The Cosmic Universe 46.13
The Cosmic Universe 46.13
The Cosmic Universe 46.13
PREFACE
Welcome to The Cosmic Universe. This textbook was written in collaboration with the OpenStax project, whose purpose
is to increase student access to high-quality learning materials, while maintaining the highest standards of academic rigor
at little to no cost. It was created by re-organizing and editing some material from the OpenStax textbooks University
Physics (http://cnx.org/contents/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@5.22) and Astronomy
(http://cnx.org/contents/2e737be8-ea65-48c3-aa0a-9f35b4c6a966@13.2) , and by the addition of new material.
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By incorporating the history of ideas, the thematic approaches lead directly to an understanding of the scientific method
- not as some dry set of steps, but as an actual, evolving human experience. As teachers at a liberal arts college, we feel
that every physics major should also understand how history, science, politics, religion, and ethics interact as parts of that
experience.
And, we also feel that it is important for physics majors to encounter, early in their college careers, the major unanswered
questions in physics, many of which are part of the study of astrophysics and cosmology - e.g. dark matter and dark energy.
The Cosmic Universe textbook contains material from various subfields of physics, from classical mechanics to optics to
relativity to quantum mechanics. The choice of topics and of the astrophysics theme were made to provide a first-semester
experience that is:
Interesting, because it deals with a very active field of current study in physics
New to virtually all of the students, because it involves topics not taught in most high-school physics courses
Rigorously mathematical in its approach, at the level of algebra and introductory calculus, more so than a
traditional college textbook in introductory astronomy
Not crucially dependent upon previous fluency in the use of calculus
Presented using a coherent story line
Calculus in The Cosmic Universe
The fact that The Cosmic Universe course is taken mostly by students in their first semester of college strongly influences
our use of calculus in this book. Many introductory calculus-based physics texts will, in an early chapter, derive the
equations of one-dimensional kinematics using integrals. For our student audience, where up to 50% are simultaneously
enrolled in their first semester of college calculus, such an approach can be discouraging and therefore counterproductive.
Knowing that they are studying (first) limits and (next) derivatives and then (perhaps by the end of the semester) integration
influences our use of calculus. We attempt to include calculus, conceptually at first, and hope that its physical significance
(of the derivative in particular) and practical applications can enhance the students' understanding of both the physics and
the math. As we are fond of asking our students, "Why did Newton invent the calculus in the first place?"
The book is organized as follows:
Introduction
Chapter 1: Introducing Physics
Chapter 2: The Universe at Its Limits
Chapter 3: Straight-Line Motion
Chapter 4: Circular Motion as One-Dimensional Motion
Chapter 5: Relativistic Kinematics in One Dimension
Chapter 6: Introduction to Vectors
Chapter 7: Kinematics in Two Dimensions
Chapter 8: Overview of the Solar System and Kepler's Laws
Chapter 9: The Newtonian Synthesis
Chapter 10: Extension of Newton's Laws to Rotations
Chapter 11: Work and Energy
Chapter 12: Momentum in One Dimension
Chapter 13: Angular Momentum
Chapter 14: Reflection and Refraction
Chapter 15: Dispersion, Total Internal Reflection, Luminosity and Apparent Brightness
Chapter 16: Image Formation
1 | INTRODUCING PHYSICS
Chapter Outline
1.1 Introducing Astrophysics
1.2 The Scope and Scale of Physics
1.3 Units and Standards
1.4 Unit Conversion
1.5 Dimensional Analysis
1.6 Estimates and Fermi Calculations
1.7 Significant Figures
1.8 Solving Problems in Physics
Introduction
You are hopefully reading this book (and taking the course with which it is associated) because you want to get a
mathematically rigorous introduction to the science of physics. To physicists, this means you want to learn to understand
how the world works (and how the things in it work). You likely want to do so because your eventual goal is a career in
physics, engineering, or a related field.
This book is like no other textbook in terms of its approach to introductory physics. The traditional undergraduate physics
course sequence is quite historical. It begins with the 17th and 18th-century development of ideas about motion from
Galileo and Newton, continues with the 19th-century study of energy and heat (by Joule and Carnot), moves on to the
study of electricity and magnetism (by Faraday and Maxwell), and follows through with the 20th-century ideas of Einstein
(especially in relativity) and of Planck, Bohr, Schrodinger and Heisenberg (in quantum physics).
Such an historical approach can have the strength of helping the reader to appreciate the process of science – how each new
discovery is built upon those that came before. A typical textbook’s ordering of the major subdisciplines in physics might
be:
• Classical Mechanics
• Waves and Sound
• Heat and Thermodynamics
• Electricity and Magnetism
• Optics
• Relativity
• Quantum Physics
But, because a rigorous, introductory study of physics can only realistically take place over a protracted period of time
(typically three or four semesters), this traditional, topical approach can sometimes lose track of the big picture. There are
important ideas that cross multiple subdisciplines, have taken many centuries to develop and, in fact, continue to be refined
even in the 21st century.
This book is fashioned around one such big-picture “story line” – our understanding of the Universe. Where did the
Universe (and we) come from? Where are we going? It is the oldest and, to us, the most interesting story in the history of
humankind. It is also multidisciplinary within physics, involving almost all of the traditional subtopics listed above.
Why begin your study of physics with astrophysics? For one thing, in the 21st century, no branch of physics is more active
and dynamic in its pursuit of scientific knowledge. Chances are you may not yet have spent much time in your formal
education studying this subject, but you can surely read almost weekly about new discoveries being made about the nature
and evolution of the Universe. New planets in astonishing numbers have been detected orbiting distant stars. Some are
considered “Earth-like” – are they possibly homes to life? Even some bodies in our own solar system are being revealed
to be very unlike our previous ideas about them. The recent NASA New Horizons mission revealed information about the
dwarf planet Pluto (see Figure 1.1) that has required astronomers to completely rethink the theories of its formation and
6 Chapter 1 | Introducing Physics
evolution.
Even more recently, NASA's Juno mission making close approaches to Jupiter has made exquisitely precise images of
magnetic storms there (see Figure 1.2). Those discoveries not only make us re-think our understanding of the planet, but
may in fact lead to new models for the basic formation of planetary magnetic fields.
No questions in science are older than those from astronomy. The recorded history of science begins with astronomy.
Thousands of years ago, from the West (the ancient Greeks) and the East (the Chinese), we find written records of
astronomical data and theories of the Universe. But even before that, it isn’t hard to speculate that human beings’ earliest
questions about their world must have included:
• What causes the regular cycles of the Sun and Moon?
• What are we seeing when we look up in the sky on a clear night?
• Are all those “dots” the same kind of thing? What kind of thing?
• How far away are they?
Indeed, astronomy has been a primary scientific endeavor for well over 2000 years. In the 20th and 21st centuries, the search
for answers to these questions has become “astrophysics”, involving multiple areas of physics. Our story line will include
the study of:
I. Mechanics (the study of motion)
A. Kinematics (a description of motion)
B. Dynamics (an explanation of why things move as they do)
1. Mass, Forces and Newton’s Laws
2. Energy (and its conservation)
a. Kinetic energy
b. Potential energy
3. Momentum (and its conservation)
4. Angular Momentum (and its conservation)
II. Relativity (at high speeds)
A. Time dilation and length contraction
B. Mass-energy conversion: E = mc2
III. Thermodynamics
A. Temperature
B. Heat as transfer of thermal energy
C. Thermal energy as the internal kinetic energy of molecules
D. Phase changes with temperature
1. Liquid-Solid transition: the melting point
2. Gas-Liquid transition: the boiling point
IV. Optics (Everything that we know about what’s “out there” comes from studying the light that reaches us here on
Earth.)
A. Geometric (ray) optics (used to construct lenses, mirrors and telescopes)
B. Physical (wave) optics
C. Electromagnetic waves and the electromagnetic spectrum – from radio waves to gamma rays
D. Types of light (continuous sources vs. line sources)
E. Interference
F. Diffraction
G. The Doppler shift
V. Quantum Physics (After 1900, we learned that the physics that applies at very small scales is not Newtonian.)
A. The ultimate sources of light are molecules, atoms, electrons, and nuclei.
B. Are things in our Universe ultimately particles or waves? (Yes!)
As you can see, if we followed this story line through a typical introductory physics textbook, it would take the whole book
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to complete it – insofar as we can ever say it is “complete”. (Perhaps a better way to put it would be “up to its present state
of understanding.”)
It may sound a bit overdramatic when we say that the ultimate goal of this book (and this course) is to understand the
Universe in its entirety. From an astrophysical point of view, we will work from the inside out – beginning with our own
solar system, moving on to consider stars other than our own Sun, and then to study galaxies of stars and what lies beyond,
the large-scale structures that form out of clusters of galaxies.
We will also be faced with the fact the telescopes are time machines, i.e. the farther out in space we look at objects, the
farther back in time we see them. The term cosmology means the study of the history of the Universe, from its beginning to
now and into the future beyond. Our outward journey, then, will help us to paint a picture of that history.
We will study the evidence for the beginning of our Universe in a “Big Bang”, and trace the evolution of stars, solar systems
and galaxies over the roughly 14 billion years of its existence. Through our understanding of the processes that have shaped
the past and the present, we will be able to examine and discuss possible fates (or futures) of the Universe. Does the future
depend upon unseen “stuff” referred to as dark matter? Does it involve an unexplained repulsive force (like an anti-gravity)
called dark energy?
Whether or not the preceding material is enough to convince you that a course in astrophysics is an interesting place to begin
your study of physics, we hope you will leave this chapter with one important take-home thought. We paraphrase the late
astronomer Carl Sagan who, through his ground-breaking television series, Cosmos, did so much to popularize and explain
astrophysics to the general public. Just think about this: “Every atom inside your body, at this instant, once lived inside a
star.”
Does that statement make sense to you? Because it’s absolutely a true, scientific fact. By the end of “The Cosmic Universe”,
you will know both why that is true and how we know it to be so.
Figure 1.3 This image might be showing any number of things. It might be a whirlpool in a tank of
water or perhaps a collage of paint and shiny beads done for art class. Without knowing the size of the
object in units we all recognize, such as meters or inches, it is difficult to know what we’re looking at. In
fact, this image shows the Whirlpool Galaxy (and its companion galaxy), which is about 60,000 light-
years in diameter (about 6 × 10 17 km across). (credit: S. Beckwith (STScI) Hubble Heritage Team,
(STScI/AURA), ESA, NASA)
As noted in the figure caption, this image is of the Whirlpool Galaxy. Galaxies are as immense as atoms are small, yet the
same laws of physics describe both, along with all the rest of nature—an indication of the underlying unity in the universe.
The laws of physics are surprisingly few, implying an underlying simplicity to nature’s apparent complexity. In this text,
you learn about the laws of physics. Galaxies and atoms may seem far removed from your daily life, but as you begin to
explore this broad-ranging subject, you may soon come to realize that physics plays a much larger role in your life than you
first thought, no matter your life goals or career choice.
Figure 1.4 These two interacting islands of stars (galaxies) are so far away that their light takes hundreds of millions of years
to reach us on Earth (photographed with the Hubble Space Telescope). (credit: modification of work by NASA, ESA, the Hubble
Heritage (STScl/AURA)-ESA/Hubble Collaboration, and K. Noll (STScl))
We invite you to come along on a series of voyages to explore the universe as astronomers and physicists understand
it today. Beyond Earth are vast and magnificent realms full of objects that have no counterpart on our home planet.
Nevertheless, we hope to show you that the evolution of the universe has been directly responsible for your presence on
10 Chapter 1 | Introducing Physics
Earth today.
Along your journey, you will encounter:
• a canyon system so large that, on Earth, it would stretch from Los Angeles to Washington, DC (Figure 1.5).
Figure 1.5 This image of Mars is centered on the Valles Marineris (Mariner Valley) complex of canyons, which is as long as
the United States is wide. (credit: modification of work by NASA)
• a crater and other evidence on Earth that tell us that the dinosaurs (and many other creatures) died because of a
cosmic collision.
• a tiny moon whose gravity is so weak that one good throw from its surface could put a baseball into orbit.
• a collapsed star so dense that to duplicate its interior we would have to squeeze every human being on Earth into a
single raindrop.
• exploding stars whose violent end could wipe clean all of the life-forms on a planet orbiting a neighboring star
(Figure 1.6).
• a “cannibal galaxy” that has already consumed a number of its smaller galaxy neighbors and is not yet finished
finding new victims.
• a radio echo that is the faint but unmistakable signal of the creation event for our universe.
Such discoveries are what make astronomy such an exciting field for scientists and many others—but you will explore much
more than just the objects in our universe and the latest discoveries about them. We will pay equal attention to the process
by which we have come to understand the realms beyond Earth and the tools we use to increase that understanding.
We gather information about the cosmos from the messages the universe sends our way. Because the stars are the
fundamental building blocks of the universe, decoding the message of starlight has been a central challenge and triumph of
modern astronomy. By the time you have finished reading this text, you will know a bit about how to read that message and
how to understand what it is telling us.
Physics is devoted to the understanding of all natural phenomena. In physics, we try to understand physical phenomena at all
scales—from the world of subatomic particles to the entire universe. Despite the breadth of the subject, the various subfields
of physics share a common core. The same basic training in physics will prepare you to work in any area of physics and the
related areas of science and engineering. In this section, we investigate the scope of physics; the scales of length, mass, and
time over which the laws of physics have been shown to be applicable; and the process by which science in general, and
physics in particular, operates.
huge clouds of gas and dust. Its companion galaxy is also visible to the right. This pair of galaxies lies a staggering billion
trillion miles (1.4 × 10 21 mi) from our own galaxy (which is called the Milky Way). The stars and planets that make up
the Whirlpool Galaxy might seem to be the furthest thing from most people’s everyday lives, but the Whirlpool is a great
starting point to think about the forces that hold the universe together. The forces that cause the Whirlpool Galaxy to act as
it does are thought to be the same forces we contend with here on Earth, whether we are planning to send a rocket into space
or simply planning to raise the walls for a new home. The gravity that causes the stars of the Whirlpool Galaxy to rotate and
revolve is thought to be the same as what causes water to flow over hydroelectric dams here on Earth. When you look up
at the stars, realize the forces out there are the same as the ones here on Earth. Through a study of physics, you may gain a
greater understanding of the interconnectedness of everything we can see and know in this universe.
Think, now, about all the technological devices you use on a regular basis. Computers, smartphones, global positioning
systems (GPSs), MP3 players, and satellite radio might come to mind. Then, think about the most exciting modern
technologies you have heard about in the news, such as trains that levitate above tracks, “invisibility cloaks” that bend light
around them, and microscopic robots that fight cancer cells in our bodies. All these groundbreaking advances, commonplace
or unbelievable, rely on the principles of physics. Aside from playing a significant role in technology, professionals such
as engineers, pilots, physicians, physical therapists, electricians, and computer programmers apply physics concepts in their
daily work. For example, a pilot must understand how wind forces affect a flight path; a physical therapist must understand
how the muscles in the body experience forces as they move and bend. As you will learn in this text, the principles of
physics are propelling new, exciting technologies, and these principles are applied in a wide range of careers.
The underlying order of nature makes science in general, and physics in particular, interesting and enjoyable to study. For
example, what do a bag of chips and a car battery have in common? Both contain energy that can be converted to other
forms. The law of conservation of energy (which says that energy can change form but is never lost) ties together such
topics as food calories, batteries, heat, light, and watch springs. Understanding this law makes it easier to learn about the
various forms energy takes and how they relate to one another. Apparently unrelated topics are connected through broadly
applicable physical laws, permitting an understanding beyond just the memorization of lists of facts.
Science consists of theories and laws that are the general truths of nature, as well as the body of knowledge they encompass.
Scientists are continuously trying to expand this body of knowledge and to perfect the expression of the laws that describe
it. Physics, which comes from the Greek phúsis, meaning “nature,” is concerned with describing the interactions of
energy, matter, space, and time to uncover the fundamental mechanisms that underlie every phenomenon. This concern for
describing the basic phenomena in nature essentially defines the scope of physics.
Physics aims to understand the world around us at the most basic level. It emphasizes the use of a small number of
quantitative laws to do this, which can be useful to other fields pushing the performance boundaries of existing technologies.
Consider a smartphone (Figure 1.7). Physics describes how electricity interacts with the various circuits inside the device.
This knowledge helps engineers select the appropriate materials and circuit layout when building a smartphone. Knowledge
of the physics underlying these devices is required to shrink their size or increase their processing speed. Or, think about a
GPS. Physics describes the relationship between the speed of an object, the distance over which it travels, and the time it
takes to travel that distance. When you use a GPS in a vehicle, it relies on physics equations to determine the travel time
from one location to another.
Knowledge of physics is useful in everyday situations as well as in nonscientific professions. It can help you understand
how microwave ovens work, why metals should not be put into them, and why they might affect pacemakers. Physics allows
you to understand the hazards of radiation and to evaluate these hazards rationally and more easily. Physics also explains the
reason why a black car radiator helps remove heat in a car engine, and it explains why a white roof helps keep the inside of
a house cool. Similarly, the operation of a car’s ignition system as well as the transmission of electrical signals throughout
our body’s nervous system are much easier to understand when you think about them in terms of basic physics.
Physics is a key element of many important disciplines and contributes directly to others. Chemistry, for example—since
it deals with the interactions of atoms and molecules—has close ties to atomic and molecular physics. Most branches of
engineering are concerned with designing new technologies, processes, or structures within the constraints set by the laws
of physics. In architecture, physics is at the heart of structural stability and is involved in the acoustics, heating, lighting, and
cooling of buildings. Parts of geology rely heavily on physics, such as radioactive dating of rocks, earthquake analysis, and
heat transfer within Earth. Some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines.
Physics has many applications in the biological sciences. On the microscopic level, it helps describe the properties of
cells and their environments. On the macroscopic level, it explains the heat, work, and power associated with the human
body and its various organ systems. Physics is involved in medical diagnostics, such as radiographs, magnetic resonance
imaging, and ultrasonic blood flow measurements. Medical therapy sometimes involves physics directly; for example,
cancer radiotherapy uses ionizing radiation. Physics also explains sensory phenomena, such as how musical instruments
make sound, how the eye detects color, and how lasers transmit information.
It is not necessary to study all applications of physics formally. What is most useful is knowing the basic laws of physics
and developing skills in the analytical methods for applying them. The study of physics also can improve your problem-
solving skills. Furthermore, physics retains the most basic aspects of science, so it is used by all the sciences, and the study
of physics makes other sciences easier to understand.
is easy enough to say, but to come to grips with what it really means, we need to get a little bit quantitative. So, before
surveying the various scales that physics allows us to explore, let’s first look at the concept of “order of magnitude,” which
we use to come to terms with the vast ranges of length, mass, and time that we consider in this text (Figure 1.8).
Figure 1.8 (a) Using a scanning tunneling microscope, scientists can see the individual atoms (diameters around 10 –10 m) that
compose this sheet of gold. (b) Tiny phytoplankton swim among crystals of ice in the Antarctic Sea. They range from a few
micrometers (1 μm is 10–6 m) to as much as 2 mm (1 mm is 10–3 m) in length. (c) These two colliding galaxies, known as NGC
4676A (right) and NGC 4676B (left), are nicknamed “The Mice” because of the tail of gas emanating from each one. They are
located 300 million light-years from Earth in the constellation Coma Berenices. Eventually, these two galaxies will merge into
one. (credit a: modification of work by Erwinrossen; credit b: modification of work by Prof. Gordon T. Taylor, Stony Brook
University; NOAA Corps Collections; credit c: modification of work by NASA, H. Ford (JHU), G. Illingworth (UCSC/LO), M.
Clampin (STScI), G. Hartig (STScI), the ACS Science Team, and ESA)
Order of magnitude
The order of magnitude of a number is the power of 10 that most closely approximates it. Thus, the order of magnitude
refers to the scale (or size) of a value. Each power of 10 represents a different order of magnitude. For example,
10 1, 10 2, 10 3, and so forth, are all different orders of magnitude, as are 10 0 = 1, 10 −1, 10 −2, and 10 −3. To find the
order of magnitude of a number, take the base-10 logarithm of the number and round it to the nearest integer, then the order
of magnitude of the number is simply the resulting power of 10. For example, the order of magnitude of 800 is 103 because
log 10 800 ≈ 2.903, which rounds to 3. Similarly, the order of magnitude of 450 is 103 because log 10 450 ≈ 2.653,
which rounds to 3 as well. Thus, we say the numbers 800 and 450 are of the same order of magnitude: 103. However, the
order of magnitude of 250 is 102 because log 10 250 ≈ 2.397, which rounds to 2.
An equivalent but quicker way to find the order of magnitude of a number is first to write it in scientific notation and then
check to see whether the first factor is greater than or less than 10 = 10 0.5 ≈ 3. The idea is that 10 = 10 0.5 is halfway
between 1 = 10 0 and 10 = 10 1 on a log base-10 scale. Thus, if the first factor is less than 10, then we round it down
to 1 and the order of magnitude is simply whatever power of 10 is required to write the number in scientific notation. On
the other hand, if the first factor is greater than 10, then we round it up to 10 and the order of magnitude is one power of
10 higher than the power needed to write the number in scientific notation. For example, the number 800 can be written in
scientific notation as 8 × 10 2. Because 8 is bigger than 10 ≈ 3, we say the order of magnitude of 800 is 10 2 + 1 = 10 3.
The number 450 can be written as 4.5 × 10 2, so its order of magnitude is also 103 because 4.5 is greater than 3. However,
250 written in scientific notation is 2.5 × 10 2 and 2.5 is less than 3, so its order of magnitude is 10 2.
The order of magnitude of a number is designed to be a ballpark estimate for the scale (or size) of its value. It is simply a
way of rounding numbers consistently to the nearest power of 10. This makes doing rough mental math with very big and
very small numbers easier. For example, the diameter of a hydrogen atom is on the order of 10−10 m, whereas the diameter
of the Sun is on the order of 109 m, so it would take roughly 10 9 /10 −10 = 10 19 hydrogen atoms to stretch across the
diameter of the Sun. This is much easier to do in your head than using the more precise values of 1.06 × 10 −10 m for a
hydrogen atom diameter and 1.39 × 10 9 m for the Sun’s diameter, to find that it would take 1.31 × 10 19 hydrogen atoms
to stretch across the Sun’s diameter. In addition to being easier, the rough estimate is also nearly as informative as the precise
calculation.
Figure 1.9 This table shows the orders of magnitude of length, mass, and time.
16 Chapter 1 | Introducing Physics
Visit this site (https://openstaxcollege.org/l/21scaleuniv) to explore interactively the vast range of length
scales in our universe. Scroll down and up the scale to view hundreds of organisms and objects, and click on the
individual objects to learn more about each one.
Building Models
How did we come to know the laws governing natural phenomena? What we refer to as the laws of nature are concise
descriptions of the universe around us. They are human statements of the underlying laws or rules that all natural processes
follow. Such laws are intrinsic to the universe; humans did not create them and cannot change them. We can only discover
and understand them. Their discovery is a very human endeavor, with all the elements of mystery, imagination, struggle,
triumph, and disappointment inherent in any creative effort (Figure 1.10). The cornerstone of discovering natural laws is
observation; scientists must describe the universe as it is, not as we imagine it to be.
Figure 1.10 (a) Enrico Fermi (1901–1954) was born in Italy. On accepting the Nobel Prize in
Stockholm in 1938 for his work on artificial radioactivity produced by neutrons, he took his
family to America rather than return home to the government in power at the time. He became an
American citizen and was a leading participant in the Manhattan Project. (b) Marie Curie
(1867–1934) sacrificed monetary assets to help finance her early research and damaged her
physical well-being with radiation exposure. She is the only person to win Nobel prizes in both
physics and chemistry. One of her daughters also won a Nobel Prize. (credit a: United States
Department of Energy)
A model is a representation of something that is often too difficult (or impossible) to display directly. Although a model
is justified by experimental tests, it is only accurate in describing certain aspects of a physical system. An example is the
Bohr model of single-electron atoms, in which the electron is pictured as orbiting the nucleus, analogous to the way planets
orbit the Sun (Figure 1.11). We cannot observe electron orbits directly, but the mental image helps explain some of the
observations we can make, such as the emission of light from hot gases (atomic spectra). However, other observations
show that the picture in the Bohr model is not really what atoms look like. The model is “wrong,” but is still useful for
some purposes. Physicists use models for a variety of purposes. For example, models can help physicists analyze a scenario
and perform a calculation or models can be used to represent a situation in the form of a computer simulation. Ultimately,
however, the results of these calculations and simulations need to be double-checked by other means—namely, observation
and experimentation.
The word theory means something different to scientists than what is often meant when the word is used in everyday
conversation. In particular, to a scientist a theory is not the same as a “guess” or an “idea” or even a “hypothesis.” The
phrase “it’s just a theory” seems meaningless and silly to scientists because science is founded on the notion of theories. To
a scientist, a theory is a testable explanation for patterns in nature supported by scientific evidence and verified multiple
times by various groups of researchers. Some theories include models to help visualize phenomena whereas others do not.
Newton’s theory of gravity, for example, does not require a model or mental image, because we can observe the objects
directly with our own senses. The kinetic theory of gases, on the other hand, is a model in which a gas is viewed as being
composed of atoms and molecules. Atoms and molecules are too small to be observed directly with our senses—thus, we
picture them mentally to understand what the instruments tell us about the behavior of gases. Although models are meant
only to describe certain aspects of a physical system accurately, a theory should describe all aspects of any system that falls
within its domain of applicability. In particular, any experimentally testable implication of a theory should be verified. If an
experiment ever shows an implication of a theory to be false, then the theory is either thrown out or modified suitably (for
example, by limiting its domain of applicability).
A law uses concise language to describe a generalized pattern in nature supported by scientific evidence and repeated
experiments. Often, a law can be expressed in the form of a single mathematical equation. Laws and theories are similar
in that they are both scientific statements that result from a tested hypothesis and are supported by scientific evidence.
However, the designation law is usually reserved for a concise and very general statement that describes phenomena in
nature, such as the law that energy is conserved during any process, or Newton’s second law of motion, which relates force
(F), mass (m), and acceleration (a) by the simple equation F = ma. A theory, in contrast, is a less concise statement of
observed behavior. For example, the theory of evolution and the theory of relativity cannot be expressed concisely enough to
be considered laws. The biggest difference between a law and a theory is that a theory is much more complex and dynamic.
A law describes a single action whereas a theory explains an entire group of related phenomena. Less broadly applicable
statements are usually called principles (such as Pascal’s principle, which is applicable only in fluids), but the distinction
between laws and principles often is not made carefully.
The models, theories, and laws we devise sometimes imply the existence of objects or phenomena that are as yet
unobserved. These predictions are remarkable triumphs and tributes to the power of science. It is the underlying order in
the universe that enables scientists to make such spectacular predictions. However, if experimentation does not verify our
predictions, then the theory or law is wrong, no matter how elegant or convenient it is. Laws can never be known with
absolute certainty because it is impossible to perform every imaginable experiment to confirm a law for every possible
scenario. Physicists operate under the assumption that all scientific laws and theories are valid until a counterexample is
observed. If a good-quality, verifiable experiment contradicts a well-established law or theory, then the law or theory must
be modified or overthrown completely.
The study of science in general, and physics in particular, is an adventure much like the exploration of an uncharted ocean.
Discoveries are made; models, theories, and laws are formulated; and the beauty of the physical universe is made more
sublime for the insights gained.
18 Chapter 1 | Introducing Physics
As we saw previously, the range of objects and phenomena studied in physics is immense. From the incredibly short lifetime
of a nucleus to the age of Earth, from the tiny sizes of subnuclear particles to the vast distance to the edges of the known
universe, from the force exerted by a jumping flea to the force between Earth and the Sun, there are enough factors of 10
to challenge the imagination of even the most experienced scientist. Giving numerical values for physical quantities and
equations for physical principles allows us to understand nature much more deeply than qualitative descriptions alone. To
comprehend these vast ranges, we must also have accepted units in which to express them. We shall find that even in the
potentially mundane discussion of meters, kilograms, and seconds, a profound simplicity of nature appears: all physical
quantities can be expressed as combinations of only seven base physical quantities.
We define a physical quantity either by specifying how it is measured or by stating how it is calculated from other
measurements. For example, we might define distance and time by specifying methods for measuring them, such as using
a meter stick and a stopwatch. Then, we could define average speed by stating that it is calculated as the total distance
traveled divided by time of travel.
Measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the length
of a race, which is a physical quantity, can be expressed in units of meters (for sprinters) or kilometers (for distance runners).
Without standardized units, it would be extremely difficult for scientists to express and compare measured values in a
meaningful way (Figure 1.12).
Two major systems of units are used in the world: SI units (for the French Système International d’Unités), also known
as the metric system, and English units (also known as the customary or imperial system). English units were historically
used in nations once ruled by the British Empire and are still widely used in the United States. English units may also be
referred to as the foot–pound–second (fps) system, as opposed to the centimeter–gram–second (cgs) system. You may also
encounter the term SAE units, named after the Society of Automotive Engineers. Products such as fasteners and automotive
tools (for example, wrenches) that are measured in inches rather than metric units are referred to as SAE fasteners or SAE
wrenches.
Virtually every other country in the world (except the United States) now uses SI units as the standard. The metric system
is also the standard system agreed on by scientists and mathematicians.
You are probably already familiar with some derived quantities that can be formed from the base quantities in Table 1.1.
For example, the geometric concept of area is always calculated as the product of two lengths. Thus, area is a derived
quantity that can be expressed in terms of SI base units using square meters (m × m = m 2). Similarly, volume is a derived
quantity that can be expressed in cubic meters (m 3). Speed is length per time; so in terms of SI base units, we could
measure it in meters per second (m/s). Volume mass density (or just density) is mass per volume, which is expressed in
terms of SI base units such as kilograms per cubic meter (kg/m3). Angles can also be thought of as derived quantities
because they can be defined as the ratio of the arc length subtended by two radii of a circle to the radius of the circle. This
is how the radian is defined. Depending on your background and interests, you may be able to come up with other derived
quantities, such as the mass flow rate (kg/s) or volume flow rate (m3/s) of a fluid, electric charge (A · s), mass flux density
[kg/(m 2 · s)], and so on. We will see many more examples throughout this text. For now, the point is that every physical
quantity can be derived from the seven base quantities in Table 1.1, and the units of every physical quantity can be derived
from the seven SI base units.
For the most part, we use SI units in this text. Non-SI units are used in a few applications in which they are in very common
use, such as the measurement of temperature in degrees Celsius (°C), the measurement of fluid volume in liters (L), and
the measurement of energies of elementary particles in electron-volts (eV). Whenever non-SI units are discussed, they are
tied to SI units through conversions. For example, 1 L is 10 −3 m 3.
Check out a comprehensive source of information on SI units (https://openstaxcollege.org/l/21SIUnits) at
the National Institute of Standards and Technology (NIST) Reference on Constants, Units, and Uncertainty.
Units of Time, Length, and Mass: The Second, Meter, and Kilogram
The initial chapters in this textbook are concerned with mechanics, fluids, and waves. In these subjects all pertinent physical
quantities can be expressed in terms of the base units of length, mass, and time. Therefore, we now turn to a discussion of
these three base units, leaving discussion of the others until they are needed later.
20 Chapter 1 | Introducing Physics
The second
The SI unit for time, the second (abbreviated s), has a long history. For many years it was defined as 1/86,400 of a mean
solar day. More recently, a new standard was adopted to gain greater accuracy and to define the second in terms of a
nonvarying or constant physical phenomenon (because the solar day is getting longer as a result of the very gradual slowing
of Earth’s rotation). Cesium atoms can be made to vibrate in a very steady way, and these vibrations can be readily observed
and counted. In 1967, the second was redefined as the time required for 9,192,631,770 of these vibrations to occur (Figure
1.13). Note that this may seem like more precision than you would ever need, but it isn’t—GPSs rely on the precision of
atomic clocks to be able to give you turn-by-turn directions on the surface of Earth, far from the satellites broadcasting their
location.
The meter
The SI unit for length is the meter (abbreviated m); its definition has also changed over time to become more precise. The
meter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This measurement was
improved in 1889 by redefining the meter to be the distance between two engraved lines on a platinum–iridium bar now
kept near Paris. By 1960, it had become possible to define the meter even more accurately in terms of the wavelength of
light, so it was again redefined as 1,650,763.73 wavelengths of orange light emitted by krypton atoms. In 1983, the meter
was given its current definition (in part for greater accuracy) as the distance light travels in a vacuum in 1/299,792,458 of a
second (Figure 1.14). This change came after knowing the speed of light to be exactly 299,792,458 m/s. The length of the
meter will change if the speed of light is someday measured with greater accuracy.
Figure 1.14 The meter is defined to be the distance light travels in 1/299,792,458 of a second in a vacuum. Distance
traveled is speed multiplied by time.
The kilogram
The SI unit for mass is the kilogram (abbreviated kg); it is defined to be the mass of a platinum–iridium cylinder
kept with the old meter standard at the International Bureau of Weights and Measures near Paris. Exact replicas of the
standard kilogram are also kept at the U.S. National Institute of Standards and Technology (NIST), located in Gaithersburg,
Maryland, outside of Washington, DC, and at other locations around the world. Scientists at NIST are currently investigating
two complementary methods of redefining the kilogram (see Figure 1.15). The determination of all other masses can be
Figure 1.15 Redefining the SI unit of mass. Complementary methods are being investigated for use in an upcoming
redefinition of the SI unit of mass. (a) The U.S. National Institute of Standards and Technology’s watt balance is a machine that
balances the weight of a test mass against the current and voltage (the “watt”) produced by a strong system of magnets. (b) The
International Avogadro Project is working to redefine the kilogram based on the dimensions, mass, and other known properties of
a silicon sphere. (credit a and credit b: National Institute of Standards and Technology)
Metric Prefixes
SI units are part of the metric system, which is convenient for scientific and engineering calculations because the units are
categorized by factors of 10. Table 1.2 lists the metric prefixes and symbols used to denote various factors of 10 in SI units.
For example, a centimeter is one-hundredth of a meter (in symbols, 1 cm = 10–2 m) and a kilometer is a thousand meters (1
km = 103 m). Similarly, a megagram is a million grams (1 Mg = 106 g), a nanosecond is a billionth of a second (1 ns = 10–9
s), and a terameter is a trillion meters (1 Tm = 1012 m).
The only rule when using metric prefixes is that you cannot “double them up.” For example, if you have measurements in
petameters (1 Pm = 1015 m), it is not proper to talk about megagigameters, although 10 6 × 10 9 = 10 15. In practice, the
only time this becomes a bit confusing is when discussing masses. As we have seen, the base SI unit of mass is the kilogram
(kg), but metric prefixes need to be applied to the gram (g), because we are not allowed to “double-up” prefixes. Thus, a
thousand kilograms (103 kg) is written as a megagram (1 Mg) since
10 3 kg = 10 3 × 10 3 g = 10 6 g = 1 Mg.
Incidentally, 103 kg is also called a metric ton, abbreviated t. This is one of the units outside the SI system considered
acceptable for use with SI units.
As we see in the next section, metric systems have the advantage that conversions of units involve only powers of 10. There
are 100 cm in 1 m, 1000 m in 1 km, and so on. In nonmetric systems, such as the English system of units, the relationships
are not as simple—there are 12 in. in 1 ft, 5280 ft in 1 mi, and so on.
Another advantage of metric systems is that the same unit can be used over extremely large ranges of values simply by
scaling it with an appropriate metric prefix. The prefix is chosen by the order of magnitude of physical quantities commonly
found in the task at hand. For example, distances in meters are suitable in construction, whereas distances in kilometers are
appropriate for air travel, and nanometers are convenient in optical design. With the metric system there is no need to invent
new units for particular applications. Instead, we rescale the units with which we are already familiar.
Example 1.1
From Table 1.2, we see that 1016 is between “peta-” (1015) and “exa-” (1018). If we use the “peta-” prefix, then
we find that 1.93 × 10 16 g = 1.93 × 10 1 Pg, since 16 = 1 + 15. Alternatively, if we use the “exa-” prefix we
find that 1.93 × 10 16 g = 1.93 × 10 −2 Eg, since 16 = −2 + 18. Because the problem asks for the numerical
value between one and 1000, we use the “peta-” prefix and the answer is 19.3 Pg.
Significance
It is easy to make silly arithmetic errors when switching from one prefix to another, so it is always a good idea to
check that our final answer matches the number we started with. An easy way to do this is to put both numbers in
scientific notation and count powers of 10, including the ones hidden in prefixes. If we did not make a mistake,
the powers of 10 should match up. In this problem, we started with 1.93 × 10 13 kg, so we have 13 + 3 = 16
powers of 10. Our final answer in scientific notation is 1.93 × 10 1 Pg, so we have 1 + 15 = 16 powers of 10. So,
1.1 Check Your Understanding Restate 4.79 × 10 5 kg using a metric prefix such that the resulting number
is bigger than one but less than 1000.
It is often necessary to convert from one unit to another. For example, if you are reading a European cookbook, some
quantities may be expressed in units of liters and you need to convert them to cups. Or perhaps you are reading walking
directions from one location to another and you are interested in how many miles you will be walking. In this case, you may
need to convert units of feet or meters to miles.
The Power of 1
Even though it will be expressed in new units, we do NOT want to change the actual value of the quantity. In algebra, there
is a safe way to keep the value of a quantity unchanged: multiply it by 1. So, if all we ever do is multiply our quantity by 1,
we are assured that we keep the same value.
The secret is in a clever use of the many ways there are in which to write the quantity 1. In particular, any fraction whose
numerator and denominator are equal does in fact have the value 1. The particular fractions we will choose are called
conversion factors.
Let’s consider a simple example of how to convert units. Suppose we want to convert 80 m to kilometers. The first thing to
do is to list the units you have and the units to which you want to convert. In this case, we have units in meters and we want
to convert to kilometers. Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor
is a ratio that expresses how many of one unit are equal to another unit. For example, there are 12 in. in 1 ft, 1609 m in 1
mi, 100 cm in 1 m, 60 s in 1 min, and so on. Refer to Appendix B for a more complete list of conversion factors. In this
case, we know that there are 1000 m in 1 km. Now we can set up our unit conversion. We write the units we have and then
multiply them by the conversion factor so the units cancel out, as shown:
80 m × 1 km = 0.080 km.
1000 m
Why did the actual quantity (the distance involved) not change? Because all we did, mathematically, was to multiply it by
1. Our conversion factor is a fraction, the value of whose numerator (1 km) is equal to the value of its denominator (1000
m). So, it is just another way to write 1.
Note that the unwanted meter unit cancels, leaving only the desired kilometer unit. You can use this method to convert
between any type of unit. Of course, the conversion of 80 m to kilometers is simply the use of a metric prefix, as we saw in
the preceding section, so we can get the same answer just as easily by noting that
80 m = 8.0 × 10 1 m = 8.0 × 10 −2 km = 0.080 km,
since “kilo-” means 103 (see Table 1.2) and 1 = −2 + 3. However, using conversion factors is handy when converting
between units that are not metric or when converting between derived units, as the following examples illustrate.
24 Chapter 1 | Introducing Physics
Example 1.2
Significance
Check the answer in the following ways:
1. Be sure that each conversion factor is a fraction whose numerator and denominator are equal. This ensures
that all you ever do is multiply your quantity by 1 (sometimes repeatedly).
2. Be sure the units in the unit conversion cancel correctly. If the unit conversion factor was written upside
down, the units do not cancel correctly in the equation. We see the “miles” in the numerator in 0.50
mi/min cancels the “mile” in the denominator in the first conversion factor. Also, the “min” in the
denominator in 0.50 mi/min cancels the “min” in the numerator in the second conversion factor.
3. Check that the units of the final answer are the desired units. The problem asked us to solve for average
speed in units of meters per second and, after the cancellations, the only units left are a meter (m) in the
numerator and a second (s) in the denominator, so we have indeed obtained these units.
1.2 Check Your Understanding Light travels about 9 Pm in a year. Given that a year is about 3 × 10 7 s,
what is the speed of light in meters per second?
Example 1.3
Strategy
We need to convert grams to kilograms and cubic centimeters to cubic meters. The conversion factors we need are
1 kg = 10 3 g and 1 cm = 10 −2 m. However, we are dealing with cubic centimeters (cm 3 = cm × cm × cm),
so we have to use the second conversion factor three times (that is, we need to cube it). The idea is still to multiply
by the conversion factors in such a way that they cancel the units we want to get rid of and introduce the units we
want to keep.
Solution
⎛ cm ⎞
3
g kg 7.86
10 3 g ⎝10 −2 m ⎠
7.86 3
× × = kg/m 3 = 7.86 × 10 3 kg/m 3
cm (10 3)(10 −6)
Significance
Remember, it’s always important to check the answer.
1. Be sure that each conversion factor is a fraction whose numerator and denominator are equal. In this case,
the first conversion factor has a numerator of 1 kg and a denominator of 103 g. The second conversion
factor has a numerator of 1 cm and a denominator of 10-2 m.
2. Be sure to cancel the units in the unit conversion correctly. We see that the gram (“g”) in the numerator
in 7.86 g/cm3 cancels the “g” in the denominator in the first conversion factor. Also, the three factors of
“cm” in the denominator in 7.86 g/cm3 cancel with the three factors of “cm” in the numerator that we get
by cubing the second conversion factor.
3. Check that the units of the final answer are the desired units. The problem asked for us to convert to
kilograms per cubic meter. After the cancellations just described, we see the only units we have left are
“kg” in the numerator and three factors of “m” in the denominator (that is, one factor of “m” cubed, or
“m3”). Therefore, the units on the final answer are correct.
1.3 Check Your Understanding We know from Figure 1.9 that the diameter of Earth is on the order of 107
m, so the order of magnitude of its surface area is 1014 m2. What is that in square kilometers (that is, km2)? (Try
doing this both by converting 107 m to km and then squaring it and then by converting 1014 m2 directly to
square kilometers. You should get the same answer both ways.)
Unit conversions may not seem very interesting, but not doing them can be costly. One famous example of this situation was
seen with the Mars Climate Orbiter. This probe was launched by NASA on December 11, 1998. On September 23, 1999,
while attempting to guide the probe into its planned orbit around Mars, NASA lost contact with it. Subsequent investigations
showed a piece of software called SM_FORCES (or “small forces”) was recording thruster performance data in the English
units of pound-seconds (lb-s). However, other pieces of software that used these values for course corrections expected them
to be recorded in the SI units of newton-seconds (N-s), as dictated in the software interface protocols. This error caused the
probe to follow a very different trajectory from what NASA thought it was following, which most likely caused the probe
either to burn up in the Martian atmosphere or to shoot out into space. This failure to pay attention to unit conversions cost
hundreds of millions of dollars, not to mention all the time invested by the scientists and engineers who worked on the
project.
1.4 Check Your Understanding Given that 1 lb (pound) is 4.45 N, were the numbers being output by
SM_FORCES too big or too small?
The dimension of any physical quantity expresses its dependence on the base quantities as a product of symbols (or powers
26 Chapter 1 | Introducing Physics
of symbols) representing the base quantities. Table 1.3 lists the base quantities and the symbols used for their dimension.
For example, a measurement of length is said to have dimension L or L1, a measurement of mass has dimension M or
M1, and a measurement of time has dimension T or T1. Like units, dimensions obey the rules of algebra. Thus, area is
the product of two lengths and so has dimension L2, or length squared. Similarly, volume is the product of three lengths
and has dimension L3, or length cubed. Speed has dimension length over time, L/T or LT–1. Volumetric mass density has
dimension M/L3 or ML–3, or mass over length cubed. In general, the dimension of any physical quantity can be written as
L a M b T c I d Θ e N f J g for some powers a, b, c, d, e, f , and g. We can write the dimensions of a length in this form with
a = 1 and the remaining six powers all set equal to zero: L 1 = L 1 M 0 T 0 I 0 Θ 0 N 0 J 0. Any quantity with a dimension that
can be written so that all seven powers are zero (that is, its dimension is L 0 M 0 T 0 I 0 Θ 0 N 0 J 0 ) is called dimensionless
(or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists often call dimensionless
quantities pure numbers.
Physicists often use square brackets around the symbol for a physical quantity to represent the dimensions of that quantity.
For example, if r is the radius of a cylinder and h is its height, then we write [r] = L and [h] = L to indicate the
dimensions of the radius and height are both those of length, or L. Similarly, if we use the symbol A for the surface area of
a cylinder and V for its volume, then [A] = L2 and [V] = L3. If we use the symbol m for the mass of the cylinder and ρ
for the density of the material from which the cylinder is made, then [m] = M and [ρ] = ML −3.
The importance of the concept of dimension arises from the fact that any mathematical equation relating physical quantities
must be dimensionally consistent, which means the equation must obey the following rules:
• Every term in an expression must have the same dimensions; it does not make sense to add or subtract quantities of
differing dimension (think of the old saying: “You can’t add apples and oranges”). In particular, the expressions on
each side of the equality in an equation must have the same dimensions.
• The arguments of any of the standard mathematical functions such as trigonometric functions (such as sine and
cosine), logarithms, or exponential functions that appear in the equation must be dimensionless. These functions
require pure numbers as inputs and give pure numbers as outputs.
If either of these rules is violated, an equation is not dimensionally consistent and cannot possibly be a correct statement
of physical law. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws of
physics, and even to suggest the form that new laws of physics might take. This last use of dimensions is beyond the scope
of this text, but is something you will undoubtedly learn later in your academic career.
Example 1.4
Strategy
One natural strategy is to look it up, but this could take time to find information from a reputable source. Besides,
even if we think the source is reputable, we shouldn’t trust everything we read. It is nice to have a way to double-
check just by thinking about it. Also, we might be in a situation in which we cannot look things up (such as during
a test). Thus, the strategy is to find the dimensions of both expressions by making use of the fact that dimensions
follow the rules of algebra. If either expression does not have the same dimensions as area, then it cannot possibly
be the correct equation for the area of a circle.
Solution
We know the dimension of area is L2. Now, the dimension of the expression πr 2 is
since the constant π is a pure number and the radius r is a length. Therefore, πr 2 has the dimension of area.
Similarly, the dimension of the expression 2πr is
[2πr] = [2] · [π] · [r] = 1 · 1 · L = L,
since the constants 2 and π are both dimensionless and the radius r is a length. We see that 2πr has the
dimension of length, which means it cannot possibly be an area.
We rule out 2πr because it is not dimensionally consistent with being an area. We see that πr 2 is dimensionally
consistent with being an area, so if we have to choose between these two expressions, πr 2 is the one to choose.
Significance
This may seem like kind of a silly example, but the ideas are very general. As long as we know the dimensions
of the individual physical quantities that appear in an equation, we can check to see whether the equation is
dimensionally consistent. On the other hand, knowing that true equations are dimensionally consistent, we can
match expressions from our imperfect memories to the quantities for which they might be expressions. Doing this
will not help us remember dimensionless factors that appear in the equations (for example, if you had accidentally
conflated the two expressions from the example into 2πr 2, then dimensional analysis is no help), but it does
help us remember the correct basic form of equations.
1.5 Check Your Understanding Suppose we want the formula for the volume of a sphere. The two
expressions commonly mentioned in elementary discussions of spheres are 4πr 2 and 4πr 3 /3. One is the
volume of a sphere of radius r and the other is its surface area. Which one is the volume?
Example 1.5
left expression and two in the expression on the right, so we look at each in turn:
[s] = L
[vt] = [v] · [t] = LT −1 · T = LT 0 = L
[0.5at 2] = [a] · [t] 2 = LT −2 · T 2 = LT 0 = L.
All three terms have the same dimension, so this equation is dimensionally consistent.
b. Again, there are no trigonometric, exponential, or logarithmic functions, so we only need to look at the
dimensions of each of the three terms appearing in the equation:
[s] = L
[vt 2] = [v] · [t] 2 = LT −1 · T 2 = LT
[at] = [a] · [t] = LT −2 · T = LT −1.
None of the three terms has the same dimension as any other, so this is about as far from being
dimensionally consistent as you can get. The technical term for an equation like this is nonsense.
c. This equation has a trigonometric function in it, so first we should check that the argument of the sine
function is dimensionless:
⎡at 2 ⎤ [a] · [t] 2 LT −2 · T 2 L
⎣ s ⎦ = [s] = L
= = 1.
L
The argument is dimensionless. So far, so good. Now we need to check the dimensions of each of the two
terms (that is, the left expression and the right expression) in the equation:
[v] = LT −1
⎡ ⎛at 2 ⎞⎤
⎣sin⎝ s ⎠⎦ = 1.
The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equation
is another example of “nonsense.”
Significance
If we are trusting people, these types of dimensional checks might seem unnecessary. But, rest assured, any
textbook on a quantitative subject such as physics (including this one) almost certainly contains some equations
with typos. Checking equations routinely by dimensional analysis save us the embarrassment of using an incorrect
equation. Also, checking the dimensions of an equation we obtain through algebraic manipulation is a great way
to make sure we did not make a mistake (or to spot a mistake, if we made one).
One further point that needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that
dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity
with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line
tangent to its graph and slopes are ratios, so for physical quantities v and t, we have that the dimension of the derivative of
v with respect to t is just the ratio of the dimension of v over that of t:
⎡ dv ⎤ [v]
⎣ dt ⎦ = [t] .
Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the
dimension of v times the dimension of t:
⎡
⎣∫ vdt⎤⎦ = [v] · [t].
By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration
or differentiation.
On many occasions, physicists, other scientists, and engineers need to make estimates for a particular quantity. Other
terms sometimes used are guesstimates, order-of-magnitude approximations, back-of-the-envelope calculations, or Fermi
calculations. (The physicist Enrico Fermi mentioned earlier was famous for his ability to estimate various kinds of data with
surprising precision.) Will that piece of equipment fit in the back of the car or do we need to rent a truck? How long will
this download take? About how large a current will there be in this circuit when it is turned on? How many houses could
a proposed power plant actually power if it is built? Note that estimating does not mean guessing a number or a formula
at random. Rather, estimation means using prior experience and sound physical reasoning to arrive at a rough idea of a
quantity’s value. Because the process of determining a reliable approximation usually involves the identification of correct
physical principles and a good guess about the relevant variables, estimating is very useful in developing physical intuition.
Estimates also allow us perform “sanity checks” on calculations or policy proposals by helping us rule out certain scenarios
or unrealistic numbers. They allow us to challenge others (as well as ourselves) in our efforts to learn truths about the world.
Many estimates are based on formulas in which the input quantities are known only to a limited precision. As you develop
physics problem-solving skills (which are applicable to a wide variety of fields), you also will develop skills at estimating.
You develop these skills by thinking more quantitatively and by being willing to take risks. As with any skill, experience
helps. Familiarity with dimensions (see Table 1.3) and units (see Table 1.1 and Table 1.2), and the scales of base
quantities (see Figure 1.9) also helps.
To make some progress in estimating, you need to have some definite ideas about how variables may be related. The
following strategies may help you in practicing the art of estimation:
• Get big lengths from smaller lengths. When estimating lengths, remember that anything can be a ruler. Thus,
imagine breaking a big thing into smaller things, estimate the length of one of the smaller things, and multiply to
get the length of the big thing. For example, to estimate the height of a building, first count how many floors it
has. Then, estimate how big a single floor is by imagining how many people would have to stand on each other’s
shoulders to reach the ceiling. Last, estimate the height of a person. The product of these three estimates is your
estimate of the height of the building. It helps to have memorized a few length scales relevant to the sorts of
problems you find yourself solving. For example, knowing some of the length scales in Figure 1.9 might come
in handy. Sometimes it also helps to do this in reverse—that is, to estimate the length of a small thing, imagine a
bunch of them making up a bigger thing. For example, to estimate the thickness of a sheet of paper, estimate the
thickness of a stack of paper and then divide by the number of pages in the stack. These same strategies of breaking
big things into smaller things or aggregating smaller things into a bigger thing can sometimes be used to estimate
other physical quantities, such as masses and times.
• Get areas and volumes from lengths. When dealing with an area or a volume of a complex object, introduce a simple
model of the object such as a sphere or a box. Then, estimate the linear dimensions (such as the radius of the sphere
or the length, width, and height of the box) first, and use your estimates to obtain the volume or area from standard
geometric formulas. If you happen to have an estimate of an object’s area or volume, you can also do the reverse;
that is, use standard geometric formulas to get an estimate of its linear dimensions.
• Get masses from volumes and densities. When estimating masses of objects, it can help first to estimate its volume
and then to estimate its mass from a rough estimate of its average density (recall, density has dimension mass over
length cubed, so mass is density times volume). For this, it helps to remember that the density of air is around 1 kg/
m3, the density of water is 103 kg/m3, and the densest everyday solids max out at around 104 kg/m3. Asking yourself
whether an object floats or sinks in either air or water gets you a ballpark estimate of its density. You can also do
this the other way around; if you have an estimate of an object’s mass and its density, you can use them to get an
estimate of its volume.
• If all else fails, bound it. For physical quantities for which you do not have a lot of intuition, sometimes the best
you can do is think something like: Well, it must be bigger than this and smaller than that. For example, suppose
30 Chapter 1 | Introducing Physics
you need to estimate the mass of a moose. Maybe you have a lot of experience with moose and know their average
mass offhand. If so, great. But for most people, the best they can do is to think something like: It must be bigger
than a person (of order 102 kg) and less than a car (of order 103 kg). If you need a single number for a subsequent
calculation, you can take the geometric mean of the upper and lower bound—that is, you multiply them together
and then take the square root. For the moose mass example, this would be
0.5
⎛ 2
⎝10 × 10 3⎞⎠ = 10 2.5 = 10 0.5 × 10 2 ≈ 3 × 10 2 kg.
The tighter the bounds, the better. Also, no rules are unbreakable when it comes to estimation. If you think the value
of the quantity is likely to be closer to the upper bound than the lower bound, then you may want to bump up your
estimate from the geometric mean by an order or two of magnitude.
• One “sig. fig.” is fine. There is no need to go beyond one significant figure when doing calculations to obtain an
estimate. In most cases, the order of magnitude is good enough. The goal is just to get in the ballpark figure, so keep
the arithmetic as simple as possible.
• Ask yourself: Does this make any sense? Last, check to see whether your answer is reasonable. How does it compare
with the values of other quantities with the same dimensions that you already know or can look up easily? If you
get some wacky answer (for example, if you estimate the mass of the Atlantic Ocean to be bigger than the mass of
Earth, or some time span to be longer than the age of the universe), first check to see whether your units are correct.
Then, check for arithmetic errors. Then, rethink the logic you used to arrive at your answer. If everything checks
out, you may have just proved that some slick new idea is actually bogus.
Example 1.6
Next, using our average depth estimate of D = 3 × 10 3 m, which was obtained by bounding, we estimate the
volume of Earth’s oceans to be
V = AD = (3 × 10 14 m 2)(3 × 10 3 m) = 9 × 10 17 m 3.
Last, we estimate the mass of the world’s oceans to be
M = ρV = (10 3 kg/m 3)(9 × 10 17 m 3) = 9 × 10 20 kg.
Thus, we estimate that the order of magnitude of the mass of the planet’s oceans is 1021 kg.
Significance
To verify our answer to the best of our ability, we first need to answer the question: Does this make any sense?
From Figure 1.9, we see the mass of Earth’s atmosphere is on the order of 1019 kg and the mass of Earth is on
the order of 1025 kg. It is reassuring that our estimate of 1021 kg for the mass of Earth’s oceans falls somewhere
between these two. So, yes, it does seem to make sense. It just so happens that we did a search on the Web for
“mass of oceans” and the top search results all said 1.4 × 10 21 kg, which is the same order of magnitude as our
estimate. Now, rather than having to trust blindly whoever first put that number up on a website (most of the other
sites probably just copied it from them, after all), we can have a little more confidence in it.
1.7 Check Your Understanding Figure 1.9 says the mass of the atmosphere is 1019 kg. Assuming the
density of the atmosphere is 1 kg/m3, estimate the height of Earth’s atmosphere. Do you think your answer is an
underestimate or an overestimate? Explain why.
How many piano tuners are there in New York City? How many leaves are on that tree? If you are studying photosynthesis
or thinking of writing a smartphone app for piano tuners, then the answers to these questions might be of great interest to
you. Otherwise, you probably couldn’t care less what the answers are. However, these are exactly the sorts of estimation
problems that people in various tech industries have been asking potential employees to evaluate their quantitative reasoning
skills. If building physical intuition and evaluating quantitative claims do not seem like sufficient reasons for you to practice
estimation problems, how about the fact that being good at them just might land you a high-paying job?
For practice estimating relative lengths, areas, and volumes, check out this PhET
(https://openstaxcollege.org/l/21lengthgame) simulation, titled “Estimation.”
Figure 1.16 shows two instruments used to measure the mass of an object. The digital scale has mostly replaced the
double-pan balance in physics labs because it gives more accurate and precise measurements. But what exactly do we
mean by accurate and precise? Aren’t they the same thing? In this section we examine in detail the process of making and
reporting a measurement.
32 Chapter 1 | Introducing Physics
Figure 1.16 (a) A double-pan mechanical balance is used to compare different masses. Usually an object with unknown mass
is placed in one pan and objects of known mass are placed in the other pan. When the bar that connects the two pans is
horizontal, then the masses in both pans are equal. The “known masses” are typically metal cylinders of standard mass such as 1
g, 10 g, and 100 g. (b) Many mechanical balances, such as double-pan balances, have been replaced by digital scales, which can
typically measure the mass of an object more precisely. A mechanical balance may read only the mass of an object to the nearest
tenth of a gram, but many digital scales can measure the mass of an object up to the nearest thousandth of a gram. (credit a:
modification of work by Serge Melki; credit b: modification of work by Karel Jakubec)
Figure 1.17 A GPS attempts to locate a restaurant at the center of the bull’s-eye. The
black dots represent each attempt to pinpoint the location of the restaurant. (a) The dots
are spread out quite far apart from one another, indicating low precision, but they are each
rather close to the actual location of the restaurant, indicating high accuracy. (b) The dots
are concentrated rather closely to one another, indicating high precision, but they are
rather far away from the actual location of the restaurant, indicating low accuracy. (credit
a and credit b: modification of works by Dark Evil)
Example 1.7
Solution
Substitute the values into the equation:
Percent Uncertainty
δA × 100% = 0.2 lb × 100% = 3.9% ≈ 4%.
A 5.1 lb
Significance
We can conclude the average weight of a bag of apples from this store is 5.1 lb ± 4%. Notice the percent
uncertainty is dimensionless because the units of weight in δA = 0.2 lb canceled those inn A = 5.1 lb when we
took the ratio.
1.8 Check Your Understanding A high school track coach has just purchased a new stopwatch. The
stopwatch manual states the stopwatch has an uncertainty of ±0.05 s. Runners on the track coach’s team
regularly clock 100-m sprints of 11.49 s to 15.01 s. At the school’s last track meet, the first-place sprinter came
in at 12.04 s and the second-place sprinter came in at 12.07 s. Will the coach’s new stopwatch be helpful in
timing the sprint team? Why or why not?
Ws ± δWs? We've already seen that the value of Ws is found by simply subtracting Wt - Wp. But what do we do with the
uncertainties? The simplest way might be to just add the uncertainties, using the logic that they both contribute to the
uncertainty in the final quantity. However more sophisticated analysis reveals that, if the uncertainties in the individual
measurements were independent of one another, we are likely overestimating the uncertainty in our final result if we simply
add them. A more accurate final answer is obtained by taking the square root of the sum of the squares of the individual
uncertainties. That is:
Adding Uncertainties in Quadrature
δW s = δW t2 + δW p2 (1.2)
So, δW s = 2 2 + 1 2 = 5 = 2.236
In our example, then, the uncertainty in the weight of the suitcase, is 2.236 lb. However, we must follow the rules for
significant figures (discussed in detail below). Since the individual weight measurements are expressed to the nearest lb, we
will round this calculated number to the nearest lb, and express the weight of the suitcase as 18 ± 2 lb.
Multiplication or Division of Quantities
Uncertainty exists in anything calculated from measured quantities. For example, the area of a floor calculated from
measurements of its length and width has an uncertainty because the length and width each have uncertainties. How big is
the uncertainty in something you calculate by multiplication or division? In this case, the measurements may not even have
the same dimensions. In this case, assuming that the uncertainties in the individual measurements are independent of one
another, the percent uncertainty in a quantity calculated by multiplication or division is the quadrature sum of the percent
uncertainties in the items used to make the calculation. Equivalently, the relative uncertainty in a quantity calculated by
multiplication or division is the quadrature sum of the relative uncertainties in the items used to make the calculation. In our
case, if A = H × W, then
Adding Relative Uncertainties in Quadrature
δA = ⎛δL ⎞ + ⎛δW ⎞
2 2 (1.3)
A ⎝L⎠ ⎝W ⎠
For example, if a floor has a length of L = 4.00 m and a width of W = 3.00 m, with uncertainties of 1% and 2%, respectively,
then the area of the floor is 12.0 m2 and has an uncertainty of 1 2 + 2 2 = 5 = 2.236 %. (Expressed as an area, this is
0.268 m2 [ 12.0 m 2 × 0.02236 ], which we round to 0.3 m2 since the area of the floor is given to a tenth of a square meter.)
So, we express the area of the floor as A = 12.0 ± 0.3 m2.
1300 in scientific notation as 1.3 × 10 3, 1.30 × 10 3, or 1.300 × 10 3, depending on whether it has two, three, or four
significant figures. Zeros are significant except when they serve only as placeholders.
Significant figures in calculations
When combining measurements with different degrees of precision, the number of significant digits in the final answer can
be no greater than the number of significant digits in the least-precise measured value. There are two different rules, one
for multiplication and division and the other for addition and subtraction.
1. For multiplication and division, the result should have the same number of significant figures as the quantity
with the least number of significant figures entering into the calculation. For example, the area of a circle can be
calculated from its radius using A = πr2. Let’s see how many significant figures the area has if the radius has only
two—say, r = 1.2 m. Using a calculator with an eight-digit output, we would calculate
A = πr 2 = (3.1415927…) × (1.2 m) 2 = 4.5238934 m 2.
But because the radius has only two significant figures, it limits the calculated quantity to two significant figures, or
A = 4.5 m 2,
7.56 kg
−6.052 kg
+13.7 kg
= 15.2 kg.
15.208 kg
Next, we identify the least-precise measurement: 13.7 kg. This measurement is expressed to the 0.1 decimal place,
so our final answer must also be expressed to the 0.1 decimal place. Thus, the answer is rounded to the tenths place,
giving us 15.2 kg.
Significant figures in this text
In this text, most numbers are assumed to have three significant figures. Furthermore, consistent numbers of significant
figures are used in all worked examples. An answer given to three digits is based on input good to at least three digits,
for example. If the input has fewer significant figures, the answer will also have fewer significant figures. Care is also
taken that the number of significant figures is reasonable for the situation posed. In some topics, particularly in optics, more
accurate numbers are needed and we use more than three significant figures. Finally, if a number is exact, such as the two
in the formula for the circumference of a circle, C = 2πr, it does not affect the number of significant figures in a calculation.
Likewise, conversion factors such as 100 cm/1 m are considered exact and do not affect the number of significant figures in
a calculation.
Figure 1.18 Problem-solving skills are essential to your success in physics. (credit:
“scui3asteveo”/Flickr)
Problem-solving skills are clearly essential to success in a quantitative course in physics. More important, the ability
to apply broad physical principles—usually represented by equations—to specific situations is a very powerful form of
knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be
applied to new situations whereas a list of facts cannot be made long enough to contain every possible circumstance. Such
analytical skills are useful both for solving problems in this text and for applying physics in everyday life.
As you are probably well aware, a certain amount of creativity and insight is required to solve problems. No rigid procedure
works every time. Creativity and insight grow with experience. With practice, the basics of problem solving become almost
automatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is to work as
many end-of-section problems as possible, starting with the easiest to build confidence and then progressing to the more
difficult. After you become involved in physics, you will see it all around you, and you can begin to apply it to situations
you encounter outside the classroom, just as is done in many of the applications in this text.
Although there is no simple step-by-step method that works for every problem, the following three-stage process facilitates
problem solving and makes it more meaningful. The three stages are strategy, solution, and significance. This process is
used in examples throughout the book. Here, we look at each stage of the process in turn.
Strategy
Strategy is the beginning stage of solving a problem. The idea is to figure out exactly what the problem is and then develop
a strategy for solving it. Some general advice for this stage is as follows:
• Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketch
at the outset. You often need to decide which direction is positive and note that on your sketch. When you have
identified the physical principles, it is much easier to find and apply the equations representing those principles.
Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of
nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical
solution is meaningless.
• Make a list of what is given or can be inferred from the problem as stated (identify the “knowns”). Many problems
are stated very succinctly and require some inspection to determine what is known. Drawing a sketch be very useful
at this point as well. Formally identifying the knowns is of particular importance in applying physics to real-world
situations. For example, the word stopped means the velocity is zero at that instant. Also, we can often take initial
time and position as zero by the appropriate choice of coordinate system.
• Identify exactly what needs to be determined in the problem (identify the unknowns). In complex problems,
especially, it is not always obvious what needs to be found or in what sequence. Making a list can help identify the
unknowns.
38 Chapter 1 | Introducing Physics
• Determine which physical principles can help you solve the problem. Since physical principles tend to be expressed
in the form of mathematical equations, a list of knowns and unknowns can help here. It is easiest if you can find
equations that contain only one unknown—that is, all the other variables are known—so you can solve for the
unknown easily. If the equation contains more than one unknown, then additional equations are needed to solve the
problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems
it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may
have to use two (or more) different equations to get the final answer.
Solution
The solution stage is when you do the math. Substitute the knowns (along with their units) into the appropriate equation
and obtain numerical solutions complete with units. That is, do the algebra, calculus, geometry, or arithmetic necessary to
find the unknown from the knowns, being sure to carry the units through the calculations. This step is clearly important
because it produces the numerical answer, along with its units. Notice, however, that this stage is only one-third of the
overall problem-solving process.
Significance
After having done the math in the solution stage of problem solving, it is tempting to think you are done. But, always
remember that physics is not math. Rather, in doing physics, we use mathematics as a tool to help us understand nature. So,
after you obtain a numerical answer, you should always assess its significance:
• Check your units. If the units of the answer are incorrect, then an error has been made and you should go back over
your previous steps to find it. One way to find the mistake is to check all the equations you derived for dimensional
consistency. However, be warned that correct units do not guarantee the numerical part of the answer is also correct.
• Check the answer to see whether it is reasonable. Does it make sense? This step is extremely important: –the goal
of physics is to describe nature accurately. To determine whether the answer is reasonable, check both its magnitude
and its sign, in addition to its units. The magnitude should be consistent with a rough estimate of what it should
be. It should also compare reasonably with magnitudes of other quantities of the same type. The sign usually tells
you about direction and should be consistent with your prior expectations. Your judgment will improve as you solve
more physics problems, and it will become possible for you to make finer judgments regarding whether nature is
described adequately by the answer to a problem. This step brings the problem back to its conceptual meaning. If
you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to
solve a problem mechanically.
• Check to see whether the answer tells you something interesting. What does it mean? This is the flip side of the
question: Does it make sense? Ultimately, physics is about understanding nature, and we solve physics problems to
learn a little something about how nature operates. Therefore, assuming the answer does make sense, you should
always take a moment to see if it tells you something about the world that you find interesting. Even if the answer to
this particular problem is not very interesting to you, what about the method you used to solve it? Could the method
be adapted to answer a question that you do find interesting? In many ways, it is in answering questions such as
these science that progresses.
CHAPTER 1 REVIEW
KEY TERMS
accuracy the degree to which a measured value agrees with an accepted reference value for that measurement
base quantity physical quantity chosen by convention and practical considerations such that all other physical quantities
can be expressed as algebraic combinations of them
base unit standard for expressing the measurement of a base quantity within a particular system of units; defined by a
particular procedure used to measure the corresponding base quantity
conversion factor a ratio that expresses how many of one unit are equal to another unit
derived quantity physical quantity defined using algebraic combinations of base quantities
derived units units that can be calculated using algebraic combinations of the fundamental units
dimension expression of the dependence of a physical quantity on the base quantities as a product of powers of symbols
representing the base quantities; in general, the dimension of a quantity has the form L a M b T c I d Θ e N f J g for some
powers a, b, c, d, e, f, and g.
dimensionally consistent equation in which every term has the same dimensions and the arguments of any
mathematical functions appearing in the equation are dimensionless
dimensionless quantity with a dimension of L 0 M 0 T 0 I 0 Θ 0 N 0 J 0 = 1; also called quantity of dimension 1 or a pure
number
discrepancy the difference between the measured value and a given standard or expected value
English units system of measurement used in the United States; includes units of measure such as feet, gallons, and
pounds
estimation using prior experience and sound physical reasoning to arrive at a rough idea of a quantity’s value; sometimes
called an “order-of-magnitude approximation,” a “guesstimate,” a “back-of-the-envelope calculation”, or a “Fermi
calculation”
kilogram SI unit for mass, abbreviated kg
law description, using concise language or a mathematical formula, of a generalized pattern in nature supported by
scientific evidence and repeated experiments
meter SI unit for length, abbreviated m
method of adding percents the percent uncertainty in a quantity calculated by multiplication or division is the sum of
the percent uncertainties in the items used to make the calculation.
metric system system in which values can be calculated in factors of 10
model representation of something often too difficult (or impossible) to display directly
order of magnitude the size of a quantity as it relates to a power of 10
percent uncertainty the ratio of the uncertainty of a measurement to the measured value, expressed as a percentage
physical quantity characteristic or property of an object that can be measured or calculated from other measurements
physics science concerned with describing the interactions of energy, matter, space, and time; especially interested in
what fundamental mechanisms underlie every phenomenon
precision the degree to which repeated measurements agree with each other
second the SI unit for time, abbreviated s
SI units the international system of units that scientists in most countries have agreed to use; includes units such as
meters, liters, and grams
significant figures used to express the precision of a measuring tool used to measure a value
40 Chapter 1 | Introducing Physics
theory testable explanation for patterns in nature supported by scientific evidence and verified multiple times by various
groups of researchers
uncertainty a quantitative measure of how much measured values deviate from one another
units standards used for expressing and comparing measurements
KEY EQUATIONS
Percent uncertainty Percent uncertainty = δA × 100%
A
2 ⎛δy ⎞ 2
⎛δx ⎞
Adding relative uncertainties in quadrature If z = xy or z = x/y then δz
z = ⎝ x +⎝y ⎠
⎠
SUMMARY
1.2 The Scope and Scale of Physics
• Physics is about trying to find the simple laws that describe all natural phenomena.
• Physics operates on a vast range of scales of length, mass, and time. Scientists use the concept of the order of
magnitude of a number to track which phenomena occur on which scales. They also use orders of magnitude to
compare the various scales.
• Scientists attempt to describe the world by formulating models, theories, and laws.
The three stages of the process for solving physics problems used in this book are as follows:
• Strategy: Determine which physical principles are involved and develop a strategy for using them to solve the
problem.
• Solution: Do the math necessary to obtain a numerical solution complete with units.
• Significance: Check the solution to make sure it makes sense (correct units, reasonable magnitude and sign) and
assess its significance.
CONCEPTUAL QUESTIONS
6. Can the validity of a model be limited or must it be
1.2 The Scope and Scale of Physics universally valid? How does this compare with the required
1. What is physics? validity of a theory or a law?
3. If two different theories describe experimental 8. What are the SI base units of length, mass, and time?
observations equally well, can one be said to be more valid
than the other (assuming both use accepted rules of logic)? 9. What is the difference between a base unit and a derived
unit? (b) What is the difference between a base quantity and
4. What determines the validity of a theory? a derived quantity? (c) What is the difference between a
base quantity and a base unit?
5. Certain criteria must be satisfied if a measurement or
observation is to be believed. Will the criteria necessarily 10. For each of the following scenarios, refer to Figure
be as strict for an expected result as for an unexpected 1.9 and Table 1.2 to determine which metric prefix on
result? the meter is most appropriate for each of the following
scenarios. (a) You want to tabulate the mean distance from
42 Chapter 1 | Introducing Physics
the Sun for each planet in the solar system. (b) You want relationship between the accuracy and the discrepancy of a
to compare the sizes of some common viruses to design a measurement?
mechanical filter capable of blocking the pathogenic ones.
(c) You want to list the diameters of all the elements on the
periodic table. (d) You want to list the distances to all the 1.8 Solving Problems in Physics
stars that have now received any radio broadcasts sent from
12. What information do you need to choose which
Earth 10 years ago.
equation or equations to use to solve a problem?
1.7 Significant Figures 13. What should you do after obtaining a numerical
answer when solving a problem?
11. (a) What is the relationship between the precision
and the uncertainty of a measurement? (b) What is the
PROBLEMS
18. Roughly how many times longer than the mean life of
1.2 The Scope and Scale of Physics an extremely unstable atomic nucleus is the lifetime of a
14. Find the order of magnitude of the following physical human?
quantities. (a) The mass of Earth’s atmosphere:
5.1 × 10 18 kg; (b) The mass of the Moon’s atmosphere: 19. Calculate the approximate number of atoms in a
bacterium. Assume the average mass of an atom in the
25,000 kg; (c) The mass of Earth’s hydrosphere: bacterium is 10 times the mass of a proton.
1.4 × 10 21 kg; (d) The mass of Earth: 5.97 × 10 24 kg;
(e) The mass of the Moon: 7.34 × 10 22 kg; (f) The 20. (a) Calculate the number of cells in a hummingbird
assuming the mass of an average cell is 10 times the mass
Earth–Moon distance (semimajor axis): 3.84 × 10 8 m; (g) of a bacterium. (b) Making the same assumption, how
The mean Earth–Sun distance: 1.5 × 10 11 m; (h) The many cells are there in a human?
equatorial radius of Earth: 6.38 × 10 6 m; (i) The mass of 21. Assuming one nerve impulse must end before another
−31 can begin, what is the maximum firing rate of a nerve in
an electron: 9.11 × 10 kg; (j) The mass of a proton:
impulses per second?
−27
1.67 × 10 kg; (k) The mass of the Sun:
1.99 × 10 30 kg. 22. About how many floating-point operations can a
supercomputer perform each year?
on the base SI unit of length: the meter. Rewrite them in 35. Soccer fields vary in size. A large soccer field is 115
scientific notation without the prefix. For example, 4.2 Pm m long and 85.0 m wide. What is its area in square feet?
would be rewritten as 4.2 × 10 15 m. (a) 89 Tm; (b) 89 pm; (Assume that 1 m = 3.281 ft.)
(c) 711 mm; (d) 0.45 μm.
36. What is the height in meters of a person who is 6 ft 1.0
in. tall?
27. The following lengths are given in meters. Use metric
prefixes to rewrite them so the numerical value is bigger
37. Mount Everest, at 29,028 ft, is the tallest mountain on
than one but less than 1000. For example, 7.9 × 10 −2 m Earth. What is its height in kilometers? (Assume that 1 m =
could be written either as 7.9 cm or 79 mm. (a) 3.281 ft.)
7.59 × 10 7 m; (b) 0.0074 m; (c) 8.8 × 10 −11 m; (d)
38. The speed of sound is measured to be 342 m/s on
1.63 × 10 13 m.
a certain day. What is this measurement in kilometers per
hour?
28. The following masses are written using metric
prefixes on the gram. Rewrite them in scientific notation 39. Tectonic plates are large segments of Earth’s crust that
in terms of the SI base unit of mass: the kilogram. For move slowly. Suppose one such plate has an average speed
example, 40 Mg would be written as 4 × 10 4 kg. (a) 23 of 4.0 cm/yr. (a) What distance does it move in 1.0 s at
mg; (b) 320 Tg; (c) 42 ng; (d) 7 g; (e) 9 Pg. this speed? (b) What is its speed in kilometers per million
years?
29. The following masses are given in kilograms. Use
metric prefixes on the gram to rewrite them so the 40. The average distance between Earth and the Sun is
numerical value is bigger than one but less than 1000. 1.5 × 10 11 m. (a) Calculate the average speed of Earth in
For example, 7 × 10 −4 kg could be written as 70 cg or its orbit (assumed to be circular) in meters per second. (b)
What is this speed in miles per hour?
700 mg. (a) 3.8 × 10 −5 kg; (b) 2.3 × 10 17 kg; (c)
2.4 × 10 −11 kg; (d) 8 × 10 15 kg; (e) 4.2 × 10 −3 kg. 41. The density of nuclear matter is about 1018 kg/m3.
Given that 1 mL is equal in volume to cm3, what is the
density of nuclear matter in megagrams per microliter (that
is, Mg/μL )?
1.4 Unit Conversion
30. The volume of Earth is on the order of 1021 m3.
(a) What is this in cubic kilometers (km3)? (b) What is it 42. The density of aluminum is 2.7 g/cm3. What is the
in cubic miles (mi3)? (c) What is it in cubic centimeters density in kilograms per cubic meter?
(cm3)?
43. A commonly used unit of mass in the English system
31. The speed limit on some interstate highways is is the pound-mass, abbreviated lbm, where 1 lbm = 0.454
roughly 100 km/h. (a) What is this in meters per second? kg. What is the density of water in pound-mass per cubic
(b) How many miles per hour is this? foot?
32. A car is traveling at a speed of 33 m/s. (a) What is its 44. A furlong is 220 yd. A fortnight is 2 weeks. Convert
speed in kilometers per hour? (b) Is it exceeding the 90 km/ a speed of one furlong per fortnight to millimeters per
h speed limit? second.
33. In SI units, speeds are measured in meters per second 45. It takes 2π radians (rad) to get around a circle, which
(m/s). But, depending on where you live, you’re probably is the same as 360°. How many radians are in 1°?
more comfortable of thinking of speeds in terms of either
kilometers per hour (km/h) or miles per hour (mi/h). In 46. Light travels a distance of about 3 × 10 8 m/s. A
this problem, you will see that 1 m/s is roughly 4 km/h
light-minute is the distance light travels in 1 min. If the Sun
or 2 mi/h, which is handy to use when developing your
physical intuition. More precisely, show that (a) is 1.5 × 10 11 m from Earth, how far away is it in light-
1.0 m/s = 3.6 km/h and (b) 1.0 m/s = 2.2 mi/h. minutes?
34. American football is played on a 100-yd-long field, 47. A light-nanosecond is the distance light travels in 1 ns.
excluding the end zones. How long is the field in meters? Convert 1 ft to light-nanoseconds.
(Assume that 1 m = 3.281 ft.)
44 Chapter 1 | Introducing Physics
49. A fluid ounce is about 30 mL. What is the volume of a 57. Assuming the human body is primarily made of water,
12 fl-oz can of soda pop in cubic meters? estimate the number of molecules in it. (Note that water has
a molecular mass of 18 g/mol and there are roughly 1024
atoms in a mole.)
1.5 Dimensional Analysis
50. A student is trying to remember some formulas from 58. Estimate the mass of air in a classroom.
geometry. In what follows, assume A is area, V is
volume, and all other variables are lengths. Determine 59. Estimate the number of molecules that make up Earth,
which formulas are dimensionally consistent. (a) assuming an average molecular mass of 30 g/mol. (Note
V = πr 2 h; (b) A = 2πr 2 + 2πrh; (c) V = 0.5bh; (d) there are on the order of 1024 objects per mole.)
51. Consider the physical quantities s, v, a, and t with 61. Roughly how many solar systems would it take to tile
dimensions [s] = L, [v] = LT −1, [a] = LT −2, and the disk of the Milky Way?
[t] = T. Determine whether each of the following
62. (a) Estimate the density of the Moon. (b) Estimate the
equations is dimensionally consistent. (a) v 2 = 2as; (b) diameter of the Moon. (c) Given that the Moon subtends
s = vt 2 + 0.5at 2; (c) v = s/t; (d) a = v/t. at an angle of about half a degree in the sky, estimate its
distance from Earth.
52. Consider the physical quantities m, s, v, a, 63. The average density of the Sun is on the order 103 kg/
and t with dimensions [m] = M, [s] = L, [v] = LT–1, m3. (a) Estimate the diameter of the Sun. (b) Given that the
[a] = LT–2, and [t] = T. Assuming each of the following Sun subtends at an angle of about half a degree in the sky,
equations is dimensionally consistent, find the dimension estimate its distance from Earth.
of the quantity on the left-hand side of the equation: (a) F =
ma; (b) K = 0.5mv2; (c) p = mv; (d) W = mas; (e) L = mvr. 64. Estimate the mass of a virus.
53. Suppose quantity s is a length and quantity t is a 65. A floating-point operation is a single arithmetic
time. Suppose the quantities v and a are defined by v operation such as addition, subtraction, multiplication, or
= ds/dt and a = dv/dt. (a) What is the dimension of v? division. (a) Estimate the maximum number of floating-
(b) What is the dimension of the quantity a? What are the point operations a human being could possibly perform in
a lifetime. (b) How long would it take a supercomputer to
dimensions of (c) ∫ vdt, (d) ∫ adt, and (e) da/dt?
perform that many floating-point operations?
55. The arc length formula says the length s of arc 67. Suppose your bathroom scale reads your mass as 65
subtended by angle Ɵ in a circle of radius r is given by kg with a 3% uncertainty. What is the uncertainty in your
the equation s = r Ɵ . What are the dimensions of (a) s, mass (in kilograms)?
(b) r, and (c) Ɵ?
68. A good-quality measuring tape can be off by 0.50 cm
over a distance of 20 m. What is its percent uncertainty?
min. What is the percent uncertainty in this measurement? Assuming the same percent uncertainty, what is the
uncertainty in a blood pressure measurement of 80 mm
70. (a) Suppose that a person has an average heart rate of Hg?
72.0 beats/min. How many beats does he or she have in 2.0
years? (b) In 2.00 years? (c) In 2.000 years? 76. A person measures his or her heart rate by counting the
number of beats in 30 s. If 40 ± 1 beats are counted in 30.0
71. A can contains 375 mL of soda. How much is left after ± 0.5 s, what is the heart rate and its uncertainty in beats per
308 mL is removed? minute?
72. State how many significant figures are proper in the 77. What is the area of a circle 3.102 cm in diameter?
results of the following calculations: (a)
(106.7)(98.2) / (46.210)(1.01); (b) (18.7) 2; (c) 78. Determine the number of significant figures in the
⎛
following measurements: (a) 0.0009, (b) 15,450.0, (c)
⎝1.60 × 10 −19⎞⎠(3712) 6×103, (d) 87.990, and (e) 30.42.
73. (a) How many significant figures are in the numbers 79. Perform the following calculations and express your
99 and 100.? (b) If the uncertainty in each number is 1, answer using the correct number of significant digits. (a)
what is the percent uncertainty in each? (c) Which is a A woman has two bags weighing 13.5 lb and one bag
more meaningful way to express the accuracy of these two with a weight of 10.2 lb. What is the total weight of the
numbers: significant figures or percent uncertainties? bags? (b) The force F on an object is equal to its mass m
multiplied by its acceleration a. If a wagon with mass 55
kg accelerates at a rate of 0.0255 m/s2, what is the force on
74. (a) If your speedometer has an uncertainty of 2.0 km/h the wagon? (The unit of force is called the newton and it is
at a speed of 90 km/h, what is the percent uncertainty? (b) expressed with the symbol N.)
If it has the same percent uncertainty when it reads 60 km/
h, what is the range of speeds you could be going?
ADDITIONAL PROBLEMS
80. Consider the equation y = mt +b, where the dimension is the uncertainty in the average speed?
of y is length and the dimension of t is time, and m and b
are constants. What are the dimensions and SI units of (a) 84. The sides of a small rectangular box are measured to
m and (b) b? be 1.80 ± 0.1 cm, 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm long.
Calculate its volume and uncertainty in cubic centimeters.
81. Consider the equation
s = s 0 + v 0 t + a 0 t /2 + j 0 t /6 + S 0 t /24 + ct 5 /120,
2 3 4
85. When nonmetric units were used in the United
Kingdom, a unit of mass called the pound-mass (lbm) was
where s is a length and t is a time. What are the dimensions
used, where 1 lbm = 0.4539 kg. (a) If there is an uncertainty
and SI units of (a) s 0, (b) v 0, (c) a 0, (d) j 0, (e) S 0,
of 0.0001 kg in the pound-mass unit, what is its percent
and (f) c? uncertainty? (b) Based on that percent uncertainty, what
mass in pound-mass has an uncertainty of 1 kg when
82. (a) A car speedometer has a 5% uncertainty. What is converted to kilograms?
the range of possible speeds when it reads 90 km/h? (b)
Convert this range to miles per hour. Note 1 km = 0.6214 86. The length and width of a rectangular room are
mi. measured to be 3.955 ± 0.005 m and 3.050 ± 0.005 m.
Calculate the area of the room and its uncertainty in square
83. A marathon runner completes a 42.188-km course in 2 meters.
h, 30 min, and 12 s. There is an uncertainty of 25 m in the
distance traveled and an uncertainty of 1 s in the elapsed 87. A car engine moves a piston with a circular cross-
time. (a) Calculate the percent uncertainty in the distance. section of 7.500 ± 0.002 cm in diameter a distance of
(b) Calculate the percent uncertainty in the elapsed time. (c) 3.250 ± 0.001 cm to compress the gas in the cylinder. (a)
What is the average speed in meters per second? (d) What By what amount is the gas decreased in volume in cubic
46 Chapter 1 | Introducing Physics
CHALLENGE PROBLEMS
88. The first atomic bomb was detonated on July 16, 1945, of TNT” (abbreviated “t TNT”), where 1 t TNT is about
at the Trinity test site about 200 mi south of Los Alamos. 4.2 GJ. Convert your answer to (b) into kilotons of TNT
In 1947, the U.S. government declassified a film reel of the (that is, kt TNT). Compare your answer with the quick-
explosion. From this film reel, British physicist G. I. Taylor and-dirty estimate of 10 kt TNT made by physicist Enrico
was able to determine the rate at which the radius of the Fermi shortly after witnessing the explosion from what was
fireball from the blast grew. Using dimensional analysis, he thought to be a safe distance. (Reportedly, Fermi made his
was then able to deduce the amount of energy released in estimate by dropping some shredded bits of paper right
the explosion, which was a closely guarded secret at the before the remnants of the shock wave hit him and looked
time. Because of this, Taylor did not publish his results until to see how far they were carried by it.)
1950. This problem challenges you to recreate this famous
calculation. (a) Using keen physical insight developed from 89. The purpose of this problem is to show the entire
years of experience, Taylor decided the radius r of the concept of dimensional consistency can be summarized
fireball should depend only on time since the explosion, by the old saying “You can’t add apples and oranges.” If
t, the density of the air, ρ, and the energy of the initial you have studied power series expansions in a calculus
explosion, E. Thus, he made the educated guess that course, you know the standard mathematical functions such
r = kE a ρ b t c for some dimensionless constant k and some as trigonometric functions, logarithms, and exponential
functions can be expressed as infinite sums of the form
unknown exponents a, b, and c. Given that [E] = ML2T–2, ∞
determine the values of the exponents necessary to make ∑ an xn = a0 + a1 x + a2 x2 + a3 x3 + ⋯ , where
this equation dimensionally consistent. (Hint: Notice the n=0
equation implies that k = rE −a ρ −b t −c and that [k] = 1. the an are dimensionless constants for all
) (b) By analyzing data from high-energy conventional n = 0, 1, 2, ⋯ and x is the argument of the function.
explosives, Taylor found the formula he derived seemed to (If you have not studied power series in calculus yet, just
be valid as long as the constant k had the value 1.03. From trust us.) Use this fact to explain why the requirement
the film reel, he was able to determine many values of r that all terms in an equation have the same dimensions is
and the corresponding values of t. For example, he found sufficient as a definition of dimensional consistency. That
that after 25.0 ms, the fireball had a radius of 130.0 m. is, it actually implies the arguments of standard
Use these values, along with an average air density of 1.25 mathematical functions must be dimensionless, so it is not
kg/m3, to calculate the initial energy release of the Trinity really necessary to make this latter condition a separate
detonation in joules (J). (Hint: To get energy in joules, requirement of the definition of dimensional consistency as
you need to make sure all the numbers you substitute in we have done in this section.
are expressed in terms of SI base units.) (c) The energy
released in large explosions is often cited in units of “tons
Introduction
In astronomy we deal with distances on a scale you may never have thought about before, with numbers larger than any
you may have encountered. We adopt two approaches that make dealing with astronomical numbers a little bit easier. First,
we use a system for writing large and small numbers called scientific notation (or sometimes powers-of-ten notation). This
system is very appealing because it eliminates the many zeros that can seem overwhelming to the reader. In scientific
notation, if you want to write a number such as 500,000,000, you express it as 5 × 108. The small raised number after
the 10, called an exponent, keeps track of the number of places we had to move the decimal point to the left to convert
500,000,000 to 5. If you are encountering this system for the first time or would like a refresher, we suggest you look back
at Section 1.3 for more information. The second way we try to keep numbers simple is to use a consistent set of units—the
metric International System of Units, or SI (from the French Système International d’Unités). The SI system is summarized
in Appendix A.
Watch this brief PBS animation (https://openstax.org/l/30scinotation) that explains how scientific notation
works and why it’s useful.
A common unit astronomers use to describe distances in the universe is a light-year, which is the distance light travels
during one year. Because light always travels at the same speed, and because its speed turns out to be the fastest possible
speed in the universe, it makes a good standard for keeping track of distances. You might be confused because a “light-year”
seems to imply that we are measuring time, but this mix-up of time and distance is common in everyday life as well. For
example, when your friend asks where the movie theater is located, you might say “about 20 minutes from downtown.”
So, how many kilometers are there in a light-year? Light travels at the amazing pace of 3 × 105 kilometers per second (km/
s), which makes a light-year 9.46 × 1012 kilometers. You might think that such a large unit would reach the nearest star
easily, but the stars are far more remote than our imaginations might lead us to believe. Even the nearest star is 4.3 light-
years away—more than 40 trillion kilometers. Other stars visible to the unaided eye are hundreds to thousands of light-years
away (Figure 2.1).
48 Chapter 2 | The Universe at Its Limits
Example 2.1.
Scientific Notation
In 2015, the richest human being on our planet had a net worth of $79.2 billion. Some might say this is an
astronomical sum of money. Express this amount in scientific notation.
Solution
$79.2 billion can be written $79,200,000,000. Expressed in scientific notation it becomes $7.92 × 1010.
Example 2.2.
And, to answer the original question precisely, we will convert that distance into units of km. The final unit
conversion gives us:
⎛ ⎞
9.46×10 15 m × ⎝ 1 km ⎠ = 9.46×10 12 km
1000 m
That’s almost 10,000,000,000,000 km that light covers in a year. To help you imagine how long this distance is,
we’ll mention that a string 1 light-year long could fit around the circumference of Earth 236 million times.
There is another reason the speed of light is such a natural unit of distance for astronomers. Information about the universe
comes to us almost exclusively through various forms of light, and all such light travels at the speed of light—that is, 1
light-year every year. This sets a limit on how quickly we can learn about events in the universe. If a star is 100 light-years
away, the light we see from it tonight left that star 100 years ago and is just now arriving in our neighborhood. The soonest
we can learn about any changes in that star is 100 years after the fact. For a star 500 light-years away, the light we detect
tonight left 500 years ago and is carrying 500-year-old news.
Because many of us are accustomed to instant news from the Internet, some might find this frustrating.
“You mean, when I see that star up there,” you ask, “I won’t know what’s actually happening there for another 500 years?”
But this isn’t the most helpful way to think about the situation. For astronomers, now is when the light reaches us here on
Earth. There is no way for us to know anything about that star (or other object) until its light reaches us.
But what at first may seem a great frustration is actually a tremendous benefit in disguise. If astronomers really want to
piece together what has happened in the universe since its beginning, they must find evidence about each epoch (or period
of time) of the past. Where can we find evidence today about cosmic events that occurred billions of years ago?
The delay in the arrival of light provides an answer to this question. The farther out in space we look, the longer the light has
taken to get here, and the longer ago it left its place of origin. By looking billions of light-years out into space, astronomers
are actually seeing billions of years into the past. In this way, we can reconstruct the history of the cosmos and get a sense
of how it has evolved over time.
This is one reason why astronomers strive to build telescopes that can collect more and more of the faint light in the
universe. The more light we collect, the fainter the objects we can observe. On average, fainter objects are farther away
and can, therefore, tell us about periods of time even deeper in the past. Instruments such as the Hubble Space Telescope
(Figure 2.2) and the Very Large Telescope in Chile (which you will learn about in the chapter on Astronomical
Instruments (https://legacy.cnx.org/content/m59802/latest/) ), are giving astronomers views of deep space and
deep time better than any we have had before.
50 Chapter 2 | The Universe at Its Limits
We can now take a brief introductory tour of the universe as astronomers understand it today to get acquainted with the
types of objects and distances you will encounter throughout the text. We begin at home with Earth, a nearly spherical planet
about 13,000 kilometers in diameter (Figure 2.3). A space traveler entering our planetary system would easily distinguish
Earth from the other planets in our solar system by the large amount of liquid water that covers some two thirds of its crust.
If the traveler had equipment to receive radio or television signals, or came close enough to see the lights of our cities at
night, she would soon find signs that this watery planet has sentient life.
Figure 2.3 This image shows the Western hemisphere as viewed from space 35,400 kilometers (about 22,000 miles) above
Earth. Data about the land surface from one satellite was combined with another satellite’s data about the clouds to create the
image. (credit: modification of work by R. Stockli, A. Nelson, F. Hasler, NASA/ GSFC/ NOAA/ USGS)
Our nearest astronomical neighbor is Earth’s satellite, commonly called the Moon. Figure 2.4 shows Earth and the Moon
drawn to scale on the same diagram. Notice how small we have to make these bodies to fit them on the page with the right
scale. The Moon’s distance from Earth is about 30 times Earth’s diameter, or approximately 384,000 kilometers, and it takes
about a month for the Moon to revolve around Earth. The Moon’s diameter is 3476 kilometers, about one fourth the size of
Earth.
Figure 2.4 This image shows Earth and the Moon shown to scale for both size and distance. (credit: modification of work by
NASA)
Light (or radio waves) takes 1.3 seconds to travel between Earth and the Moon. If you’ve seen videos of the Apollo flights
to the Moon, you may recall that there was a delay of about 3 seconds between the time Mission Control asked a question
and the time the astronauts responded. This was not because the astronomers were thinking slowly, but rather because it
took the radio waves almost 3 seconds to make the round trip.
Earth revolves around our star, the Sun, which is about 150 million kilometers away—approximately 400 times as far away
from us as the Moon. We call the average Earth–Sun distance an astronomical unit (AU) because, in the early days of
astronomy, it was the most important measuring standard. Light takes slightly more than 8 minutes to travel 1 astronomical
unit, which means the latest news we receive from the Sun is always 8 minutes old. The diameter of the Sun is about 1.5
million kilometers; Earth could fit comfortably inside one of the minor eruptions that occurs on the surface of our star. If
the Sun were reduced to the size of a basketball, Earth would be a small apple seed about 30 meters from the ball.
It takes Earth 1 year (3 × 107 seconds) to go around the Sun at our distance; to make it around, we must travel at
approximately 110,000 kilometers per hour. (If you, like many students, still prefer miles to kilometers, you might find the
following trick helpful. To convert kilometers to miles, just multiply kilometers by 0.6. Thus, 110,000 kilometers per hour
becomes 66,000 miles per hour.) Because gravity holds us firmly to Earth and there is no resistance to Earth’s motion in the
vacuum of space, we participate in this extremely fast-moving trip without being aware of it day to day.
Earth is only one of eight planets that revolve around the Sun. These planets, along with their moons and swarms of smaller
bodies such as dwarf planets, make up the solar system (Figure 2.5). A planet is defined as a body of significant size that
orbits a star and does not produce its own light. (If a large body consistently produces its own light, it is then called a star.)
Later in the book this definition will be modified a bit, but it is perfectly fine for now as you begin your voyage.
52 Chapter 2 | The Universe at Its Limits
Figure 2.5 The Sun, the planets, and some dwarf planets are shown with their sizes drawn to scale. The orbits of the planets are
much more widely separated than shown in this drawing. Notice the size of Earth compared to the giant planets. (credit:
modification of work by NASA)
We are able to see the nearby planets in our skies only because they reflect the light of our local star, the Sun. If the planets
were much farther away, the tiny amount of light they reflect would usually not be visible to us. The planets we have so
far discovered orbiting other stars were found from the pull their gravity exerts on their parent stars, or from the light they
block from their stars when they pass in front of them. We can’t see most of these planets directly, although a few are now
being imaged directly.
The Sun is our local star, and all the other stars are also enormous balls of glowing gas that generate vast amounts of energy
by nuclear reactions deep within. We will discuss the processes that cause stars to shine in more detail later in the book. The
other stars look faint only because they are so very far away. If we continue our basketball analogy, Proxima Centauri, the
nearest star beyond the Sun, which is 4.3 light-years away, would be almost 7000 kilometers from the basketball.
When you look up at a star-filled sky on a clear night, all the stars visible to the unaided eye are part of a single collection of
stars we call the Milky Way Galaxy, or simply the Galaxy. (When referring to the Milky Way, we capitalize Galaxy; when
talking about other galaxies of stars, we use lowercase galaxy.) The Sun is one of hundreds of billions of stars that make
up the Galaxy; its extent, as we will see, staggers the human imagination. Within a sphere 10 light-years in radius centered
on the Sun, we find roughly ten stars. Within a sphere 100 light-years in radius, there are roughly 10,000 (104) stars—far
too many to count or name—but we have still traversed only a tiny part of the Milky Way Galaxy. Within a 1000-light-year
sphere, we find some ten million (107) stars; within a sphere of 100,000 light-years, we finally encompass the entire Milky
Way Galaxy.
Our Galaxy looks like a giant disk with a small ball in the middle. If we could move outside our Galaxy and look down
on the disk of the Milky Way from above, it would probably resemble the galaxy in Figure 2.6, with its spiral structure
outlined by the blue light of hot adolescent stars.
The Sun is somewhat less than 30,000 light-years from the center of the Galaxy, in a location with nothing much to
distinguish it. From our position inside the Milky Way Galaxy, we cannot see through to its far rim (at least not with
ordinary light) because the space between the stars is not completely empty. It contains a sparse distribution of gas (mostly
the simplest element, hydrogen) intermixed with tiny solid particles that we call interstellar dust. This gas and dust collect
into enormous clouds in many places in the Galaxy, becoming the raw material for future generations of stars. Figure 2.7
shows an image of the disk of the Galaxy as seen from our vantage point.
54 Chapter 2 | The Universe at Its Limits
Figure 2.7 Because we are inside the Milky Way Galaxy, we see its disk in cross-section flung
across the sky like a great milky white avenue of stars with dark “rifts” of dust. In this dramatic
image, part of it is seen above Trona Pinnacles in the California desert. (credit: Ian Norman)
Typically, the interstellar material is so extremely sparse that the space between stars is a much better vacuum than anything
we can produce in terrestrial laboratories. Yet, the dust in space, building up over thousands of light-years, can block the
light of more distant stars. Like the distant buildings that disappear from our view on a smoggy day in Los Angeles, the
more distant regions of the Milky Way cannot be seen behind the layers of interstellar smog. Luckily, astronomers have
found that stars and raw material shine with various forms of light, some of which do penetrate the smog, and so we have
been able to develop a pretty good map of the Galaxy.
Recent observations, however, have also revealed a rather surprising and disturbing fact. There appears to be more—much
more—to the Galaxy than meets the eye (or the telescope). From various investigations, we have evidence that much of our
Galaxy is made of material we cannot currently observe directly with our instruments. We therefore call this component of
the Galaxy dark matter. We know the dark matter is there by the pull its gravity exerts on the stars and raw material we can
observe, but what this dark matter is made of and how much of it exists remain a mystery. Furthermore, this dark matter is
not confined to our Galaxy; it appears to be an important part of other star groupings as well.
By the way, not all stars live by themselves, as the Sun does. Many are born in double or triple systems with two, three,
or more stars revolving about each other. Because the stars influence each other in such close systems, multiple stars
allow us to measure characteristics that we cannot discern from observing single stars. In a number of places, enough stars
have formed together that we recognized them as star clusters (Figure 2.8). Some of the largest of the star clusters that
astronomers have cataloged contain hundreds of thousands of stars and take up volumes of space hundreds of light-years
across.
You may hear stars referred to as “eternal,” but in fact no star can last forever. Since the “business” of stars is making
energy, and energy production requires some sort of fuel to be used up, eventually all stars run out of fuel. This news should
not cause you to panic, though, because our Sun still has at least 5 or 6 billion years to go. Ultimately, the Sun and all stars
will die, and it is in their death throes that some of the most intriguing and important processes of the universe are revealed.
For example, we now know that many of the atoms in our bodies were once inside stars. These stars exploded at the ends of
their lives, recycling their material back into the reservoir of the Galaxy. In this sense, all of us are literally made of recycled
“star dust.”
In a very rough sense, you could think of the solar system as your house or apartment and the Galaxy as your town, made up
of many houses and buildings. In the twentieth century, astronomers were able to show that, just as our world is made up of
many, many towns, so the universe is made up of enormous numbers of galaxies. (We define the universe to be everything
that exists that is accessible to our observations.) Galaxies stretch as far into space as our telescopes can see, many billions
of them within the reach of modern instruments. When they were first discovered, some astronomers called galaxies island
universes, and the term is aptly descriptive; galaxies do look like islands of stars in the vast, dark seas of intergalactic space.
The nearest galaxy, discovered in 1993, is a small one that lies 75,000 light-years from the Sun in the direction of the
constellation Sagittarius, where the smog in our own Galaxy makes it especially difficult to discern. (A constellation, we
should note, is one of the 88 sections into which astronomers divide the sky, each named after a prominent star pattern
within it.) Beyond this Sagittarius dwarf galaxy lie two other small galaxies, about 160,000 light-years away. First recorded
by Magellan’s crew as he sailed around the world, these are called the Magellanic Clouds (Figure 2.9). All three of these
small galaxies are satellites of the Milky Way Galaxy, interacting with it through the force of gravity. Ultimately, all three
may even be swallowed by our much larger Galaxy, as other small galaxies have been over the course of cosmic time.
56 Chapter 2 | The Universe at Its Limits
Figure 2.9 This image shows both the Large Magellanic Cloud and the Small Magellanic Cloud above the telescopes of the
Atacama Large Millimeter/Submillimeter Array (ALMA) in the Atacama Desert of northern Chile. (credit: ESO, C. Malin)
The nearest large galaxy is a spiral quite similar to our own, located in the constellation of Andromeda, and is thus called
the Andromeda galaxy; it is also known by one of its catalog numbers, M31 (Figure 2.10). M31 is a little more than 2
million light-years away and, along with the Milky Way, is part of a small cluster of more than 50 galaxies referred to as the
Local Group.
Figure 2.10 The Andromeda galaxy (M31) is a spiral-shaped collection of stars similar to our own Milky Way. (credit: Adam
Evans)
At distances of 10 to 15 million light-years, we find other small galaxy groups, and then at about 50 million light-years
there are more impressive systems with thousands of member galaxies. We have discovered that galaxies occur mostly in
clusters, both large and small (Figure 2.11).
58 Chapter 2 | The Universe at Its Limits
Figure 2.11 In this image, you can see part of a cluster of galaxies located about 60 million
light-years away in the constellation of Fornax. All the objects that are not pinpoints of light in
the picture are galaxies of billions of stars. (credit: ESO, J. Emerson, VISTA. Acknowledgment:
Cambridge Astronomical Survey Unit)
Some of the clusters themselves form into larger groups called superclusters. The Local Group is part of a supercluster of
galaxies, called the Virgo Supercluster, which stretches over a diameter of 110 million light-years. We are just beginning to
explore the structure of the universe at these enormous scales and are already encountering some unexpected findings.
At even greater distances, where many ordinary galaxies are too dim to see, we find quasars. These are brilliant centers of
galaxies, glowing with the light of an extraordinarily energetic process. The enormous energy of the quasars is produced by
gas that is heated to a temperature of millions of degrees as it falls toward a massive black hole and swirls around it. The
brilliance of quasars makes them the most distant beacons we can see in the dark oceans of space. They allow us to probe
the universe 10 billion light-years away or more, and thus 10 billion years or more in the past.
With quasars we can see way back close to the Big Bang explosion that marks the beginning of time. Beyond the quasars
and the most distant visible galaxies, we have detected the feeble glow of the explosion itself, filling the universe and thus
coming to us from all directions in space. The discovery of this “afterglow of creation” is considered to be one of the most
significant events in twentieth-century science, and we are still exploring the many things it has to tell us about the earliest
times of the universe.
Measurements of the properties of galaxies and quasars in remote locations require large telescopes, sophisticated light-
amplifying devices, and painstaking labor. Every clear night, at observatories around the world, astronomers and students
are at work on such mysteries as the birth of new stars and the large-scale structure of the universe, fitting their results into
the tapestry of our understanding.
The foregoing discussion has likely impressed on you that the universe is extraordinarily large and extraordinarily empty.
On average, it is 10,000 times more empty than our Galaxy. Yet, as we have seen, even the Galaxy is mostly empty space.
The air we breathe has about 1019 atoms in each cubic centimeter—and we usually think of air as empty space. In the
interstellar gas of the Galaxy, there is about one atom in every cubic centimeter. Intergalactic space is filled so sparsely that
to find one atom, on average, we must search through a cubic meter of space. Most of the universe is fantastically empty;
places that are dense, such as the human body, are tremendously rare.
Even our most familiar solids are mostly space. If we could take apart such a solid, piece by piece, we would eventually
reach the tiny molecules from which it is formed. Molecules are the smallest particles into which any matter can be divided
while still retaining its chemical properties. A molecule of water (H2O), for example, consists of two hydrogen atoms and
one oxygen atom bonded together.
Molecules, in turn, are built of atoms, which are the smallest particles of an element that can still be identified as that
element. For example, an atom of gold is the smallest possible piece of gold. Nearly 100 different kinds of atoms (elements)
exist in nature. Most of them are rare, and only a handful account for more than 99% of everything with which we come in
contact. The most abundant elements in the cosmos today are listed in Table 2.1; think of this table as the “greatest hits”
of the universe when it comes to elements.
Table 2.1
All atoms consist of a central, positively charged nucleus surrounded by negatively charged electrons. The bulk of the matter
in each atom is found in the nucleus, which consists of positive protons and electrically neutral neutrons all bound tightly
together in a very small space. Each element is defined by the number of protons in its atoms. Thus, any atom with 6 protons
in its nucleus is called carbon, any with 50 protons is called tin, and any with 70 protons is called ytterbium. (For a list of
the elements, see Appendix F.)
1. This list of elements is arranged in order of the atomic number, which is the number of protons in each
nucleus.
60 Chapter 2 | The Universe at Its Limits
The distance from an atomic nucleus to its electrons is typically 100,000 times the size of the nucleus itself. This is why
we say that even solid matter is mostly space. The typical atom is far emptier than the solar system out to Neptune. (The
distance from Earth to the Sun, for example, is only 100 times the size of the Sun.) This is one reason atoms are not like
miniature solar systems.
Remarkably, physicists have discovered that everything that happens in the universe, from the smallest atomic nucleus to
the largest superclusters of galaxies, can be explained through the action of only four forces: gravity, electromagnetism
(which combines the actions of electricity and magnetism), and two forces that act at the nuclear level. The fact that there
are four forces (and not a million, or just one) has puzzled physicists and astronomers for many years and has led to a quest
for a unified picture of nature.
To construct an atom, particle by particle, check out this guided animation (https://openstax.org/l/
30buildanatom) for building an atom.
If you are new to astronomy, you have probably reached the end of our brief tour in this chapter with mixed emotions. On
the one hand, you may be fascinated by some of the new ideas you’ve read about and you may be eager to learn more. On
the other hand, you may be feeling a bit overwhelmed by the number of topics we have covered, and the number of new
words and ideas we have introduced. Learning astronomy is a little like learning a new language: at first it seems there are
so many new expressions that you’ll never master them all, but with practice, you soon develop facility with them.
At this point you may also feel a bit small and insignificant, dwarfed by the cosmic scales of distance and time. But, there
is another way to look at what you have learned from our first glimpses of the cosmos. Let us consider the history of the
universe from the Big Bang to today and compress it, for easy reference, into a single year. (We have borrowed this idea
from Carl Sagan’s 1997 Pulitzer Prize-winning book, The Dragons of Eden.)
On this scale, the Big Bang happened at the first moment of January 1, and this moment, when you are reading this chapter
would be the end of the very last second of December 31. When did other events in the development of the universe happen
in this “cosmic year?” Our solar system formed around September 10, and the oldest rocks we can date on Earth go back to
the third week in September (Figure 2.12).
Figure 2.12 On a cosmic calendar, where the time since the Big Bang is compressed into 1 year, creatures we would call
human do not emerge on the scene until the evening of December 31. (credit: February: modification of work by NASA, JPL-
Caltech, W. Reach (SSC/Caltech); March: modification of work by ESA, Hubble and NASA, Acknowledgement: Giles
Chapdelaine; April: modification of work by NASA, ESA, CFHT, CXO, M.J. Jee (University of California, Davis), A. Mahdavi
(San Francisco State University); May: modification of work by NASA, JPL-Caltech; June: modification of work by NASA/
ESA; July: modification of work by NASA, JPL-Caltech, Harvard-Smithsonian; August: modification of work by NASA, JPL-
Caltech, R. Hurt (SSC-Caltech); September: modification of work by NASA; October: modification of work by NASA;
November: modification of work by Dénes Emőke)
Where does the origin of human beings fall during the course of this cosmic year? The answer turns out to be the evening
of December 31. The invention of the alphabet doesn’t occur until the fiftieth second of 11:59 p.m. on December 31. And
the beginnings of modern astronomy are a mere fraction of a second before the New Year. Seen in a cosmic context, the
amount of time we have had to study the stars is minute, and our success in piecing together as much of the story as we
have is remarkable.
Certainly our attempts to understand the universe are not complete. As new technologies and new ideas allow us to gather
more and better data about the cosmos, our present picture of astronomy will very likely undergo many changes. Still, as
you read our current progress report on the exploration of the universe, take a few minutes every once in a while just to
savor how much you have already learned.
are from government-supported instruments or projects, paid for by tax dollars, and therefore are free of copyright laws.)
Here are three resources we especially like:
• Astronomy Picture of the Day: apod.nasa.gov/apod/astropix.html. Two space scientists scour the Internet and select
one beautiful astronomy image to feature each day. Their archives range widely, from images of planets and nebulae
to rockets and space instruments; they also have many photos of the night sky. The search function (see the menu
on the bottom of the page) works quite well for finding something specific among the many years’ worth of daily
images.
• Hubble Space Telescope Images: www.hubblesite.org/newscenter/archive/browse/images. Starting at this page, you
can select from among hundreds of Hubble pictures by subject or by date. Note that many of the images have
supporting pictures with them, such as diagrams, animations, or comparisons. Excellent captions and background
information are provided. Other ways to approach these images are through the more public-oriented Hubble
Gallery (www.hubblesite.org/gallery) and the European homepage (www.spacetelescope.org/images).
• National Aeronautics and Space Administration’s (NASA’s) Planetary Photojournal: photojournal.jpl.nasa.gov. This
site features thousands of images from planetary exploration, with captions of varied length. You can select images
by world, feature name, date, or catalog number, and download images in a number of popular formats. However,
only NASA mission images are included. Note the Photojournal Search option on the menu at the top of the
homepage to access ways to search their archives.
Videos
Cosmic Voyage: www.youtube.com/watch?v=qxXf7AJZ73A. This video presents a portion of Cosmic Voyage, narrated by
Morgan Freeman (8:34).
Powers of Ten: www.youtube.com/watch?v=0fKBhvDjuy0. This classic short video is a much earlier version of Powers of
Ten, narrated by Philip Morrison (9:00).
The Known Universe: www.youtube.com/watch?v=17jymDn0W6U. This video tour from the American Museum of
Natural History has realistic animation, music, and captions (6:30).
Wanderers: apod.nasa.gov/apod/ap141208.html. This video provides a tour of the solar system, with narrative by Carl
Sagan, imagining other worlds with dramatically realistic paintings (3:50).
3 | STRAIGHT-LINE MOTION
Figure 3.1 A JR Central L0 series five-car maglev (magnetic levitation) train undergoing a test run on the Yamanashi Test
Track. The maglev train’s motion can be described using kinematics, the subject of this chapter. (credit: modification of work by
“Maryland GovPics”/Flickr)
Chapter Outline
3.1 Position, Displacement and Average Velocity
3.2 Instantaneous Velocity and Speed
3.3 Average and Instantaneous Acceleration
3.4 Motion with Constant Acceleration
3.5 Free Fall
Introduction
Our universe is full of objects in motion. From the stars, planets, and galaxies; to the motion of people and animals; down
to the microscopic scale of atoms and molecules—everything in our universe is in motion. We can describe motion using
the two disciplines of kinematics and dynamics. We will eventually study dynamics, which is concerned with the causes of
motion, in Newton’s Laws of Motion (https://legacy.cnx.org/content/m58294/latest/) ; but, there is much to be
learned about motion without referring to what causes it, and this is the study of kinematics. Kinematics involves describing
motion through properties such as position, time, velocity, and acceleration.
A full treatment of kinematics considers motion in two and three dimensions. For now, we discuss motion in one dimension,
which provides us with the tools necessary to study multidimensional motion. A good example of an object undergoing one-
dimensional motion is the maglev (magnetic levitation) train depicted at the beginning of this chapter. As it travels, say, from
Tokyo to Kyoto, it is at different positions along the track at various times in its journey, and therefore has displacements,
or changes in position. It also has a variety of velocities along its path and it undergoes accelerations (changes in velocity).
With the skills learned in this chapter we can calculate these quantities and average velocity. All these quantities can be
described using kinematics, without knowing the train’s mass or the forces involved.
64 Chapter 3 | Straight-Line Motion
When you’re in motion, the basic questions to ask are: Where are you? Where are you going? How fast are you getting
there? The answers to these questions require that you specify your position, your displacement, and your average
velocity—the terms we define in this section.
Position
To describe the motion of an object, you must first be able to describe its position (x): where it is at any particular time.
More precisely, we need to specify its position relative to a convenient frame of reference. A frame of reference is an
arbitrary set of axes from which the position and motion of an object are described. Earth is often used as a frame of
reference, and we often describe the position of an object as it relates to stationary objects on Earth. For example, a rocket
launch could be described in terms of the position of the rocket with respect to Earth as a whole, whereas a cyclist’s position
could be described in terms of where she is in relation to the buildings she passes Figure 3.2. In other cases, we use
reference frames that are not stationary but are in motion relative to Earth. To describe the position of a person in an airplane,
for example, we use the airplane, not Earth, as the reference frame. To describe the position of an object undergoing one-
dimensional motion, we often use the variable x. Later in the chapter, during the discussion of free fall, we use the variable
y.
Displacement
If an object moves relative to a frame of reference—for example, if a professor moves to the right relative to a whiteboard
Figure 3.3—then the object’s position changes. This change in position is called displacement. The word displacement
implies that an object has moved, or has been displaced. Although position is the numerical value of x along a straight line
where an object might be located, displacement gives the change in position along this line. Since displacement indicates
direction, it is a vector and can be either positive or negative, depending on the choice of positive direction. Also, an analysis
of motion can have many displacements embedded in it. If right is positive and an object moves 2 m to the right, then 4 m
to the left, the individual displacements are 2 m and −4 m, respectively.
Figure 3.3 A professor paces left and right while lecturing. Her position relative to Earth
is given by x. The +2.0-m displacement of the professor relative to Earth is represented by
an arrow pointing to the right.
Displacement
Displacement Δx is the change in position of an object:
Δx = x f − x 0, (3.1)
We use the uppercase Greek letter delta (Δ) to mean “change in” whatever quantity follows it; thus, Δx means change in
position (final position less initial position). We always solve for displacement by subtracting initial position x 0 from final
position x f . Note that the SI unit for displacement is the meter, but sometimes we use kilometers or other units of length.
Keep in mind that when units other than meters are used in a problem, you may need to convert them to meters to complete
the calculation (see Appendix B).
Objects in motion can also have a series of displacements. In the previous example of the pacing professor, the individual
displacements are 2 m and −4 m, giving a total displacement of −2 m. We define total displacement Δx Total , as the sum
of the individual displacements, and express this mathematically with the equation
Δx Total = ∑ Δx i, (3.2)
Δx 1 = x 1 − x 0 = 2 − 0 = 2 m.
Similarly,
Δx 2 = x 2 − x 1 = −2 − (2) = −4 m.
66 Chapter 3 | Straight-Line Motion
Thus,
Δx Total = Δx 1 + Δx 2 = 2 − 4 = −2 m.
The total displacement is 2 − 4 = −2 m to the left, or in the negative direction. It is also useful to calculate the magnitude
of the displacement, or its size. The magnitude of the displacement is always positive. This is the absolute value of
the displacement, because displacement is a vector and cannot have a negative value of magnitude. In our example, the
magnitude of the total displacement is 2 m, whereas the magnitudes of the individual displacements are 2 m and 4 m.
The magnitude of the total displacement should not be confused with the distance traveled. Distance traveled x Total , is the
total length of the path traveled between two positions. In the previous problem, the distance traveled is the sum of the
magnitudes of the individual displacements:
x Total = |Δx 1| + |Δx 2| = 2 + 4 = 6 m.
Average Velocity
To calculate the other physical quantities in kinematics we must introduce the time variable. The time variable allows us not
only to state where the object is (its position) during its motion, but also how fast it is moving. How fast an object is moving
is given by the rate at which the position changes with time.
For each position x i , we assign a particular time t i . If the details of the motion at each instant are not important, the rate
–
is usually expressed as the average velocity v . This vector quantity is simply the total displacement between two points
divided by the time taken to travel between them. The time taken to travel between two points is called the elapsed time
Δt .
Average Velocity
If x 1 and x 2 are the positions of an object at times t 1 and t 2 , respectively, then
It is important to note that the average velocity is a vector and can be negative, depending on positions x 1 and x 2 .
Example 3.1
Delivering Flyers
Jill sets out from her home to deliver flyers for her yard sale, traveling due east along her street lined with houses.
At 0.5 km and 9 minutes later she runs out of flyers and has to retrace her steps back to her house to get more.
This takes an additional 9 minutes. After picking up more flyers, she sets out again on the same path, continuing
where she left off, and ends up 1.0 km from her house. This third leg of her trip takes 15 minutes. At this point
she turns back toward her house, heading west. After 1.75 km and 25 minutes she stops to rest.
a. What is Jill’s total displacement to the point where she stops to rest?
b. What is the magnitude of the final displacement?
c. What is the average velocity during her entire trip?
d. What is the total distance traveled?
e. Make a graph of position versus time.
A sketch of Jill’s movements is shown in Figure 3.4.
Strategy
The problem contains data on the various legs of Jill’s trip, so it would be useful to make a table of the physical
quantities. We are given position and time in the wording of the problem so we can calculate the displacements
and the elapsed time. We take east to be the positive direction. From this information we can find the total
displacement and average velocity. Jill’s home is the starting point x 0 . The following table gives Jill’s time and
position in the first two columns, and the displacements are calculated in the third column.
t0 = 0 x0 = 0 Δx 0 = 0
t1 = 9 x 1 = 0.5 Δx 1 = x 1 − x 0 = 0.5
t 2 = 18 x2 = 0 Δx 2 = x 2 − x 1 = −0.5
t 3 = 33 x 3 = 1.0 Δx 3 = x 3 − x 2 = 1.0
t 4 = 58 x 4 = −0.75 Δx 4 = x 4 − x 3 = −1.75
Solution
a. From the above table, the total displacement is
e. We can graph Jill’s position versus time as a useful aid to see the motion; the graph is shown in Figure
3.5.
68 Chapter 3 | Straight-Line Motion
Figure 3.5 This graph depicts Jill’s position versus time. The
average velocity is the slope of a line connecting the initial and
final points.
Significance
Jill’s total displacement is −0.75 km, which means at the end of her trip she ends up 0.75 km due west of her
home. The average velocity means if someone was to walk due west at 0.013 km/min starting at the same time
Jill left her home, they both would arrive at the final stopping point at the same time. Note that if Jill were to
end her trip at her house, her total displacement would be zero, as well as her average velocity. The total distance
traveled during the 58 minutes of elapsed time for her trip is 3.75 km.
3.1 Check Your Understanding A cyclist rides 3 km west and then turns around and rides 2 km east. (a)
What is his displacement? (b) What is the distance traveled? (c) What is the magnitude of his displacement?
We have now seen how to calculate the average velocity between two positions. However, since objects in the real world
move continuously through space and time, we would like to find the velocity of an object at any single point. We can find
the velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us
better insight into the physics of motion and will be useful in later chapters.
Instantaneous Velocity
The quantity that tells us how fast an object is moving anywhere along its path is the instantaneous velocity, usually
called simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore the
displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position x
as a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation
– x(t 2) − x(t 1)
is v = t 2 − t 1 . To find the instantaneous velocity at any position, we let t 1 = t and t 2 = t + Δt . After inserting
these expressions into the equation for the average velocity and taking the limit as Δt → 0 , we find the expression for the
instantaneous velocity:
x(t + Δt) − x(t) dx(t)
v(t) = lim = .
Δt → 0 Δt dt
Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at
a specific time point t 0 is the rate of change of the position function, which is the slope of the position function x(t) at
t 0 . Figure 3.6 shows how the average velocity v = Δx between two times approaches the instantaneous velocity at t 0.
–
Δt
The instantaneous velocity is shown at time t 0 , which happens to be at the maximum of the position function. The slope
of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, t 1, t 2 , and so on,
the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position
function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there
as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.
70 Chapter 3 | Straight-Line Motion
Example 3.2
Figure 3.7 The object starts out in the positive direction, stops
for a short time, and then reverses direction, heading back
toward the origin. Notice that the object comes to rest
instantaneously, which would require an infinite force. Thus, the
graph is an approximation of motion in the real world. (The
concept of force is discussed in Newton’s Laws of Motion
(https://legacy.cnx.org/content/m58294/latest/) .)
Strategy
The graph contains three straight lines during three time intervals. We find the velocity during each time interval
by taking the slope of the line using the grid.
Solution
Figure 3.8 The velocity is positive for the first part of the trip,
zero when the object is stopped, and negative when the object
reverses direction.
Significance
During the time interval between 0 s and 0.5 s, the object’s position is moving away from the origin and the
position-versus-time curve has a positive slope. At any point along the curve during this time interval, we can
find the instantaneous velocity by taking its slope, which is +1 m/s, as shown in Figure 3.8. In the subsequent
time interval, between 0.5 s and 1.0 s, the position doesn’t change and we see the slope is zero. From 1.0 s to 2.0
s, the object is moving back toward the origin and the slope is −0.5 m/s. The object has reversed direction and has
a negative velocity.
Speed
In everyday language, most people use the terms speed and velocity interchangeably. In physics, however, they do not have
the same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.
We can calculate the average speed by finding the total distance traveled divided by the elapsed time:
Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing the
magnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the total
displacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the total
72 Chapter 3 | Straight-Line Motion
distance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time,
then we would be interested in our average speed.
However, we can calculate the instantaneous speed from the magnitude of the instantaneous velocity:
If a particle is moving along the x-axis at +7.0 m/s and another particle is moving along the same axis at −7.0 m/s, they have
different velocities, but both have the same speed of 7.0 m/s. Some typical speeds are shown in the following table.
The importance of understanding acceleration spans our day-to-day experience, as well as the vast reaches of outer space
and the tiny world of subatomic physics. In everyday conversation, to accelerate means to speed up; applying the brake
pedal causes a vehicle to slow down. We are familiar with the acceleration of our car, for example. The greater the
acceleration, the greater the change in velocity over a given time. Acceleration is widely seen in experimental physics. In
linear particle accelerator experiments, for example, subatomic particles are accelerated to very high velocities in collision
experiments, which tell us information about the structure of the subatomic world as well as the origin of the universe.
In space, cosmic rays are subatomic particles that have been accelerated to very high energies in supernovas (exploding
massive stars) and active galactic nuclei. It is important to understand the processes that accelerate cosmic rays because
these rays contain highly penetrating radiation that can damage electronics flown on spacecraft, for example.
Average Acceleration
The formal definition of acceleration is consistent with these notions just described, but is more inclusive.
Average Acceleration
Average acceleration is the rate at which velocity changes:
–a = Δv = v f − v 0 , (3.7)
Δt tf − t0
where −
a is average acceleration, v is velocity, and t is time. (The bar over the a means average acceleration.)
Because acceleration is velocity in meters divided by time in seconds, the SI units for acceleration are often abbreviated m/
s2—that is, meters per second squared or meters per second per second. This literally means by how many meters per second
the velocity changes every second. Recall that velocity is a vector—it has both magnitude and direction—which means that
a change in velocity can be a change in magnitude (or speed), but it can also be a change in direction. For example, if a
runner traveling at 10 km/h due east slows to a stop, reverses direction, continues her run at 10 km/h due west, her velocity
has changed as a result of the change in direction, although the magnitude of the velocity is the same in both directions.
Thus, acceleration occurs when velocity changes in magnitude (an increase or decrease in speed) or in direction, or both.
Acceleration as a Vector
Acceleration is a vector in the same direction as the change in velocity, Δv . Since velocity is a vector, it can change
in magnitude or in direction, or both. Acceleration is, therefore, a change in speed or direction, or both.
Keep in mind that although acceleration is in the direction of the change in velocity, it is not always in the direction of
motion. When an object slows down, its acceleration is opposite to the direction of its motion. Although this is commonly
referred to as deceleration Figure 3.9, we say the train is accelerating in a direction opposite to its direction of motion.
The term deceleration can cause confusion in our analysis because it is not a vector and it does not point to a specific
direction with respect to a coordinate system, so we do not use it. Acceleration is a vector, so we must choose the appropriate
sign for it in our chosen coordinate system. In the case of the train in Figure 3.9, acceleration is in the negative direction
in the chosen coordinate system, so we say the train is undergoing negative acceleration.
If an object in motion has a velocity in the positive direction with respect to a chosen origin and it acquires a constant
negative acceleration, the object eventually comes to a rest and reverses direction. If we wait long enough, the object passes
through the origin going in the opposite direction. This is illustrated in Figure 3.10.
74 Chapter 3 | Straight-Line Motion
Figure 3.10 An object in motion with a velocity vector toward the east under
negative acceleration comes to a rest and reverses direction. It passes the origin going
in the opposite direction after a long enough time.
Example 3.3
Strategy
First we draw a sketch and assign a coordinate system to the problem Figure 3.12. This is a simple problem, but
it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we
have negative velocity.
Figure 3.12 Identify the coordinate system, the given information, and what you want to
determine.
We can solve this problem by identifying Δv and Δt from the given information, and then calculating the
v −v
average acceleration directly from the equation –
a = Δv = t f − t 0 .
Δt f 0
Solution
First, identify the knowns: v 0 = 0, v f = −15.0 m/s (the negative sign indicates direction toward the west), Δt =
1.80 s.
Second, find the change in velocity. Since the horse is going from zero to –15.0 m/s, its change in velocity equals
its final velocity:
Δv = v f − v 0 = v f = −15.0 m/s.
Last, substitute the known values ( Δv and Δt ) and solve for the unknown –
a:
–a = Δv = −15.0 m/s = −8.33m/s 2.
Δt 1.80 s
Significance
The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of 8.33 m/s2 due
west means the horse increases its velocity by 8.33 m/s due west each second; that is, 8.33 meters per second per
second, which we write as 8.33 m/s2. This is truly an average acceleration, because the ride is not smooth. We
see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his
weight.
3.2 Check Your Understanding Protons in a linear accelerator are accelerated from rest to 2.0 × 10 7 m/s
in 10–4 s. What is the average acceleration of the protons?
Instantaneous Acceleration
Instantaneous acceleration a, or acceleration at a specific instant in time, is obtained using the same process discussed
for instantaneous velocity. That is, we calculate the average velocity between two points in time separated by Δt and let
Δt approach zero. The result is the derivative of the velocity function v(t), which is instantaneous acceleration and is
expressed mathematically as
Thus, similar to velocity being the derivative of the position function, instantaneous acceleration is the derivative of the
velocity function. We can show this graphically in the same way as instantaneous velocity. In Figure 3.13, instantaneous
acceleration at time t0 is the slope of the tangent line to the velocity-versus-time graph at time t0. We see that average
acceleration –
a = Δv approaches instantaneous acceleration as Δt approaches zero. Also in part (a) of the figure, we see
Δt
that velocity has a maximum when its slope is zero. This time corresponds to the zero of the acceleration function. In
part (b), instantaneous acceleration at the minimum velocity is shown, which is also zero, since the slope of the curve is
zero there, too. Thus, for a given velocity function, the zeros of the acceleration function give either the minimum or the
maximum velocity.
76 Chapter 3 | Straight-Line Motion
Figure 3.13 In a graph of velocity versus time, instantaneous acceleration is the slope of the tangent line. (a)
v −v
Shown is average acceleration –
a = Δv = t f − t i between times Δt = t 6 − t 1, Δt = t 5 − t 2 , and
Δt f i
Δt = t 4 − t 3 . When Δt → 0 , the average acceleration approaches instantaneous acceleration at time t0. In view
(a), instantaneous acceleration is shown for the point on the velocity curve at maximum velocity. At this point,
instantaneous acceleration is the slope of the tangent line, which is zero. At any other time, the slope of the tangent
line—and thus instantaneous acceleration—would not be zero. (b) Same as (a) but shown for instantaneous
acceleration at minimum velocity.
To illustrate this concept, let’s look at two examples. First, a simple example is shown of a velocity-versus-time graph, to
find acceleration graphically. This graph is depicted in Figure 3.14(a), which is a straight line. The corresponding graph of
acceleration versus time is found from the slope of velocity and is shown in Figure 3.14(b). In this example, the velocity
function is a straight line with a constant slope, thus acceleration is a constant.
Figure 3.14 (a, b) The velocity-versus-time graph is linear and has a negative constant slope (a) that is equal to
acceleration, shown in (b).
3.3 Check Your Understanding An airplane lands on a runway traveling east. Describe its acceleration.
However, acceleration is happening to many other objects in our universe with which we don’t have direct contact. Table
3.2 presents the acceleration of various objects. We can see the magnitudes of the accelerations extend over many orders of
magnitude.
In this table, we see that typical accelerations vary widely with different objects and have nothing to do with object size
or how massive it is. Acceleration can also vary widely with time during the motion of an object. A drag racer has a large
acceleration just after its start, but then it tapers off as the vehicle reaches a constant velocity. Its average acceleration can be
quite different from its instantaneous acceleration at a particular time during its motion. Figure 3.15 compares graphically
average acceleration with instantaneous acceleration for two very different motions.
78 Chapter 3 | Straight-Line Motion
Figure 3.15 Graphs of instantaneous acceleration versus time for two different one-dimensional motions. (a) Acceleration
varies only slightly and is always in the same direction, since it is positive. The average over the interval is nearly the same as
the acceleration at any given time. (b) Acceleration varies greatly, perhaps representing a package on a post office conveyor
belt that is accelerated forward and backward as it bumps along. It is necessary to consider small time intervals (such as from
0–1.0 s) with constant or nearly constant acceleration in such a situation.
Learn about position, velocity, and acceleration graphs. Move the little man back and forth with a mouse and plot
his motion. Set the position, velocity, or acceleration and let the simulation move the man for you. Visit this link
(https://openstaxcollege.org/l/21movmansimul) to use the moving man simulation.
You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car’s
displacement in a given time. But, we have not developed a specific equation that relates acceleration and displacement. In
this section, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement,
velocity, and acceleration. We first investigate a single object in motion, called single-body motion. Then we investigate the
motion of two objects, called two-body pursuit problems.
Notation
First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch,
is a great simplification. Since elapsed time is Δt = t f − t 0 , taking t 0 = 0 means that Δt = t f , the final time on the
stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity.
That is, x 0 is the initial position and v 0 is the initial velocity. We put no subscripts on the final values. That is, t is the
final time, x is the final position, and v is the final velocity. This gives a simpler expression for elapsed time, Δt = t . It also
simplifies the expression for x displacement, which is now Δx = x − x 0 . Also, it simplifies the expression for change in
velocity, which is now Δv = v − v 0 . To summarize, using the simplified notation, with the initial time taken to be zero,
Δt = t
Δx = x − x 0
Δv = v − v 0,
where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is
under consideration.
We now make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus to
find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal—that
is,
–a = a = constant.
Thus, we can use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit
the situations we can study nor does it degrade the accuracy of our treatment. For one thing, acceleration is constant in
a great number of situations. Furthermore, in many other situations we can describe motion accurately by assuming a
constant acceleration equal to the average acceleration for that motion. Lastly, for motion during which acceleration changes
drastically, such as a car accelerating to top speed and then braking to a stop, motion can be considered in separate parts,
each of which has its own constant acceleration.
v = Δx .
–
Δt
Substituting the simplified notation for Δx and Δt yields
– x − x0
v= t .
Solving for x gives us
–
x = x 0 + vt, (3.9)
– v0 + v (3.10)
v= .
2
– v +v
The equation v = 0 reflects the fact that when acceleration is constant, v is just the simple average of the initial and
2
final velocities. Figure 3.16 illustrates this concept graphically. In part (a) of the figure, acceleration is constant, with
velocity increasing at a constant rate. The average velocity during the 1-h interval from 40 km/h to 80 km/h is 60 km/h:
– v 0 + v 40 km/h + 80 km/h
v= = = 60 km/h.
2 2
In part (b), acceleration is not constant. During the 1-h interval, velocity is closer to 80 km/h than 40 km/h. Thus, the average
velocity is greater than in part (a).
80 Chapter 3 | Straight-Line Motion
Figure 3.16 (a) Velocity-versus-time graph with constant acceleration showing the initial and final velocities v 0 and v .
The average velocity is 1 (v 0 + v) = 60km/h . (b) Velocity-versus-time graph with an acceleration that changes with time.
2
The average velocity is not given by 1 (v 0 + v) , but is greater than 60 km/h.
2
a = Δv .
Δt
Substituting the simplified notation for Δv and Δt gives us
v − v0
a= t (constant a).
Example 3.4
Last, we determine which equation to use. To do this we figure out which kinematic equation gives the unknown
in terms of the knowns. We calculate the final velocity using Equation 3.11, v = v 0 + at .
Solution
Substitute the known values and solve:
Figure 3.17 is a sketch that shows the acceleration and velocity vectors.
Figure 3.17 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before
heading for the terminal. Note the acceleration is negative because its direction is opposite to its velocity, which is
positive.
Significance
The final velocity is much less than the initial velocity, as desired when slowing down, but is still positive (see
figure). With jet engines, reverse thrust can be maintained long enough to stop the plane and start moving it
backward, which is indicated by a negative final velocity, but is not the case here.
In addition to being useful in problem solving, the equation v = v 0 + at gives us insight into the relationships among
velocity, acceleration, and time. We can see, for example, that
• Final velocity depends on how large the acceleration is and how long it lasts
• If the acceleration is zero, then the final velocity equals the initial velocity (v = v0), as expected (in other words,
velocity is constant)
• If a is negative, then the final velocity is less than the initial velocity
All these observations fit our intuition. Note that it is always useful to examine basic equations in light of our intuition and
experience to check that they do indeed describe nature accurately.
v0 + v
= v 0 + 1 at.
2 2
v0 + v –
Since = v for constant acceleration, we have
2
v = v 0 + 1 at.
–
2
– –
Now we substitute this expression for v into the equation for displacement, x = x 0 + vt , yielding
Example 3.5
Figure 3.18 U.S. Army Top Fuel pilot Tony “The Sarge”
Schumacher begins a race with a controlled burnout. (credit: Lt.
Col. William Thurmond. Photo Courtesy of U.S. Army.)
Strategy
First, let’s draw a sketch Figure 3.19. We are asked to find displacement, which is x if we take x 0 to be zero.
(Think about x 0 as the starting line of a race. It can be anywhere, but we call it zero and measure all other
positions relative to it.) We can use the equation x = x 0 + v 0 t + 1 at 2 when we identify v 0 , a , and t from the
2
statement of the problem.
Solution
First, we need to identify the knowns. Starting from rest means that v 0 = 0 , a is given as 26.0 m/s2 and t is given
as 5.56 s.
Second, we substitute the known values into the equation to solve for the unknown:
x = x 0 + v 0 t + 1 at 2.
2
Since the initial position and velocity are both zero, this equation simplifies to
x = 1 at 2.
2
Substituting the identified values of a and t gives
What else can we learn by examining the equation x = x 0 + v 0 t + 1 at 2 ? We can see the following relationships:
2
• Displacement depends on the square of the elapsed time when acceleration is not zero. In Example 3.5, the
dragster covers only one-fourth of the total distance in the first half of the elapsed time.
–
• If acceleration is zero, then initial velocity equals average velocity (v 0 = v) , and
x = x 0 + v 0 t + 1 at 2 becomes x = x 0 + v 0 t.
2
Example 3.6
Thus,
v 2 = 2.09 × 10 4 m
/s 2
An examination of the equation v 2 = v 20 + 2a(x − x 0) can produce additional insights into the general relationships among
physical quantities:
• The final velocity depends on how large the acceleration is and the distance over which it acts.
• For a fixed acceleration, a car that is going twice as fast doesn’t simply stop in twice the distance. It takes much
farther to stop. (This is why we have reduced speed zones near schools.)
Before we get into the examples, let’s look at some of the equations more closely to see the behavior of acceleration at
extreme values. Rearranging Equation 3.11, we have
v − v0
a= t .
From this we see that, for a finite time, if the difference between the initial and final velocities is small, the acceleration is
small, approaching zero in the limit that the initial and final velocities are equal. On the contrary, in the limit t → 0 for a
finite difference between the initial and final velocities, acceleration becomes infinite.
Similarly, rearranging Equation 3.13, we can express acceleration in terms of velocities and displacement:
v 2 − v 20
a= .
2(x − x 0)
Thus, for a finite difference between the initial and final velocities acceleration becomes infinite in the limit the
displacement approaches zero. Acceleration approaches zero in the limit the difference in initial and final velocities
approaches zero for a finite displacement.
Example 3.7
Figure 3.20 Sample sketch to visualize deceleration and stopping distance of a car.
Solution
a. First, we need to identify the knowns and what we want to solve for. We know that v0 = 30.0 m/s, v = 0,
and a = −7.00 m/s2 (a is negative because it is in a direction opposite to velocity). We take x0 to be zero.
We are looking for displacement Δx , or x − x0.
Second, we identify the equation that will help us solve the problem. The best equation to use is
v 2 = v 20 + 2a(x − x 0).
This equation is best because it includes only one unknown, x. We know the values of all the other
variables in this equation. (Other equations would allow us to solve for x, but they require us to know the
stopping time, t, which we do not know. We could use them, but it would entail additional calculations.)
Third, we rearrange the equation to solve for x:
v 2 − v 20
x − x0 =
2a
Thus,
x = 64.3 m on dry concrete.
b. This part can be solved in exactly the same manner as (a). The only difference is that the acceleration is
−5.00 m/s2. The result is
x wet = 90.0 m on wet concrete.
c. When the driver reacts, the stopping distance is the same as it is in (a) and (b) for dry and wet concrete.
So, to answer this question, we need to calculate how far the car travels during the reaction time, and then
86 Chapter 3 | Straight-Line Motion
add that to the stopping time. It is reasonable to assume the velocity remains constant during the driver’s
reaction time.
–
To do this, we, again, identify the knowns and what we want to solve for. We know that v = 30.0 m/s ,
t reaction = 0.500 s , and a reaction = 0 . We take x 0-reaction to be zero. We are looking for x reaction .
–
Second, as before, we identify the best equation to use. In this case, x = x 0 + vt works well because the
only unknown value is x, which is what we want to solve for.
Third, we substitute the knowns to solve the equation:
x = 0 + (30.0 m/s)(0.500 s) = 15.0 m.
This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases
of dry and wet concrete 15.0 m greater than if he reacted instantly.
Last, we then add the displacement during the reaction time to the displacement when braking (Figure
3.21),
x braking + x reaction = x total,
and find (a) to be 64.3 m + 15.0 m = 79.3 m when dry and (b) to be 90.0 m + 15.0 m = 105 m when wet.
Figure 3.21 The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction
time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car
traveling initially at 30.0 m/s. Also shown are the total distances traveled from the point when the driver first sees a
light turn red, assuming a 0.500-s reaction time.
Significance
The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer
to stop a car on wet pavement than dry. It is interesting that reaction time adds significantly to the displacements,
but more important is the general approach to solving problems. We identify the knowns and the quantities to be
determined, then find an appropriate equation. If there is more than one unknown, we need as many independent
equations as there are unknowns to solve. There is often more than one way to solve a problem. The various parts
of this example can, in fact, be solved by other methods, but the solutions presented here are the shortest.
Example 3.8
Calculating Time
Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates
at 2.00 m/s2, how long does it take the car to travel the 200 m up the ramp? (Such information might be useful to
a traffic engineer.)
Strategy
First, we draw a sketch Figure 3.22. We are asked to solve for time t. As before, we identify the known quantities
to choose a convenient physical relationship (that is, an equation with one unknown, t.)
Solution
Again, we identify the knowns and what we want to solve for. We know that x 0 = 0,
v 0 = 10 m/s, a = 2.00 m/s 2 , and x = 200 m.
We need to solve for t. The equation x = x 0 + v 0 t + 1 at 2 works best because the only unknown in the equation
2
is the variable t, for which we need to solve. From this insight we see that when we input the knowns into the
equation, we end up with a quadratic equation.
We need to rearrange the equation to solve for t, then substituting the knowns into the equation:
which yields two solutions: t = 10.0 and t = −20.0. A negative value for time is unreasonable, since it would mean
the event happened 20 s before the motion began. We can discard that solution. Thus,
t = 10.0 s.
Significance
Whenever an equation contains an unknown squared, there are two solutions. In some problems both solutions are
meaningful; in others, only one solution is reasonable. The 10.0-s answer seems reasonable for a typical freeway
on-ramp.
3.4 Check Your Understanding A manned rocket accelerates at a rate of 20 m/s2 during launch. How long
does it take the rocket to reach a velocity of 400 m/s?
88 Chapter 3 | Straight-Line Motion
Example 3.9
Acceleration of a Spaceship
A spaceship has left Earth’s orbit and is on its way to the Moon. It accelerates at 20 m/s2 for 2 min and covers a
distance of 1000 km. What are the initial and final velocities of the spaceship?
Strategy
We are asked to find the initial and final velocities of the spaceship. Looking at the kinematic equations, we see
that one equation will not give the answer. We must use one kinematic equation to solve for one of the velocities
and substitute it into another kinematic equation to get the second velocity. Thus, we solve two of the kinematic
equations simultaneously.
Solution
x − x 0 = v 0 t + 1 at
2
6 1
1.0 × 10 m = v 0(120.0 s) + (20.0m/s 2)(120.0 s) 2
2
v 0 = 7133.3 m/s.
Significance
There are six variables in displacement, time, velocity, and acceleration that describe motion in one dimension.
The initial conditions of a given problem can be many combinations of these variables. Because of this diversity,
solutions may not be easy as simple substitutions into one of the equations. This example illustrates that solutions
to kinematics may require solving two simultaneous kinematic equations.
With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process
of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers
and insights into physical relationships. The next level of complexity in our kinematics problems involves the motion of
two interrelated bodies, called two-body pursuit problems.
Figure 3.23 A two-body pursuit scenario where car 2 has a constant velocity and car 1 is
behind with a constant acceleration. Car 1 catches up with car 2 at a later time.
The time and distance required for car 1 to catch car 2 depends on the initial distance car 1 is from car 2 as well as the
velocities of both cars and the acceleration of car 1. The kinematic equations describing the motion of both cars must be
solved to find these unknowns.
Example 3.10
Equation for the cheetah: The cheetah is accelerating from rest, so we use Equation 3.12 with x 0 = 0
and v 0 = 0 :
x = x 0 + v 0 t + 1 at 2 = 1 at 2.
2 2
Now we have an equation of motion for each animal with a common parameter, which can be eliminated
to find the solution. In this case, we solve for t:
x = vt = 1 at 2
–
2
–
2v
t= a.
The gazelle has a constant velocity of 10 m/s, which is its average velocity. The acceleration of the
cheetah is 4 m/s2. Evaluating t, the time for the cheetah to reach the gazelle, we have
– 2(10)
t = 2v
a = 4 = 5 s.
b. To get the displacement, we use either the equation of motion for the cheetah or the gazelle, since they
should both give the same answer.
Displacement of the cheetah:
x = 1 at 2 = 1 (4)(5) 2 = 50 m.
2 2
individual motion. It is also important to have a good visual perspective of the two-body pursuit problem to see
the common parameter that links the motion of both objects.
3.5 Check Your Understanding A bicycle has a constant velocity of 10 m/s. A person starts from rest and
runs to catch up to the bicycle in 30 s. What is the acceleration of the person?
An interesting application of Equation 3.4 through Equation 3.13 is called free fall, which describes the motion of an
object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s
assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we
can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But
“falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height.
If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.
Gravity
The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in
a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass.
This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction
that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people
believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of
air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height
Figure 3.24.
Figure 3.24 A hammer and a feather fall with the same constant acceleration if air resistance is negligible. This is a general
characteristic of gravity not unique to Earth, as astronaut David R. Scott demonstrated in 1971 on the Moon, where the
acceleration from gravity is only 1.67 m/s2 and there is no atmosphere.
In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball
reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is
not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between
clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.
For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in
free fall. The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects
is therefore called acceleration due to gravity. Acceleration due to gravity is constant, which means we can apply the
kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of
interesting situations.
Acceleration due to gravity is so important that its magnitude is given its own symbol, g. It is constant at any given location
on Earth and has the average value
g = 9.81 m/s 2 (or 32.2 ft/s 2).
Although g varies from 9.78 m/s2 to 9.83 m/s2, depending on latitude, altitude, underlying geological formations, and
local topography, let’s use an average value of 9.8 m/s2 rounded to two significant figures in this text unless specified
otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting
from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth).
In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the
value +g or −g depends on how we define our coordinate system. If we define the upward direction as positive, then
a = −g = −9.8 m/s 2, and if we define the downward direction as positive, then a = g = 9.8 m/s 2 .
y = y 0 + v 0 t − 1 gt 2 (3.15)
2
v 2 = v 20 − 2g(y − y 0) (3.16)
Example 3.11
Strategy
Choose the origin at the top of the building with the positive direction upward and the negative direction
downward. To find the time when the position is −98 m, we use Equation 3.15, with
y 0 = 0, v 0 = −4.9 m/s, and g = 9.8 m/s 2 .
Solution
a. Substitute the given values into the equation:
y = y 0 + v 0 t − 1 gt 2
2
−98.0 m = 0 − (4.9 m/s)t − 1 (9.8 m/s 2)t 2.
2
This simplifies to
t 2 + t − 20 = 0.
This is a quadratic equation with roots t = −5.0s and t = 4.0s . The positive root is the one we are
interested in, since time t = 0 is the time when the ball is released at the top of the building. (The time
t = −5.0s represents the fact that a ball thrown upward from the ground would have been in the air for
5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
b. Using Equation 3.14, we have
v = v 0 − gt = −4.9 m/s − (9.8m/s 2)(4.0 s) = −44.1 m/s.
Significance
For situations when two roots are obtained from a quadratic equation in the time variable, we must look at the
physical significance of both roots to determine which is correct. Since t = 0 corresponds to the time when
the ball was released, the negative root would correspond to a time before the ball was released, which is not
physically meaningful. When the ball hits the ground, its velocity is not immediately zero, but as soon as the ball
interacts with the ground, its acceleration is not g and it accelerates with a different value over a short time to zero
velocity. This problem shows how important it is to establish the correct coordinate system and to keep the signs
of g in the kinematic equations consistent.
Example 3.12
Figure 3.26 A baseball hit straight up is caught by the catcher 5.0 s later.
94 Chapter 3 | Straight-Line Motion
Strategy
Choose a coordinate system with a positive y-axis that is straight up and with an origin that is at the spot where
the ball is hit and caught.
Solution
a. Equation 3.15 gives
y = y 0 + v 0 t − 1 gt 2
2
v 2 = v 20 − 2g(y − y 0)
or
y = 30.6 m.
c. To find the time when v = 0 , we use Equation 3.14:
v = v 0 − gt
This gives t = 2.5 s . Since the ball rises for 2.5 s, the time to fall is 2.5 s.
d. The acceleration is 9.8 m/s2 everywhere, even when the velocity is zero at the top of the path. Although
the velocity is zero at the top, it is changing at the rate of 9.8 m/s2 downward.
e. The velocity at t = 5.0s can be determined with Equation 3.14:
v = v 0 − gt
= 24.5 m/s − 9.8m/s 2(5.0 s)
= −24.5 m/s.
Significance
The ball returns with the speed it had when it left. This is a general property of free fall for any initial velocity. We
used a single equation to go from throw to catch, and did not have to break the motion into two segments, upward
and downward. We are used to thinking of the effect of gravity is to create free fall downward toward Earth. It is
important to understand, as illustrated in this example, that objects moving upward away from Earth are also in a
state of free fall.
3.6 Check Your Understanding A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water.
Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity
increases faster, the speed of the ice chunk or its distance traveled?
Example 3.13
Rocket Booster
A small rocket with a booster blasts off and heads straight upward. When at a height of 5.0 km and velocity of
200.0 m/s, it releases its booster. (a) What is the maximum height the booster attains? (b) What is the velocity of
the booster at a height of 6.0 km? Neglect air resistance.
Strategy
We need to select the coordinate system for the acceleration of gravity, which we take as negative downward. We
are given the initial velocity of the booster and its height. We consider the point of release as the origin. We know
the velocity is zero at the maximum position within the acceleration interval; thus, the velocity of the booster is
zero at its maximum height, so we can use this information as well. From these observations, we use Equation
3.16, which gives us the maximum height of the booster. We also use Equation 3.16 to give the velocity at 6.0
km. The initial velocity of the booster is 200.0 m/s.
Solution
a. From Equation 3.16, v 2 = v 20 − 2g(y − y 0) . With v = 0 and y 0 = 0 , we can solve for y:
v 20 (2.0 × 10 2 m/s) 2
y= = = 2040.8 m.
−2g −2(9.8 m/s 2)
This solution gives the maximum height of the booster in our coordinate system, which has its origin at
the point of release, so the maximum height of the booster is roughly 7.0 km.
b. An altitude of 6.0 km corresponds to y = 1.0 × 10 3 m in the coordinate system we are using. The other
96 Chapter 3 | Straight-Line Motion
CHAPTER 3 REVIEW
KEY TERMS
acceleration due to gravity acceleration of an object as a result of gravity
average acceleration the rate of change in velocity; the change in velocity over time
average speed the total distance traveled divided by elapsed time
average velocity the displacement divided by the time over which displacement occurs
displacement the change in position of an object
distance traveled the total length of the path traveled between two positions
elapsed time the difference between the ending time and the beginning time
free fall the state of movement that results from gravitational force only
instantaneous acceleration acceleration at a specific point in time
instantaneous speed the absolute value of the instantaneous velocity
instantaneous velocity the velocity at a specific instant or time point
kinematics the description of motion through properties such as position, time, velocity, and acceleration
position the location of an object at a particular time
total displacement the sum of individual displacements over a given time period
two-body pursuit problem a kinematics problem in which the unknowns are calculated by solving the kinematic
equations simultaneously for two moving objects
KEY EQUATIONS
Displacement Δx = x f − x i
–v = Δx = x 2 − x 1
Average velocity Δt t2 − t1
dx(t)
Instantaneous velocity v(t) =
dt
–a = Δv = v f − v 0
Average acceleration Δt t f − t0
dv(t)
Instantaneous acceleration a(t) =
dt
–v = v 0 + v
Average velocity 2
SUMMARY
3.1 Position, Displacement and Average Velocity
• Kinematics is the description of motion without considering its causes. In this chapter, it is limited to motion along
a straight line, called one-dimensional motion.
• Displacement is the change in position of an object. The SI unit for displacement is the meter. Displacement has
direction as well as magnitude.
• Distance traveled is the total length of the path traveled between two positions.
• Time is measured in terms of change. The time between two position points x 1 and x 2 is Δt = t 2 − t 1 . Elapsed
time for an event is Δt = t f − t 0 , where t f is the final time and t 0 is the initial time. The initial time is often taken
to be zero.
–
• Average velocity v is defined as displacement divided by elapsed time. If x 1, t 1 and x 2, t 2 are two position time
points, the average velocity between these points is
x −x
v = Δx = t 2 − t 1 .
–
Δt 2 1
• Two-body pursuit problems always require two equations to be solved simultaneously for the unknowns.
• For objects in free fall, the upward direction is normally taken as positive for displacement, velocity, and
acceleration.
CONCEPTUAL QUESTIONS
quantities the same?
3.1 Position, Displacement and Average
Velocity 10. How are instantaneous velocity and instantaneous
speed related to one another? How do they differ?
1. Give an example in which there are clear distinctions
among distance traveled, displacement, and magnitude of
displacement. Identify each quantity in your example
specifically. 3.3 Average and Instantaneous Acceleration
11. Is it possible for speed to be constant while
2. Under what circumstances does distance traveled equal acceleration is not zero?
magnitude of displacement? What is the only case in which
magnitude of displacement and displacement are exactly 12. Is it possible for velocity to be constant while
the same? acceleration is not zero? Explain.
3. Bacteria move back and forth using their flagella 13. Give an example in which velocity is zero yet
(structures that look like little tails). Speeds of up to 50 μm/ acceleration is not.
s (50 × 10−6 m/s) have been observed. The total distance
traveled by a bacterium is large for its size, whereas its 14. If a subway train is moving to the left (has a negative
displacement is small. Why is this? velocity) and then comes to a stop, what is the direction of
its acceleration? Is the acceleration positive or negative?
4. Give an example of a device used to measure time and
identify what change in that device indicates a change in 15. Plus and minus signs are used in one-dimensional
time. motion to indicate direction. What is the sign of an
acceleration that reduces the magnitude of a negative
5. Does a car’s odometer measure distance traveled or velocity? Of a positive velocity?
displacement?
6. During a given time interval the average velocity of 3.4 Motion with Constant Acceleration
an object is zero. What can you say conclude about its 16. When analyzing the motion of a single object, what
displacement over the time interval? is the required number of known physical variables that
are needed to solve for the unknown quantities using the
kinematic equations?
3.2 Instantaneous Velocity and Speed
7. There is a distinction between average speed and the 17. State two scenarios of the kinematics of single object
magnitude of average velocity. Give an example that where three known quantities require two kinematic
illustrates the difference between these two quantities. equations to solve for the unknowns.
This is one-dimensional motion. (a) When is its velocity 21. The severity of a fall depends on your speed when
zero? (b) Does its velocity change direction? (c) Does the you strike the ground. All factors but the acceleration from
acceleration have the same sign on the way up as on the gravity being the same, how many times higher could a safe
way down? fall on the Moon than on Earth (gravitational acceleration
on the Moon is about one-sixth that of the Earth)?
20. Suppose you throw a rock nearly straight up at a
coconut in a palm tree and the rock just misses the coconut 22. How many times higher could an astronaut jump on
on the way up but hits the coconut on the way down. the Moon than on Earth if her takeoff speed is the same
Neglecting air resistance and the slight horizontal variation in both locations (gravitational acceleration on the Moon is
in motion to account for the hit and miss of the coconut, about on-sixth of that on Earth)?
how does the speed of the rock when it hits the coconut
on the way down compare with what it would have been if 23. When given the acceleration function, what additional
it had hit the coconut on the way up? Is it more likely to information is needed to find the velocity function and
dislodge the coconut on the way up or down? Explain. position function?
PROBLEMS
average velocity of the blast wave? b) Compare this with
3.1 Position, Displacement and Average the speed of sound, which is 343 m/s at sea level.
Velocity
24. Consider a coordinate system in which the positive x
3.2 Instantaneous Velocity and Speed
axis is directed upward vertically. What are the positions of
a particle (a) 5.0 m directly above the origin and (b) 2.0 m 30. A woodchuck runs 20 m to the right in 5 s, then turns
below the origin? and runs 10 m to the left in 3 s. (a) What is the average
velocity of the woodchuck? (b) What is its average speed?
25. A car is 2.0 km west of a traffic light at t = 0 and 5.0
km east of the light at t = 6.0 min. Assume the origin of the 31. Sketch the velocity-versus-time graph from the
coordinate system is the light and the positive x direction following position-versus-time graph.
is eastward. (a) What are the car’s position vectors at these
two times? (b) What is the car’s displacement between 0
min and 6.0 min?
35. Dr. John Paul Stapp was a U.S. Air Force officer who
studied the effects of extreme acceleration on the human 3.4 Motion with Constant Acceleration
body. On December 10, 1954, Stapp rode a rocket sled, 40. A particle moves in a straight line at a constant
accelerating from rest to a top speed of 282 m/s (1015 velocity of 30 m/s. What is its displacement between t = 0
km/h) in 5.00 s and was brought jarringly back to rest in and t = 5.0 s?
only 1.40 s. Calculate his (a) acceleration in his direction
of motion and (b) acceleration opposite to his direction 41. A particle moves in a straight line with an initial
of motion. Express each in multiples of g (9.80 m/s2) by velocity of 30 m/s and a constant acceleration of 30 m/s2. If
taking its ratio to the acceleration of gravity. at t = 0, x = 0 and v = 0 , what is the particle’s position
at t = 5 s?
36. Sketch the acceleration-versus-time graph from the
following velocity-versus-time graph.
42. A particle moves in a straight line with an initial
velocity of 30 m/s and constant acceleration 30 m/s2. (a)
What is its displacement at t = 5 s? (b) What is its velocity
at this same time?
which times is it negative? 49. (a) A light-rail commuter train accelerates at a rate of
1.35 m/s2. How long does it take to reach its top speed of
80.0 km/h, starting from rest? (b) The same train ordinarily
decelerates at a rate of 1.65 m/s2. How long does it take
to come to a stop from its top speed? (c) In emergencies,
the train can decelerate more rapidly, coming to rest from
80.0 km/h in 8.30 s. What is its emergency acceleration in
meters per second squared?
70. A very strong, but inept, shot putter puts the shot
straight up vertically with an initial velocity of 11.0 m/s.
104 Chapter 3 | Straight-Line Motion
How long a time does he have to get out of the way if the in Victoria, Australia, a hiker hears a rock break loose from
shot was released at a height of 2.20 m and he is 1.80 m a height of 105.0 m. He can’t see the rock right away, but
tall? then does, 1.50 s later. (a) How far above the hiker is the
rock when he can see it? (b) How much time does he have
71. You throw a ball straight up with an initial velocity of to move before the rock hits his head?
15.0 m/s. It passes a tree branch on the way up at a height
of 7.0 m. How much additional time elapses before the ball 74. There is a 250-m-high cliff at Half Dome in Yosemite
passes the tree branch on the way back down? National Park in California. Suppose a boulder breaks loose
from the top of this cliff. (a) How fast will it be going when
72. A kangaroo can jump over an object 2.50 m high. (a) it strikes the ground? (b) Assuming a reaction time of 0.300
Considering just its vertical motion, calculate its vertical s, how long a time will a tourist at the bottom have to get
speed when it leaves the ground. (b) How long a time is it out of the way after hearing the sound of the rock breaking
in the air? loose (neglecting the height of the tourist, which would
become negligible anyway if hit)? The speed of sound is
335.0 m/s on this day.
73. Standing at the base of one of the cliffs of Mt. Arapiles
ADDITIONAL PROBLEMS
75. Professional baseball player Nolan Ryan could pitch of 4.0 × 10 5 m/s. It enters a region 5.0 cm long where it
a baseball at approximately 160.0 km/h. At that average
velocity, how long did it take a ball thrown by Ryan to undergoes an acceleration of 6.0 × 10 12 m/s 2 along the
reach home plate, which is 18.4 m from the pitcher’s same straight line. (a) What is the electron’s velocity when
mound? Compare this with the average reaction time of a it emerges from this region? b) How long does the electron
human to a visual stimulus, which is 0.25 s. take to cross the region?
76. An airplane leaves Chicago and makes the 3000-km 82. An ambulance driver is rushing a patient to the
trip to Los Angeles in 5.0 h. A second plane leaves Chicago hospital. While traveling at 72 km/h, she notices the traffic
one-half hour later and arrives in Los Angeles at the same light at the upcoming intersections has turned amber. To
time. Compare the average velocities of the two planes. reach the intersection before the light turns red, she must
Ignore the curvature of Earth and the difference in altitude travel 50 m in 2.0 s. (a) What minimum acceleration must
between the two cities. the ambulance have to reach the intersection before the
light turns red? (b) What is the speed of the ambulance
77. Unreasonable Results A cyclist rides 16.0 km east, when it reaches the intersection?
then 8.0 km west, then 8.0 km east, then 32.0 km west,
and finally 11.2 km east. If his average velocity is 24 km/ 83. A motorcycle that is slowing down uniformly covers
h, how long did it take him to complete the trip? Is this a 2.0 successive km in 80 s and 120 s, respectively. Calculate
reasonable time? (a) the acceleration of the motorcycle and (b) its velocity at
the beginning and end of the 2-km trip.
78. An object has an acceleration of +1.2 cm/s 2 . At
84. A cyclist travels from point A to point B in 10 min.
t = 4.0 s , its velocity is −3.4 cm/s . Determine the During the first 2.0 min of her trip, she maintains a uniform
object’s velocities at t = 1.0 s and t = 6.0 s .
acceleration of 0.090 m/s 2 . She then travels at constant
velocity for the next 5.0 min. Next, she decelerates at a
79. A particle moving at constant acceleration has constant rate so that she comes to a rest at point B 3.0 min
velocities of 2.0 m/s at t = 2.0 s and −7.6 m/s at later. (a) Sketch the velocity-versus-time graph for the trip.
t = 5.2 s. What is the acceleration of the particle? (b) What is the acceleration during the last 3 min? (c) How
far does the cyclist travel?
80. A train is moving up a steep grade at constant velocity
(see following figure) when its caboose breaks loose and 85. Two trains are moving at 30 m/s in opposite directions
starts rolling freely along the track. After 5.0 s, the caboose on the same track. The engineers see simultaneously that
is 30 m behind the train. What is the acceleration of the they are on a collision course and apply the brakes when
caboose? they are 1000 m apart. Assuming both trains have the same
Figure shows a train moving up a hill. acceleration, what must this acceleration be if the trains are
to stop just short of colliding?
81. An electron is moving in a straight line with a velocity
86. A 10.0-m-long truck moving with a constant velocity 94. A coin is dropped from a hot-air balloon that is 300
of 97.0 km/h passes a 3.0-m-long car moving with a m above the ground and rising at 10.0 m/s upward. For the
constant velocity of 80.0 km/h. How much time elapses coin, find (a) the maximum height reached, (b) its position
between the moment the front of the truck is even with the and velocity 4.00 s after being released, and (c) the time
back of the car and the moment the back of the truck is even before it hits the ground.
with the front of the car?
Top drawing shows passenger car with a speed of 80 95. A soft tennis ball is dropped onto a hard floor from
kilometers per hour in front of the truck with the speed of a height of 1.50 m and rebounds to a height of 1.10 m.
97 kilometers per hour. Middle drawing shows passenger (a) Calculate its velocity just before it strikes the floor. (b)
car with a speed of 80 kilometers per hour parallel to the Calculate its velocity just after it leaves the floor on its way
truck with the speed of 97 kilometers per hour. Bottom back up. (c) Calculate its acceleration during contact with
drawing shows passenger car with a speed of 80 kilometers
the floor if that contact lasts 3.50 ms (3.50 × 10 −3 s) (d)
per hour behind the truck with a speed of 97 kilometers per
hour. How much did the ball compress during its collision with
the floor, assuming the floor is absolutely rigid?
87. A police car waits in hiding slightly off the highway.
A speeding car is spotted by the police car doing 40 m/s. At 96. Unreasonable results. A raindrop falls from a cloud
the instant the speeding car passes the police car, the police 100 m above the ground. Neglect air resistance. What is
car accelerates from rest at 4 m/s2 to catch the speeding car. the speed of the raindrop when it hits the ground? Is this a
How long does it take the police car to catch the speeding reasonable number?
car?
97. Compare the time in the air of a basketball player who
88. Pablo is running in a half marathon at a velocity of jumps 1.0 m vertically off the floor with that of a player
3 m/s. Another runner, Jacob, is 50 meters behind Pablo who jumps 0.3 m vertically.
with the same velocity. Jacob begins to accelerate at 0.05
m/s2. (a) How long does it take Jacob to catch Pablo? (b) 98. Suppose that a person takes 0.5 s to react and move his
What is the distance covered by Jacob? (c) What is the final hand to catch an object he has dropped. (a) How far does
velocity of the Jacob? the object fall on Earth, where g = 9.8 m/s 2 ? (b) How far
does the object fall on the Moon, where the acceleration
89. Unreasonable results A runner approaches the finish due to gravity is 1/6 of that on Earth?
line and is 75 m away; her average speed at this position is
8 m/s. She decelerates at this point at 0.5 m/s2. How long
does it take her to cross the finish line from 75 m away? Is 99. A hot-air balloon rises from ground level at a constant
this reasonable? velocity of 3.0 m/s. One minute after liftoff, a sandbag is
dropped accidentally from the balloon. Calculate (a) the
time it takes for the sandbag to reach the ground and (b) the
90. An airplane accelerates at 5.0 m/s2 for 30.0 s. During velocity of the sandbag when it hits the ground.
this time, it covers a distance of 10.0 km. What are the
initial and final velocities of the airplane?
100. (a) A world record was set for the men’s 100-m dash
in the 2008 Olympic Games in Beijing by Usain Bolt of
91. Compare the distance traveled of an object that Jamaica. Bolt “coasted” across the finish line with a time
undergoes a change in velocity that is twice its initial of 9.69 s. If we assume that Bolt accelerated for 3.00 s to
velocity with an object that changes its velocity by four reach his maximum speed, and maintained that speed for
times its initial velocity over the same time period. The the rest of the race, calculate his maximum speed and his
accelerations of both objects are constant. acceleration. (b) During the same Olympics, Bolt also set
the world record in the 200-m dash with a time of 19.30 s.
92. An object is moving east with a constant velocity and Using the same assumptions as for the 100-m dash, what
is at position x 0 at time t 0 = 0 . (a) With what acceleration was his maximum speed for this race?
must the object have for its total displacement to be zero at
a later time t ? (b) What is the physical interpretation of the 101. An object is dropped from a height of 75.0 m above
solution in the case for t → ∞ ? ground level. (a) Determine the distance traveled during the
first second. (b) Determine the final velocity at which the
object hits the ground. (c) Determine the distance traveled
93. A ball is thrown straight up. It passes a 2.00-m-high
during the last second of motion before hitting the ground.
window 7.50 m off the ground on its path up and takes
1.30 s to go past the window. What was the ball’s initial
velocity? 102. A steel ball is dropped onto a hard floor from a height
of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate
its velocity just before it strikes the floor. (b) Calculate its
106 Chapter 3 | Straight-Line Motion
velocity just after it leaves the floor on its way back up. 103. An object is dropped from a roof of a building of
(c) Calculate its acceleration during contact with the floor height h. During the last second of its descent, it drops a
if that contact lasts 0.0800 ms (8.00 × 10 −5 s) (d) How distance h/3. Calculate the height of the building.
much did the ball compress during its collision with the
floor, assuming the floor is absolutely rigid?
CHALLENGE PROBLEMS
104. In a 100-m race, the winner is timed at 11.2 s. The traveled at 11.8 m/s until the finish line. What was the
second-place finisher’s time is 11.6 s. How far is the difference in finish time in seconds between the winner
second-place finisher behind the winner when she crosses and runner-up? How far back was the runner-up when the
the finish line? Assume the velocity of each runner is winner crossed the finish line?
constant throughout the race.
106. In 1967, New Zealander Burt Munro set the world
105. A cyclist sprints at the end of a race to clinch a record for an Indian motorcycle, on the Bonneville Salt
victory. She has an initial velocity of 11.5 m/s and Flats in Utah, of 295.38 km/h. The one-way course was
accelerates at a rate of 0.500 m/s2 for 7.00 s. (a) What is her 8.00 km long. Acceleration rates are often described by the
final velocity? (b) The cyclist continues at this velocity to time it takes to reach 96.0 km/h from rest. If this time was
the finish line. If she is 300 m from the finish line when she 4.00 s and Burt accelerated at this rate until he reached his
starts to accelerate, how much time did she save? (c) The maximum speed, how long did it take Burt to complete the
second-place winner was 5.00 m ahead when the winner course?
started to accelerate, but he was unable to accelerate, and
4 | CIRCULAR MOTION AS
ONE-DIMENSIONAL
MOTION
Figure 4.1 Brazos wind farm in west Texas. As of 2012, wind farms in the US had a power output of 60 gigawatts, enough
capacity to power 15 million homes for a year. (credit: modification of work by “ENERGY.GOV”/Flickr)
Chapter Outline
4.1 Rotational Variables
4.2 Rotation with Constant Angular Acceleration
4.3 Relating Angular and Translational Quantities
Introduction
In Straight-Line Motion, we described motion (kinematics) in one dimension, and introduced important concepts like
displacement, velocity and acceleration. The type of motion we considered there is called translational motion, because the
moving object undergoes a translation from place to place in space. However, we know from everyday life that rotational
motion is also very important and that many objects that move have both translation and rotation. The wind turbines in our
chapter opening image are a prime example of how rotational motion impacts our daily lives, as the market for clean energy
sources continues to grow.
As we shall discover in Section 8.4, the objects that we study in astronomy often move (translate) in paths that follow
the shapes of conic sections (ellipses, parabolas, hyperbolas or circles). Those paths are the mathematical solutions to the
equations of motion which arise in the study of objects subjected to the force of gravity. A circular path is the simplest,
special case of this kind of motion.
Before considering rotating objects, we will begin by examining the idealization of a single point object moving in a path
that is shaped as a perfect circle. Of course, such an object is actually translating in a two dimensional manner. But, as we
will see, a mathematical analogy will allow us to describe that object using the kinematic equations of one dimensional
motion.
108 Chapter 4 | Circular Motion as One-Dimensional Motion
Figure 4.2 shows the orbital paths of some objects in our solar system. The motions of the planets around the Sun, although
they actually move along elliptical paths, can (at least initially) be approximated as being circular motion.
Figure 4.2 We see the orbits of typical comets and asteroids compared with those of the
planets Mercury, Venus, Earth, Mars, and Jupiter (shown in black). Shown in red are three
comets: Halley, Kopff, and Encke. In blue are the four largest asteroids: Ceres, Pallas, Vesta,
and Hygeia. While the orbits of the comets and asteroids are obviously elliptical, those of the
planets can be well approximated as circular.
Finally, to conclude the chapter, we will address the fixed-axis rotation of an extended object. Fixed-axis rotation describes
the rotation around a fixed axis of a rigid body; that is, an object that does not deform as it moves. We will show how to
apply all of the ideas we’ve developed up to this point about rotational and translational motion to such an object rotating
around a fixed axis.
So far in this text, we have only studied translational motion, including the variables that describe it: displacement, velocity,
and acceleration. Now we expand our description of motion to rotation—specifically, rotational motion about a fixed axis.
We will find that rotational motion is described by a set of related variables similar to those we used in translational motion.
Angular Velocity
Circular motion is, simply, motion in a perfectly circular path. Although this is the simplest case of rotational motion, it is
very useful for many situations, and we use it here to introduce rotational variables.
In Figure 4.3, we show (in red) a particle moving in a circle. The coordinate system is fixed and serves as a frame of
reference to define the particle’s position. Its position from the origin of the circle to the particle sweeps out the angle θ
, which increases in the counterclockwise direction as the particle moves along its circular path. The angle θ is called the
angular position of the particle. As the particle moves in its circular path, it also traces an arc length s.
The angle is related to the radius of the circle and the arc length by
θ = rs . (4.1)
The angle θ , the angular position of the particle along its path, has units of radians (rad). There are 2π radians in 360°.
Note that the radian measure is a ratio of length measurements, and therefore is a dimensionless quantity. As the particle
moves along its circular path, its angular position changes and it undergoes angular displacements Δθ.
Note again that the angle θ is measured counterclockwise starting from the positive x axis.
Now, let's note an interesting bit of mathematical simplification. Although this particle is moving in two dimensions, x and
y, we know that those coordinates can be written as:
x = rcosθ (4.2)
y = rsinθ (4.3)
However, the fact that the particle is constrained to move along a circular path, where r remains constant, means that its
only degree of motional freedom is the angle θ. So, in a mathematical sense, it is only moving in one dimension, and that
dimension is represented by the coordinate θ.
We can thus describe its location completely by simply stating the value of the coordinate θ, its angular position. Its motion
will be completely described by determining its angular displacements, Δθ , as time goes on. This means that we have
a system that is mathematically analogous to the one studied in Chapter 3, the only difference being that, there, the
dimension was called x (or y). Here, the dimension is called θ.
110 Chapter 4 | Circular Motion as One-Dimensional Motion
The magnitude of the angular velocity, denoted by ω , is the time rate of change of the angle θ as the particle moves in its
circular path. The instantaneous angular velocity is defined as the limit in which Δt → 0 in the average angular velocity
– = Δθ :
ω
Δt
ω = lim Δθ = dθ , (4.4)
Δt → 0 Δt dt
where θ is the angle of rotation (Figure 4.3). The units of angular velocity are radians per second (rad/s). Angular velocity
can also be referred to as the rotation rate in radians per second. In many situations, we are given the rotation rate in
revolutions/s or cycles/s. To find the angular velocity, we must multiply revolutions/s by 2π , since there are 2π radians in
one complete revolution. Since the direction of a positive angle in a circle is counterclockwise, we take counterclockwise
rotations as being positive and clockwise rotations as negative. (This choice is analogous to the one we made for horizontal,
translational motion. In that situation, moving from left to right was defined as moving in the positive motion, but moving
from right to left was in the negative direction.)
Which properties of motion are the same for these two points, and which are different? If you've ever ridden on a merry-go-
round, you know that the speed, v 2 , of a person located at point r 2 will be greater than the speed, v 1 , of a person located
at r 1 . Your experience tells you that, the farther you are from the axis of the rotating object, the faster you are moving. We
call v 1 and v 2 the tangential speeds, because they are moving instantaneously in a direction that is tangent to the circle
of their motion.
But, how do the angular speeds of the two particles (or people) compare? Since it will take each one exactly the same
amount of time to complete one rotation, their angular speed, ω , must be the same.
We can see how angular velocity is related to the tangential speed of the particle by going back to the definition of the
tangential displacement or arc length
s = rθ.
Noting that the radius r is a constant, we have
v t = Δs = rΔθ = r Δθ = rω
Δt Δt Δt
So, simply put, for any point located at a distance r from the axis of rotation, the tangential speed
v t = rω. (4.5)
That is, the tangential speed of the particle is its angular velocity times the radius of the circle traced out by its motion.
From Equation 4.5, we see that the tangential speed of the particle increases with its distance from the axis of rotation
for a constant angular velocity. This effect is shown in Figure 4.4. So it is for our two particles placed at different radii on
a rotating disk with a constant angular velocity. As the disk rotates, the tangential speed increases linearly with the radius
from the axis of rotation. In Figure 4.4, we see that v 1 = r 1 ω 1 and v 2 = r 2 ω 2 . But the disk has a constant angular
v v ⎛r ⎞
velocity, so ω 1 = ω 2 . This means r 1 = r 2 or v 2 = ⎝r 2 ⎠v 1 . Thus, since r 2 > r 1 , v 2 > v 1 .
1 2 1
Example 4.1
Rotation of a Flywheel
A flywheel rotates such that it sweeps out an angle at the rate of θ = ωt = (45.0 rad/s)t radians. The wheel
rotates counterclockwise when viewed in the plane of the page. (a) What is the angular velocity of the flywheel?
(b) How many radians does the flywheel rotate through in 30 s? (c) What is the tangential speed of a point on the
flywheel 10 cm from the axis of rotation?
Strategy
The functional form of the angular position of the flywheel is given in the problem as θ(t) = ωt , so we can find
the angular speed by inspection. It is just 45 rad/s. To find the angular displacement of the flywheel during 30
s, we seek the angular displacement Δθ , where the change in angular position is between 0 and 30 s. To find
the tangential speed of a point at a distance from the axis of rotation, we multiply its distance times the angular
velocity of the flywheel.
Solution
a. ω = 45 rad/s . We see that the angular velocity is a constant.
b. Δθ = θ(30 s) − θ(0 s) = 45.0(30.0 s) − 45.0(0 s) = 1350.0 rad .
c. v t = rω = (0.1 m)(45.0 rad/s) = 4.5 m/s .
Significance
In 30 s, the flywheel has rotated through quite a number of revolutions, about 215 if we divide the angular
displacement by 2π . A massive flywheel can be used to store energy in this way, if the losses due to friction are
minimal. Recent research has considered superconducting bearings on which the flywheel rests, with zero energy
loss due to friction.
Angular Acceleration
We have just discussed angular velocity for circular motion, but not all circular motion is at a uniform velocity. Envision
an ice skater spinning with his arms outstretched—when he pulls his arms inward, his angular velocity increases. Or think
about a computer’s hard disk slowing to a halt as the angular velocity decreases. We will explore these situations later, but
we can already see a need to define an angular acceleration for describing situations where ω changes. The faster the
change in ω , the greater the angular acceleration. We define the instantaneous angular acceleration α as the derivative
of angular velocity with respect to time:
2 (4.6)
α = lim Δω = dω = d 2θ ,
Δt → 0 Δt dt dt
112 Chapter 4 | Circular Motion as One-Dimensional Motion
– = Δω as Δt → 0 .
where we have taken the limit of the average angular acceleration, α
Δt
a t = rα. (4.8)
Let’s apply these ideas to the analysis of a few simple fixed-axis rotation scenarios. Before doing so, we present a problem-
solving strategy that can be applied to rotational kinematics: the description of rotational motion.
Example 4.2
Δt = Δω
α .
Solution
a. Entering known information into the definition of angular acceleration, we get
– = Δω = 250 rpm .
α
Δt 5.00 s
Because Δω is in revolutions per minute (rpm) and we want the standard units of rad/s 2 for angular
acceleration, we need to convert from rpm to rad/s:
4.1 Check Your Understanding The fan blades on a turbofan jet engine (shown below) accelerate from rest
up to a rotation rate of 40.0 rev/s in 20 s. The increase in angular velocity of the fan is constant in time. (The
GE90-110B1 turbofan engine mounted on a Boeing 777, as shown, is currently the largest turbofan engine in
the world, capable of thrusts of 330–510 kN.)
(a) What is the average angular acceleration?
(b) What is the instantaneous angular acceleration at any time during the first 20 s?
We now have a basic vocabulary for discussing fixed-axis rotational kinematics and relationships between rotational
variables. We discuss more definitions and connections in the next section.
114 Chapter 4 | Circular Motion as One-Dimensional Motion
In the preceding section, we defined the rotational variables of angular displacement, angular velocity, and angular
acceleration. In this section, we work with these definitions to derive relationships among these variables and use these
relationships to analyze rotational motion for a rigid body about a fixed axis under a constant angular acceleration. (By
the analogy previously established, these relationships also hold for any point-like object undergoing circular motion.)
This analysis forms the basis for rotational kinematics. If the angular acceleration is constant, the equations of rotational
kinematics simplify, similar to the equations of linear kinematics discussed in Section 3.4. We can then use this simplified
set of equations to describe many applications in physics and engineering where the angular acceleration of the system is
constant.
– = ω0 + ωf .
ω (4.9)
2
From the definition of the average angular velocity, we can find an equation that relates the angular position, average
angular velocity, and time:
– = Δθ .
ω
Δt
Solving for θ , we have
– t,
θf = θ0 + ω (4.10)
where we have set t 0 = 0 . This equation can be very useful if we know the average angular velocity of the system. Then we
could find the angular displacement over a given time period. Next, we find an equation relating ω , α , and t. To determine
this equation, we start with the definition of angular acceleration:
α = Δω .
Δt
We rearrange this to get αΔt = Δω . In uniform rotational motion, the angular acceleration is constant so this becomes:
α(t − t 0) = ω f − ω 0 (4.11)
Setting t 0 = 0 , we have
αt = ω f − ω 0.
ω f = ω 0 + αt, (4.12)
where ω 0 is the initial angular velocity. Equation 4.12 is the rotational counterpart to the linear kinematics equation
v f = v 0 + at . With Equation 4.12, we can find the angular velocity of an object at any specified time t given the initial
angular velocity and the angular acceleration.
Again by analogy to the discussion in Section 3.4, where we have set the initial time t 0 = 0 , we can obtain another
equation:
θ f = θ 0 + ω 0 t + 1 αt 2. (4.13)
2
Equation 4.13 is the rotational counterpart to the linear kinematics equation found in Section 3.4 for position as a
function of time. This equation gives us the angular position of a rotating rigid body at any time t given the initial conditions
(initial angular position and initial angular velocity) and the angular acceleration.
We can find an equation that is independent of time by solving for t in Equation 4.12 and substituting into Equation
4.13. Equation 4.13 becomes
⎛ω f − ω 0 ⎞ 1 ⎛ω f − ω 0 ⎞2
θf = θ0 + ω0 ⎝ α ⎠ + 2 α⎝ α ⎠
ω 0 ω f ω 20 1 ω 2f ω 0 ω f 1 ω 20
= θ0 + α − α +2 α − α +2 α
ω2 ω2
= θ 0 + 1 αf − 1 α0 ,
2 2
ω 2f − ω 20
θf − θ0 =
2α
or
ω 2f = ω 20 + 2α(Δθ). (4.14)
Equation 4.10 through Equation 4.14 describe fixed-axis rotation for constant acceleration and are summarized in
Table 4.1.
Example 4.3
Strategy
Identify the knowns and compare with the kinematic equations for constant acceleration. Look for the appropriate
equation that can be solved for the unknown, using the knowns given in the problem description.
Solution
a. We are given α and t and want to determine ω . The most straightforward equation to use is
ω f = ω 0 + αt , since all terms are known besides the unknown variable we are looking for. We are given
that ω 0 = 0 (it starts from rest), so
θ f = θ i + ω i t + 1 αt 2
2
= 0 + 0 + (0.500)⎛⎝110 rad/s 2⎞⎠(2.00 s) 2 = 220 rad.
Significance
This example illustrates that relationships among rotational quantities are highly analogous to those among linear
quantities. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at
220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.)
In the preceding example, we considered a fishing reel with a positive angular acceleration. Now let us consider what
happens with a negative angular acceleration.
Example 4.4
Calculating the Duration When the Fishing Reel Slows Down and Stops
Now the fisherman applies a brake to the spinning reel, achieving an angular acceleration of −300 rad/s 2 . How
long does it take the reel to come to a stop?
Strategy
We are asked to find the time t for the reel to come to a stop. The initial and final conditions are different from
those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is
ω 0 = 220 rad/s and the final angular velocity ω is zero. The angular acceleration is given as α = −300 rad/s 2.
Examining the available equations, we see all quantities but t are known in ω f = ω 0 + αt , making it easiest to
use this equation.
Solution
The equation states
ω f = ω 0 + αt.
We solve the equation algebraically for t and then substitute the known values as usual, yielding
ω f − ω 0 0 − 220.0 rad/s
t= α = = 0.733 s.
−300.0 rad/s 2
Significance
Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the
time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because
of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel.
A tired fish is slower, requiring a smaller acceleration.
4.2 Check Your Understanding A centrifuge used in DNA extraction spins at a maximum rate of 7000 rpm,
producing a “g-force” on the sample that is 6000 times the force of gravity. If the centrifuge takes 10 seconds to
come to rest from the maximum spin rate: (a) What is the angular acceleration of the centrifuge? (b) What is the
angular displacement of the centrifuge during this time?
Example 4.5
Strategy
a. Since the angular velocity varies linearly with time, we know that the angular acceleration is constant and
does not depend on the time variable. The angular acceleration is the slope of the angular velocity vs. time
graph, α = dω . To calculate the slope, we read directly from Figure 4.7, and see that ω 0 = 30 rad/s
dt
at t = 0 s and ω f = 0 rad/s at t = 5 s .
Solution
a. Calculating the slope, we get
ω−ω (0 − 30.0) rad/s
α= t−t 0 = = −6.0 rad/s 2.
0 (5.0 − 0) s
b. θ f = θ 0 + ω 0 t + 1 αt 2.
2
Setting θ 0 = 0 , we have
θ 0 = (30.0 rad/s)(5.0 s) + 1 (−6.0 rad/s 2)(5.0 rad/s) 2 = 150.0 − 75.0 = 75.0 rad.
2
In this section, as we consider the motion of some point on a rotating object, we will relate each of the rotational variables
to the translational variables defined in Straight-Line Motion. This will complete our ability to describe rigid-body
rotations.
Linear Rotational
Position x θ
Velocity v = dx ω = dθ
dt dt
Acceleration a = dv α = dω
dt dt
Let’s compare the linear and rotational variables individually. The linear variable of position has physical units of meters,
whereas the angular position variable has dimensionless units of radians, as can be seen from the definition of θ = rs , which
is the ratio of two lengths. The linear velocity has units of m/s, and its counterpart, the angular velocity, has units of rad/
s. In Section 4.1, we saw in the case of circular motion that the linear tangential speed of a particle at a radius r from the
axis of rotation is related to the angular velocity by the relation v t = rω . This could also apply to points on a rigid body
rotating about a fixed axis. Here, we consider only circular motion. (In circular motion, both uniform and nonuniform, there
exists another acceleration, the centripetal acceleration, which will be discussed in Section 7.4.)
Rotational Translational
θ =θ + ω
f
–t
0 x = x + –v t
0
ω f = ω 0 + αt v f = v 0 + at
θ f = θ 0 + ω 0 t + 1 αt 2 x f = x 0 + v 0 t + 1 at 2
2 2
ω 2f = ω 20 + 2α(Δθ) v 2f = v 20 + 2a(Δx)
2. The second correspondence has to do with relating linear and rotational variables in the special case of circular
motion. Importantly, any object moving in a circular path possesses both an angular velocity and a tangential linear
velocity. It possesses both an angular acceleration and a tangential linear acceleration. This is shown in Table 4.3,
where in the third column, we have listed the connecting equation that relates the linear variable to the rotational
variable for a rotating point located a distance r from its axis of rotation. The rotational variables of angular
velocity and acceleration have subscripts that indicate their definition in circular motion.
120 Chapter 4 | Circular Motion as One-Dimensional Motion
θ s θ = rs
ω vt v
ω = rt
α at a
α = rt
Example 4.6
4.3 Check Your Understanding A boy jumps on a merry-go-round with a radius of 5 m that is at rest. It
starts accelerating at a constant rate up to an angular velocity of 5 rad/s in 20 seconds. What is the distance
travelled by the boy?
CHAPTER 4 REVIEW
KEY TERMS
angular acceleration time rate of change of angular velocity
angular position angle a body has rotated through in a fixed coordinate system
angular velocity time rate of change of angular position
instantaneous angular acceleration derivative of angular velocity with respect to time
instantaneous angular velocity derivative of angular position with respect to time
kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular
acceleration, and time
SUMMARY
4.1 Rotational Variables
• The angular position θ of a rotating body is the angle the body has rotated through in a fixed coordinate system,
which serves as a frame of reference.
• The angular velocity of a rotating body about a fixed axis is defined as ω(rad/s) , the rotational rate of the body in
radians per second. The instantaneous angular velocity of a rotating body ω = lim Δω = dθ is the derivative
Δt → 0 Δt dt
with respect to time of the angular position θ , found by taking the limit Δt → 0 in the average angular velocity
– = Δθ . The angular velocity relates v to the tangential speed of a point on the rotating body through the
ω
Δt t
relation v t = rω , where r is the radius to the point and v t is the tangential speed at the given point.
• If the system’s angular velocity is not constant, then the system has an angular acceleration. The average angular
– = Δω . The
acceleration over a given time interval is the change in angular velocity over this time interval, α
Δt
instantaneous angular acceleration is the time derivative of angular velocity, α = lim Δω = dω . The sign of the
Δt → 0 Δt dt
angular acceleration α is found by examining the angular velocity. If a rotation rate of a rotating body is decreasing,
the angular acceleration is in the opposite direction to ω . If the rotation rate is increasing, the angular acceleration
is in the same direction as ω .
• The tangential acceleration of a point at a radius from the axis of rotation is the angular acceleration times the radius
to the point.
Table 4.4 Comparison of the Kinematic Equations for Rotational and Translational Motion
CONCEPTUAL QUESTIONS
time variable?
4.1 Rotational Variables
1. A clock is mounted on the wall. As you look at it, what 5. If a rigid body has a constant angular acceleration, what
is the direction of the angular velocity vector of the second is the functional form of the angular position?
hand?
6. If the angular acceleration of a rigid body is zero, what
2. What is the value of the angular acceleration of the is the functional form of the angular velocity?
second hand of the clock on the wall?
7. A massless tether with a masses tied to both ends
3. A baseball bat is swung. Do all points on the bat have rotates about a fixed axis through the center. Can the total
the same angular velocity? The same tangential speed? acceleration of the tether/mass combination be zero if the
angular velocity is constant?
PROBLEMS
11. A particle moves 3.0 m along a circle of radius 1.5
4.1 Rotational Variables m. (a) Through what angle does it rotate? (b) If the particle
8. Calculate the angular velocity of Earth. makes this trip in 1.0 s at a constant speed, what is its
angular velocity?
9. A track star runs a 400-m race on a 400-m circular track
in 45 s. What is his angular velocity assuming a constant 12. A compact disc rotates at 500 rev/min. If the diameter
speed? of the disc is 120 mm, (a) what is the tangential speed of a
point at the edge of the disc? (b) At a point halfway to the
center of the disc?
10. A wheel rotates at a constant rate of
2.0 × 10 3 rev/min . (a) What is its angular velocity in 13. Unreasonable results. The propeller of an aircraft is
radians per second? (b) Through what angle does it turn in spinning at 10 rev/s when the pilot shuts off the engine.
10 s? Express the solution in radians and degrees. The propeller reduces its angular velocity at a constant
2.0 rad/s 2 for a time period of 40 s. What is the rotation acceleration of 1.0 rad/s 2 ; at t = 0 its angular velocity
rate of the propeller in 40 s? Is this a reasonable situation? is 2.0 rad/s. (a) Determine the disk’s angular velocity at
t = 5.0 s . (b) What is the angle it has rotated through
14. A gyroscope slows from an initial rate of 32.0 rad/s at during this time? (c) What is the tangential acceleration of
a rate of 0.700 rad/s 2 . How long does it take to come to a point on the disk at t = 5.0 s ?
rest?
23. The angular velocity vs. time for a fan on a hovercraft
15. On takeoff, the propellers on a UAV (unmanned aerial is shown below. (a) What is the angle through which the fan
vehicle) increase their angular velocity from rest at a rate of blades rotate in the first 8 seconds? (b) Verify your result
ω = (25.0t) rad/s for 3.0 s. (a) What is the instantaneous using the kinematic equations.
angular velocity of the propellers at t = 2.0 s ? (b) What is
the angular acceleration?
21. A vertical wheel with a diameter of 50 cm starts 28. What is (a) the angular speed and (b) the linear speed
from rest and rotates with a constant angular acceleration of a point on Earth’s surface at latitude 30° N. Take the
of 5.0 rad/s 2 around a fixed axis through its center radius of the Earth to be 6309 km. (c) At what latitude
counterclockwise. (a) Where is the point that is initially at would your linear speed be 10 m/s?
the bottom of the wheel at t = 10 s? (b) What is the point’s
linear acceleration at this instant? 29. A bicycle wheel with radius 0.3m rotates from rest
to 3 rev/s in 5 s. What is the magnitude of the tangential
22. A circular disk of radius 10 cm has a constant angular acceleration of a point on the outside edge of the wheel?
124 Chapter 4 | Circular Motion as One-Dimensional Motion
5 | RELATIVISTIC
KINEMATICS IN ONE
DIMENSION
Chapter Outline
5.1 Invariance of Physical Laws
5.2 Relativity of Simultaneity
5.3 Time Dilation
5.4 Length Contraction
5.5 The Lorentz Transformation
5.6 Relativistic Velocity Transformation
Introduction
Figure 5.1 Special relativity explains how time passes slightly differently on Earth and within the rapidly moving global
positioning satellite (GPS). GPS units in vehicles could not find their correct location on Earth without taking this correction into
account. (credit: USAF)
Our description of motion (kinematics), thus far, has hopefully obeyed your common sense ideas about such things. For
example, when we use the equation
t = Δx
v
it hopefully seems sensible to you that, if you are moving at a constant speed, v , then the time t it takes you to travel some
distance Δx is directly proportional to that distance, but inversely proportional to your speed.
Here's another bit of common-sense kinematics: Suppose that on a winter day you are pulling a child on a sled, which is
moving to the right at a speed of v = 1 m/s. The child takes a snowball, and throws it forward at a speed u' = 1.5 m/s
126 Chapter 5 | Relativistic Kinematics in One Dimension
relative to herself, you and the sled. She's trying to hit her friend, who is standing at rest just in front of you (see Figure 1).
At what speed, u , does her friend observe the snowball to be moving?
Figure 5.2 Classically, velocities add like ordinary numbers in one-dimensional motion.
Here the girl throws a snowball forward and then backward from a sled. The velocity of the
sled relative to the Earth is v=1.0 m/s . The velocity of the snowball relative to the truck is
u′ , while its velocity relative to the Earth is u . Classically, u=v+u′ .
(Credit: OpenStax College Physics. "Relativistic Addition of Velocities." OpenStax-CNX.
September 12, 2013. https://legacy.cnx.org/content/m42540/1.6/)
The answer, of course, is child's play! (Sorry...) But your common sense tells you that it's moving at 2.5 m/s, and if you
think about an equation that tells that story, it's just
u = u' + v
We can see that this equation is still valid if the child throws the snowball in the opposite direction. In that case, the value
of u' = -1.5 m/s, with the minus sign indicating that the direction of her throw is in the negative x direction. The equation
yields u = -0.5 m/s. Thus, to the stationary friend ahead, the snowball appears to be moving away to the left at a speed of
0.5 m/s.
This classical addition of velocities may be common sense, but early in the 20th century it was shown to be wrong for
objects that travel at or near the speed of light. And, although our everyday experience does not seem to include objects that
move that fast, as Figure 5.1 shows, there are examples where the so-called relativistic corrections do matter. In particular,
as we study astrophysics (where the distance scales can be enormous) and the motion of elementary particles (which do
move near the speed of light), we must incorporate the theory of relativity.
The special theory of relativity was proposed in 1905 by Albert Einstein (1879–1955). It describes how time, space, and
physical phenomena appear in different frames of reference that are moving at constant velocity with respect to each other.
This differs from Einstein’s later work on general relativity, which deals with any frame of reference, including accelerated
frames.
The theory of relativity led to a profound change in the way we perceive space and time. The “common sense” rules that
we use to relate space and time measurements in the Newtonian worldview differ seriously from the correct rules at speeds
near the speed of light. For example, the special theory of relativity tells us that measurements of length and time intervals
are not the same in reference frames moving relative to one another. A particle might be observed to have a lifetime of
1.0 × 10 −8 s in one reference frame, but a lifetime of 2.0 × 10 −8 s in another; and an object might be measured to be
2.0 m long in one frame and 3.0 m long in another frame. These effects are usually significant only at speeds comparable to
the speed of light, but even at the much lower speeds of the global positioning satellite, which requires extremely accurate
time measurements to function, the different lengths of the same distance in different frames of reference are significant
enough that they need to be taken into account.
The modifications of kinematics in special relativity do not invalidate classical theories or require their replacement. Instead,
the equations of relativistic mechanics differ meaningfully from those of classical mechanics only for objects moving at
relativistic speeds (i.e., speeds less than, but comparable to, the speed of light). In the macroscopic world that you encounter
in your daily life, the relativistic equations reduce to classical equations, and the predictions of classical mechanics agree
closely enough with experimental results to disregard relativistic corrections.
Suppose you calculate the hypotenuse of a right triangle given the base angles and adjacent sides. Whether you calculate
the hypotenuse from one of the sides and the cosine of the base angle, or from the Pythagorean theorem, the results
should agree. Predictions based on different principles of physics must also agree, whether we consider them principles of
mechanics or principles of electromagnetism.
Albert Einstein pondered a disagreement between predictions based on electromagnetism and on assumptions made in
classical mechanics. Specifically, suppose an observer measures the velocity of a light pulse in the observer’s own rest
frame; that is, in the frame of reference in which the observer is at rest. According to the assumptions long considered
obvious in classical mechanics, if an observer measures a velocity → v in one frame of reference, and that frame of
reference is moving with velocity →
u past a second reference frame, an observer in the second frame measures the original
→
velocity as v′ = → v + → u . This sum of velocities is often referred to as Galilean relativity. If this principle is correct,
the pulse of light that the observer measures as traveling with speed c travels at speed c + u measured in the frame of the
second observer. If we reasonably assume that the laws of electrodynamics are the same in both frames of reference, then the
predicted speed of light (in vacuum) in both frames should be c = 1/ ε 0 μ 0. Each observer should measure the same speed
of the light pulse with respect to that observer’s own rest frame. To reconcile difficulties of this kind, Einstein constructed
his special theory of relativity, which introduced radical new ideas about time and space that have since been confirmed
experimentally.
Inertial Frames
All velocities are measured relative to some frame of reference. For example, a car’s motion is measured relative to its
starting position on the road it travels on; a projectile’s motion is measured relative to the surface from which it is launched;
and a planet’s orbital motion is measured relative to the star it orbits. The frames of reference in which mechanics takes the
simplest form are those that are not accelerating. Newton’s first law, the law of inertia, holds exactly in such a frame.
For example, to a passenger inside a plane flying at constant speed and constant altitude, physics seems to work exactly
the same as when the passenger is standing on the surface of Earth. When the plane is taking off, however, matters are
somewhat more complicated. In this case, the passenger at rest inside the plane concludes that a net force F on an object
is not equal to the product of mass and acceleration, ma. Instead, F is equal to ma plus a fictitious force. This situation is
128 Chapter 5 | Relativistic Kinematics in One Dimension
not as simple as in an inertial frame. The term “special” in “special relativity” refers to dealing only with inertial frames of
reference. Einstein’s later theory of general relativity deals with all kinds of reference frames, including accelerating, and
therefore non-inertial, reference frames.
This postulate denies the existence of a special or preferred inertial frame. The laws of nature do not give us a way to
endow any one inertial frame with special properties. For example, we cannot identify any inertial frame as being in a state
of “absolute rest.” We can only determine the relative motion of one frame with respect to another.
There is, however, more to this postulate than meets the eye. The laws of physics include only those that satisfy this
postulate. We will see that the definitions of energy and momentum must be altered to fit this postulate. Another outcome
of this postulate is the famous equation E = mc 2, which relates energy to mass.
Michelson-Morley Experiment
The Michelson-Morley experiment demonstrated that the speed of light in a vacuum is independent of the motion of
Earth about the Sun.
The eventual conclusion derived from this result is that light, unlike mechanical waves such as sound, does not need a
medium to carry it. Furthermore, the Michelson-Morley results implied that the speed of light c is independent of the motion
of the source relative to the observer. That is, everyone observes light to move at speed c regardless of how they move
relative to the light source or to one another. For several years, many scientists tried unsuccessfully to explain these results
within the framework of Newton’s laws.
In addition, there was a contradiction between the principles of electromagnetism and the assumption made in Newton’s
laws about relative velocity. Classically, the velocity of an object in one frame of reference and the velocity of that object in
a second frame of reference relative to the first should combine like simple vectors to give the velocity seen in the second
frame. If that were correct, then two observers moving at different speeds would see light traveling at different speeds.
Imagine what a light wave would look like to a person traveling along with it (in vacuum) at a speed c. If such a motion
were possible, then the wave would be stationary relative to the observer. It would have electric and magnetic fields whose
strengths varied with position but were constant in time. This is not allowed by Maxwell’s equations. So either Maxwell’s
equations are different in different inertial frames, or an object with mass cannot travel at speed c. Einstein concluded that
the latter is true: An object with mass cannot travel at speed c. Maxwell’s equations are correct, but Newton’s addition of
velocities is not correct for light.
Not until 1905, when Einstein published his first paper on special relativity, was the currently accepted conclusion reached.
Based mostly on his analysis that the laws of electricity and magnetism would not allow another speed for light, and only
slightly aware of the Michelson-Morley experiment, Einstein detailed his second postulate of special relativity.
In other words, the speed of light has the same definite speed for any observer, regardless of the relative motion of the
source. This deceptively simple and counterintuitive postulate, along with the first postulate, leave all else open for change.
Among the changes are the loss of agreement on the time between events, the variation of distance with speed, and the
realization that matter and energy can be converted into one another. We describe these concepts in the following sections.
5.1 Check Your Understanding Explain how special relativity differs from general relativity.
Do time intervals depend on who observes them? Intuitively, it seems that the time for a process, such as the elapsed time
for a foot race (Figure 5.3), should be the same for all observers. In everyday experiences, disagreements over elapsed time
have to do with the accuracy of measuring time. No one would be likely to argue that the actual time interval was different
for the moving runner and for the stationary clock displayed. Carefully considering just how time is measured, however,
shows that elapsed time does depends on the relative motion of an observer with respect to the process being measured.
130 Chapter 5 | Relativistic Kinematics in One Dimension
Figure 5.3 Elapsed time for a foot race is the same for all observers, but at relativistic speeds,
elapsed time depends on the motion of the observer relative to the location where the process
being timed occurs. (credit: "Jason Edward Scott Bain"/Flickr)
Consider how we measure elapsed time. If we use a stopwatch, for example, how do we know when to start and stop the
watch? One method is to use the arrival of light from the event. For example, if you’re in a moving car and observe the light
arriving from a traffic signal change from green to red, you know it’s time to step on the brake pedal. The timing is more
accurate if some sort of electronic detection is used, avoiding human reaction times and other complications.
Now suppose two observers use this method to measure the time interval between two flashes of light from flash lamps
that are a distance apart (Figure 5.4). An observer A is seated midway on a rail car with two flash lamps at opposite sides
equidistant from her. A pulse of light is emitted from each flash lamp and moves toward observer A, shown in frame (a)
of the figure. The rail car is moving rapidly in the direction indicated by the velocity vector in the diagram. An observer B
standing on the platform is facing the rail car as it passes and observes both flashes of light reach him simultaneously, as
shown in frame (c). He measures the distances from where he saw the pulses originate, finds them equal, and concludes that
the pulses were emitted simultaneously.
However, because of Observer A’s motion, the pulse from the right of the railcar, from the direction the car is moving,
reaches her before the pulse from the left, as shown in frame (b). She also measures the distances from within her frame of
reference, finds them equal, and concludes that the pulses were not emitted simultaneously.
The two observers reach conflicting conclusions about whether the two events at well-separated locations were
simultaneous. Both frames of reference are valid, and both conclusions are valid. Whether two events at separate locations
are simultaneous depends on the motion of the observer relative to the locations of the events.
Figure 5.4 (a) Two pulses of light are emitted simultaneously relative to observer B. (c) The pulses reach
observer B’s position simultaneously. (b) Because of A’s motion, she sees the pulse from the right first and
concludes the bulbs did not flash simultaneously. Both conclusions are correct.
Here, the relative velocity between observers affects whether two events a distance apart are observed to be simultaneous.
Simultaneity is not absolute. We might have guessed (incorrectly) that if light is emitted simultaneously, then two observers
halfway between the sources would see the flashes simultaneously. But careful analysis shows this cannot be the case if the
speed of light is the same in all inertial frames.
This type of thought experiment (in German, “Gedankenexperiment”) shows that seemingly obvious conclusions must be
changed to agree with the postulates of relativity. The validity of thought experiments can only be determined by actual
observation, and careful experiments have repeatedly confirmed Einstein’s theory of relativity.
132 Chapter 5 | Relativistic Kinematics in One Dimension
The analysis of simultaneity shows that Einstein’s postulates imply an important effect: Time intervals have different values
when measured in different inertial frames. Suppose, for example, an astronaut measures the time it takes for a pulse of
light to travel a distance perpendicular to the direction of his ship’s motion (relative to an earthbound observer), bounce off
a mirror, and return (Figure 5.5). How does the elapsed time that the astronaut measures in the spacecraft compare with
the elapsed time that an earthbound observer measures by observing what is happening in the spacecraft?
Examining this question leads to a profound result. The elapsed time for a process depends on which observer is measuring
it. In this case, the time measured by the astronaut (within the spaceship where the astronaut is at rest) is smaller than
the time measured by the earthbound observer (to whom the astronaut is moving). The time elapsed for the same process
is different for the observers, because the distance the light pulse travels in the astronaut’s frame is smaller than in the
earthbound frame, as seen in Figure 5.5. Light travels at the same speed in each frame, so it takes more time to travel the
greater distance in the earthbound frame.
Figure 5.5 (a) An astronaut measures the time Δτ for light to travel distance 2D in the astronaut’s frame. (b) A NASA
scientist on Earth sees the light follow the longer path 2s and take a longer time Δt. (c) These triangles are used to find the
relationship between the two distances D and s.
Time Dilation
Time dilation is the lengthening of the time interval between two events for an observer in an inertial frame that is
moving with respect to the rest frame of the events (in which the events occur at the same location).
To quantitatively compare the time measurements in the two inertial frames, we can relate the distances in Figure 5.5 to
each other, then express each distance in terms of the time of travel (respectively either Δt or Δτ ) of the pulse in the
corresponding reference frame. The resulting equation can then be solved for Δt in terms of Δτ.
The lengths D and L in Figure 5.5 are the sides of a right triangle with hypotenuse s. From the Pythagorean theorem,
s 2 = D 2 + L 2.
The lengths 2s and 2L are, respectively, the distances that the pulse of light and the spacecraft travel in time Δt in the
earthbound observer’s frame. The length D is the distance that the light pulse travels in time Δτ in the astronaut’s frame.
This gives us three equations:
2s = cΔt; 2L = vΔt; 2D = cΔτ.
134 Chapter 5 | Relativistic Kinematics in One Dimension
Note that we used Einstein’s second postulate by taking the speed of light to be c in both inertial frames. We substitute these
results into the previous expression from the Pythagorean theorem:
s2 = D2 + L2
⎛ Δt ⎞ ⎛ ⎞ ⎛ ⎞
2 2 2
⎝c 2 ⎠ = ⎝c Δτ ⎠ + ⎝v Δt ⎠ .
2 2
Then we rearrange to obtain
(cΔt) 2 − (vΔt) 2 = (cΔτ) 2.
Finally, solving for Δt in terms of Δτ gives us
Δt = Δτ . (5.1)
1 − (v/c) 2
This is equivalent to
Δt = γΔτ,
where γ is the relativistic factor (often called the Lorentz factor) given by
γ= 1 (5.2)
1− v2
c2
and v and c are the speeds of the moving observer and light, respectively.
Note the asymmetry between the two measurements. Only one of them is a measurement of the time interval between two
events—the emission and arrival of the light pulse—at the same position. It is a measurement of the time interval in the
rest frame of a single clock. The measurement in the earthbound frame involves comparing the time interval between two
events that occur at different locations. The time interval between events that occur at a single location has a separate name
to distinguish it from the time measured by the earthbound observer, and we use the separate symbol Δτ to refer to it
throughout this chapter.
Proper Time
The proper time interval Δτ between two events is the time interval measured by an observer for whom both events
occur at the same location.
The equation relating Δt and Δτ is truly remarkable. First, as stated earlier, elapsed time is not the same for different
observers moving relative to one another, even though both are in inertial frames. A proper time interval Δτ for an
observer who, like the astronaut, is moving with the apparatus, is smaller than the time interval for other observers. It is the
smallest possible measured time between two events. The earthbound observer sees time intervals within the moving system
as dilated (i.e., lengthened) relative to how the observer moving relative to Earth sees them within the moving system.
Alternatively, according to the earthbound observer, less time passes between events within the moving frame. Note that the
shortest elapsed time between events is in the inertial frame in which the observer sees the events (e.g., the emission and
arrival of the light signal) occur at the same point.
This time effect is real and is not caused by inaccurate clocks or improper measurements. Time-interval measurements of
the same event differ for observers in relative motion. The dilation of time is an intrinsic property of time itself. All clocks
moving relative to an observer, including biological clocks, such as a person’s heartbeat, or aging, are observed to run more
slowly compared with a clock that is stationary relative to the observer.
Note that if the relative velocity is much less than the speed of light (v<<c), then v 2 /c 2 is extremely small, and the
elapsed times Δt and Δτ are nearly equal. At low velocities, physics based on modern relativity approaches classical
physics—everyday experiences involve very small relativistic effects. However, for speeds near the speed of light, v 2 /c 2
is close to one, so 1 − v 2/c 2 is very small and Δt becomes significantly larger than Δτ.
Half-Life of a Muon
There is considerable experimental evidence that the equation Δt = γΔτ is correct. One example is found in cosmic ray
particles that continuously rain down on Earth from deep space. Some collisions of these particles with nuclei in the upper
atmosphere result in short-lived particles called muons. The half-life (amount of time for half of a material to decay) of a
muon is 1.52 μs when it is at rest relative to the observer who measures the half-life. This is the proper time interval Δτ.
This short time allows very few muons to reach Earth’s surface and be detected if Newtonian assumptions about time and
space were correct. However, muons produced by cosmic ray particles have a range of velocities, with some moving near
the speed of light. It has been found that the muon’s half-life as measured by an earthbound observer ( Δt ) varies with
velocity exactly as predicted by the equation Δt = γΔτ. The faster the muon moves, the longer it lives. We on Earth see
the muon last much longer than its half-life predicts within its own rest frame. As viewed from our frame, the muon decays
more slowly than it does when at rest relative to us. A far larger fraction of muons reach the ground as a result.
Before we present the first example of solving a problem in relativity, we state a strategy you can use as a guideline for
these calculations.
Note that you should not round off during the calculation. As noted in the text, you must often perform your calculations to
many digits to see the desired effect. You may round off at the very end of the problem solution, but do not use a rounded
number in a subsequent calculation. Also, check the answer to see if it is reasonable: Does it make sense? This may be more
difficult for relativity, which has few everyday examples to provide experience with what is reasonable. But you can look
for velocities greater than c or relativistic effects that are in the wrong direction (such as a time contraction where a dilation
was expected).
136 Chapter 5 | Relativistic Kinematics in One Dimension
Example 5.1
Δt = γΔτ = Δτ .
2
1 − v2
c
d. Do the calculation. Use the expression for γ to determine Δt from Δτ :
Δt = 1s
⎛ 5830 m/s ⎞
2
1−⎝
3.00 × 10 8 m/s ⎠
= 1.000000000189 s
= 1 s + 1.89 × 10 −10 s.
Significance
The very high speed of the HTV-2 is still only 10-5 times the speed of light. Relativistic effects for the HTV-2 are
negligible for almost all purposes, but are not zero.
Example 5.2
Δt = γΔτ = 1 Δτ
1 − v 2/c 2
Δτ = 1 − v 2/c 2
Δt
⎛Δτ ⎞
2 2
⎝ Δt ⎠ = 1 − v2
c
v = 1 − (Δτ/Δt) 2.
c
d. Do the calculation:
v = 1 − (1/1.01) 2
c
= 0.14.
Significance
The result shows that an object must travel at very roughly 10% of the speed of light for its motion to produce
significant relativistic time dilation effects.
Example 5.3
As we will discuss later, in the muon’s reference frame, it travels a shorter distance than measured in Earth’s
reference frame.
Strategy
A clock moving with the muon measures the proper time of its decay process, so the time we are given is
Δτ = 2.20μs. The earthbound observer measures Δt as given by the equation Δt = γΔτ. Because the velocity
is given, we can calculate the time in Earth’s frame of reference.
138 Chapter 5 | Relativistic Kinematics in One Dimension
Solution
a. Identify the knowns: v = 0.950c, Δτ = 2.20μs.
with
γ= 1 .
2
1 − v2
c
d. Do the calculation. Use the expression for γ to determine Δt from Δτ :
Δt = γΔτ
= 1 Δτ
2
1 − v2
c
2.20μs
=
1 − (0.950) 2
= 7.05 μs.
Example 5.4
Relativistic Television
A non-flat screen, older-style television display (Figure 5.7) works by accelerating electrons over a short
distance to relativistic speed, and then using electromagnetic fields to control where the electron beam strikes a
fluorescent layer at the front of the tube. Suppose the electrons travel at 6.00 × 10 7 m/s through a distance of
0.200 m from the start of the beam to the screen. (a) What is the time of travel of an electron in the rest frame of
the television set? (b) What is the electron’s time of travel in its own rest frame?
Figure 5.7 The electron beam in a cathode ray tube television display.
Δt = dv .
d. Do the calculation:
t = 0.200 m
6.00 × 10 7 m/s
= 3.33 × 10 −9 s.
Significance
The time of travel is extremely short, as expected. Because the calculation is entirely within a single frame of
reference, relativity is not involved, even though the electron speed is close to c.
Strategy for (b)
(b) In the frame of reference of the electron, the vacuum tube is moving and the electron is stationary. The
electron-emitting cathode leaves the electron and the front of the vacuum tube strikes the electron with the
electron at the same location. Therefore we use the time dilation formula to relate the proper time in the electron
rest frame to the time in the television frame.
140 Chapter 5 | Relativistic Kinematics in One Dimension
Solution
a. Identify the knowns (from part a):
Δt = 3.33 × 10 −9 s; v = 6.00 × 10 7 m/s; d = 0.200 m.
b. Identify the unknown: τ.
c. Express the answer as an equation:
Δt = γΔτ = Δτ
1 − v 2/c 2
Δτ = Δt 1 − v 2/c 2.
d. Do the calculation:
⎛ ⎞
2
7
Δτ = ⎛⎝3.33 × 10 −9 s⎞⎠ 1 − 6.00 × 108 m/s
⎝3.00 × 10 m/s ⎠
= 3.26 × 10 −9 s.
Significance
The time of travel is shorter in the electron frame of reference. Because the problem requires finding the time
interval measured in different reference frames for the same process, relativity is involved. If we had tried to
calculate the time in the electron rest frame by simply dividing the 0.200 m by the speed, the result would be
slightly incorrect because of the relativistic speed of the electron.
The paradox here is that the two twins cannot both be correct. As with all paradoxes, conflicting conclusions come from
a false premise. In fact, the astronaut’s motion is significantly different from that of the earthbound twin. The astronaut
accelerates to a high velocity and then decelerates to view the star system. To return to Earth, she again accelerates and
decelerates. The spacecraft is not in a single inertial frame to which the time dilation formula can be directly applied. That
is, the astronaut twin changes inertial references. The earthbound twin does not experience these accelerations and remains
in the same inertial frame. Thus, the situation is not symmetric, and it is incorrect to claim that the astronaut observes the
same effects as her twin. The lack of symmetry between the twins will be still more evident when we analyze the journey
later in this chapter in terms of the path the astronaut follows through four-dimensional space-time.
In 1971, American physicists Joseph Hafele and Richard Keating verified time dilation at low relative velocities by flying
extremely accurate atomic clocks around the world on commercial aircraft. They measured elapsed time to an accuracy of
a few nanoseconds and compared it with the time measured by clocks left behind. Hafele and Keating’s results were within
experimental uncertainties of the predictions of relativity. Both special and general relativity had to be taken into account,
because gravity and accelerations were involved as well as relative motion.
5.3 Check Your Understanding a. A particle travels at 1.90 × 10 8 m/s and lives 2.10 × 10 −8 s when at
rest relative to an observer. How long does the particle live as viewed in the laboratory?
b. Spacecraft A and B pass in opposite directions at a relative speed of 4.00 × 10 7 m/s. An internal clock in
spacecraft A causes it to emit a radio signal for 1.00 s. The computer in spacecraft B corrects for the beginning
and end of the signal having traveled different distances, to calculate the time interval during which ship A was
emitting the signal. What is the time interval that the computer in spacecraft B calculates?
142 Chapter 5 | Relativistic Kinematics in One Dimension
The length of the train car in Figure 5.9 is the same for all the passengers. All of them would agree on the simultaneous
location of the two ends of the car and obtain the same result for the distance between them. But simultaneous events in one
inertial frame need not be simultaneous in another. If the train could travel at relativistic speeds, an observer on the ground
would see the simultaneous locations of the two endpoints of the car at a different distance apart than observers inside the
car. Measured distances need not be the same for different observers when relativistic speeds are involved.
Proper Length
Two observers passing each other always see the same value of their relative speed. Even though time dilation implies that
the train passenger and the observer standing alongside the tracks measure different times for the train to pass, they still
agree that relative speed, which is distance divided by elapsed time, is the same. If an observer on the ground and one on the
train measure a different time for the length of the train to pass the ground observer, agreeing on their relative speed means
they must also see different distances traveled.
The muon discussed in m58563 (https://legacy.cnx.org/content/m58563/latest/#fs-id1167793912924) illustrates
this concept (Figure 5.10). To an observer on Earth, the muon travels at 0.950c for 7.05 μs from the time it is produced
until it decays. Therefore, it travels a distance relative to Earth of:
L 0 = vΔt = (0.950)(3.00 × 10 8 m/s)(7.05 × 10 −6 s⎞⎠ = 2.01 km.
In the muon frame, the lifetime of the muon is 2.20 μs. In this frame of reference, the Earth, air, and ground have only
enough time to travel:
L = vΔτ = (0.950)(3.00 × 10 8 m/s)(2.20 × 10 −6 s) km = 0.627 km.
The distance between the same two events (production and decay of a muon) depends on who measures it and how they are
moving relative to it.
Proper Length
Proper length L 0 is the distance between two points measured by an observer who is at rest relative to both of the
points.
The earthbound observer measures the proper length L 0 because the points at which the muon is produced and decays are
stationary relative to Earth. To the muon, Earth, air, and clouds are moving, so the distance L it sees is not the proper length.
Figure 5.10 (a) The earthbound observer sees the muon travel 2.01 km. (b) The same path has length 0.627 km seen from the
muon’s frame of reference. The Earth, air, and clouds are moving relative to the muon in its frame, and have smaller lengths
along the direction of travel.
Length Contraction
To relate distances measured by different observers, note that the velocity relative to the earthbound observer in our muon
example is given by
L0
v= .
Δt
The time relative to the earthbound observer is Δt, because the object being timed is moving relative to this observer. The
velocity relative to the moving observer is given by
v= L.
Δτ
The moving observer travels with the muon and therefore observes the proper time Δτ. The two velocities are identical;
thus,
L0
= L.
Δt Δτ
We know that Δt = γΔτ. Substituting this equation into the relationship above gives
L (5.3)
L = γ0 .
Substituting for γ gives an equation relating the distances measured by different observers.
Length Contraction
Length contraction is the decrease in the measured length of an object from its proper length when measured in a
reference frame that is moving with respect to the object:
2 (5.4)
L = L0 1 − v2
c
where L 0 is the length of the object in its rest frame, and L is the length in the frame moving with velocity v.
If we measure the length of anything moving relative to our frame, we find its length L to be smaller than the proper length
144 Chapter 5 | Relativistic Kinematics in One Dimension
L 0 that would be measured if the object were stationary. For example, in the muon’s rest frame, the distance Earth moves
between where the muon was produced and where it decayed is shorter than the distance traveled as seen from the Earth’s
frame. Those points are fixed relative to Earth but are moving relative to the muon. Clouds and other objects are also
contracted along the direction of motion as seen from muon’s rest frame.
Thus, two observers measure different distances along their direction of relative motion, depending on which one is
measuring distances between objects at rest.
But what about distances measured in a direction perpendicular to the relative motion? Imagine two observers moving along
their x-axes and passing each other while holding meter sticks vertically in the y-direction. Figure 5.11 shows two meter
sticks M and M′ that are at rest in the reference frames of two boys S and S′, respectively. A small paintbrush is attached
to the top (the 100-cm mark) of stick M′. Suppose that S′ is moving to the right at a very high speed v relative to S, and the
sticks are oriented so that they are perpendicular, or transverse, to their relative velocity vector. The sticks are held so that
as they pass each other, their lower ends (the 0-cm marks) coincide. Assume that when S looks at his stick M afterwards,
he finds a line painted on it, just below the top of the stick. Because the brush is attached to the top of the other boy’s stick
M′, S can only conclude that stick M′ is less than 1.0 m long.
Now when the boys approach each other, S′, like S, sees a meter stick moving toward him with speed v. Because their
situations are symmetric, each boy must make the same measurement of the stick in the other frame. So, if S measures stick
M′ to be less than 1.0 m long, S′ must measure stick M to be also less than 1.0 m long, and S′ must see his paintbrush
pass over the top of stick M and not paint a line on it. In other words, after the same event, one boy sees a painted line on a
stick, while the other does not see such a line on that same stick!
Einstein’s first postulate requires that the laws of physics (as, for example, applied to painting) predict that S and S′, who
are both in inertial frames, make the same observations; that is, S and S′ must either both see a line painted on stick M, or
both not see that line. We are therefore forced to conclude our original assumption that S saw a line painted below the top
of his stick was wrong! Instead, S finds the line painted right at the 100-cm mark on M. Then both boys will agree that a
line is painted on M, and they will also agree that both sticks are exactly 1 m long. We conclude then that measurements of
a transverse length must be the same in different inertial frames.
Example 5.5
Figure 5.12 (a) The earthbound observer measures the proper distance between Earth and Alpha Centauri. (b) The
astronaut observes a length contraction because Earth and Alpha Centauri move relative to her ship. She can travel this
shorter distance in a smaller time (her proper time) without exceeding the speed of light.
Strategy
First, note that a light year (ly) is a convenient unit of distance on an astronomical scale—it is the distance light
travels in a year. For part (a), the 4.300-ly distance between Alpha Centauri and Earth is the proper distance
L 0, because it is measured by an earthbound observer to whom both stars are (approximately) stationary. To
the astronaut, Earth and Alpha Centauri are moving past at the same velocity, so the distance between them is the
contracted length L. In part (b), we are given γ, so we can find v by rearranging the definition of γ to express v
in terms of c.
Solution for (a)
For part (a):
a. Identify the knowns: L 0 = 4.300 ly; γ = 30.00.
d. Do the calculation:
L
L = γ0
4.300 ly
=
30.00
= 0.1433 ly.
146 Chapter 5 | Relativistic Kinematics in One Dimension
γ= 1 .
2
1 − v2
c
Then solve for the unknown v/c by first squaring both sides and then rearranging:
γ2 = 1
2
1 − v2
c
v2 = 1 − 1
c2 γ2
v = 1 − 12 .
c
γ
d. Do the calculation:
v = 1− 1
c
γ2
= 1− 1
(30.00) 2
= 0.99944
or
v = 0.9994 c.
Significance
Remember not to round off calculations until the final answer, or you could get erroneous results. This is
especially true for special relativity calculations, where the differences might only be revealed after several
decimal places. The relativistic effect is large here ⎛⎝γ = 30.00⎞⎠, and we see that v is approaching (not equaling)
the speed of light. Because the distance as measured by the astronaut is so much smaller, the astronaut can travel
it in much less time in her frame.
People traveling at extremely high velocities could cover very large distances (thousands or even millions of light years) and
age only a few years on the way. However, like emigrants in past centuries who left their home, these people would leave the
Earth they know forever. Even if they returned, thousands to millions of years would have passed on Earth, obliterating most
of what now exists. There is also a more serious practical obstacle to traveling at such velocities; immensely greater energies
would be needed to achieve such high velocities than classical physics predicts can be attained. This will be discussed later
in the chapter.
Why don’t we notice length contraction in everyday life? The distance to the grocery store does not seem to depend on
2
whether we are moving or not. Examining the equation L = L 0 1 − v 2 , we see that at low velocities (v<<c), the
c
lengths are nearly equal, which is the classical expectation. But length contraction is real, if not commonly experienced. For
example, a charged particle such as an electron traveling at relativistic velocity has electric field lines that are compressed
along the direction of motion as seen by a stationary observer (Figure 5.13). As the electron passes a detector, such as a
coil of wire, its field interacts much more briefly, an effect observed at particle accelerators such as the 3-km-long Stanford
Linear Accelerator (SLAC). In fact, to an electron traveling down the beam pipe at SLAC, the accelerator and Earth are all
moving by and are length contracted. The relativistic effect is so great that the accelerator is only 0.5 m long to the electron.
It is actually easier to get the electron beam down the pipe, because the beam does not have to be as precisely aimed to get
down a short pipe as it would to get down a pipe 3 km long. This, again, is an experimental verification of the special theory
of relativity.
5.4 Check Your Understanding A particle is traveling through Earth’s atmosphere at a speed of 0.750c. To
an earthbound observer, the distance it travels is 2.50 km. How far does the particle travel as viewed from the
particle’s reference frame?
We have used the postulates of relativity to examine, in particular examples, how observers in different frames of reference
measure different values for lengths and the time intervals. We can gain further insight into how the postulates of relativity
change the Newtonian view of time and space by examining the transformation equations that give the space and time
coordinates of events in one inertial reference frame in terms of those in another. We first examine how position and
time coordinates transform between inertial frames according to the view in Newtonian physics. Then we examine how
this has to be changed to agree with the postulates of relativity. Finally, we examine the resulting Lorentz transformation
equations and some of their consequences in terms of four-dimensional space-time diagrams, to support the view that the
consequences of special relativity result from the properties of time and space itself, rather than electromagnetism.
Implicit in these equations is the assumption that time measurements made by observers in both S and S′ are the same.
That is,
t = t′.
148 Chapter 5 | Relativistic Kinematics in One Dimension
We denote the velocity of the particle by u rather than v to avoid confusion with the velocity v of one frame of reference
with respect to the other. Velocities in each frame differ by the velocity that one frame has as seen from the other frame.
The left-hand sides of the two expressions can be set equal because both are zero. Because y = y′ and z = z′, we obtain
x 2 − c 2 t 2 = x′ 2 − c 2 t′ 2. (5.5)
This cannot be satisfied for nonzero relative velocity v of the two frames if we assume the Galilean transformation results
in t = t′ with x = x′ + vt′.
To find the correct set of transformation equations, assume the two coordinate systems S and S′ in Figure 5.14. First
suppose that an event occurs at (x′, 0, 0, t′) in S′ and at (x, 0, 0, t) in S, as depicted in the figure.
Suppose that at the instant that the origins of the coordinate systems in S and S′ coincide, a flash bulb emits a spherically
spreading pulse of light starting from the origin. At time t, an observer in S finds the origin of S′ to be at x = vt. With
the help of a friend in S, the S′ observer also measures the distance from the event to the origin of S′ and finds it to be
x′ 1 − v 2 /c 2. This follows because we have already shown the postulates of relativity to imply length contraction. Thus
the position of the event in S is
x = vt + x′ 1 − v 2 /c 2
and
x′ = x − vt .
1 − v 2 /c 2
The postulates of relativity imply that the equation relating distance and time of the spherical wave front:
x2 + y2 + z2 − c2 t2 = 0
must apply both in terms of primed and unprimed coordinates, which was shown above to lead to Equation 5.5:
x 2 − c 2 t 2 = x′ 2 − c 2 t′ 2.
We combine this with the equation relating x and x′ to obtain the relation between t and t′ :
2
t′ = t − vx/c .
2 2
1 − v /c
The equations relating the time and position of the events as seen in S are then
2
t = t′ + vx′/c
2 2
1 − v /c
x = x′ + vt′
1 − v 2 /c 2
y = y′
z = z′.
This set of equations, relating the position and time in the two inertial frames, is known as the Lorentz transformation.
They are named in honor of H.A. Lorentz (1853–1928), who first proposed them. Interestingly, he justified the
transformation on what was eventually discovered to be a fallacious hypothesis. The correct theoretical basis is Einstein’s
special theory of relativity.
The reverse transformation expresses the variables in S in terms of those in S′. Simply interchanging the primed and
unprimed variables and substituting gives:
t′ = t − vx/c 2
1 − v 2 /c 2
x′ = x − vt
1 − v 2 /c 2
y′ = y
z′ = z.
Example 5.6
c. Express the answer as an equation. The time signal starts as ⎛⎝x′, t 1′⎞⎠ and stops at ⎛⎝x′, t 2′⎞⎠. Note
that the x′ coordinate of both events is the same because the clock is at rest in S′. Write the first
Lorentz transformation equation in terms of Δt = t 2 − t 1, Δx = x 2 − x 1, and similarly for the primed
coordinates, as:
150 Chapter 5 | Relativistic Kinematics in One Dimension
2
Δt = Δt′ + vΔx′/c .
2
1 − v2
c
Because the position of the clock in S′ is fixed, Δx′ = 0, and the time interval Δt becomes:
Δt = Δt′ .
2
1 − v2
c
d. Do the calculation.
With Δt′ = 1.2 s this gives:
Δt = 1.2 s = 1.6 s.
⎛1 ⎞
2
1− ⎝2 ⎠
Note that the Lorentz transformation reproduces the time dilation equation.
Example 5.7
L′ = (100 m) 1 − v 2/c 2
= (100 m) 1 − (0.20) 2
= 98.0 m.
Note that the Lorentz transformation gave the length contraction equation for the street.
Example 5.8
both ends of the 26 m long passenger car when the middle of the car passes him at a speed of c/2. Find the
separation in time between when the bulbs flashed as seen by the train passenger seated in the middle of the car.
Figure 5.15 An person watching a train go by observes two bulbs flash simultaneously at opposite ends
of a passenger car. There is another passenger inside of the car observing the same flashes but from a
different perspective.
Solution
a. Identify the known: Δt = 0.
Note that the spatial separation of the two events is between the two lamps, not the distance of the lamp
to the passenger.
b. Identify the unknown: Δt′ = t′2 − t′1.
Again, note that the time interval is between the flashes of the lamps, not between arrival times for
reaching the passenger.
c. Express the answer as an equation:
2
Δt = Δt′ + vΔx′/c .
1 − v 2 /c 2
d. Do the calculation:
Δt′ + 2c (26 m)/c 2
0 =
1 − v 2 /c 2
Δt′ = − 26 m/s = − ⎛ 26 m/s8
2c 2⎝3.00×10 m/s⎞⎠
Δt′ = −4.33×10 −8 s.
Significance
The sign indicates that the event with the larger x 2′, namely, the flash from the right, is seen to occur first in the
S′ frame, as found earlier for this example, so that t 2 < t 1.
Space-time
Relativistic phenomena can be analyzed in terms of events in a four-dimensional space-time. When phenomena such as the
twin paradox, time dilation, length contraction, and the dependence of simultaneity on relative motion are viewed in this
way, they are seen to be characteristic of the nature of space and time, rather than specific aspects of electromagnetism.
In three-dimensional space, positions are specified by three coordinates on a set of Cartesian axes, and the displacement of
one point from another is given by:
⎛
Δx, Δy, Δz⎞⎠ = (x 2 − x 1, y 2 − y 1, z 2 − z 1).
⎝
The distance Δr is invariant under a rotation of axes. If a new set of Cartesian axes rotated around the origin relative to the
original axes are used, each point in space will have new coordinates in terms of the new axes, but the distance Δr′ given
by
Δr′ 2 = (Δx′) 2 + ⎛⎝Δy′⎞⎠ 2 + (Δz′) 2.
That has the same value that Δr 2 had. Something similar happens with the Lorentz transformation in space-time.
Define the separation between two events, each given by a set of x, y, z¸ and ct along a four-dimensional Cartesian system
of axes in space-time, as
Δx, Δy, Δz, cΔt⎞⎠ = ⎛⎝x 2 − x 1, y 2 − y 1, z 2 − z 1, c(t 2 − t 1)⎞⎠.
⎛
⎝
The space-time interval is invariant under the Lorentz transformation. This follows from the postulates of relativity, and can
be seen also by substitution of the previous Lorentz transformation equations into the expression for the space-time interval:
Δs 2 = (Δx) 2 + ⎛⎝Δy⎞⎠ 2 + (Δz) 2 − (cΔt) 2
⎛ Δt′ + vΔx′ ⎞2
⎛ ⎞
2
⎜ c2 ⎟
= ⎜ Δx′ + vΔt′ ⎟ + ⎛⎝Δy′⎞⎠ 2 + (Δz′) 2 − ⎜c
⎝ 1 − v /c ⎠
2 2 2 2⎟
⎝ 1 − v /c ⎠
= (Δx′) 2 + ⎛⎝Δy′⎞⎠ 2 + (Δz′) 2 − (cΔt′) 2
= Δs′ 2.
5.5 Check Your Understanding Show that if a time increment dt elapses for an observer who sees the
particle moving with velocity v, it corresponds to a proper time particle increment for the particle of dτ = γdt.
Figure 5.16 The light cone consists of all the world lines
followed by light from the event A at the vertex of the cone.
Consider now the world line of a particle through space-time. Any world line outside of the cone, such as one passing from
A through C, would involve speeds greater than c, and would therefore not be possible.
The twin paradox seen in space-time
The twin paradox discussed earlier involves an astronaut twin traveling at near light speed to a distant star system, and
returning to Earth. Because of time dilation, the space twin is predicted to age much less than the earthbound twin. This
seems paradoxical because we might have expected at first glance for the relative motion to be symmetrical and naively
thought it possible to also argue that the earthbound twin should age less.
To analyze this in terms of a space-time diagram, assume that the origin of the axes used is fixed in Earth. The world line of
the earthbound twin is then along the time axis.
The world line of the astronaut twin, who travels to the distant star and then returns, must deviate from a straight line path
in order to allow a return trip. As seen in Figure 5.17, the circumstances of the two twins are not at all symmetrical. Their
paths in space-time are of manifestly different length. Specifically, the world line of the earthbound twin has length 2cΔt,
which then gives the proper time that elapses for the earthbound twin as 2Δt. The distance to the distant star system is
Δx = vΔt. The proper time that elapses for the space twin is 2Δτ where
c 2 Δτ 2 = (cΔt) 2 − (Δx) 2.
This is considerably shorter than the proper time for the earthbound twin by the ratio
2 2 2 2
cΔτ = (cΔt) − (Δx) = (cΔt) − (vΔt)
cΔt (cΔt) 2 (cΔt) 2
2
= 1 − v 2 = 1γ .
c
consistent with the time dilation formula. The twin paradox is therefore seen to be no paradox at all. The situation of the
two twins is not symmetrical in the space-time diagram. The only surprise is perhaps that the seemingly longer path on the
space-time diagram corresponds to the smaller proper time interval.
154 Chapter 5 | Relativistic Kinematics in One Dimension
Figure 5.17 The space twin and the earthbound twin, in the
twin paradox example, follow world lines of different length
through space-time.
unchanged and corresponds to a rotation of axes in the four-dimensional space-time. If the S and S′ frames are in relative
motion along their shared x-direction the space and time axes of S′ are rotated by an angle α as seen from S, in the way
shown in shown in Figure 5.18, where:
tanα = vc = β.
The line labeled “v = c” at 45° to the x-axis corresponds to the edge of the light cone, and is unaffected by the Lorentz
transformation, in accordance with the second postulate of relativity. The “v = c” line, and the light cone it represents, are
the same for both the S and S′ frame of reference.
Simultaneity
Simultaneity of events at separated locations depends on the frame of reference used to describe them, as given by the
scissors-like “rotation” to new time and space coordinates as described. If two events have the same t values in the unprimed
frame of reference, they need not have the same values measured along the ct′-axis, and would then not be simultaneous
in the primed frame.
As a specific example, consider the near-light-speed train in which flash lamps at the two ends of the car have flashed
simultaneously in the frame of reference of an observer on the ground. The space-time graph is shown Figure 5.19. The
flashes of the two lamps are represented by the dots labeled “Left flash lamp” and “Right flash lamp” that lie on the light
cone in the past. The world line of both pulses travel along the edge of the light cone to arrive at the observer on the ground
simultaneously. Their arrival is the event at the origin. They therefore had to be emitted simultaneously in the unprimed
frame, as represented by the point labeled as t(both). But time is measured along the ct′-axis in the frame of reference of
the observer seated in the middle of the train car. So in her frame of reference, the emission event of the bulbs labeled as t′
(left) and t′ (right) were not simultaneous.
Figure 5.19 The train example revisited. The flashes occur at the same time
t(both) along the time axis of the ground observer, but at different times, along
the t′ time axis of the passenger.
In terms of the space-time diagram, the two observers are merely using different time axes for the same events because
they are in different inertial frames, and the conclusions of both observers are equally valid. As the analysis in terms of the
space-time diagrams further suggests, the property of how simultaneity of events depends on the frame of reference results
from the properties of space and time itself, rather than from anything specifically about electromagnetism.
Do you remember the snowball-throwing child we discussed in the Introduction to this chapter? In that example, we
added velocities using the common-sense method known as the Galilean velocity transformation. However, the relativistic
156 Chapter 5 | Relativistic Kinematics in One Dimension
addition of velocities is quite different. Before explaining how velocities add in special relativity, it is important to briefly
consider a classical example where the motion is not confined to one dimension. (While a complete discussion of the
classical addition of velocities in more than one dimension will be presented in Section 7.5, a simple example here will
help us understand what it is about relativity that is so fundamentally different.)
Classical Relativity
Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a
sailor at the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the
mast, or will it hit behind the mast because the ship is moving forward? The answer is that if air resistance is negligible,
the binoculars will hit at the base of the mast at a point directly below its point of release. Now let us consider what two
different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars have no
horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See Figure 5.20.)
To the observer on shore, the binoculars and the ship have the same horizontal velocity, so both move the same distance
forward while the binoculars are falling. This observer sees the curved path shown in Figure 5.20. Although the paths look
different to the different observers, each sees the same result—the binoculars hit at the base of the mast and not behind it.
To get the correct description, it is crucial to correctly specify the velocities relative to the observer.
If we analyze this situation classically, it is obvious that the two observers will measure different horizontal velocities for
the binoculars. After all, the boat is moving horizontally. But, classically, how would the two observers compare the vertical
velocity of the binoculars. And how long would each observer think that it took the binoculars to travel from the sailor's
hand to the deck of the boat?
The classical answer is that the two observers would measure exactly the same vertical motion (free fall) and would measure
the same time of fall from the sailor's hand to the deck of the boat. This is common sense! The purely horizontal motion of
the boat does not affect the vertical motion of the binoculars.
But, remember that the Lorentz transformation shows that even the measurement of time will be different for two observers
in the relativistic case. So, as we will see, even velocities in directions other than the direction of the observers' relative
motion will be measured differently by two different observers. This is definitely not common sense. But it is the truth of
special relativity.
Velocity Transformations
Imagine a car traveling at night along a straight road, as in Figure 5.21. The driver sees the light leaving the headlights
at speed c within the car’s frame of reference. If the Galilean transformation applied to light, then the light from the car’s
headlights would approach the pedestrian at a speed u = v + c, contrary to Einstein’s postulates.
Figure 5.21 According to experimental results and the second postulate of relativity, light from the car’s headlights moves
away from the car at speed c and toward the observer on the sidewalk at speed c.
Both the distance traveled and the time of travel are different in the two frames of reference, and they must differ in a way
that makes the speed of light the same in all inertial frames. The correct rules for transforming velocities from one frame to
another can be obtained from the Lorentz transformation equations.
We thus obtain the equations for the velocity components of the object as seen in frame S:
⎛ u′x + v ⎞⎟ ⎛ u′ / γ ⎞ ⎛ u′ / γ ⎞
ux = ⎜ ⎜ ⎟, u z = ⎜ ⎟.
y z
, u y =
⎝1 + vu′x / c 2 ⎠ ⎝1 + vu′x / c 2 ⎠ ⎝1 + vu′x / c 2 ⎠
Compare this with how the Galilean transformation of classical mechanics says the velocities transform, by adding simply
as vectors:
u x = u′x + u, u y = u′y, u z = u′.
z
When the relative velocity of the frames is much smaller than the speed of light, that is, when v ≪ c, the special relativity
velocity addition law reduces to the Galilean velocity law. When the speed v of S′ relative to S is comparable to the speed
of light, the relativistic velocity addition law gives a much smaller result than the classical (Galilean) velocity addition
does.
Example 5.9
Figure 5.22 How fast does a light signal approach Earth if sent from a
spaceship traveling at 0.500c?
Strategy
Because the light and the spaceship are moving at relativistic speeds, we cannot use simple velocity addition.
Instead, we determine the speed at which the light approaches Earth using relativistic velocity addition.
Solution
a. Identify the knowns: v = 0.500c; u′ = c.
d. Do the calculation:
u = v + vu′
u′
1+ 2
c
= 0.500c +c
1 + (0.500c)(c)
2 c
(0.500 + 1)c
=
⎛c 2 + 0.500c 2 ⎞
⎝ c2 ⎠
= c.
Significance
Relativistic velocity addition gives the correct result. Light leaves the ship at speed c and approaches Earth at
speed c. The speed of light is independent of the relative motion of source and observer, whether the observer is
on the ship or earthbound.
Velocities cannot add to greater than the speed of light, provided that v is less than c and u′ does not exceed c. The
following example illustrates that relativistic velocity addition is not as symmetric as classical velocity addition.
Example 5.10
Figure 5.23 A canister is fired at 0.7500c toward Earth or away from Earth.
Strategy
Because the canister and the spaceship are moving at relativistic speeds, we must determine the speed of the
canister by an earthbound observer using relativistic velocity addition instead of simple velocity addition.
Solution for (a)
a. Identify the knowns: v = 0.500c; u′ = 0.750c.
d. Do the calculation:
u = v + vu′
u′
1+ 2
c
= 0.500c + 0.750c
1 + (0.500c)(0.750c)
2
c
= 0.909c.
Solution for (b)
a. Identify the knowns: v = 0.500c; u′ = −0.750c.
d. Do the calculation:
160 Chapter 5 | Relativistic Kinematics in One Dimension
u = v + vu′
u′
1+ 2
c
0.500c + (−0.750c)
=
1 + (0.500c)(−0.750c)
2 c
= −0.400c.
Significance
The minus sign indicates a velocity away from Earth (in the opposite direction from v), which means the canister
is heading toward Earth in part (a) and away in part (b), as expected. But relativistic velocities do not add as
simply as they do classically. In part (a), the canister does approach Earth faster, but at less than the vector
sum of the velocities, which would give 1.250c. In part (b), the canister moves away from Earth at a velocity
of −0.400c, which is faster than the −0.250c expected classically. The differences in velocities are not even
symmetric: In part (a), an observer on Earth sees the canister and the ship moving apart at a speed of 0.409c, and
at a speed of 0.900c in part (b).
5.6 Check Your Understanding Distances along a direction perpendicular to the relative motion of the two
frames are the same in both frames. Why then are velocities perpendicular to the x-direction different in the two
frames?
CHAPTER 5 REVIEW
KEY TERMS
classical (Galilean) velocity addition method of adding velocities when v<<c; velocities add like regular numbers
in one-dimensional motion: u = v + u′, where v is the velocity between two observers, u is the velocity of an object
relative to one observer, and u′ is the velocity relative to the other observer
event occurrence in space and time specified by its position and time coordinates (x, y, z, t) measured relative to a frame
of reference
first postulate of special relativity laws of physics are the same in all inertial frames of reference
Galilean relativity if an observer measures a velocity in one frame of reference, and that frame of reference is moving
with a velocity past a second reference frame, an observer in the second frame measures the original velocity as the
vector sum of these velocities
Galilean transformation relation between position and time coordinates of the same events as seen in different
reference frames, according to classical mechanics
inertial frame of reference reference frame in which a body at rest remains at rest and a body in motion moves at a
constant speed in a straight line unless acted on by an outside force
length contraction decrease in observed length of an object from its proper length L 0 to length L when its length is
observed in a reference frame where it is traveling at speed v
Lorentz transformation relation between position and time coordinates of the same events as seen in different
reference frames, according to the special theory of relativity
Michelson-Morley experiment investigation performed in 1887 that showed that the speed of light in a vacuum is the
same in all frames of reference from which it is viewed
proper length L 0; the distance between two points measured by an observer who is at rest relative to both of the points;
for example, earthbound observers measure proper length when measuring the distance between two points that are
stationary relative to Earth
proper time Δτ is the time interval measured by an observer who sees the beginning and end of the process that the
time interval measures occur at the same location
relativistic velocity addition method of adding velocities of an object moving at a relativistic speeds
rest frame frame of reference in which the observer is at rest
second postulate of special relativity light travels in a vacuum with the same speed c in any direction in all inertial
frames
special theory of relativity theory that Albert Einstein proposed in 1905 that assumes all the laws of physics have the
same form in every inertial frame of reference, and that the speed of light is the same within all inertial frames
time dilation lengthening of the time interval between two events when seen in a moving inertial frame rather than the
rest frame of the events (in which the events occur at the same location)
world line path through space-time
KEY EQUATIONS
Δt = Δτ = γτ
2
Time dilation 1 − v2
c
γ= 1
2
Lorentz factor 1 − v2
c
162 Chapter 5 | Relativistic Kinematics in One Dimension
2 L
Length contraction L = L 0 1 − v 2 = γ0
c
x= x′ + vt′
1 − v 2 /c 2
y = y′
z = z′
2
Inverse Lorentz transformation t′ = t − vx/c
2 2
1 − v /c
x′ = x − vt
1 − v 2 /c 2
y′ = y
z′ = z
⎛ u′x + v ⎞⎟ ⎛ u′ /γ ⎞ ⎛ u′ /γ ⎞
ux = ⎜ ⎜ ⎟, u z = ⎜ ⎟
y z
Relativistic velocity addition , u y =
⎝1 + vu′x /c 2 ⎠ ⎝1 + vu′x /c 2 ⎠ ⎝1 + vu′x /c 2 ⎠
SUMMARY
5.1 Invariance of Physical Laws
• Relativity is the study of how observers in different reference frames measure the same event.
• Modern relativity is divided into two parts. Special relativity deals with observers in uniform (unaccelerated)
motion, whereas general relativity includes accelerated relative motion and gravity. Modern relativity is consistent
with all empirical evidence thus far and, in the limit of low velocity and weak gravitation, gives close agreement
with the predictions of classical (Galilean) relativity.
• An inertial frame of reference is a reference frame in which a body at rest remains at rest and a body in motion
moves at a constant speed in a straight line unless acted upon by an outside force.
• Modern relativity is based on Einstein’s two postulates. The first postulate of special relativity is that the laws of
physics are the same in all inertial frames of reference. The second postulate of special relativity is that the speed of
light c is the same in all inertial frames of reference, independent of the relative motion of the observer and the light
source.
• The Michelson-Morley experiment demonstrated that the speed of light in a vacuum is independent of the motion
of Earth about the sun.
Δt = Δτ = γΔτ,
2
1 − v2
c
where
γ= 1 .
2
1 − v2
c
• The premise of the twin paradox is faulty because the traveling twin is accelerating. The journey is not symmetrical
for the two twins.
• Time dilation is usually negligible at low relative velocities, but it does occur, and it has been verified by
experiment.
• The proper time is the shortest measure of any time interval. Any observer who is moving relative to the system
being observed measures a time interval longer than the proper time.
The concept that times and distances are the same in all inertial frames in the Galilean transformation, however, is
inconsistent with the postulates of special relativity.
• The relativistically correct Lorentz transformation equations are
164 Chapter 5 | Relativistic Kinematics in One Dimension
x= x′ + vt′ x′ = x − vt
1 − v 2 /c 2 1 − v 2 /c 2
y = y′ y′ = y
z = z′ z′ = z
We can obtain these equations by requiring an expanding spherical light signal to have the same shape and speed of
growth, c, in both reference frames.
• Relativistic phenomena can be explained in terms of the geometrical properties of four-dimensional space-time, in
which Lorentz transformations correspond to rotations of axes.
• The Lorentz transformation corresponds to a space-time axis rotation, similar in some ways to a rotation of space
axes, but in which the invariant spatial separation is given by Δs rather than distances Δr, and that the Lorentz
transformation involving the time axis does not preserve perpendicularity of axes or the scales along the axes.
• The analysis of relativistic phenomena in terms of space-time diagrams supports the conclusion that these
phenomena result from properties of space and time itself, rather than from the laws of electromagnetism.
CONCEPTUAL QUESTIONS
6. (a) How could you travel far into the future of Earth
5.1 Invariance of Physical Laws without aging significantly? (b) Could this method also
1. Which of Einstein’s postulates of special relativity allow you to travel into the past?
includes a concept that does not fit with the ideas of
classical physics? Explain.
5.4 Length Contraction
2. Is Earth an inertial frame of reference? Is the sun? 7. To whom does an object seem greater in length, an
Justify your response. observer moving with the object or an observer moving
relative to the object? Which observer measures the
3. When you are flying in a commercial jet, it may appear object’s proper length?
to you that the airplane is stationary and Earth is moving
beneath you. Is this point of view valid? Discuss briefly. 8. Relativistic effects such as time dilation and length
contraction are present for cars and airplanes. Why do these
effects seem strange to us?
5.3 Time Dilation
4. (a) Does motion affect the rate of a clock as measured 9. Suppose an astronaut is moving relative to Earth at
by an observer moving with it? (b) Does motion affect how a significant fraction of the speed of light. (a) Does he
an observer moving relative to a clock measures its rate? observe the rate of his clocks to have slowed? (b) What
change in the rate of earthbound clocks does he see? (c)
Does his ship seem to him to shorten? (d) What about the
5. To whom does the elapsed time for a process seem
distance between two stars that lie in the direction of his
to be longer, an observer moving relative to the process
motion? (e) Do he and an earthbound observer agree on his
or an observer moving with the process? Which observer
velocity relative to Earth?
measures the interval of proper time?
PROBLEMS
observer moving with it? Base your calculation on its
5.3 Time Dilation velocity relative to the Earth and the time it lives (proper
10. (a) What is γ if v = 0.250c ? (b) If v = 0.500c ? time). (c) Verify that these two distances are related through
length contraction γ = 3.20.
Earth and Mars appear to move in opposite directions with S′ is moving in the positive direction along the positive
speeds 108,000 km/h and 86,871 km/h, respectively. What x-axis with a constant speed v and observes the same two
is the speed of Mars at this instant when observed from events in his frame. The origins of the two frames coincide
Earth? at t = t′ = 0. (a) Find the positions and timings of these
two events in the frame S′ (a) according to the Galilean
28. A man is running on a straight road perpendicular to transformation, and (b) according to the Lorentz
a train track and away from the track at a speed of 12 m/ transformation.
s. The train is moving with a speed of 30 m/s with respect
to the track. What is the speed of the man with respect to a
passenger sitting at rest in the train?
5.6 Relativistic Velocity Transformation
29. A man is running on a straight road that makes 30° 33. If two spaceships are heading directly toward each
other at 0.800c, at what speed must a canister be shot from
with the train track. The man is running in the direction on
the first ship to approach the other at 0.999c as seen by the
the road that is away from the track at a speed of 12 m/s.
second ship?
The train is moving with a speed of 30 m/s with respect to
the track. What is the speed of the man with respect to a
passenger sitting at rest in the train? 34. Two planets are on a collision course, heading directly
toward each other at 0.250c. A spaceship sent from one
planet approaches the second at 0.750c as seen by the
30. In a frame at rest with respect to the billiard table, a
second planet. What is the velocity of the ship relative to
billiard ball of mass m moving with speed v strikes another
the first planet?
billiard ball of mass m at rest. The first ball comes to rest
after the collision while the second ball takes off with speed
v in the original direction of the motion of the first ball. 35. When a missile is shot from one spaceship toward
This shows that momentum is conserved in this frame. (a) another, it leaves the first at 0.950c and approaches the
Now, describe the same collision from the perspective of a other at 0.750c. What is the relative velocity of the two
frame that is moving with speed v in the direction of the ships?
motion of the first ball. (b) Is the momentum conserved in
this frame? 36. What is the relative velocity of two spaceships if one
fires a missile at the other at 0.750c and the other observes
31. In a frame at rest with respect to the billiard table, it to approach at 0.950c?
two billiard balls of same mass m are moving toward each
other with the same speed v. After the collision, the two 37. Prove that for any relative velocity v between two
balls come to rest. (a) Show that momentum is conserved observers, a beam of light sent from one to the other will
in this frame. (b) Now, describe the same collision from approach at speed c (provided that v is less than c, of
the perspective of a frame that is moving with speed v course).
in the direction of the motion of the first ball. (c) Is the
momentum conserved in this frame? 38. Show that for any relative velocity v between two
observers, a beam of light projected by one directly away
32. In a frame S, two events are observed: event 1: a from the other will move away at the speed of light
pion is created at rest at the origin and event 2: the pion (provided that v is less than c, of course).
disintegrates after time τ . Another observer in a frame
ADDITIONAL PROBLEMS
39. (a) At what relative velocity is γ = 1.50 ? (b) At what a high-velocity space probe indicate that 24.0 h have passed
relative velocity is γ = 100 ? on board. (b) What is unreasonable about this result? (c)
Which assumptions are unreasonable or inconsistent?
40. (a) At what relative velocity is γ = 2.00 ? (b) At what 42. (a) How long does it take the astronaut in Example
relative velocity is γ = 10.0 ? 5.5 to travel 4.30 ly at 0.99944c (as measured by the
earthbound observer)? (b) How long does it take according
to the astronaut? (c) Verify that these two times are related
41. Unreasonable Results (a) Find the value of γ
through time dilation with γ = 30.00 as given.
required for the following situation. An earthbound
observer measures 23.9 h to have passed while signals from
43. (a) How fast would an athlete need to be running
for a 100- m race to look 100 yd long? (b) Is the answer 51. An observer sees two events 1.5 × 10 −8 s apart at
consistent with the fact that relativistic effects are difficult a separation of 800 m. How fast must a second observer
to observe in ordinary circumstances? Explain. be moving relative to the first to see the two events occur
simultaneously?
44. (a) Find the value of γ for the following situation. An
astronaut measures the length of his spaceship to be 100 m, 52. An observer standing by the railroad tracks sees two
while an earthbound observer measures it to be 25.0 m. (b) bolts of lightning strike the ends of a 500-m-long train
What is the speed of the spaceship relative to Earth? simultaneously at the instant the middle of the train passes
him at 50 m/s. Use the Lorentz transformation to find the
45. A clock in a spaceship runs one-tenth the rate at which time between the lightning strikes as measured by a
an identical clock on Earth runs. What is the speed of the passenger seated in the middle of the train.
spaceship?
53. Two astronomical events are observed from Earth to
46. An astronaut has a heartbeat rate of 66 beats per occur at a time of 1 s apart and a distance separation of
minute as measured during his physical exam on Earth. The 1.5 × 10 9 m from each other. (a) Determine whether
heartbeat rate of the astronaut is measured when he is in separation of the two events is space like or time like. (b)
a spaceship traveling at 0.5c with respect to Earth by an State what this implies about whether it is consistent with
observer (A) in the ship and by an observer (B) on Earth. special relativity for one event to have caused the other?
(a) Describe an experimental method by which observer B
on Earth will be able to determine the heartbeat rate of the 54. Two astronomical events are observed from Earth to
astronaut when the astronaut is in the spaceship. (b) What occur at a time of 0.30 s apart and a distance separation of
will be the heartbeat rate(s) of the astronaut reported by
observers A and B?
2.0 × 10 9 m from each other. How fast must a spacecraft
travel from the site of one event toward the other to make
the events occur at the same time when measured in the
47. A spaceship (A) is moving at speed c/2 with respect
frame of reference of the spacecraft?
to another spaceship (B). Observers in A and B set their
clocks so that the event at (x, y, z, t) of turning on a laser in
spaceship B has coordinates (0, 0, 0, 0) in A and also (0, 0, 55. A spacecraft starts from being at rest at the origin and
0, 0) in B. An observer at the origin of B turns on the laser accelerates at a constant rate g, as seen from Earth, taken to
at t = 0 and turns it off at t = τ in his time. What is the be an inertial frame, until it reaches a speed of c/2. (a) Show
time duration between on and off as seen by an observer in that the increment of proper time is related to the elapsed
A? time in Earth’s frame by: dτ = 1 − v 2/c 2dt.
(b) Find an expression for the elapsed time to reach speed
48. Same two observers as in the preceding exercise, but c/2 as seen in Earth’s frame. (c) Use the relationship in (a)
now we look at two events occurring in spaceship A. A to obtain a similar expression for the elapsed proper time to
photon arrives at the origin of A at its time t = 0 and reach c/2 as seen in the spacecraft, and determine the ratio
another photon arrives at (x = 1.00 m, 0, 0) at t = 0 in of the time seen from Earth with that on the spacecraft to
the frame of ship A. (a) Find the coordinates and times of reach the final speed.
the two events as seen by an observer in frame B. (b) In
which frame are the two events simultaneous and in which 56. (a) All but the closest galaxies are receding from our
frame are they are not simultaneous? own Milky Way Galaxy. If a galaxy 12.0 × 10 9 ly away
is receding from us at 0.900c, at what velocity relative to
49. Same two observers as in the preceding exercises. A us must we send an exploratory probe to approach the other
rod of length 1 m is laid out on the x-axis in the frame of galaxy at 0.990c as measured from that galaxy? (b) How
B from origin to (x = 1.00 m, 0, 0). What is the length of long will it take the probe to reach the other galaxy as
the rod observed by an observer in the frame of spaceship measured from Earth? You may assume that the velocity of
A? the other galaxy remains constant. (c) How long will it then
take for a radio signal to be beamed back? (All of this is
50. An observer at origin of inertial frame S sees a possible in principle, but not practical.)
flashbulb go off at x = 150 km, y = 15.0 km, and
57. Suppose a spaceship heading straight toward the Earth
z = 1.00 km at time t = 4.5 × 10 −4 s. At what time and
at 0.750c can shoot a canister at 0.500c relative to the ship.
position in the S ′ system did the flash occur, if S ′ is (a) What is the velocity of the canister relative to Earth, if it
moving along shared x-direction with S at a velocity is shot directly at Earth? (b) If it is shot directly away from
v = 0.6c ? Earth?
168 Chapter 5 | Relativistic Kinematics in One Dimension
58. Repeat the preceding problem with the ship heading 60. (a) Suppose the speed of light were only 3000 m/s.
directly away from Earth. A jet fighter moving toward a target on the ground at 800
m/s shoots bullets, each having a muzzle velocity of 1000
59. If a spaceship is approaching the Earth at 0.100c and m/s. What are the bullets’ velocity relative to the target?
a message capsule is sent toward it at 0.100c relative to (b) If the speed of light was this small, would you observe
Earth, what is the speed of the capsule relative to the ship? relativistic effects in everyday life? Discuss.
Time Dilation; an experiment with mu-mesons (1960 Time Dilation - Albert Einstein and the Theory of
Educational Physics) https://www.youtube.com/ Relativity: https://www.youtube.com/
watch?v=tbsdrHlLfVQ&t=4s A classic 1960 educational watch?v=KHjpBjgIMVk&t=13s YouTube animation
video showing the verification of time dilation in the muon- explaining time dilation (1:22)
lifetime experiment (35:53)
Simultaneity - Albert Einstein and the Theory of Relativity
6 | INTRODUCTION TO
VECTORS
Figure 6.1 A signpost gives information about distances and directions to towns or to other locations relative to the location of
the signpost. Distance is a scalar quantity. Knowing the distance alone is not enough to get to the town; we must also know the
direction from the signpost to the town. The direction, together with the distance, is a vector quantity commonly called the
displacement vector. A signpost, therefore, gives information about displacement vectors from the signpost to towns. (credit:
modification of work by “studio tdes”/Flickr)
Chapter Outline
6.1 Scalars and Vectors
6.2 Coordinate Systems and Components of a Vector
6.3 Algebra of Vectors
Introduction
Vectors are essential to physics and engineering. Many fundamental physical quantities are vectors, including displacement,
velocity, force, and electric and magnetic vector fields. Scalar products of vectors define other fundamental scalar physical
quantities, such as energy. Vector products of vectors define still other fundamental vector physical quantities, such as torque
and angular momentum. In other words, vectors are a component part of physics in much the same way as sentences are a
component part of literature.
In introductory physics, vectors are Euclidean quantities that have geometric representations as arrows in one dimension (in
a line), in two dimensions (in a plane), or in three dimensions (in space). They can be added, subtracted, or multiplied. In
this chapter, we explore elements of vector algebra for applications in mechanics and in electricity and magnetism. Vector
operations also have numerous generalizations in other branches of physics.
170 Chapter 6 | Introduction to Vectors
Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. For
example, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100
m.” A physical quantity that can be specified completely in this manner is called a scalar quantity. Scalar is a synonym of
“number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.
Scalar quantities that have the same physical units can be added or subtracted according to the usual rules of algebra for
numbers. For example, a class ending 10 min earlier than 50 min lasts 50 min − 10 min = 40 min . Similarly, a 60-cal
serving of corn followed by a 200-cal serving of donuts gives 60 cal + 200 cal = 260 cal of energy. When we multiply
a scalar quantity by a number, we obtain the same scalar quantity but with a larger (or smaller) value. For example, if
yesterday’s breakfast had 200 cal of energy and today’s breakfast has four times as much energy as it had yesterday, then
today’s breakfast has 4(200 cal) = 800 cal of energy. Two scalar quantities can also be multiplied or divided by each other
to form a derived scalar quantity. For example, if a train covers a distance of 100 km in 1.0 h, its speed is 100.0 km/1.0 h =
27.8 m/s, where the speed is a derived scalar quantity obtained by dividing distance by time.
Many physical quantities, however, cannot be described completely by just a single number of physical units. For example,
when the U.S. Coast Guard dispatches a ship or a helicopter for a rescue mission, the rescue team must know not only the
distance to the distress signal, but also the direction from which the signal is coming so they can get to its origin as quickly
as possible. Physical quantities specified completely by giving a number of units (magnitude) and a direction are called
vector quantities. Examples of vector quantities include displacement, velocity, position, force, and torque. In the language
of mathematics, physical vector quantities are represented by mathematical objects called vectors (Figure 6.2). We can
add or subtract two vectors, and we can multiply a vector by a scalar or by another vector, but we cannot divide by a vector.
The operation of division by a vector is not defined.
Let’s examine vector algebra using a graphical method to be aware of basic terms and to develop a qualitative
understanding. In practice, however, when it comes to solving physics problems, we use analytical methods, which we’ll
see in the next section. Analytical methods are more simple computationally and more accurate than graphical methods.
From now on, to distinguish between a vector and a scalar quantity, we adopt the common convention that a letter in bold
type with an arrow above it denotes a vector, and a letter without an arrow denotes a scalar. For example, a distance of 2.0
km, which is a scalar quantity, is denoted by d = 2.0 km, whereas a displacement of 2.0 km in some direction, which is a
→
vector quantity, is denoted by d .
Suppose you tell a friend on a camping trip that you have discovered a terrific fishing hole 6 km from your tent. It is unlikely
your friend would be able to find the hole easily unless you also communicate the direction in which it can be found with
respect to your campsite. You may say, for example, “Walk about 6 km northeast from my tent.” The key concept here is
that you have to give not one but two pieces of information—namely, the distance or magnitude (6 km) and the direction
(northeast).
Displacement is a general term used to describe a change in position, such as during a trip from the tent to the fishing hole.
Displacement is an example of a vector quantity. If you walk from the tent (location A) to the hole (location B), as shown
→
in Figure 6.3, the vector D , representing your displacement, is drawn as the arrow that originates at point A and ends
→
at point B. The arrowhead marks the end of the vector. The direction of the displacement vector D is the direction of the
→
arrow. The length of the arrow represents the magnitude D of vector D . Here, D = 6 km. Since the magnitude of a vector
is its length, which is a positive number, the magnitude is also indicated by placing the absolute value notation around the
| |
→
symbol that denotes the vector; so, we can write equivalently that D ≡ D . To solve a vector problem graphically, we
→
need to draw the vector D to scale. For example, if we assume 1 unit of distance (1 km) is represented in the drawing
by a line segment of length u = 2 cm, then the total displacement in this example is represented by a vector of length
d = 6u = 6(2 cm) = 12 cm , as shown in Figure 6.4. Notice that here, to avoid confusion, we used D = 6 km to denote
the magnitude of the actual displacement and d = 12 cm to denote the length of its representation in the drawing.
→
Figure 6.4 A displacement D of magnitude 6 km is drawn
to scale as a vector of length 12 cm when the length of 2 cm
represents 1 unit of displacement (which in this case is 1 km).
Suppose your friend walks from the campsite at A to the fishing pond at B and then walks back: from the fishing pond at
→
B to the campsite at A. The magnitude of the displacement vector D AB from A to B is the same as the magnitude of the
→
displacement vector D BA from B to A (it equals 6 km in both cases), so we can write D AB = D BA . However, vector
→ → → →
D AB is not equal to vector D BA because these two vectors have different directions: D AB ≠ D BA . In Figure
→
6.3, vector D BA would be represented by a vector with an origin at point B and an end at point A, indicating vector
→ →
D BA points to the southwest, which is exactly 180° opposite to the direction of vector D AB . We say that vector
→ → → →
D BA is antiparallel to vector D AB and write D AB = − D BA , where the minus sign indicates the antiparallel
direction.
Two vectors that have identical directions are said to be parallel vectors—meaning, they are parallel to each other. Two
→ → → →
parallel vectors A and B are equal, denoted by A = B , if and only if they have equal magnitudes
| | | |
→ →
A = B .
Two vectors with directions perpendicular to each other are said to be orthogonal vectors. These relations between vectors
are illustrated in Figure 6.5.
→ →
Figure 6.5 Various relations between two vectors A and B . (a)
→ → → →
A ≠ B because A ≠ B . (b) A ≠ B because they are not
→ →
parallel and A ≠ B . (c) A ≠ − A because they have different
| | |
→ →
| →
directions (even though A = − A = A) . (d) A = B
→
6.1 Check Your Understanding Two motorboats named Alice and Bob are moving on a lake. Given the
information about their velocity vectors in each of the following situations, indicate whether their velocity
vectors are equal or otherwise. (a) Alice moves north at 6 knots and Bob moves west at 6 knots. (b) Alice moves
west at 6 knots and Bob moves west at 3 knots. (c) Alice moves northeast at 6 knots and Bob moves south at 3
knots. (d) Alice moves northeast at 6 knots and Bob moves southwest at 6 knots. (e) Alice moves northeast at 2
knots and Bob moves closer to the shore northeast at 2 knots.
Figure 6.6 Displacement vectors for a fishing trip. (a) Stopping to rest at point C while walking from camp (point A) to the
pond (point B). (b) Going back for the dropped tackle box (point D). (c) Finishing up at the fishing pond.
Suppose your friend departs from point A (the campsite) and walks in the direction to point B (the fishing pond), but,
along the way, stops to rest at some point C located three-quarters of the distance between A and B, beginning from
→
point A (Figure 6.6(a)). What is his displacement vector D AC when he reaches point C? We know that if he walks
→
all the way to B, his displacement vector relative to A is D AB , which has magnitude D AB = 6 km and a direction
of northeast. If he walks only a 0.75 fraction of the total distance, maintaining the northeasterly direction, at point C he
must be 0.75D AB = 4.5 km away from the campsite at A. So, his displacement vector at the rest point C has magnitude
→
D AC = 4.5 km = 0.75D AB and is parallel to the displacement vector D AB . All of this can be stated succinctly in the
form of the following vector equation:
→ →
D AC = 0.75 D AB.
In a vector equation, both sides of the equation are vectors. The previous equation is an example of a vector multiplied by a
→
positive scalar (number) α = 0.75 . The result, D AC , of such a multiplication is a new vector with a direction parallel to
→
the direction of the original vector D AB .
→ → →
In general, when a vector A is multiplied by a positive scalar α , the result is a new vector B that is parallel to A :
→ → (6.1)
B =α A .
The magnitude
| |
→ →
| |
B of this new vector is obtained by multiplying the magnitude A of the original vector, as expressed
by the scalar equation:
B = |α| A. (6.2)
In a scalar equation, both sides of the equation are numbers. Equation 6.2 is a scalar equation because the magnitudes
of vectors are scalar quantities (and positive numbers). If the scalar α is negative in the vector equation Equation 6.1,
then the magnitude
| |
→ →
B of the new vector is still given by Equation 6.2, but the direction of the new vector B is
→
antiparallel to the direction of A . These principles are illustrated in Figure 6.7(a) by two examples where the length
→ → →
of vector A is 1.5 units. When α = 2 , the new vector B = 2 A has length B = 2A = 3.0 units (twice as long
→ →
as the original vector) and is parallel to the original vector. When α = −2 , the new vector C = −2 A has length
C = | − 2| A = 3.0 units (twice as long as the original vector) and is antiparallel to the original vector.
Now suppose your fishing buddy departs from point A (the campsite), walking in the direction to point B (the fishing
hole), but he realizes he lost his tackle box when he stopped to rest at point C (located three-quarters of the distance
between A and B, beginning from point A). So, he turns back and retraces his steps in the direction toward the campsite
and finds the box lying on the path at some point D only 1.2 km away from point C (see Figure 6.6(b)). What is his
→ →
displacement vector D AD when he finds the box at point D? What is his displacement vector D DB from point D to the
→ →
hole? We have already established that at rest point C his displacement vector is D AC = 0.75 D AB . Starting at point
→
C, he walks southwest (toward the campsite), which means his new displacement vector D CD from point C to point
→ →
→
D is antiparallel to D AB . Its magnitude
| →
D CD is
→
| D CD = 1.2 km = 0.2D AB , so his second displacement vector is
D CD = −0.2 D AB . His total displacement D AD relative to the campsite is the vector sum of the two displacement
→ →
vectors: vector D AC (from the campsite to the rest point) and vector D CD (from the rest point to the point where he
finds his box):
→ → → (6.3)
D AD = D AC + D CD.
The vector sum of two (or more) vectors is called the resultant vector or, for short, the resultant. When the vectors on the
→
right-hand-side of Equation 6.3 are known, we can find the resultant D AD as follows:
→ → → → → → → (6.4)
D AD = D AC + D CD = 0.75 D AB − 0.2 D AB = (0.75 − 0.2) D AB = 0.55 D AB.
→
When your friend finally reaches the pond at B, his displacement vector D AB from point A is the vector sum of his
176 Chapter 6 | Introduction to Vectors
→ →
displacement vector D AD from point A to point D and his displacement vector D DB from point D to the fishing hole:
→ → → →
D AB = D AD + D DB (see Figure 6.6(c)). This means his displacement vector D DB is the difference of two
vectors:
→ → → → → (6.5)
D DB = D AB − D AD = D AB + (− D AD).
Notice that a difference of two vectors is nothing more than a vector sum of two vectors because the second term in
→ →
Equation 6.5 is vector − D AD (which is antiparallel to D AD) . When we substitute Equation 6.4 into Equation
6.5, we obtain the second displacement vector:
→ → → → → → → (6.6)
D DB = D AB − D AD = D AB − 0.55 D AB = (1.0 − 0.55) D AB = 0.45 D AB.
This result means your friend walked D DB = 0.45D AB = 0.45(6.0 km) = 2.7 km from the point where he finds his tackle
box to the fishing hole.
→ →
When vectors A and B lie along a line (that is, in one dimension), such as in the camping example, their resultant
→ → → → → →
R = A + B and their difference D = A − B both lie along the same direction. We can illustrate the addition
or subtraction of vectors by drawing the corresponding vectors to scale in one dimension, as shown in Figure 6.7.
→ →
To illustrate the resultant when A and B are two parallel vectors, we draw them along one line by placing the origin
of one vector at the end of the other vector in head-to-tail fashion (see Figure 6.7(b)). The magnitude of this resultant
→
is the sum of their magnitudes: R = A + B. The direction of the resultant is parallel to both vectors. When vector A
→
is antiparallel to vector B , we draw them along one line in either head-to-head fashion (Figure 6.7(c)) or tail-to-
tail fashion. The magnitude of the vector difference, then, is the absolute value D = | A − B| of the difference of their
→
magnitudes. The direction of the difference vector D is parallel to the direction of the longer vector.
In general, in one dimension—as well as in higher dimensions, such as in a plane or in space—we can add any number of
vectors and we can do so in any order because the addition of vectors is commutative,
→ → → → (6.7)
A + B = B + A ,
and associative,
→ → → → → → (6.8)
( A + B ) + C = A + ( B + C ).
→ → → (6.9)
α 1 A + α 2 A = (α 1 + α 2) A .
||
^ →
u ≡ u = 1 . The only role of a unit vector is to specify direction. For example, instead of saying vector D AB has a
Example 6.1
A Ladybug Walker
A long measuring stick rests against a wall in a physics laboratory with its 200-cm end at the floor. A ladybug
lands on the 100-cm mark and crawls randomly along the stick. It first walks 15 cm toward the floor, then it walks
56 cm toward the wall, then it walks 3 cm toward the floor again. Then, after a brief stop, it continues for 25 cm
toward the floor and then, again, it crawls up 19 cm toward the wall before coming to a complete rest (Figure
6.8). Find the vector of its total displacement and its final resting position on the stick.
Strategy
If we choose the direction along the stick toward the floor as the direction of unit vector ^
u , then the direction
toward the floor is + ^
u and the direction toward the wall is − ^
u . The ladybug makes a total of five
displacements:
→
D 1 = (15 cm)( + ^
u ),
→
D 2 = (56 cm)(− ^
u ),
→
D 3 = (3 cm)( + ^
u ),
→ ^
D 4 = (25 cm)( + u ), and
→
D 5 = (19 cm)(− ^
u ).
→
The total displacement D is the resultant of all its displacement vectors.
Figure 6.8 Five displacements of the ladybug. Note that in this schematic drawing,
magnitudes of displacements are not drawn to scale. (credit: modification of work by
“Persian Poet Gal”/Wikimedia Commons)
178 Chapter 6 | Introduction to Vectors
Solution
The resultant of all the displacement vectors is
→ → → → → →
D = D 1+ D 2+ D 3+ D 4+ D 5
= (15 cm)( + ^
u ) + (56 cm)(− ^
u ) + (3 cm)( + ^
u ) + (25 cm)( + ^
u ) + (19 cm)(− ^
u)
= (15 − 56 + 3 + 25 − 19)cm ^u
= −32 cm ^
u.
In this calculation, we use the distributive law given by Equation 6.9. The result reads that the total
displacement vector points away from the 100-cm mark (initial landing site) toward the end of the meter stick
that touches the wall. The end that touches the wall is marked 0 cm, so the final position of the ladybug is at the
(100 – 32)cm = 68-cm mark.
6.2 Check Your Understanding A cave diver enters a long underwater tunnel. When her displacement with
respect to the entry point is 20 m, she accidentally drops her camera, but she doesn’t notice it missing until she
is some 6 m farther into the tunnel. She swims back 10 m but cannot find the camera, so she decides to end the
dive. How far from the entry point is she? Taking the positive direction out of the tunnel, what is her
displacement vector relative to the entry point?
Figure 6.9 In navigation, the laws of geometry are used to draw resultant displacements on
nautical maps.
For a geometric construction of the sum of two vectors in a plane, we follow the parallelogram rule. Suppose two vectors
→ →
A and B are at the arbitrary positions shown in Figure 6.10. Translate either one of them in parallel to the beginning
of the other vector, so that after the translation, both vectors have their origins at the same point. Now, at the end of vector
→ → → →
A we draw a line parallel to vector B and at the end of vector B we draw a line parallel to vector A (the dashed
lines in Figure 6.10). In this way, we obtain a parallelogram. From the origin of the two vectors we draw a diagonal that is
→ → → →
the resultant R of the two vectors: R = A + B (Figure 6.10(a)). The other diagonal of this parallelogram is the
→ → →
vector difference of the two vectors D = A − B , as shown in Figure 6.10(b). Notice that the end of the difference
→
vector is placed at the end of vector A .
Figure 6.10 The parallelogram rule for the addition of two vectors. Make the parallel translation of each vector to a point
where their origins (marked by the dot) coincide and construct a parallelogram with two sides on the vectors and the other
→
two sides (indicated by dashed lines) parallel to the vectors. (a) Draw the resultant vector R along the diagonal of the
parallelogram from the common point to the opposite corner. Length R of the resultant vector is not equal to the sum of the
→ → →
magnitudes of the two vectors. (b) Draw the difference vector D = A − B along the diagonal connecting the ends of
→ → →
the vectors. Place the origin of vector D at the end of vector B and the end (arrowhead) of vector D at the end of
→
vector A . Length D of the difference vector is not equal to the difference of magnitudes of the two vectors.
It follows from the parallelogram rule that neither the magnitude of the resultant vector nor the magnitude of the difference
vector can be expressed as a simple sum or difference of magnitudes A and B, because the length of a diagonal cannot be
expressed as a simple sum of side lengths. When using a geometric construction to find magnitudes
| | | |
→ →
R and D , we
have to use trigonometry laws for triangles, which may lead to complicated algebra. There are two ways to circumvent this
algebraic complexity. One way is to use the method of components, which we examine in the next section. The other way is
to draw the vectors to scale, as is done in navigation, and read approximate vector lengths and angles (directions) from the
graphs. In this section we examine the second approach.
If we need to add three or more vectors, we repeat the parallelogram rule for the pairs of vectors until we find the resultant
of all of the resultants. For three vectors, for example, we first find the resultant of vector 1 and vector 2, and then we
find the resultant of this resultant and vector 3. The order in which we select the pairs of vectors does not matter because
the operation of vector addition is commutative and associative (see Equation 6.7 and Equation 6.8). Before we state a
general rule that follows from repetitive applications of the parallelogram rule, let’s look at the following example.
Suppose you plan a vacation trip in Florida. Departing from Tallahassee, the state capital, you plan to visit your uncle
Joe in Jacksonville, see your cousin Vinny in Daytona Beach, stop for a little fun in Orlando, see a circus performance
in Tampa, and visit the University of Florida in Gainesville. Your route may be represented by five displacement vectors
→ → → → →
A , B , C , D , and E , which are indicated by the red vectors in Figure 6.11. What is your total displacement
when you reach Gainesville? The total displacement is the vector sum of all five displacement vectors, which may be
found by using the parallelogram rule four times. Alternatively, recall that the displacement vector has its beginning at
the initial position (Tallahassee) and its end at the final position (Gainesville), so the total displacement vector can be
drawn directly as an arrow connecting Tallahassee with Gainesville (see the green vector in Figure 6.11). When we use
→
the parallelogram rule four times, the resultant R we obtain is exactly this green vector connecting Tallahassee with
→ → → → → →
Gainesville: R = A + B + C + D + E .
180 Chapter 6 | Introduction to Vectors
Figure 6.11 When we use the parallelogram rule four times, we obtain the resultant vector
→ → → → → →
R = A + B + C + D + E , which is the green vector connecting Tallahassee with Gainesville.
Drawing the resultant vector of many vectors can be generalized by using the following tail-to-head geometric
→ → → → →
construction. Suppose we want to draw the resultant vector R of four vectors A , B , C , and D (Figure
6.12(a)). We select any one of the vectors as the first vector and make a parallel translation of a second vector to a position
where the origin (“tail”) of the second vector coincides with the end (“head”) of the first vector. Then, we select a third
vector and make a parallel translation of the third vector to a position where the origin of the third vector coincides with
the end of the second vector. We repeat this procedure until all the vectors are in a head-to-tail arrangement like the one
→
shown in Figure 6.12. We draw the resultant vector R by connecting the origin (“tail”) of the first vector with the end
(“head”) of the last vector. The end of the resultant vector is at the end of the last vector. Because the addition of vectors is
associative and commutative, we obtain the same resultant vector regardless of which vector we choose to be first, second,
third, or fourth in this construction.
Example 6.2
Figure 6.13 Vectors used in Example 6.2 and in the Check Your Understanding feature that follows.
Strategy
In geometric construction, to find a vector means to find its magnitude and its direction angle with the horizontal
direction. The strategy is to draw to scale the vectors that appear on the right-hand side of the equation and
construct the resultant vector. Then, use a ruler and a protractor to read the magnitude of the resultant and the
direction angle. For parts (a) and (b) we use the parallelogram rule. For (c) we use the tail-to-head method.
Solution
→ →
For parts (a) and (b), we attach the origin of vector B to the origin of vector A , as shown in Figure 6.14,
→ →
and construct a parallelogram. The shorter diagonal of this parallelogram is the sum A + B . The longer of
182 Chapter 6 | Introduction to Vectors
→ →
the diagonals is the difference A − B . We use a ruler to measure the lengths of the diagonals, and a protractor
→
to measure the angles with the horizontal. For the resultant R , we obtain R = 5.8 cm and θ R ≈ 0° . For the
→
difference D , we obtain D = 16.2 cm and θ D = 49.3° , which are shown in Figure 6.14.
Figure 6.14 Using the parallelogram rule to solve (a) (finding the resultant, red) and (b) (finding
the difference, blue).
→
For (c), we can start with vector −3 B and draw the remaining vectors tail-to-head as shown in Figure 6.15.
In vector addition, the order in which we draw the vectors is unimportant, but drawing the vectors to scale is very
→
important. Next, we draw vector S from the origin of the first vector to the end of the last vector and place the
→ →
arrowhead at the end of S . We use a ruler to measure the length of S , and find that its magnitude is
S = 36.9 cm. We use a protractor and find that its direction angle is θ S = 52.9° . This solution is shown in Figure
6.15.
6.3 → → →
Check Your Understanding Using the three displacement vectors A , B , and F in Figure
→
6.13, choose a convenient scale, and use a ruler and a protractor to find vector G given by the vector
→ → → →
equation G = A + 2 B − F .
Observe the addition of vectors in a plane by visiting this vector calculator (https://openstaxcollege.org/l/
21compveccalc) and this Phet simulation (https://openstaxcollege.org/l/21phetvecaddsim) .
Vectors are usually described in terms of their components in a coordinate system. Even in everyday life we naturally invoke
the concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions to
a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction 37° north
of east.
In a rectangular (Cartesian) xy-coordinate system in a plane, a point in a plane is described by a pair of coordinates (x, y).
→
In a similar fashion, a vector A in a plane is described by a pair of its vector coordinates. The x-coordinate of vector
→ →
A is called its x-component and the y-coordinate of vector A is called its y-component. The vector x-component is
→ →
a vector denoted by A x . The vector y-component is a vector denoted by A y . In the Cartesian system, the x and y
vector components of a vector are the orthogonal projections of this vector onto the x- and y-axes, respectively. In this way,
following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of
its vector components:
→ → → (6.10)
A = A x + A y.
→ →
As illustrated in Figure 6.16, vector A is the diagonal of the rectangle where the x-component A x is the side parallel
→ →
to the x-axis and the y-component A y is the side parallel to the y-axis. Vector component A x is orthogonal to vector
→
component A y .
184 Chapter 6 | Introduction to Vectors
→
Figure 6.16 Vector A in a plane in the Cartesian coordinate
system is the vector sum of its vector x- and y-components. The
→
x-vector component A x is the orthogonal projection of vector
→ →
A onto the x-axis. The y-vector component A y is the
→
orthogonal projection of vector A onto the y-axis. The numbers
A x and A y that multiply the unit vectors are the scalar components
of the vector.
^
It is customary to denote the positive direction on the x-axis by the unit vector i and the positive direction on the y-axis
^ ^ ^
by the unit vector j . Unit vectors of the axes, i and j , define two orthogonal directions in the plane. As shown in
Figure 6.16, the x- and y- components of a vector can now be written in terms of the unit vectors of the axes:
⎧→ ^ (6.11)
⎨→
A x = Ax i
⎩ A y = Ay j .
^
→ → →
The vectors A x and A y defined by Equation 6.11 are the vector components of vector A . The numbers A x and
→
A y that define the vector components in Equation 6.11 are the scalar components of vector A . Combining Equation
6.10 with Equation 6.11, we obtain the component form of a vector:
→ ^ ^ (6.12)
A = Ax i + Ay j .
If we know the coordinates b(x b, y b) of the origin point of a vector (where b stands for “beginning”) and the coordinates
e(x e, y e) of the end point of a vector (where e stands for “end”), we can obtain the scalar components of a vector simply
by subtracting the origin point coordinates from the end point coordinates:
⎧ A x = xe − xb (6.13)
⎨
⎩A y = y e − y b.
Example 6.3
Figure 6.17 The graph of the displacement vector. The vector points from
the origin point at b to the end point at e.
Significance
Notice that the physical unit—here, 1 cm—can be placed either with each component immediately before the unit
vector or globally for both components, as in Equation 6.14. Often, the latter way is more convenient because
it is simpler.
186 Chapter 6 | Introduction to Vectors
→ ^ ^
The vector x-component D x = −4.0 i = 4.0(− i ) of the displacement vector has the magnitude
| || ||
→
D x
^ ^
||
= − 4.0 i = 4.0 because the magnitude of the unit vector is i = 1 . Notice, too, that the direction
^
of the x-component is − i , which is antiparallel to the direction of the +x-axis; hence, the x-component vector
→ →
D x points to the left, as shown in Figure 6.17. The scalar x-component of vector D is D x = −4.0 .
→ ^
Similarly, the vector y-component D y = + 2.9 j of the displacement vector has magnitude
| ||||
→
D
^
y
^ ^
||
= 2.9 j = 2.9 because the magnitude of the unit vector is j = 1 . The direction of the y-component
→
is + j , which is parallel to the direction of the +y-axis. Therefore, the y-component vector D y points up, as
→ →
seen in Figure 6.17. The scalar y-component of vector D is D y = + 2.9 . The displacement vector D is
the resultant of its two vector components.
The vector component form of the displacement vector Equation 6.14 tells us that the mouse pointer has been
moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.
6.4 Check Your Understanding A blue fly lands on a sheet of graph paper at a point located 10.0 cm to the
right of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the left
edge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lower
left-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing.
→
When we know the scalar components A x and A y of a vector A , we can find its magnitude A and its direction angle
θ A . The direction angle—or direction, for short—is the angle the vector forms with the positive direction on the x-axis.
The angle θ A is measured in the counterclockwise direction from the +x-axis to the vector (Figure 6.18). Because the
lengths A, A x , and A y form a right triangle, they are related by the Pythagorean theorem:
A 2 = A 2x + A 2y ⇔ A = A 2x + A 2y. (6.15)
This equation works even if the scalar components of a vector are negative. The direction angle θ A of a vector is defined
via the tangent function of angle θ A in the triangle shown in Figure 6.18:
Ay ⎛A y⎞ (6.16)
tan θ A =
Ax
⇒ θ A = tan −1
⎝ A x ⎠.
→
Figure 6.18 For vector A , its magnitude A and its direction
angle θ A are related to the magnitudes of its scalar components
because A, A x , and A y form a right triangle.
When the vector lies either in the first quadrant or in the fourth quadrant, where component A x is positive (Figure 6.19),
the angle θ in Equation 6.16) is identical to the direction angle θ A . For vectors in the fourth quadrant, angle θ is
negative, which means that for these vectors, direction angle θ A is measured clockwise from the positive x-axis. Similarly,
for vectors in the second quadrant, angle θ is negative. When the vector lies in either the second or third quadrant, where
component A x is negative, the direction angle is θ A = θ + 180° (Figure 6.19).
Example 6.4
Solution
→
The magnitude of vector D is
6.5 Check Your Understanding If the displacement vector of a blue fly walking on a sheet of graph paper is
→ ^ ^
D = (−5.00 i − 3.00 j )cm , find its magnitude and direction.
In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of
many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each
car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly
be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding
the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading
to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach
is to find vector components when the direction and magnitude of a vector are known.
Let us return to the right triangle in Figure 6.18. The quotient of the adjacent side A x to the hypotenuse A is the cosine
function of direction angle θ A , A x/A = cos θ A , and the quotient of the opposite side A y to the hypotenuse A is the sine
function of θ A , A y/A = sin θ A . When magnitude A and direction θ A are known, we can solve these relations for the
scalar components:
⎧A x = A cos θ A (6.17)
⎨ .
⎩ A y = A sin θ A
When calculating vector components with Equation 6.17, care must be taken with the angle. The direction angle θ A
of a vector is the angle measured counterclockwise from the positive direction on the x-axis to the vector. The clockwise
measurement gives a negative angle.
Example 6.5
On the second leg of Trooper’s wanderings, the magnitude of the displacement is L 2 = 300.0 m and the
direction is north. The direction angle is θ 2 = + 90° . We obtain the following results:
On the third leg, the displacement magnitude is L 3 = 50.0 m and the direction is 30° west of north. The
direction angle measured counterclockwise from the eastern direction is θ 3 = 30° + 90° = + 120° . This gives
the following answers:
L 3x = L 3 cos θ 3 = (50.0 m) cos 120° = −25.0 m,
L 3y = L 3 sin θ 3 = (50.0 m) sin 120° = + 43.3 m,
→ ^ ^ ^ ^
L 3 = L 3x i + L 3y j = (−25.0 i + 43.3 j )m.
On the fourth leg of the excursion, the displacement magnitude is L 4 = 80.0 m and the direction is south. The
direction angle can be taken as either θ 4 = −90° or θ 4 = + 270° . We obtain
190 Chapter 6 | Introduction to Vectors
On the last leg, the magnitude is L 5 = 150.0 m and the angle is θ 5 = −23° + 270° = + 247° (23° west of
south), which gives
L 5x = L 5 cos θ 5 = (150.0 m) cos 247° = −58.6 m,
L 5y = L 5 sin θ 5 = (150.0 m) sin 247° = −138.1 m,
→ ^ ^ ^ ^
L 5 = L 5x i + L 5y j = (−58.6 i − 138.1 j )m.
6.6 Check Your Understanding If Trooper runs 20 m west before taking a rest, what is his displacement
vector?
Polar Coordinates
To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system
^ ^
these directions are given by unit vectors i and j along the x-axis and the y-axis, respectively. The Cartesian coordinate
system is very convenient to use in describing displacements and velocities of objects and the forces acting on them.
However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usually
work in the polar coordinate system.
In the polar coordinate system, the location of point P in a plane is given by two polar coordinates (Figure 6.20). The first
polar coordinate is the radial coordinate r, which is the distance of point P from the origin. The second polar coordinate is
an angle φ that the radial vector makes with some chosen direction, usually the positive x-direction. In polar coordinates,
angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point
^
P. This radial direction is described by a unit radial vector ^
r . The second unit vector t is a vector orthogonal to the
^
radial direction ^
r . The positive + t direction indicates how the angle φ changes in the counterclockwise direction. In
this way, a point P that has coordinates (x, y) in the rectangular system can be described equivalently in the polar coordinate
system by the two polar coordinates (r, φ) . Equation 6.17 is valid for any vector, so we can use it to express the x-
and y-coordinates of vector →
r . In this way, we obtain the connection between the polar coordinates and rectangular
coordinates of point P:
⎧x = r cos φ (6.18)
⎨ .
⎩ y = r sin φ
^
Figure 6.20 Using polar coordinates, the unit vector r defines
the positive direction along the radius r (radial direction) and,
^
orthogonal to it, the unit vector t defines the positive direction of
rotation by the angle φ .
Example 6.6
Polar Coordinates
A treasure hunter finds one silver coin at a location 20.0 m away from a dry well in the direction 20° north of
east and finds one gold coin at a location 10.0 m away from the well in the direction 20° north of west. What are
the polar and rectangular coordinates of these findings with respect to the well?
Strategy
The well marks the origin of the coordinate system and east is the +x-direction. We identify radial distances from
the locations to the origin, which are r S = 20.0 m (for the silver coin) and r G = 10.0 m (for the gold coin). To
find the angular coordinates, we convert 20° to radians: 20° = π20/180 = π/9 . We use Equation 6.18 to find
the x- and y-coordinates of the coins.
Solution
The angular coordinate of the silver coin is φ S = π/9 , whereas the angular coordinate of the gold coin is
φ G = π − π/9 = 8π/9 . Hence, the polar coordinates of the silver coin are (r S, φ S) = (20.0 m, π/9) and those
of the gold coin are (r G, φ G) = (10.0 m, 8π/9) . We substitute these coordinates into Equation 6.18 to obtain
rectangular coordinates. For the gold coin, the coordinates are
⎧x G = r G cos φ G = (10.0 m) cos 8π/9 = −9.4 m
⎨ ⇒ (x G, y G) = (−9.4 m, 3.4 m).
⎩y G = r G sin φ G = (10.0 m) sin 8π/9 = 3.4 m
directions, so we need not two but three unit vectors to define a three-dimensional coordinate system. In the Cartesian
^ ^
coordinate system, the first two unit vectors are the unit vector of the x-axis i and the unit vector of the y-axis j . The
^
third unit vector k is the direction of the z-axis (Figure 6.21). The order in which the axes are labeled, which is the
order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The
^ ^ ^
order x-y-z, which is equivalent to the order i - j - k , defines the standard right-handed coordinate system (positive
orientation).
→ → ^
In three-dimensional space, vector A has three vector components: the x-component A x = A x i , which is the part
→ → ^ →
of vector A along the x-axis; the y-component A y = A y j , which is the part of A along the y-axis; and the
→ ^
z-component A z = A z k , which is the part of the vector along the z-axis. A vector in three-dimensional space is the
vector sum of its three vector components (Figure 6.22):
→ ^ ^ ^ (6.19)
A = A x i + Ay j + Az k .
If we know the coordinates of its origin b(x b, y b, z b) and of its end e(x e, y e, z e) , its scalar components are obtained by
taking their differences: A x and A y are given by Equation 6.13 and the z-component is given by
A z = z e − z b. (6.20)
A = A 2x + A 2y + A 2z . (6.21)
This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in Figure 6.22,
the diagonal in the xy-plane has length A 2x + A 2y and its square adds to the square A 2z to give A 2 . Note that when the
z-component is zero, the vector lies entirely in the xy-plane and its description is reduced to two dimensions.
Example 6.7
Takeoff of a Drone
During a takeoff of IAI Heron (Figure 6.23), its position with respect to a control tower is 100 m above the
ground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200
m to the east, and 2100 m to the north. What is the drone’s displacement vector with respect to the control tower?
What is the magnitude of its displacement vector?
Strategy
We take the origin of the Cartesian coordinate system as the control tower. The direction of the +x-axis is given
194 Chapter 6 | Introduction to Vectors
^ ^
by unit vector i to the east, the direction of the +y-axis is given by unit vector j to the north, and the direction
^
of the +z-axis is given by unit vector k , which points up from the ground. The drone’s first position is the origin
(or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacement
vector.
Solution
We identify b(300.0 m, 200.0 m, 100.0 m) and e(480.0 m, 370.0 m, 250.0m), and use Equation 6.13 and
Equation 6.20 to find the scalar components of the drone’s displacement vector:
⎧D x = x e − x b = 1200.0 m − 300.0 m = 900.0 m,
⎨D y = y e − y b = 2100.0 m − 200.0 m = 1900.0 m,
⎩D z = z e − z b = 250.0 m − 100.0 m = 150.0 m.
We substitute these components into Equation 6.19 to find the displacement vector:
→ ^ ^ ^ ^ ^ ^ ^ ^ ^
D = D x i + D y j + D z k = 900.0 m i + 1900.0 m j + 150.0 m k = (0.90 i + 1.90 j + 0.15 k ) km.
6.7 Check Your Understanding If the average velocity vector of the drone in the displacement in Example
^ ^ ^
6.7 is →
u = (15.0 i + 31.7 j + 2.5 k )m/s , what is the magnitude of the drone’s velocity vector?
Vectors can be added together and multiplied by scalars. Vector addition is associative (Equation 6.8) and commutative
(Equation 6.7), and vector multiplication by a sum of scalars is distributive (Equation 6.9). Also, scalar multiplication
by a sum of vectors is distributive:
→ → → → (6.22)
α( A + B ) = α A + α B .
→ ^ ^ ^
In this equation, α is any number (a scalar). For example, a vector antiparallel to vector A = A x i + A y j + A z k can
→
be expressed simply by multiplying A by the scalar α = −1 :
→ ^ ^ ^ (6.23)
− A = −A x i − A y j − A z k .
Example 6.8
Direction of Motion
^ ^ ^
In a Cartesian coordinate system where i denotes geographic east, j denotes geographic north, and k
denotes altitude above sea level, a military convoy advances its position through unknown territory with velocity
→ ^ ^ ^
v = (4.0 i + 3.0 j + 0.1 k )km/h . If the convoy had to retreat, in what geographic direction would it be
moving?
Solution
^
The velocity vector has the third component → v z = ( + 0.1km/h) k , which says the convoy is climbing at a rate
of 100 m/h through mountainous terrain. At the same time, its velocity is 4.0 km/h to the east and 3.0 km/h to
the north, so it moves on the ground in direction tan −1(3 /4) ≈ 37° north of east. If the convoy had to retreat,
its new velocity vector →
u would have to be antiparallel to →
v and be in the form →
u = −α →
v , where
^ ^ ^
α is a positive number. Thus, the velocity of the retreat would be →
u = α(−4.0 i − 3.0 j − 0.1 k )km/h . The
negative sign of the third component indicates the convoy would be descending. The direction angle of the retreat
velocity is tan −1(−3α/ − 4α) ≈ 37° south of west. Therefore, the convoy would be moving on the ground in
direction 37° south of west while descending on its way back.
→
The generalization of the number zero to vector algebra is called the null vector, denoted by 0 . All components of the
→ ^ ^ ^
null vector are zero, 0 = 0 i + 0 j + 0 k , so the null vector has no length and no direction.
→ →
Two vectors A and B are equal vectors if and only if their difference is the null vector:
→ → → ^ ^ ^ ^ ^ ^ ^ ^ ^
0 = A − B = (A x i + A y j + A z k ) − (B x i + B y j + B z k ) = (A x − B x) i + (A y − B y) j + (A z − B z) k .
This vector equation means we must have simultaneously A x − B x = 0 , A y − B y = 0 , and A z − B z = 0 . Hence, we can
→ → → →
write A = B if and only if the corresponding components of vectors A and B are equal:
⎧A x = B x (6.24)
⎨A y = B y.
→ →
A = B ⇔
⎩A z = B z
Two vectors are equal when their corresponding scalar components are equal.
Resolving vectors into their scalar components (i.e., finding their scalar components) and expressing them analytically in
vector component form (given by Equation 6.19) allows us to use vector algebra to find sums or differences of many
→ →
vectors analytically (i.e., without using graphical methods). For example, to find the resultant of two vectors A and B
, we simply add them component by component, as follows:
→ → → ^ ^ ^ ^ ^ ^ ^ ^ ^
R = A + B = (A x i + A y j + A z k ) + (B x i + B y j + B z k ) = (A x + B x) i + (A y + B y) j + (A z + B z) k .
→ ^ ^ ^
In this way, using Equation 6.24, scalar components of the resultant vector R = R x i + R y j + R z k are the sums of
→ →
corresponding scalar components of vectors A and B :
196 Chapter 6 | Introduction to Vectors
⎧R x = A x + B x,
⎨R y = A y + B y,
⎩ R z = A z + B z.
Analytical methods can be used to find components of a resultant of many vectors. For example, if we are to sum up
→ → → → → ^ ^ ^
N vectors F 1, F 2, F 3, … , F N , where each vector is F k = F kx i + F ky j + F kz k , the resultant vector
→
F R is
⎛ ^ ^⎞
N N
→ → → → → → ^
F R= F 1+ F 2+ F 3+…+ F N = ∑ F k= ∑ ⎝F kx i + F ky j + F kz k
⎠
k=1 k=1
⎛ N ⎞^ ⎛ N ⎞^ ⎛ N ⎞^
= ⎜ ∑ F kx⎟ i + ⎜ ∑ F ky⎟ j + ⎜ ∑ F kz⎟ k .
⎝k = 1 ⎠ ⎝k = 1 ⎠ ⎝k = 1 ⎠
⎧ N (6.25)
⎪F Rx = ∑ F kx = F 1x + F 2x + … + F Nx
⎪ k=1
⎪ N
⎨F Ry = ∑ F ky = F 1y + F 2y + … + F Ny
⎪ k=1
⎪ N
⎪F = ∑ F = F + F + … + F .
⎩ Rz k = 1 kz 1z 2z Nz
Having found the scalar components, we can write the resultant in vector component form:
→ ^ ^ ^
F R = F Rx i + F Ry j + F Rz k .
Analytical methods for finding the resultant and, in general, for solving vector equations are very important in physics
because many physical quantities are vectors. For example, we use this method in kinematics to find resultant displacement
vectors and resultant velocity vectors, in mechanics to find resultant force vectors and the resultants of many derived vector
quantities, and in electricity and magnetism to find resultant electric or magnetic vector fields.
Example 6.9
For (a) we may substitute directly into Equation 6.24 to find the scalar components of the resultant:
⎧R x = A x + B x + C x = 8.19 cm − 2.39 cm + 6.93 cm = 12.73 cm
⎨ .
⎩R y = A y + B y + C y = 5.73 cm − 6.58 cm + 4.00 cm = 3.15 cm
→ ^ ^ ^ ^
Therefore, the resultant vector is R = R x i + R y j = (12.7 i + 3.1 j )cm .
→ ^ ^ ^ ^
Hence, the difference vector is D = D x i + D y j = (10.6 i + 12.3 j )cm .
→
For (c), we can write vector S in the following explicit form:
→ → → → ^ ^ ^ ^ ^ ^
S = A − 3 B + C = (A x i + A y j ) − 3(B x i + B y j ) + (C x i + C y j )
^ ^
= (A x − 3B x + C x) i + (A y − 3B y + C y) j .
→
Then, the scalar components of S are
⎧S x = A x − 3B x + C x = 8.19 cm − 3(−2.39 cm) + 6.93 cm = 22.29 cm
⎨ .
⎩S y = A y − 3B y + C y = 5.73 cm − 3(−6.58 cm) + 4.00 cm = 29.47 cm
→ ^ ^ ^ ^
The vector is S = S x i + S y j = (22.3 i + 29.5 j )cm .
Significance
Having found the vector components, we can illustrate the vectors by graphing or we can compute magnitudes
and direction angles, as shown in Figure 6.24. Results for the magnitudes in (b) and (c) can be compared with
results for the same problems obtained with the graphical method, shown in Figure 6.14 and Figure 6.15.
Notice that the analytical method produces exact results and its accuracy is not limited by the resolution of a ruler
or a protractor, as it was with the graphical method used in Example 6.2 for finding this same resultant.
198 Chapter 6 | Introduction to Vectors
Figure 6.24 Graphical illustration of the solutions obtained analytically in Example 6.9.
6.8 → → →
Check Your Understanding Three displacement vectors A , B , and F (Figure 6.13) are
specified by their magnitudes A = 10.00, B = 7.00, and F = 20.00, respectively, and by their respective direction
angles with the horizontal direction α = 35° , β = −110° , and φ = 110° . The physical units of the
→ → → →
magnitudes are centimeters. Use the analytical method to find vector G = A + 2 B − F . Verify that
G = 28.15 cm and that θ G = −68.65° .
Example 6.10
Strategy
We assume that east is the direction of the positive x-axis and north is the direction of the positive y-axis. As
→ →
in Example 6.9, we have to resolve the three given forces— A (the pull from Ang), B (the pull from
→
Bing), and C (the pull from Chang)—into their scalar components and then find the scalar components of the
→ → → → →
resultant vector R = A + B + C . When the pulling force D from Dong balances out this resultant,
→ → → → → → →
the sum of D and R must give the null vector D + R = 0 . This means that D = − R , so the
→
pull from Dong must be antiparallel to R .
Solution
The direction angles are θ A = −α = −55° , θ B = 90° − β = 30° , and θ C = 90° + γ = 145° , and substituting
them into Equation 6.17 gives the scalar components of the three given forces:
⎧A x = A cos θ A = (160.0 N) cos (−55°) = + 91.8 N
⎨
⎩A y = A sin θ A = (160.0 N) sin (−55°) = −131.1 N
⎧B x = B cos θ B = (200.0 N) cos 30° = + 173.2 N
⎨ .
⎩B y = B sin θ B = (200.0 N) sin 30° = + 100.0 N
⎧C x = C cos θ C = (140.0 N) cos 145° = −114.7 N
⎨
⎩C y = C sin θ C = (140.0 N) sin 145° = + 80.3 N
→ → → →
Now we compute scalar components of the resultant vector R = A + B + C :
⎧R x = A x + B x + C x = + 91.8 N + 173.2 N − 114.7 N = + 150.3 N
⎨ .
⎩R y = A y + B y + C y = −131.1 N + 100.0 N + 80.3 N = + 49.2 N
→
The antiparallel vector to the resultant R is
→ → ^ ^ ^ ^
D = − R = −R x i − R y j = (−150.3 i − 49.2 j ) N.
6.9 Check Your Understanding Suppose that Bing in Example 6.10 leaves the game to attend to more
important matters, but Ang, Chang, and Dong continue playing. Ang and Chang’s pull on the toy does not
change, but Dong runs around and bites on the toy in a different place. With how big a force and in what
direction must Dong pull on the toy now to balance out the combined pulls from Chang and Ang? Illustrate this
situation by drawing a vector diagram indicating all forces involved.
200 Chapter 6 | Introduction to Vectors
Example 6.11
Vector Algebra
→ → → → ^
Find the magnitude of the vector C that satisfies the equation 2 A − 6 B + 3 C = 2 j , where
→ ^ ^ → ^ ^
A = i − 2 k and B = − j + k /2 .
Strategy
→ → →
We first solve the given equation for the unknown vector C . Then we substitute A and B ; group the
^ ^ ^
terms along each of the three directions i , j , and k ; and identify the scalar components C x , C y , and C z .
Finally, we substitute into Equation 6.21 to find magnitude C.
Solution
→ → → ^
2 A −6 B +3 C = 2j
→ ^ → →
3 C = 2j −2 A +6 B
→ ^ → →
C = 2 j − 2 A +2 B
3 3
^ ^ ^ ⎛ ^ ^⎞ ^ ^ ^ ^ ^
= 2 j − 2 ( i − 2 k ) + 2⎜− j + k ⎟ = 2 j − 2 i + 4 k − 2 j + k
3 3 ⎝ 2 ⎠ 3 3 3
^ ⎛ ⎞^ ⎛ ⎞^
= − 2 i + ⎝2 − 2⎠ j + ⎝4 + 1⎠ k
3 3 3
^ ^ ^
= −2 i − 4 j + 7 k.
3 3 3
The components are C x = −2 /3 , C y = −4/3 , and C z = 7 /3 , and substituting into Equation 6.21 gives
Example 6.12
Displacement of a Skier
Starting at a ski lodge, a cross-country skier goes 5.0 km north, then 3.0 km west, and finally 4.0 km southwest
before taking a rest. Find his total displacement vector relative to the lodge when he is at the rest point. How far
and in what direction must he ski from the rest point to return directly to the lodge?
Strategy
^
We assume a rectangular coordinate system with the origin at the ski lodge and with the unit vector i pointing
^ → → →
east and the unit vector j pointing north. There are three displacements: D 1 , D 2 , and D 3 . We
identify their magnitudes as D 1 = 5.0 km , D 2 = 3.0 km , and D 3 = 4.0 km . We identify their directions are
the angles θ 1 = 90° , θ 2 = 180° , and θ 3 = 180° + 45° = 225° . We resolve each displacement vector to its
scalar components and substitute the components into Equation 6.24 to obtain the scalar components of the
→
resultant displacement D from the lodge to the rest point. On the way back from the rest point to the lodge, the
→ → →
displacement is B = − D . Finally, we find the magnitude and direction of B .
Solution
Scalar components of the displacement vectors are
→ ^ ^ ^ ^
Hence, the skier’s net displacement vector is D = D x i + D y j = (−5.8 i + 2.2 j )km . On the way back
→ → ^ ^ ^ ^
to the lodge, his displacement is B = − D = −(−5.8 i + 2.2 j )km = (5.8 i − 2.2 j )km . Its magnitude is
B = B 2x + B 2y = (5.8) 2 + (−2.2) 2 km = 6.2 km and its direction angle is θ = tan −1(−2.2/5.8) = −20.8° .
Therefore, to return to the lodge, he must go 6.2 km in a direction about 21° south of east.
Significance
Notice that no figure is needed to solve this problem by the analytical method. Figures are required when using a
graphical method; however, we can check if our solution makes sense by sketching it, which is a useful final step
in solving any vector problem.
Example 6.13
Displacement of a Jogger
A jogger runs up a flight of 200 identical steps to the top of a hill and then runs along the top of the hill 50.0 m
before he stops at a drinking fountain (Figure 6.26). His displacement vector from point A at the bottom of the
→ ^ ^
steps to point B at the fountain is D AB = (−90.0 i + 30.0 j )m . What is the height and width of each step in
the flight? What is the actual distance the jogger covers? If he makes a loop and returns to point A, what is his net
displacement vector?
Strategy
→ →
The displacement vector D AB is the vector sum of the jogger’s displacement vector D AT along the stairs
→
(from point A at the bottom of the stairs to point T at the top of the stairs) and his displacement vector D TB on
the top of the hill (from point T at the top of the stairs to the fountain at point B). We must find the horizontal and
→ →
the vertical components of D TB . If each step has width w and height h, the horizontal component of D TB
must have a length of 200w and the vertical component must have a length of 200h. The actual distance the jogger
covers is the sum of the distance he runs up the stairs and the distance of 50.0 m that he runs along the top of the
hill.
Solution
In the coordinate system indicated in Figure 6.26, the jogger’s displacement vector on the top of the hill is
→ ^
D TB = (−50.0 m) i . His net displacement vector is
→ → →
D AB = D AT + D TB.
→
Therefore, his displacement vector D TB along the stairs is
→ → → ^ ^ ^ ^ ^
D AT = D AB − D = (−90.0 i + 30.0 j )m − (−50.0 m) i = [(−90.0 + 50.0) i + 30.0 j )]m
TB
^ ^
= (−40.0 i + 30.0 j )m.
Its scalar components are D AT x = −40.0 m and D ATy = 30.0 m . Therefore, we must have
Thus, the actual distance he runs is D AT + D TB = 50.0 m + 50.0 m = 100.0 m . When he makes a loop and
comes back from the fountain to his initial position at point A, the total distance he covers is twice this distance,
or 200.0 m. However, his net displacement vector is zero, because when his final position is the same as his initial
position, the scalar components of his net displacement vector are zero (Equation 6.13).
In many physical situations, we often need to know the direction of a vector. For example, we may want to know the
direction of a magnetic field vector at some point or the direction of motion of an object. We have already said direction is
given by a unit vector, which is a dimensionless entity—that is, it has no physical units associated with it. When the vector
in question lies along one of the axes in a Cartesian system of coordinates, the answer is simple, because then its unit vector
of direction is either parallel or antiparallel to the direction of the unit vector of an axis. For example, the direction of vector
→ ^ ^ ^ ^ →
d = −5 m i is unit vector d = − i . The general rule of finding the unit vector V of direction for any vector V is
to divide it by its magnitude V:
^ → (6.26)
V= V .
V
We see from this expression that the unit vector of direction is indeed dimensionless because the numerator and the
denominator in Equation 6.26 have the same physical unit. In this way, Equation 6.26 allows us to express the unit
vector of direction in terms of unit vectors of the axes. The following example illustrates this principle.
Example 6.14
The unit vector of the convoy’s direction of motion is the unit vector ^
v that is parallel to the velocity vector. The
unit vector is obtained by dividing a vector by its magnitude, in accordance with Equation 6.26.
Solution
The magnitude of the vector →
v is
→ ^ ^ ^
^ (4.000 i + 3.000 j + 0.100 k )km/h
v = vv =
5.001km/h
^ ^ ^
(4.000 i + 3.000 j + 0.100 k )
=
5.001
4.000 ^ 3.000 ^ 0.100 ^
= i + j + k
5.001 5.001 5.001
^ ^ ^
= (79.98 i + 59.99 j + 2.00 k ) × 10 −2.
Significance
Note that when using the analytical method with a calculator, it is advisable to carry out your calculations to at
least three decimal places and then round off the final answer to the required number of significant figures, which
is the way we performed calculations in this example. If you round off your partial answer too early, you risk your
final answer having a huge numerical error, and it may be far off from the exact answer or from a value measured
in an experiment.
6.10 Check Your Understanding Verify that vector ^ v obtained in Example 6.14 is indeed a unit vector
by computing its magnitude. If the convoy in Example 6.8 was moving across a desert flatland—that is, if the
third component of its velocity was zero—what is the unit vector of its direction of motion? Which geographic
direction does it represent?
204 Chapter 6 | Introduction to Vectors
CHAPTER 6 REVIEW
KEY TERMS
antiparallel vectors two vectors with directions that differ by 180°
parallel vectors two vectors with exactly the same direction angles
parallelogram rule geometric construction of the vector sum in a plane
polar coordinate system an orthogonal coordinate system where location in a plane is given by polar coordinates
polar coordinates a radial coordinate and an angle
radial coordinate distance to the origin in a polar coordinate system
resultant vector vector sum of two (or more) vectors
scalar a number, synonymous with a scalar quantity in physics
scalar component a number that multiplies a unit vector in a vector component of a vector
scalar equation equation in which the left-hand and right-hand sides are numbers
scalar quantity quantity that can be specified completely by a single number with an appropriate physical unit
tail-to-head geometric construction geometric construction for drawing the resultant vector of many vectors
unit vector vector of a unit magnitude that specifies direction; has no physical unit
unit vectors of the axes unit vectors that define orthogonal directions in a plane or in space
vector mathematical object with magnitude and direction
vector components orthogonal components of a vector; a vector is the vector sum of its vector components.
vector equation equation in which the left-hand and right-hand sides are vectors
vector quantity physical quantity described by a mathematical vector—that is, by specifying both its magnitude and its
direction; synonymous with a vector in physics
vector sum resultant of the combination of two (or more) vectors
KEY EQUATIONS
→ →
Multiplication by a scalar (vector equation) B =α A
→ → → →
Commutative law A + B = B + A
→ → → → → →
Associative law ( A + B )+ C = A +( B + C )
→ → →
Distributive law α 1 A + α 2 A = (α 1 + α 2) A
→ ^ ^
The component form of a vector in two dimensions A = Ax i + Ay j
⎧A x = xe − xb
Scalar components of a vector in two dimensions ⎨
⎩A y = ye − yb
⎛A y⎞
The direction angle of a vector in a plane θ A = tan −1
⎝A x ⎠
⎧A x = A cos θ A
Scalar components of a vector in a plane ⎨
⎩A y = A sin θ A
⎧x = r cos φ
Polar coordinates in a plane ⎨
⎩y = r sin φ
→ ^ ^ ^
The component form of a vector in three dimensions A = A x i + Ay j + Az k
→ → → →
Distributive property α( A + B ) = α A + α B
→ → ^ ^ ^
Antiparallel vector to A − A = −A x i − A y j − A z k
⎧A x = B x
⎨A y = B y
→ →
A = B ⇔
⎩A z = B z
Equal vectors
⎧ N
⎪F Rx = ∑ F kx = F 1x + F 2x + … + F Nx
⎪ k=1
⎪ N
⎨F Ry = ∑ F ky = F 1y + F 2y + … + F Ny
⎪
Components of the resultant of N vectors
k=1
⎪ N
⎪F = ∑ F = F + F + … + F
⎩ Rz k = 1 kz 1z 2z Nz
^ →
General unit vector V= V
V
206 Chapter 6 | Introduction to Vectors
SUMMARY
6.1 Scalars and Vectors
• A vector quantity is any quantity that has magnitude and direction, such as displacement or velocity. Vector
quantities are represented by mathematical objects called vectors.
• Geometrically, vectors are represented by arrows, with the end marked by an arrowhead. The length of the vector is
its magnitude, which is a positive scalar. On a plane, the direction of a vector is given by the angle the vector makes
with a reference direction, often an angle with the horizontal. The direction angle of a vector is a scalar.
• Two vectors are equal if and only if they have the same magnitudes and directions. Parallel vectors have the same
direction angles but may have different magnitudes. Antiparallel vectors have direction angles that differ by 180° .
Orthogonal vectors have direction angles that differ by 90° .
• When a vector is multiplied by a scalar, the result is another vector of a different length than the length of the original
vector. Multiplication by a positive scalar does not change the original direction; only the magnitude is affected.
Multiplication by a negative scalar reverses the original direction. The resulting vector is antiparallel to the original
vector. Multiplication by a scalar is distributive. Vectors can be divided by nonzero scalars but cannot be divided by
vectors.
• Two or more vectors can be added to form another vector. The vector sum is called the resultant vector. We can add
vectors to vectors or scalars to scalars, but we cannot add scalars to vectors. Vector addition is commutative and
associative.
• To construct a resultant vector of two vectors in a plane geometrically, we use the parallelogram rule. To construct
a resultant vector of many vectors in a plane geometrically, we use the tail-to-head method.
CONCEPTUAL QUESTIONS
11. Can a magnitude of a vector be negative?
6.1 Scalars and Vectors
1. A weather forecast states the temperature is predicted to 12. Can the magnitude of a particle’s displacement be
be −5 °C the following day. Is this temperature a vector or greater that the distance traveled?
a scalar quantity? Explain.
13. If two vectors are equal, what can you say about their
2. Which of the following is a vector: a person’s height, components? What can you say about their magnitudes?
the altitude on Mt. Everest, the velocity of a fly, the age of What can you say about their directions?
Earth, the boiling point of water, the cost of a book, Earth’s
population, or the acceleration of gravity? 14. If three vectors sum up to zero, what geometric
condition do they satisfy?
3. Give a specific example of a vector, stating its
magnitude, units, and direction.
6.2 Coordinate Systems and Components of a
4. What do vectors and scalars have in common? How do Vector
they differ?
15. Give an example of a nonzero vector that has a
component of zero.
→ →
5. Suppose you add two vectors A and B . What
relative direction between them produces the resultant with 16. Explain why a vector cannot have a component greater
the greatest magnitude? What is the maximum magnitude? than its own magnitude.
What relative direction between them produces the
resultant with the smallest magnitude? What is the 17. If two vectors are equal, what can you say about their
minimum magnitude? components?
9. When a 10,000-m runner competing on a 400-m track 20. If two vectors have the same magnitude, do their
crosses the finish line, what is the runner’s net components have to be the same?
displacement? Can this displacement be zero? Explain.
PROBLEMS
ascends again for 7 m and descends again for 24.0 m,
6.1 Scalars and Vectors where he makes a stop, waiting for his buddy. Assuming the
21. A scuba diver makes a slow descent into the depths positive direction up to the surface, express his net vertical
of the ocean. His vertical position with respect to a boat displacement vector in terms of the unit vector. What is his
on the surface changes several times. He makes the first distance to the boat?
stop 9.0 m from the boat but has a problem with equalizing
the pressure, so he ascends 3.0 m and then continues 22. In a tug-of-war game on one campus, 15 students pull
descending for another 12.0 m to the second stop. From on a rope at both ends in an effort to displace the central
there, he ascends 4 m and then descends for 18.0 m, knot to one side or the other. Two students pull with force
208 Chapter 6 | Introduction to Vectors
196 N each to the right, four students pull with force 98 deal during the day and he is blown along the following
N each to the left, five students pull with force 62 N each directions: 2.50 km and 45.0° north of west, then 4.70 km
to the left, three students pull with force 150 N each to the and 60.0° south of east, then 1.30 km and 25.0° south of
right, and one student pulls with force 250 N to the left. west, then 5.10 km straight east, then 1.70 km and 5.00°
Assuming the positive direction to the right, express the net
east of north, then 7.20 km and 55.0° south of west, and
pull on the knot in terms of the unit vector. How big is the
net pull on the knot? In what direction? finally 2.80 km and 10.0° north of east. Use a graphical
method to find the castaway’s final position relative to the
23. Suppose you walk 18.0 m straight west and then 25.0 island.
m straight north. How far are you from your starting point
and what is the compass direction of a line connecting 28. A small plane flies 40.0 km in a direction 60° north
your starting point to your final position? Use a graphical of east and then flies 30.0 km in a direction 15° north of
method. east. Use a graphical method to find the total distance the
plane covers from the starting point and the direction of the
24. For the vectors given in the following figure, use path to the final position.
a graphical method to find the following resultants: (a)
→ → → → → →
A + B , (b) C + B , (c) D + F , (d) 29. A trapper walks a 5.0-km straight-line distance from
→ → → → → → his cabin to the lake, as shown in the following figure. Use
A − B , (e) D − F , (f) A + 2 F , (g); and (h) a graphical method (the parallelogram rule) to determine
→ → → the trapper’s displacement directly to the east and
A −4 D +2 F .
displacement directly to the north that sum up to his
resultant displacement vector. If the trapper walked only in
directions east and north, zigzagging his way to the lake,
how many kilometers would he have to walk to get to the
lake?
nearest centimeter.
6.2 Coordinate Systems and Components of a
Vector 40. A chameleon is resting quietly on a lanai screen,
waiting for an insect to come by. Assume the origin of a
33. Assuming the +x-axis is horizontal and points to the
Cartesian coordinate system at the lower left-hand corner
right, resolve the vectors given in the following figure to
of the screen and the horizontal direction to the right as the
their scalar components and express them in vector
+x-direction. If its coordinates are (2.000 m, 1.000 m), (a)
component form.
how far is it from the corner of the screen? (b) What is its
location in polar coordinates?
34. Suppose you walk 18.0 m straight west and then 25.0
m straight north. How far are you from your starting point?
41. Two points in the Cartesian plane are A(2.00 m, −4.00
What is your displacement vector? What is the direction of
m) and B(−3.00 m, 3.00 m). Find the distance between
your displacement? Assume the +x-axis is horizontal to the
them and their polar coordinates.
right.
39. Two points in a plane have polar coordinates 46. A small plane flies 40.0 km in a direction 60° north
P 1(2.500 m, π/6) and P 2(3.800 m, 2π/3) . Determine of east and then flies 30.0 km in a direction 15° north
their Cartesian coordinates and the distance between them of east. Use the analytical method to find the total distance
in the Cartesian coordinate system. Round the distance to a the plane covers from the starting point, and the geographic
210 Chapter 6 | Introduction to Vectors
direction of its displacement vector. What is its vector. (b) How far is the restaurant from the post office?
displacement vector? (c) If he returns directly from the restaurant to the post
office, what is his displacement vector on the return trip?
47. In an attempt to escape a desert island, a castaway (d) What is his compass heading on the return trip? Assume
builds a raft and sets out to sea. The wind shifts a great the +x-axis is to the east.
deal during the day, and she is blown along the following
straight lines: 2.50 km and 45.0° north of west, then 4.70 51. An adventurous dog strays from home, runs three
km and 60.0° south of east, then 1.30 km and 25.0° south blocks east, two blocks north, and one block east, one block
of west, then 5.10 km due east, then 1.70 km and 5.00° north, and two blocks west. Assuming that each block is
about a 100 yd, use the analytical method to find the dog’s
east of north, then 7.20 km and 55.0° south of west, and
net displacement vector, its magnitude, and its direction.
finally 2.80 km and 10.0° north of east. Use the analytical Assume the +x-axis is to the east. How would your answer
method to find the resultant vector of all her displacement be affected if each block was about 100 m?
vectors. What is its magnitude and direction?
→ ^ ^
48. Assuming the +x-axis is horizontal to the right for 52. If D = (6.00 i − 8.00 j )m ,
the vectors given in the following figure, use the analytical → ^ ^
→ → B = (−8.00 i + 3.00 j )m , and
method to find the following resultants: (a) A + B ,
→ ^ ^
→ → → → → → A = (26.0 i + 19.0 j )m , find the unknown constants a
(b) C + B , (c) D + F , (d) A − B , (e)
→ → → → → → → → → → →
D − F , (f) A + 2 F , (g) C −2 D +3 F , and b such that a D + b B + A = 0 .
→ → →
and (h) A − 4 D + 2 F .
→ ^ ^
53. Given the displacement vector D = (3 i − 4 j )m,
→
find the displacement vector R so that
→ → ^
D + R = −4D j .
49. Given the vectors in the preceding figure, find vector 56. A barge is pulled by the two tugboats shown in the
→ → → → following figure. One tugboat pulls on the barge with a
R that solves equations (a) D + R = F and (b) force of magnitude 4000 units of force at 15° above the
→ → → →
C − 2 D + 5 R = 3 F . Assume the +x-axis is line AB (see the figure and the other tugboat pulls on the
horizontal to the right. barge with a force of magnitude 5000 units of force at 12°
below the line AB. Resolve the pulling forces to their scalar
components and find the components of the resultant force
50. A delivery man starts at the post office, drives 40 km
pulling on the barge. What is the magnitude of the resultant
north, then 20 km west, then 60 km northeast, and finally
pull? What is its direction relative to the line AB?
50 km north to stop for lunch. Use the analytical method
to determine the following: (a) Find his net displacement
Figure 6.28
ADDITIONAL PROBLEMS
58. You fly 32.0 km in a straight line in still air in the coordinates of the following points: (a) (−x, y), (b) (−2x,
direction 35.0° south of west. (a) Find the distances you −2y), and (c) (3x, −3y).
would have to fly due south and then due west to arrive
at the same point. (b) Find the distances you would have → →
61. Vectors A and B have identical magnitudes
to fly first in a direction 45.0° south of west and then
of 5.0 units. Find the angle between them if
in a direction 45.0° west of north. Note these are the → → ^
components of the displacement along a different set of A + B =5 2j .
axes—namely, the one rotated by 45° with respect to the
axes in (a). 62. Starting at the island of Moi in an unknown
archipelago, a fishing boat makes a round trip with two
59. Rectangular coordinates of a point are given by (2, y) stops at the islands of Noi and Poi. It sails from Moi for
and its polar coordinates are given by (r, π/6) . Find y and 4.76 nautical miles (nmi) in a direction 37° north of east
r. to Noi. From Noi, it sails 69° west of north to Poi. On
its return leg from Poi, it sails 28° east of south. What
60. If the polar coordinates of a point are (r, φ) and distance does the boat sail between Noi and Poi? What
distance does it sail between Moi and Poi? Express your
its rectangular coordinates are (x, y) , determine the polar
answer both in nautical miles and in kilometers. Note: 1
212 Chapter 6 | Introduction to Vectors
nmi = 1852 m. → ^ ^
u = (−18.0 i − 13.0 j )km/h , how fast and in what
63. An air traffic controller notices two signals from two ^ ^
geographic direction is it heading? Here, i and j are
planes on the radar monitor. One plane is at altitude 800
m and in a 19.2-km horizontal distance to the tower in a directions to geographic east and north, respectively.
direction 25° south of west. The second plane is at altitude
1100 m and its horizontal distance is 17.6 km and 20° 69. Find the scalar components of three-dimensional
→ →
south of west. What is the distance between these planes? vectors G and H in the following figure and write
the vectors in vector component form in terms of the unit
→ → → vectors of the axes.
64. Show that when A + B = C , then
Vector G has magnitude 10.0. Its projection in the x y plane
2 2 2
C = A + B + 2AB cos φ , where φ is the angle is between the positive x and positive y directions, at an
→ → angle of 45 degrees from the positive x direction. The angle
between vectors A and B . between vector G and the positive z direction is 60 degrees.
Vector H has magnitude 15.0. Its projection in the x y
65. Four force vectors each have the same magnitude f. plane is between the negative x and positive y directions,
What is the largest magnitude the resultant force vector at an angle of 30 degrees from the positive y direction. The
may have when these forces are added? What is the angle between vector H and the positive z direction is 450
smallest magnitude of the resultant? Make a graph of both degrees.
situations.
70. A diver explores a shallow reef off the coast of Belize.
66. A skater glides along a circular path of radius 5.00 She initially swims 90.0 m north, makes a turn to the east
m in clockwise direction. When he coasts around one- and continues for 200.0 m, then follows a big grouper for
half of the circle, starting from the west point, find (a) 80.0 m in the direction 30° north of east. In the meantime,
the magnitude of his displacement vector and (b) how far a local current displaces her by 150.0 m south. Assuming
he actually skated. (c) What is the magnitude of his the current is no longer present, in what direction and how
displacement vector when he skates all the way around the far should she now swim to come back to the point where
circle and comes back to the west point? she started?
7 | KINEMATICS IN TWO
AND THREE DIMENSIONS
Figure 7.1 The Red Arrows is the aerobatics display team of Britain’s Royal Air Force. Based in Lincolnshire, England, they
perform precision flying shows at high speeds, which requires accurate measurement of position, velocity, and acceleration in
three dimensions. (credit: modification of work by Phil Long)
Chapter Outline
7.1 Displacement and Velocity Vectors
7.2 Acceleration Vector
7.3 Projectile Motion
7.4 Uniform Circular Motion
7.5 Relative Motion in One and Two Dimensions
Introduction
To give a complete description of kinematics, we must explore motion in two and three dimensions. After all, most objects
in our universe do not move in straight lines; rather, they follow curved paths. From kicked footballs to the flight paths of
birds to the orbital motions of celestial bodies and down to the flow of blood plasma in your veins, most motion follows
curved trajectories.
Fortunately, the treatment of motion in one dimension in the previous chapter has given us a foundation on which to build,
as the concepts of position, displacement, velocity, and acceleration defined in one dimension can be expanded to two and
three dimensions. Consider the Red Arrows, also known as the Royal Air Force Aerobatic team of the United Kingdom.
Each jet follows a unique curved trajectory in three-dimensional airspace, as well as has a unique velocity and acceleration.
Thus, to describe the motion of any of the jets accurately, we must assign to each jet a unique position vector in three
dimensions as well as a unique velocity and acceleration vector. We can apply the same basic equations for displacement,
velocity, and acceleration we derived in Motion Along a Straight Line to describe the motion of the jets in two and three
dimensions, but with some modifications—in particular, the inclusion of vectors.
In this chapter we also explore two special types of motion in two dimensions: projectile motion and circular motion. Last,
we conclude with a discussion of relative motion. In the chapter-opening picture, each jet has a relative motion with respect
214 Chapter 7 | Kinematics in Two and Three Dimensions
to any other jet in the group or to the people observing the air show on the ground.
Displacement and velocity in two or three dimensions are straightforward extensions of the one-dimensional definitions.
However, now they are vector quantities, so calculations with them have to follow the rules of vector algebra, not scalar
algebra.
Displacement Vector
To describe motion in two and three dimensions, we must first establish a coordinate system and a convention for the axes.
We generally use the coordinates x, y, and z to locate a particle at point P(x, y, z) in three dimensions. If the particle is
moving, the variables x, y, and z are functions of time (t):
x = x(t) y = y(t) z = z(t). (7.1)
The position vector from the origin of the coordinate system to point P is →
r (t). In unit vector notation, introduced in
Coordinate Systems and Components of a Vector, →
r (t) is
→ ^ ^ ^ (7.2)
r (t) = x(t) i + y(t) j + z(t) k .
Figure 7.2 shows the coordinate system and the vector to point P, where a particle could be located at a particular time
t. Note the orientation of the x, y, and z axes. This orientation is called a right-handed coordinate system (Coordinate
Systems and Components of a Vector) and it is used throughout the chapter.
With our definition of the position of a particle in three-dimensional space, we can formulate the three-dimensional
displacement. Figure 7.3 shows a particle at time t 1 located at P 1 with position vector →
r (t 1). At a later time t 2, the
Δ→
r = →
r (t 2) − →
r (t 1). (7.3)
Vector addition is discussed in Algebra of Vectors. Note that this is the same operation we did in one dimension, but now
the vectors are in three-dimensional space.
→ → →
Figure 7.3 The displacement Δ r = r (t 2) − r (t 1) is
the vector from P 1 to P 2 .
Example 7.1
Figure 7.4 Two position vectors are drawn from the center of Earth,
which is the origin of the coordinate system, with the y-axis as north and the
x-axis as east. The vector between them is the displacement of the satellite.
→ ^
r (t 1) = 6770. km j
→ ^ ^
r (t 2) = 6770. km (cos 45°) i + 6770. km (sin(−45°)) j .
→ ^
r (t 1) = 6770. j
→ ^ ^
r (t 2) = 4787 i − 4787 j .
Note that the satellite took a curved path along its circular orbit to get from its initial position to its final position in this
example. It also could have traveled 4787 km east, then 11,557 km south to arrive at the same location. Both of these paths
are longer than the length of the displacement vector. In fact, the displacement vector gives the shortest path between two
points in one, two, or three dimensions.
Many applications in physics can have a series of displacements, as discussed in the previous chapter. The total
displacement is the sum of the individual displacements, only this time, we need to be careful, because we are adding
vectors. We illustrate this concept with an example of Brownian motion.
Example 7.2
Brownian Motion
Brownian motion is a chaotic random motion of particles suspended in a fluid, resulting from collisions with
the molecules of the fluid. This motion is three-dimensional. The displacements in numerical order of a particle
undergoing Brownian motion could look like the following, in micrometers (Figure 7.6):
^ ^ ^
Δ→
r 1 = 2.0 i + j + 3.0 k
^ ^
Δ→
r 2 = − i + 3.0 k
^ ^ ^
Δ→
r 3 = 4.0 i − 2.0 j + k
^ ^ ^
Δ→
r 4 = −3.0 i + j + 2.0 k .
Solution
We form the sum of the displacements and add them as vectors:
Δ→
r Total =∑Δ→
r i=Δ→
r 1+Δ→
r 2+Δ→
r 3+Δ→
r 4
^ ^ ^
= (2.0 − 1.0 + 4.0 − 3.0) i + (1.0 + 0 − 2.0 + 1.0) j + (3.0 + 3.0 + 1.0 + 2.0) k
^ ^ ^
= 2.0 i + 0 j + 9.0 k μm.
Velocity Vector
In the previous chapter we found the instantaneous velocity by calculating the derivative of the position function with
respect to time. We can do the same operation in two and three dimensions, but we use vectors. The instantaneous velocity
vector is now
→
r (t + Δt) − →
r (t) d → (7.4)
→
v (t) = lim = r .
Δt → 0 Δt dt
Let’s look at the relative orientation of the position vector and velocity vector graphically. In Figure 7.7 we show the
vectors → r (t) and → r (t + Δt), which give the position of a particle moving along a path represented by the gray line.
As Δt goes to zero, the velocity vector, given by Equation 7.4, becomes tangent to the path of the particle at time t.
→ ^ ^ ^
r (t) = x(t) i + y(t) j + z(t) k ,
we can write
→ ^ ^ ^ (7.5)
v (t) = v x(t) i + v y(t) j + v z(t) k
where
If only the average velocity is of concern, we have the vector equivalent of the one-dimensional average velocity for two
and three dimensions:
→
r (t 2) − → r (t 1) (7.7)
→
v = .
avg t2 − t1
Example 7.3
dr(t) ^ ^ ^
(a) v(t) = = 4.0t i + 3.0 j + 5.0 k m/s
dt
→ ^ ^ ^
v (2.0s) = 8.0 i + 3.0 j + 5.0 k m/s
Speed → | |
v (2.0 s) = 8 2 + 3 2 + 5 2 = 9.9 m/s.
^ ^
Check Your Understanding The position function of a particle is →
7.1
r (t) = 3.0t 3 i + 4.0 j . (a) What is
the instantaneous velocity at t = 3 s? (b) Is the average velocity between 2 s and 4 s equal to the instantaneous
velocity at t = 3 s?
Independence of Motion
In the kinematic description of motion, we are able to treat the horizontal and vertical components of motion separately.
In many cases, motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.
An example illustrating the independence of vertical and horizontal motions is given by two baseballs. One baseball is
dropped from rest. At the same instant, another is thrown horizontally from the same height and it follows a curved path. A
stroboscope captures the positions of the balls at fixed time intervals as they fall (Figure 7.8).
It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity implies
vertical motion is independent of whether the ball is moving horizontally. (Assuming no air resistance, the vertical motion
of a falling object is influenced by gravity only, not by any horizontal forces.) Careful examination of the ball thrown
horizontally shows it travels the same horizontal distance between flashes. This is because there are no additional forces on
the ball in the horizontal direction after it is thrown. This result means horizontal velocity is constant and is affected neither
by vertical motion nor by gravity (which is vertical). Note this case is true for ideal conditions only. In the real world, air
resistance affects the speed of the balls in both directions.
The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional
motions (horizontal and vertical). The key to analyzing such motion, called projectile motion, is to resolve it into motions
along perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because the
components are independent.
222 Chapter 7 | Kinematics in Two and Three Dimensions
Instantaneous Acceleration
In addition to obtaining the displacement and velocity vectors of an object in motion, we often want to know its acceleration
vector at any point in time along its trajectory. This acceleration vector is the instantaneous acceleration and it can be
obtained from the derivative with respect to time of the velocity function, as we have seen in a previous chapter. The only
difference in two or three dimensions is that these are now vector quantities. Taking the derivative with respect to time
→v (t), we find
→
v (t + Δt) − →
v (t) d →
v (t) (7.8)
→
a (t) = lim = .
t→0 Δt dt
Also, since the velocity is the derivative of the position function, we can write the acceleration in terms of the second
derivative of the position function:
Example 7.4
^ ^ ^
(b) Evaluating →
a (2.0 s) = 5.0 i + 4.0 j − 24.0 k m/s 2 gives us the direction in unit vector notation. The
magnitude of the acceleration is → | |
a (2.0 s) = 5.0 2 + 4.0 2 + (−24.0) 2 = 24.8 m/s 2.
Significance
In this example we find that acceleration has a time dependence and is changing throughout the motion. Let’s
consider a different velocity function for the particle.
Example 7.5
The velocity function is linear in time in the x direction and is constant in the y and z directions.
(b) Taking the derivative of the velocity function, we find
→ ^
a (t) = −2 i m/s 2.
The acceleration vector is a constant in the negative x-direction.
(c) The trajectory of the particle can be seen in Figure 7.9. Let’s look in the y and z directions first. The particle’s
position increases steadily as a function of time with a constant velocity in these directions. In the x direction,
however, the particle follows a path in positive x until t = 5 s, when it reverses direction. We know this from
looking at the velocity function, which becomes zero at this time and negative thereafter. We also know this
because the acceleration is negative and constant—meaning, the particle is decelerating, or accelerating in the
negative direction. The particle’s position reaches 25 m, where it then reverses direction and begins to accelerate
in the negative x direction. The position reaches zero at t = 10 s.
224 Chapter 7 | Kinematics in Two and Three Dimensions
Figure 7.9 The particle starts at point (x, y, z) = (0, 0, 0) with position vector
→
r = 0. The projection of the trajectory onto the xy-plane is shown. The
values of y and z increase linearly as a function of time, whereas x has a turning
point at t = 5 s and 25 m, when it reverses direction. At this point, the x
component of the velocity becomes negative. At t = 10 s, the particle is back to 0
m in the x direction.
Significance
By graphing the trajectory of the particle, we can better understand its motion, given by the numerical results of
the kinematic equations.
7.2 Check Your Understanding Suppose the acceleration function has the form
→ ^ ^ ^
a (t) = a i + b j + c k m/s 2, where a, b, and c are constants. What can be said about the functional form of
the velocity function?
Constant Acceleration
Multidimensional motion with constant acceleration can be treated the same way as shown in the previous chapter for
one-dimensional motion. Earlier we showed that three-dimensional motion is equivalent to three one-dimensional motions,
each along an axis perpendicular to the others. To develop the relevant equations in each direction, let’s consider the two-
dimensional problem of a particle moving in the xy plane with constant acceleration, ignoring the z-component for the
moment. The acceleration vector is
→ ^ ^
a = a 0x i + a 0y j .
Each component of the motion has a separate set of equations similar to Equation 3.9–Equation 3.13 of the
Section 3.4. We show only the equations for position and velocity in the x- and y-directions. A similar set of kinematic
equations could be written for motion in the z-direction:
x(t) = x 0 + v 0x t + 1 a x t 2 (7.13)
2
v 2x(t) = v 20x + 2a x(x − x 0) (7.14)
y(t) = y 0 + v 0y t + 1 a y t 2 (7.17)
2
v 2y(t) = v 20y + 2a y(y − y 0). (7.18)
Here the subscript 0 denotes the initial position or velocity. Equation 7.11 to Equation 7.18 can be substituted into
Equation 7.2 and Equation 7.5 without the z-component to obtain the position vector and velocity vector as a function
of time in two dimensions:
→ ^ ^ ^ ^
r (t) = x(t) i + y(t) j and →
v (t) = v x(t) i + v y(t) j .
The following example illustrates a practical use of the kinematic equations in two dimensions.
Example 7.6
A Skier
Figure 7.10 shows a skier moving with an acceleration of 2.1 m/s 2 down a slope of 15° at t = 0. With the
origin of the coordinate system at the front of the lodge, her initial position and velocity are
→ ^ ^
r (0) = (75.0 i − 50.0 j ) m
and
→ ^ ^
v (0) = (4.1 i − 1.1 j ) m/s.
(a) What are the x- and y-components of the skier’s position and velocity as functions of time? (b) What are her
position and velocity at t = 10.0 s?
226 Chapter 7 | Kinematics in Two and Three Dimensions
Figure 7.10 A skier has an acceleration of 2.1 m/s 2 down a slope of 15°. The origin of the coordinate system is at
the ski lodge.
Strategy
Since we are evaluating the components of the motion equations in the x and y directions, we need to find the
components of the acceleration and put them into the kinematic equations. The components of the acceleration
are found by referring to the coordinate system in Figure 7.10. Then, by inserting the components of the initial
position and velocity into the motion equations, we can solve for her position and velocity at a later time t.
Solution
(a) The origin of the coordinate system is at the top of the hill with y-axis vertically upward and the x-axis
horizontal. By looking at the trajectory of the skier, the x-component of the acceleration is positive and the
y-component is negative. Since the angle is 15° down the slope, we find
Inserting the initial position and velocity into Equation 7.12 and Equation 7.13 for x, we have
(b) Now that we have the equations of motion for x and y as functions of time, we can evaluate them at t = 10.0 s:
The magnitude of the velocity of the skier at 10.0 s is 25 m/s, which is 60 mi/h.
Significance
It is useful to know that, given the initial conditions of position, velocity, and acceleration of an object, we can
find the position, velocity, and acceleration at any later time.
With Equation 7.8 through Equation 7.10 we have completed the set of expressions for the position, velocity, and
acceleration of an object moving in two or three dimensions. If the trajectories of the objects look something like the “Red
Arrows” in the opening picture for the chapter, then the expressions for the position, velocity, and acceleration can be quite
complicated. In the sections to follow we examine two special cases of motion in two and three dimensions by looking at
projectile motion and circular motion.
At this University of Colorado Boulder website (https://openstaxcollege.org/l/21phetmotladyb) ,
you can explore the position velocity and acceleration of a ladybug with an interactive simulation that allows you
to change these parameters.
Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of
gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as
they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their
path is called a trajectory. The motion of falling objects as discussed in Section 3.5 is a simple one-dimensional type of
projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion,
and our treatment neglects the effects of air resistance.
The most important fact to remember here is that motions along perpendicular axes are independent and thus can be
analyzed separately. We discussed this fact in Section 7.1, where we saw that vertical and horizontal motions are
independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: one along the
horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from
gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary,
we call the horizontal axis the x-axis and the vertical axis the y-axis. It is not required that we use this choice of axes; it is
simply convenient in the case of gravitational acceleration. In other cases we may choose a different set of axes. Figure
7.11 illustrates the notation for displacement, where we define → s to be the total displacement, and → x and → y are its
component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y.
228 Chapter 7 | Kinematics in Two and Three Dimensions
→
s has
Figure 7.11 The total displacement s of a soccer ball at a point along its path. The vector
→ →
components x and y along the horizontal and vertical axes. Its magnitude is s and it makes an angle
θ with the horizontal.
To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find
their components along the x- and y-axes. Let’s assume all forces except gravity (such as air resistance and friction, for
example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:
a y = −g = −9.8 m/s 2 ( − 32 ft/s 2).
Because gravity is vertical, a x = 0. If a x = 0, this means the initial velocity in the x direction is equal to the final velocity
in the x direction, or v x = v 0x. With these conditions on acceleration and velocity, we can write the kinematic Equation
7.11 through Equation 7.18 for motion in a uniform gravitational field, including the rest of the kinematic equations
for a constant acceleration from Section 3.4. The kinematic equations for motion in a uniform gravitational field become
kinematic equations with a y = −g, a x = 0 :
Horizontal Motion
v 0x = v x, x = x 0 + v x t (7.19)
Vertical Motion
y = y 0 + 1 (v 0y + v y)t (7.20)
2
v y = v 0y − gt (7.21)
y = y 0 + v 0y t − 1 gt 2 (7.22)
2
v 2y = v 20y − 2g(y − y 0) (7.23)
Using this set of equations, we can analyze projectile motion, keeping in mind some important points.
Figure 7.12 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions
along the vertical and horizontal axes. (b) The horizontal motion is simple, because a x = 0 and v x is a constant. (c) The
velocity in the vertical direction begins to decrease as the object rises. At its highest point, the vertical velocity is zero. As the
object falls toward Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial
vertical velocity. (d) The x and y motions are recombined to give the total velocity at any given point on the trajectory.
230 Chapter 7 | Kinematics in Two and Three Dimensions
Example 7.7
Strategy
The motion can be broken into horizontal and vertical motions in which a x = 0 and a y = −g. We can then
define x 0 and y 0 to be zero and solve for the desired quantities.
Solution
(a) By “height” we mean the altitude or vertical position y above the starting point. The highest point in any
trajectory, called the apex, is reached when v y = 0. Since we know the initial and final velocities, as well as the
initial position, we use the following equation to find y:
v 2y = v 20y − 2g(y − y 0).
0 = v 20y − 2gy.
Now we must find v 0y, the component of the initial velocity in the y direction. It is given by v 0y = v 0 sinθ 0,
where v 0 is the initial velocity of 70.0 m/s and θ 0 = 75° is the initial angle. Thus,
and y is
(67.6 m/s) 2
y= .
2(9.80 m/s 2)
Thus, we have
y = 233 m.
Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the
acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical
component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity
reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for
large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is
not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the
same height.
(b) As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest
point. In this case, the easiest method is to use v y = v 0y − gt. Because v y = 0 at the apex, this equation reduces
to simply
0 = v 0y − gt
or
v 0y
t = g = 67.6 m/s2 = 6.90s.
9.80 m/s
This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several
seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + 1 (v 0y + v y)t. This
2
is left for you as an exercise to complete.
(c) Because air resistance is negligible, a x = 0 and the horizontal velocity is constant, as discussed earlier. The
horizontal displacement is the horizontal velocity multiplied by time as given by x = x 0 + v x t, where x 0 is
equal to zero. Thus,
x = v x t,
where v x is the x-component of the velocity, which is given by
When solving Example 7.7(a), the expression we found for y is valid for any projectile motion when air resistance is
negligible. Call the maximum height y = h. Then,
v 20y
h= .
2g
This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical
component of the initial velocity.
7.3 Check Your Understanding A rock is thrown horizontally off a cliff 100.0 m high with a velocity of
15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion?
(c) Which equations describe the vertical motion? (d) What is the rock’s velocity at the point of impact?
Example 7.8
Figure 7.14 The trajectory of a tennis ball hit into the stands.
Strategy
Again, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solve
for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, we solve
for t first. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity.
This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain → v at
final time t, determined in the first part of the example.
Solution
(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We
can find the time for this by using Equation 7.22:
y = y 0 + v 0y t − 1 gt 2.
2
If we take the initial position y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the
vertical component of the initial velocity:
we can combine them to find the magnitude of the total velocity vector → v and the angle θ it makes with
the horizontal. Since v x is constant, we can solve for it at any horizontal location. We choose the starting point
because we know both the initial velocity and the initial angle. Therefore,
v x = v 0 cosθ 0 = (30 m/s)cos 45° = 21.2 m/s.
⎛v y ⎞ ⎛ ⎞
θ v = tan −1⎝v ⎠ = tan −1 ⎝ 21.2 ⎠ = −53.1°.
x −15.9
Significance
(a) As mentioned earlier, the time for projectile motion is determined completely by the vertical motion. Thus,
any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude spends
3.79 s in the air. (b) The negative angle means the velocity is 53.1° below the horizontal at the point of impact.
This result is consistent with the fact that the ball is impacting at a point on the other side of the apex of the
trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the
magnitude of the initial velocity we expect since it is impacting 10.0 m above the launch elevation.
y − y 0 = v 0y t − 1 gt 2 = (v 0 sinθ 0)t − 1 gt 2 = 0.
2 2
Factoring, we have
⎛ gt ⎞
t⎝v 0 sinθ 0 − = 0.
2⎠
Solving for t gives us
This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation 7.24 does not
apply when the projectile lands at a different elevation than it was launched, as we saw in Example 7.8 of the tennis player
hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly
proportional to the initial velocity in the y direction and inversely proportional to g. Thus, on the Moon, where gravity is
one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.
Trajectory
The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and
solving for y(x). We take x 0 = y 0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives
x = v 0x t ⇒ t = vx = x .
0x v 0 cosθ 0
Substituting the expression for t into the equation for the position y = (v 0 sinθ 0)t − 1 gt 2 gives
2
⎛ x ⎞ 1 ⎛ x ⎞
2
y = (v 0 sinθ 0)
⎝v 0 cosθ 0 ⎠ − 2 g⎝v 0 cosθ 0 ⎠ .
Rearranging terms, we have
⎡ g ⎤ (7.25)
y = (tanθ 0)x − ⎢ ⎥x 2.
⎣2(v 0 cosθ 0) ⎦
2
This trajectory equation is of the form y = ax + bx 2, which is an equation of a parabola with coefficients
g
a = tanθ 0, b = − .
2(v 0 cosθ 0) 2
Range
From the trajectory equation we can also find the range, or the horizontal distance traveled by the projectile. Factoring
Equation 7.25, we have
⎡ g ⎤
y = x⎢tanθ 0 − x ⎥.
⎣ 2(v 0 cosθ 0) 2 ⎦
The position y is zero for both the launch point and the impact point, since we are again considering only a flat horizontal
surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch point, and
2v 20 sinθ 0 cosθ 0
x= g ,
corresponding to the impact point. Using the trigonometric identity 2sinθcosθ = sin2θ and setting x = R for range, we
find
v 20 sin2θ 0 (7.26)
R= g .
Note particularly that Equation 7.26 is valid only for launch and impact on a horizontal surface. We see the range is
directly proportional to the square of the initial speed v 0 and sin2θ 0 , and it is inversely proportional to the acceleration of
gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, we
see from the factor sin2θ 0 that the range is maximum at 45°. These results are shown in Figure 7.15. In (a) we see that
the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at 45°. This is true only for
conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting
that the same range is found for two initial launch angles that sum to 90°. The projectile launched with the smaller angle
has a lower apex than the higher angle, but they both have the same range.
Example 7.9
the ball drop with a minimum amount of rolling after impact. Here, he angles the shot at 70° to the horizontal to
minimize rolling after impact. Both shots are hit and impacted on a level surface.
(a) What is the initial speed of the ball at the second hole?
(b) What is the initial speed of the ball at the fourth hole?
(c) Write the trajectory equation for both cases.
(d) Graph the trajectories.
Strategy
We see that the range equation has the initial speed and angle, so we can solve for the initial speed for both (a)
and (b). When we have the initial speed, we can use this value to write the trajectory equation.
Solution
2
v sin2θ Rg 90.0 m(9.8 m/s 2)
(a) R = 0 g 0 ⇒ v 0 = = = 37.0 m/s
sin2θ 0 sin(2(70°))
2
v sin2θ Rg 90.0 m(9.8 m/s 2)
(b) R = 0 g 0 ⇒ v 0 = = = 31.9 m/s
sin2θ 0 sin(2(30°))
(c)
⎡ g ⎤
y = x⎢tanθ 0 − x ⎥
⎣ 2(v 0 cosθ 0) 2 ⎦
⎡ ⎤
Second hole: y = x⎢tan 70° − 9.8 m/s 2 x⎥ = 2.75x − 0.0306x 2
⎣ 2[(37.0 m/s)(cos 70°)] 2 ⎦
⎡ ⎤
Fourth hole: y = x⎢tan 30° − 9.8 m/s 2 x⎥ = 0.58x − 0.0064x 2
⎣ 2[(31.9 m/s)(cos30°)] 2 ⎦
(d) Using a graphing utility, we can compare the two trajectories, which are shown in Figure 7.16.
Significance
The initial speed for the shot at 70° is greater than the initial speed of the shot at 30°. Note from Figure 7.16
that two projectiles launched at the same speed but at different angles have the same range if the launch angles
add to 90°. The launch angles in this example add to give a number greater than 90°. Thus, the shot at 70° has
to have a greater launch speed to reach 90 m, otherwise it would land at a shorter distance.
7.4 Check Your Understanding If the two golf shots in Example 7.9 were launched at the same speed,
which shot would have the greatest range?
When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference
of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity
changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile
has farther to fall than it would on level ground, as shown in Figure 7.17, which is based on a drawing in Newton’s
Principia . If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 s
an object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000mi/hr)
near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. This
is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit.
These and other aspects of orbital motion, such as Earth’s rotation, are covered in greater depth in Newton's Synthesis
(https://legacy.cnx.org/content/m58344/latest/) .
Centripetal Acceleration
In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-
dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory
such as a circle. In this case the velocity vector is changing, or d →
v /dt ≠ 0. This is shown in Figure 7.18. As the particle
moves counterclockwise in time Δt on the circular path, its position vector moves from →
r (t) to →
r (t + Δt). The
velocity vector has constant magnitude and is tangent to the path as it changes from →
v (t) to →
v (t + Δt), changing its
v (t) is perpendicular to the position vector →
direction only. Since the velocity vector → r (t), the triangles formed by the
position vectors and Δ → v are similar. Furthermore, since →
r , and the velocity vectors and Δ → r (t) = →
r (t + Δt) | | | |
and → |
v (t) = → | | |
v (t + Δt) , the two triangles are isosceles. From these facts we can make the assertion
Δv = Δr or Δv = v Δr.
v r r
Figure 7.18 (a) A particle is moving in a circle at a constant speed, with position and velocity vectors at times t
→
and t + Δt. (b) Velocity vectors forming a triangle. The two triangles in the figure are similar. The vector Δ v
points toward the center of the circle in the limit Δt → 0.
The direction of the acceleration can also be found by noting that as Δt and therefore Δθ approach zero, the vector Δ →
v
approaches a direction perpendicular to →
v . In the limit Δt → 0, Δ →
v is perpendicular to →
v . Since →
v is tangent
to the circle, the acceleration d →
v /dt points toward the center of the circle. Summarizing, a particle moving in a circle at
a constant speed has an acceleration with magnitude
2 (7.27)
a C = vr .
The direction of the acceleration vector is toward the center of the circle (Figure 7.19). This is a radial acceleration and
is called the centripetal acceleration, which is why we give it the subscript c. The word centripetal comes from the Latin
words centrum (meaning “center”) and petere (meaning to seek”), and thus takes the meaning “center seeking.”
240 Chapter 7 | Kinematics in Two and Three Dimensions
Let’s investigate some examples that illustrate the relative magnitudes of the velocity, radius, and centripetal acceleration.
Example 7.10
Creating an Acceleration of 1 g
A jet is flying at 134.1 m/s along a straight line and makes a turn along a circular path level with the ground. What
does the radius of the circle have to be to produce a centripetal acceleration of 1 g on the pilot and jet toward the
center of the circular trajectory?
Strategy
Given the speed of the jet, we can solve for the radius of the circle in the expression for the centripetal
acceleration.
Solution
Set the centripetal acceleration equal to the acceleration of gravity: 9.8 m/s 2 = v 2/r.
Solving for the radius, we find
(134.1 m/s) 2
r= = 1835 m = 1.835 km.
9.8 m/s 2
Significance
To create a greater acceleration than g on the pilot, the jet would either have to decrease the radius of its circular
trajectory or increase its speed on its existing trajectory or both.
7.5 Check Your Understanding A flywheel has a radius of 20.0 cm. What is the speed of a point on the edge
of the flywheel if it experiences a centripetal acceleration of 900.0 cm/s 2 ?
Centripetal acceleration can have a wide range of values, depending on the speed and radius of curvature of the circular
path. Typical centripetal accelerations are given in the following table.
→ ^ ^ (7.28)
r (t) = Acosωt i + Asin ωt j .
Here, ω is a constant called the angular frequency of the particle. The angular frequency has units of radians (rad) per
second and is simply the number of radians of angular measure through which the particle passes per second. The angle θ
that the position vector has at any particular time is ωt .
If T is the period of motion, or the time to complete one revolution ( 2π rad), then
ω = 2π .
T
Velocity and acceleration can be obtained from the position function by differentiation:
242 Chapter 7 | Kinematics in Two and Three Dimensions
→ d→
r (t) ^ ^ (7.29)
v (t) = = −Aωsin ωt i + Aωcos ωt j .
dt
It can be shown from Figure 7.20 that the velocity vector is tangential to the circle at the location of the particle, with
magnitude Aω. Similarly, the acceleration vector is found by differentiating the velocity:
→ d→
v (t) ^ ^ (7.30)
a (t) = = −Aω 2 cos ωt i − Aω 2 sin ωt j .
dt
From this equation we see that the acceleration vector has magnitude Aω 2 and is directed opposite the position vector,
toward the origin, because →
a (t) = −ω 2 →
r (t).
Example 7.11
→ ^ ^
r (2.0 × 10 −7 s) = A cos ω(2.0 × 10 −7 s) i + A sin ω(2.0 × 10 −7 s) j m
^
= 0.175cos[(2.856 × 10 7 rad/s)(2.0 × 10 −7 s)] i
^
+0.175sin[(2.856 × 10 7 rad/s)(2.0 × 10 −7 s)] j m
^ ^ ^ ^
= 0.175cos(5.712 rad) i + 0.175sin(5.712 rad) j = 0.147 i − 0.095 j m.
From this result we see that the proton is located slightly below the x-axis. This is shown in Figure 7.21.
Significance
We picked the initial position of the particle to be on the x-axis. This was completely arbitrary. If a different
starting position were given, we would have a different final position at t = 200 ns.
aT = | |
d →v
.
(7.31)
dt
The direction of tangential acceleration is tangent to the circle whereas the direction of centripetal acceleration is radially
inward toward the center of the circle. Thus, a particle in circular motion with a tangential acceleration has a total
acceleration that is the vector sum of the centripetal and tangential accelerations:
→
a = →
a → (7.32)
C+ a T.
244 Chapter 7 | Kinematics in Two and Three Dimensions
The acceleration vectors are shown in Figure 7.22. Note that the two acceleration vectors →
a C and →
a T are
perpendicular to each other, with →
a C in the radial direction and →
a T in the tangential direction. The total acceleration
→
a points at an angle between →
a →
a T.
C and
Example 7.12
| |
→ 2c
a T = d v = 32 = 12.03 m/s 2 = 1.5 m/s 2.
dt t (2.0)
Total acceleration is
and θ = tan −1 3.1 = 64° from the tangent to the circle. See Figure 7.23.
1.5
Significance
The directions of centripetal and tangential accelerations can be described more conveniently in terms of a polar
coordinate system, with unit vectors in the radial and tangential directions. This coordinate system, which is used
for motion along curved paths, is discussed in detail later in the book.
We have already discussed (in Section 5.6) how velocities transform non-intuitively in the world of relativity. The
discussion in this chapter will assume that the velocities of all the objects involved are much, much less than the speed
of light, so that the relativistic corrections may be ignored. However, from the extensive discussion in the chapter on
Relativistic Kinematics in One Dimension, we already understand the importance in kinematics of asking the
question, "Who is the observer?"
Motion Relative to What?
Motion does not happen in isolation. If you’re riding in a train moving at 10 m/s east, this velocity is measured relative to
the ground on which you’re traveling. However, if another train passes you at 15 m/s east, your velocity relative to this other
train is different from your velocity relative to the ground. Your velocity relative to the other train is 5 m/s west. To explore
this idea further, we first need to establish some terminology.
Reference Frames
To discuss relative motion in one or more dimensions, we return to the concept of reference frames. When we say an object
has a certain velocity, we must state it has a velocity with respect to a given reference frame. In most examples we have
examined so far, this reference frame has been Earth. If you say a person is sitting in a train moving at 10 m/s east, then you
246 Chapter 7 | Kinematics in Two and Three Dimensions
imply the person on the train is moving relative to the surface of Earth at this velocity, and Earth is the reference frame. We
can expand our view of the motion of the person on the train and say Earth is spinning in its orbit around the Sun, in which
case the motion becomes more complicated. In this case, the solar system is the reference frame. In summary, all discussion
of relative motion must define the reference frames involved. We now develop a method to refer to reference frames in
relative motion.
Note the ordering of the subscripts for the various reference frames in Equation 7.33. The subscripts for the coupling
reference frame, which is the train, appear consecutively in the right-hand side of the equation. Figure 7.24 shows the
correct order of subscripts when forming the vector equation.
^
Adding the vectors, we find →
v PE = 8 m/s i , so the person is moving 8 m/s east with respect to Earth. Graphically, this
is shown in Figure 7.25.
Figure 7.25 Velocity vectors of the train with respect to Earth, person with respect to the
train, and person with respect to Earth.
→
r = →
r → (7.34)
PS PS′ + r S′ S.
The relative velocities are the time derivatives of the position vectors. Therefore,
→
v = →
v → (7.35)
PS PS′ + v S′ S.
The velocity of a particle relative to S is equal to its velocity relative to S′ plus the velocity of S′ relative to S.
We can extend Equation 7.35 to any number of reference frames. For particle P with velocities
→
v PA, →
v PB , and →
v PC in frames A, B, and C,
→
v = →
v → → (7.36)
PC PA + v AB + v BC.
We can also see how the accelerations are related as observed in two reference frames by differentiating Equation 7.35:
→
a = →
a → (7.37)
PS PS′ + a S′ S.
This says the acceleration of a particle is the same as measured by two observers moving at a constant velocity relative to
each other.
248 Chapter 7 | Kinematics in Two and Three Dimensions
Example 7.13
Strategy
First, we must establish the reference frame common to both vehicles, which is Earth. Then, we write the
velocities of each with respect to the reference frame of Earth, which enables us to form a vector equation that
links the car, the truck, and Earth to solve for the velocity of the car with respect to the truck.
Solution
^
The velocity of the car with respect to Earth is →
v CE = 80 km/h i . The velocity of the truck with respect to
^
Earth is →
v TE = −70 km/h j . Using the velocity addition rule, the relative motion equation we are seeking is
→
v = →
v →
CT CE + v ET.
Here, → v CT is the velocity of the car with respect to the truck, and Earth is the connecting reference frame.
Since we have the velocity of the truck with respect to Earth, the negative of this vector is the velocity of Earth
with respect to the truck: →
v ET = − → v TE. The vector diagram of this equation is shown in Figure 7.28.
We can now solve for the velocity of the car with respect to the truck:
and
⎛ ⎞
θ = tan −1 ⎝70.0 ⎠ = 41.2° north of east.
80.0
Significance
Drawing a vector diagram showing the velocity vectors can help in understanding the relative velocity of the two
objects.
7.6 Check Your Understanding A boat heads north in still water at 4.5 m/s directly across a river that is
running east at 3.0 m/s. What is the velocity of the boat with respect to Earth?
Example 7.14
CHAPTER 7 REVIEW
KEY TERMS
acceleration vector instantaneous acceleration found by taking the derivative of the velocity function with respect to
time in unit vector notation
angular frequency ω, rate of change of an angle with which an object that is moving on a circular path
centripetal acceleration component of acceleration of an object moving in a circle that is directed radially inward
toward the center of the circle
displacement vector vector from the initial position to a final position on a trajectory of a particle
position vector vector from the origin of a chosen coordinate system to the position of a particle in two- or three-
dimensional space
projectile motion motion of an object subject only to the acceleration of gravity
range maximum horizontal distance a projectile travels
reference frame coordinate system in which the position, velocity, and acceleration of an object at rest or moving is
measured
relative velocity velocity of an object as observed from a particular reference frame, or the velocity of one reference
frame with respect to another reference frame
tangential acceleration magnitude of which is the time rate of change of speed. Its direction is tangent to the circle.
time of flight elapsed time a projectile is in the air
total acceleration vector sum of centripetal and tangential accelerations
trajectory path of a projectile through the air
velocity vector vector that gives the instantaneous speed and direction of a particle; tangent to the trajectory
KEY EQUATIONS
→ ^ ^ ^
Position vector r (t) = x(t) i + y(t) j + z(t) k
Displacement vector Δ→
r = →
r (t 2) − →
r (t 1)
→
r (t + Δt) − →
r (t) d →
→
v (t) = lim = r
Velocity vector
Δt → 0 Δt dt
→ ^ ^ ^
Velocity in terms of components v (t) = v x(t) i + v y(t) j + v z(t) k
2(v 0 sinθ)
Time of flight T tof = g
⎡ g ⎤
Trajectory y = (tanθ 0)x − ⎢ ⎥x 2
⎣2(v 0 cosθ 0) 2 ⎦
v 20 sin 2θ 0
Range R= g
2
Centripetal acceleration a C = vr
→ ^ ^
Position vector, uniform circular motion r (t) = Acos ωt i + A sin ωt j
→ d→
r (t) ^ ^
Velocity vector, uniform circular motion v (t) = = −Aω sin ωt i + Aω cos ωt j
dt
→ d→
v (t) ^ ^
Acceleration vector, uniform circular motion a (t) = = −Aω 2 cos ωt i − Aω 2 sin ωt j
dt
Tangential acceleration aT =
d →v | |
dt
→
a = →
a →
Total acceleration C+ a T
SUMMARY
7.1 Displacement and Velocity Vectors
• The position function →
r (t) gives the position as a function of time of a particle moving in two or three
dimensions. Graphically, it is a vector from the origin of a chosen coordinate system to the point where the particle
is located at a specific time.
• The displacement vector Δ →
r gives the shortest distance between any two points on the trajectory of a particle in
two or three dimensions.
• Instantaneous velocity gives the speed and direction of a particle at a specific time on its trajectory in two or three
dimensions, and is a vector in two and three dimensions.
• The velocity vector is tangent to the trajectory of the particle.
• Displacement →
r (t) can be written as a vector sum of the one-dimensional displacements →
x (t), →
y (t), →
z (t)
along the x, y, and z directions.
• Velocity →
v (t) can be written as a vector sum of the one-dimensional velocities v x(t), v y(t), v z(t) along the x, y,
and z directions.
• The kinematic equations for constant acceleration can be written as the vector sum of the constant acceleration
equations in the x, y, and z directions.
2(v 0 sinθ)
T to f = g .
This equation is valid only when the projectile lands at the same elevation from which it was launched.
• The maximum horizontal distance traveled by a projectile is called the range. Again, the equation for range is valid
only when the projectile lands at the same elevation from which it was launched.
• Nonuniform circular motion occurs when there is tangential acceleration of an object executing circular motion such
that the speed of the object is changing. This acceleration is called tangential acceleration →
a T. The magnitude of
tangential acceleration is the time rate of change of the magnitude of the velocity. The tangential acceleration vector
is tangential to the circle, whereas the centripetal acceleration vector points radially inward toward the center of the
circle. The total acceleration is the vector sum of tangential and centripetal accelerations.
• An object executing uniform circular motion can be described with equations of motion. The position vector of the
^ ^
object is →
r (t) = A cos ωt i + A sin ωt j , where A is the magnitude | →r (t)|, which is also the radius of the
circle, and ω is the angular frequency.
• If S and S′ are two reference frames moving relative to each other at a constant velocity, then the velocity of an
object relative to S is equal to its velocity relative to S′ plus the velocity of S′ relative to S.
• If two reference frames are moving relative to each other at a constant velocity, then the accelerations of an object
as observed in both reference frames are equal.
CONCEPTUAL QUESTIONS
9. A dime is placed at the edge of a table so it hangs over
7.1 Displacement and Velocity Vectors slightly. A quarter is slid horizontally on the table surface
1. What form does the trajectory of a particle have if perpendicular to the edge and hits the dime head on. Which
the distance from any point A to point B is equal to the coin hits the ground first?
magnitude of the displacement from A to B?
PROBLEMS
23. The position of a particle is
7.1 Displacement and Velocity Vectors → ^ ^ ^
r (t) = 4.0t 2 i − 3.0 j + 2.0t 3 k m. (a) What is the
17. The coordinates of a particle in a rectangular
coordinate system are (1.0, –4.0, 6.0). What is the position velocity of the particle at 0 s and at 1.0 s? (b) What is the
vector of the particle? average velocity between 0 s and 1.0 s?
18. The position of a particle changes from 24. Clay Matthews, a linebacker for the Green Bay
→ ^ ^ Packers, can reach a speed of 10.0 m/s. At the start of a
r 1 = (2.0 i + 3.0 j )cm to play, Matthews runs downfield at 45° with respect to the
→ ^ ^ 50-yard line and covers 8.0 m in 1 s. He then runs straight
r 2 = (−4.0 i + 3.0 j ) cm. What is the particle’s
down the field at 90° with respect to the 50-yard line for
displacement? 12 m, with an elapsed time of 1.2 s. (a) What is Matthews’
final displacement from the start of the play? (b) What is
19. The 18th hole at Pebble Beach Golf Course is a dogleg his average velocity?
to the left of length 496.0 m. The fairway off the tee is
taken to be the x direction. A golfer hits his tee shot a 25. The F-35B Lighting II is a short-takeoff and vertical
distance of 300.0 m, corresponding to a displacement landing fighter jet. If it does a vertical takeoff to 20.00-m
^
Δ→
r 1 = 300.0 m i , and hits his second shot 189.0 m height above the ground and then follows a flight path
angled at 30° with respect to the ground for 20.00 km,
^ ^
with a displacement Δ → r 2 = 172.0 m i + 80.3 m j . what is the final displacement?
What is the final displacement of the golf ball from the tee?
of Mars? Note that Mars has an acceleration of gravity of its equator just above its surface. At what speed must the jet
2 travel if the magnitude of its acceleration is g?
3.7 m/s .
57. In 1999, Robbie Knievel was the first to jump the 67. A fan is rotating at a constant 360.0 rev/min. What is
Grand Canyon on a motorcycle. At a narrow part of the the magnitude of the acceleration of a point on one of its
canyon (69.0 m wide) and traveling 35.8 m/s off the takeoff blades 10.0 cm from the axis of rotation?
ramp, he reached the other side. What was his launch
angle? 68. A point located on the second hand of a large clock
has a radial acceleration of 0.1cm/s 2. How far is the point
58. You throw a baseball at an initial speed of 15.0 m/s at from the axis of rotation of the second hand?
an angle of 30° with respect to the horizontal. What would
the ball’s initial speed have to be at 30° on a planet that
has twice the acceleration of gravity as Earth to achieve the 7.5 Relative Motion in One and Two
same range? Consider launch and impact on a horizontal Dimensions
surface.
69. The coordinate axes of the reference frame S′ remain
59. Aaron Rogers throws a football at 20.0 m/s to his wide parallel to those of S, as S′ moves away from S at a
receiver, who runs straight down the field at 9.4 m/s for → ^ ^ ^
constant velocity v S′ = (4.0 i + 3.0 j + 5.0 k ) m/s.
20.0 m. If Aaron throws the football when the wide receiver
has reached 10.0 m, what angle does Aaron have to launch (a) If at time t = 0 the origins coincide, what is the position
the ball so the receiver catches it at the 20.0 m mark? of the origin O′ in the S frame as a function of time?
(b) How is particle position for →
r (t) and →
r ′(t), as
measured in S and S′, respectively, related? (c) What
7.4 Uniform Circular Motion
is the relationship between particle velocities
60. A flywheel is rotating at 30 rev/s. What is the total →
v (t) and →
v ′(t) ? (d) How are accelerations
angle, in radians, through which a point on the flywheel
rotates in 40 s? →
a (t) and →
a ′ (t) related?
61. A particle travels in a circle of radius 10 m at a 70. The coordinate axes of the reference frame S′ remain
constant speed of 20 m/s. What is the magnitude of the
parallel to those of S, as S′ moves away from S at a
acceleration?
^ ^ ^
constant velocity → v S′ S = (1.0 i + 2.0 j + 3.0 k )t m/s
62. Cam Newton of the Carolina Panthers throws a perfect
. (a) If at time t = 0 the origins coincide, what is the
football spiral at 8.0 rev/s. The radius of a pro football is 8.5
position of origin O′ in the S frame as a function of time?
cm at the middle of the short side. What is the centripetal
acceleration of the laces on the football? (b) How is particle position for →
r (t) and →
r ′(t) , as
measured in S and S′, respectively, related? (c) What
63. A fairground ride spins its occupants inside a flying is the relationship between particle velocities
saucer-shaped container. If the horizontal circular path the →
v (t) and →
v ′(t) ? (d) How are accelerations
riders follow has an 8.00-m radius, at how many
revolutions per minute are the riders subjected to a
→
a (t) and →
a ′ (t) related?
centripetal acceleration equal to that of gravity?
71. The velocity of a particle in reference frame A is
64. A runner taking part in the 200-m dash must run ^ ^
around the end of a track that has a circular arc with a (2.0 i + 3.0 j ) m/s. The velocity of reference frame A
radius of curvature of 30.0 m. The runner starts the race at ^
a constant speed. If she completes the 200-m dash in 23.2 with respect to reference frame B is 4.0 k m/s, and the
s and runs at constant speed throughout the race, what is velocity of reference frame B with respect to C is
her centripetal acceleration as she runs the curved portion ^
of the track? 2.0 j m/s. What is the velocity of the particle in reference
frame C?
65. What is the acceleration of Venus toward the Sun,
assuming a circular orbit? 72. Raindrops fall vertically at 4.5 m/s relative to the
earth. What does an observer in a car moving at 22.0 m/s in
66. An experimental jet rocket travels around Earth along a straight line measure as the velocity of the raindrops?
73. A seagull can fly at a velocity of 9.00 m/s in still air. to get across and how far downstream is the boat when it
(a) If it takes the bird 20.0 min to travel 6.00 km straight reaches the opposite shore?
into an oncoming wind, what is the velocity of the wind?
(b) If the bird turns around and flies with the wind, how 76. A small plane flies at 200 km/h in still air. If the
long will it take the bird to return 6.00 km? wind blows directly out of the west at 50 km/h, (a) in what
direction must the pilot head her plane to move directly
74. A ship sets sail from Rotterdam, heading due north at north across land and (b) how long does it take her to reach
7.00 m/s relative to the water. The local ocean current is a point 300 km directly north of her starting point?
1.50 m/s in a direction 40.0° north of east. What is the
velocity of the ship relative to Earth? 77. A cyclist traveling southeast along a road at 15 km/h
feels a wind blowing from the southwest at 25 km/h. To a
75. A boat can be rowed at 8.0 km/h in still water. (a) How stationary observer, what are the speed and direction of the
much time is required to row 1.5 km downstream in a river wind?
moving 3.0 km/h relative to the shore? (b) How much time
is required for the return trip? (c) In what direction must the 78. A river is moving east at 4 m/s. A boat starts from
boat be aimed to row straight across the river? (d) Suppose the dock heading 30° north of west at 7 m/s. If the river
the river is 0.8 km wide. What is the velocity of the boat is 1800 m wide, (a) what is the velocity of the boat with
with respect to Earth and how much time is required to respect to Earth and (b) how long does it take the boat to
get to the opposite shore? (e) Suppose, instead, the boat is cross the river?
aimed straight across the river. How much time is required
ADDITIONAL PROBLEMS
79. A Formula One race car is traveling at 89.0 m/s along
a straight track enters a turn on the race track with radius
of curvature of 200.0 m. What centripetal acceleration must
the car have to stay on the track?
85. A propeller blade at rest starts to rotate from t = 0 s to t velocity 5 s after the rockets fire?
= 5.0 s with a tangential acceleration of the tip of the blade
at 3.00 m/s 2. The tip of the blade is 1.5 m from the axis of 92. A crossbow is aimed horizontally at a target 40 m
rotation. At t = 5.0 s, what is the total acceleration of the tip away. The arrow hits 30 cm below the spot at which it was
of the blade? aimed. What is the initial velocity of the arrow?
86. A particle is executing circular motion with a constant 93. A long jumper can jump a distance of 8.0 m when he
angular frequency of ω = 4.00 rad/s. If time t = 0 takes off at an angle of 45° with respect to the horizontal.
corresponds to the position of the particle being located at Assuming he can jump with the same initial speed at all
y = 0 m and x = 5 m, (a) what is the position of the particle angles, how much distance does he lose by taking off at
at t = 10 s? (b) What is its velocity at this time? (c) What is 30° ?
its acceleration?
94. On planet Arcon, the maximum horizontal range of
87. A particle’s centripetal acceleration is a C = 4.0 m/s 2 a projectile launched at 10 m/s is 20 m. What is the
acceleration of gravity on this planet?
at t = 0 s. It is executing uniform circular motion about an
axis at a distance of 5.0 m. What is its velocity at t = 10 s?
95. A mountain biker encounters a jump on a race course
that sends him into the air at 60° to the horizontal. If he
88. A rod 3.0 m in length is rotating at 2.0 rev/s about
an axis at one end. Compare the centripetal accelerations at lands at a horizontal distance of 45.0 m and 20 m below his
radii of (a) 1.0 m, (b) 2.0 m, and (c) 3.0 m. launch point, what is his initial speed?
CHALLENGE PROBLEMS
99. World’s Longest Par 3. The tee of the world’s longest (b) In addition to clearing the crossbar, the football must
par 3 sits atop South Africa’s Hanglip Mountain at 400.0 be high enough in the air early during its flight to clear the
m above the green and can only be reached by helicopter. reach of the onrushing defensive lineman. If the lineman is
The horizontal distance to the green is 359.0 m. Neglect 4.6 m away and has a vertical reach of 2.5 m, can he block
air resistance and answer the following questions. (a) If the 45.7-m field goal attempt? (c) What if the lineman is 1.0
a golfer launches a shot that is 40° with respect to the m away?
horizontal, what initial velocity must she give the ball? (b)
What is the time to reach the green?
8 | OVERVIEW OF THE
SOLAR SYSTEM
Chapter Outline
8.1 Overview of Our Planetary System
8.2 Composition and Structure of Planets
8.3 Origin of the Solar System
8.4 Kepler's Laws of Planetary Motion
Introduction
Figure 8.1 This picture was taken by the Curiosity Rover on Mars in 2012. The image is reconstructed digitally from 55
different images taken by a camera on the rover’s extended mast, so that the many positions of the mast (which acted like a selfie
stick) are edited out. (credit: modification of work by NASA/JPL-Caltech/MSSS)
Surrounding the Sun is a complex system of worlds with a wide range of conditions: eight major planets, many dwarf
planets, hundreds of moons, and countless smaller objects. Thanks largely to visits by spacecraft, we can now envision the
members of the solar system as other worlds like our own, each with its own chemical and geological history, and unique
sights that interplanetary tourists may someday visit. Some have called these past few decades the “golden age of planetary
exploration,” comparable to the golden age of exploration in the fifteenth century, when great sailing ships plied Earth’s
oceans and humanity became familiar with our own planet’s surface.
In this chapter, we discuss our planetary system and introduce the idea of comparative planetology—studying how the
planets work by comparing them with one another. We want to get to know the planets not only for what we can learn
about them, but also to see what they can tell us about the origin and evolution of the entire solar system. In the upcoming
chapters, we describe the better-known members of the solar system and begin to compare them to the thousands of planets
that have been discovered recently, orbiting other stars.
264 Chapter 8 | Overview of the Solar System
The solar system[1] consists of the Sun and many smaller objects: the planets, their moons and rings, and such “debris”
as asteroids, comets, and dust. Decades of observation and spacecraft exploration have revealed that most of these objects
formed together with the Sun about 4.5 billion years ago. They represent clumps of material that condensed from an
enormous cloud of gas and dust. The central part of this cloud became the Sun, and a small fraction of the material in the
outer parts eventually formed the other objects.
During the past 50 years, we have learned more about the solar system than anyone imagined before the space age. In
addition to gathering information with powerful new telescopes, we have sent spacecraft directly to many members of the
planetary system. (Planetary astronomy is the only branch of our science in which we can, at least vicariously, travel to the
objects we want to study.) With evocative names such as Voyager, Pioneer, Curiosity, and Pathfinder, our robot explorers
have flown past, orbited, or landed on every planet, returning images and data that have dazzled both astronomers and the
public. In the process, we have also investigated two dwarf planets, hundreds of fascinating moons, four ring systems, a
dozen asteroids, and several comets (smaller members of our solar system that we will discuss later).
Our probes have penetrated the atmosphere of Jupiter and landed on the surfaces of Venus, Mars, our Moon, Saturn’s moon
Titan, the asteroids Eros and Itokawa, and the Comet Churyumov-Gerasimenko (usually referred to as 67P). Humans have
set foot on the Moon and returned samples of its surface soil for laboratory analysis (Figure 8.2). We have even discovered
other places in our solar system that might be able to support some kind of life.
Figure 8.2 The lunar lander and surface rover from the Apollo 15 mission are seen in this view of the one place beyond Earth
that has been explored directly by humans. (credit: modification of work by David R. Scott, NASA)
View this gallery of NASA images (https://openstaxcollege.org/l/30projapolloarc) that trace the history of
1. The generic term for a group of planets and other bodies circling a star is planetary system. Ours is called the solar
system because our Sun is sometimes called Sol. Strictly speaking, then, there is only one solar system; planets orbiting
other stars are in planetary systems.
An Inventory
The Sun, a star that is brighter than about 80% of the stars in the Galaxy, is by far the most massive member of the solar
system, as shown in Table 8.1. It is an enormous ball about 1.4 million kilometers in diameter, with surface layers of
incandescent gas and an interior temperature of millions of degrees. The Sun will be discussed in later chapters as our first,
and best-studied, example of a star.
Table 8.1
Table 8.1 also shows that most of the material of the planets is actually concentrated in the largest one, Jupiter, which
is more massive than all the rest of the planets combined. Astronomers were able to determine the masses of the planets
centuries ago using Kepler’s laws of planetary motion and Newton’s law of gravity to measure the planets’ gravitational
effects on one another or on moons that orbit them (see Orbits and Gravity (https://legacy.cnx.org/content/
m59773/latest/) ). Today, we make even more precise measurements of their masses by tracking their gravitational effects
on the motion of spacecraft that pass near them.
Beside Earth, five other planets were known to the ancients—Mercury, Venus, Mars, Jupiter, and Saturn—and two were
discovered after the invention of the telescope: Uranus and Neptune. The eight planets all revolve in the same direction
around the Sun. They orbit in approximately the same plane, like cars traveling on concentric tracks on a giant, flat
racecourse. Each planet stays in its own “traffic lane,” following a nearly circular orbit about the Sun and obeying the
“traffic” laws discovered by Galileo, Kepler, and Newton. Besides these planets, we have also been discovering smaller
worlds beyond Neptune that are called trans-Neptunian objects or TNOs (see Figure 8.3). The first to be found, in 1930,
was Pluto, but others have been discovered during the twenty-first century. One of them, Eris, is about the same size as
Pluto and has at least one moon (Pluto has five known moons.) The largest TNOs are also classed as dwarf planets, as
is the largest asteroid, Ceres. (Dwarf planets will be discussed further in the chapter on Rings, Moons, and Pluto
(https://legacy.cnx.org/content/m59860/latest/) ). To date, more than 1750 of these TNOs have been discovered.
266 Chapter 8 | Overview of the Solar System
Figure 8.3 All eight major planets orbit the Sun in roughly the same plane. The five currently known dwarf planets are also
shown: Eris, Haumea, Pluto, Ceres, and Makemake. Note that Pluto’s orbit is not in the plane of the planets.
Each of the planets and dwarf planets also rotates (spins) about an axis running through it, and in most cases the direction
of rotation is the same as the direction of revolution about the Sun. The exceptions are Venus, which rotates backward very
slowly (that is, in a retrograde direction), and Uranus and Pluto, which also have strange rotations, each spinning about an
axis tipped nearly on its side. We do not yet know the spin orientations of Eris, Haumea, and Makemake.
The four planets closest to the Sun (Mercury through Mars) are called the inner or terrestrial planets. Often, the Moon is
also discussed as a part of this group, bringing the total of terrestrial objects to five. (We generally call Earth’s satellite “the
Moon,” with a capital M, and the other satellites “moons,” with lowercase m’s.) The terrestrial planets are relatively small
worlds, composed primarily of rock and metal. All of them have solid surfaces that bear the records of their geological
history in the forms of craters, mountains, and volcanoes (Figure 8.4).
Figure 8.4 The pockmarked face of the terrestrial world of Mercury is more typical of the inner planets than the watery surface
of Earth. This black-and-white image, taken with the Mariner 10 spacecraft, shows a region more than 400 kilometers wide.
(credit: modification of work by NASA/John Hopkins University Applied Physics Laboratory/Carnegie Institution of
Washington)
The next four planets (Jupiter through Neptune) are much larger and are composed primarily of lighter ices, liquids, and
gases. We call these four the jovian planets (after “Jove,” another name for Jupiter in mythology) or giant planets—a name
they richly deserve (Figure 8.5). More than 1400 Earths could fit inside Jupiter, for example. These planets do not have
solid surfaces on which future explorers might land. They are more like vast, spherical oceans with much smaller, dense
cores.
Figure 8.5 This montage shows the four giant planets: Jupiter, Saturn, Uranus, and Neptune. Below them, Earth is shown to
scale. (credit: modification of work by NASA, Solar System Exploration)
Near the outer edge of the system lies Pluto, which was the first of the distant icy worlds to be discovered beyond Neptune
(Pluto was visited by a spacecraft, the NASA New Horizons mission, in 2015 [see Figure 8.6]). Table 8.2 summarizes
some of the main facts about the planets.
Figure 8.6 This intriguing image from the New Horizons spacecraft, taken when it flew by the dwarf planet in July 2015,
shows some of its complex surface features. The rounded white area is temporarily being called the Sputnik Plain, after
humanity’s first spacecraft. (credit: modification of work by NASA/Johns Hopkins University Applied Physics Laboratory/
Southwest Research Institute)
The Planets
Name Distance from Sun Revolution Period Diameter Mass Density
(AU)[2] (y) (km) (1023 kg) (g/cm3)[3]
Mercury 0.39 0.24 4,878 3.3 5.4
Table 8.2
The Planets
Name Distance from Sun Revolution Period Diameter Mass Density
(AU) (y) (km) (1023 kg) (g/cm3)
Venus 0.72 0.62 12,120 48.7 5.2
Earth 1.00 1.00 12,756 59.8 5.5
Mars 1.52 1.88 6,787 6.4 3.9
Jupiter 5.20 11.86 142,984 18,991 1.3
Saturn 9.54 29.46 120,536 5686 0.7
Uranus 19.18 84.07 51,118 866 1.3
Neptune 30.06 164.82 49,660 1030 1.6
Table 8.2
Example 8.1
Comparing Densities
Let’s compare the densities of several members of the solar system. The density of an object equals its mass
divided by its volume. The volume (V) of a sphere (like a planet) is calculated using the equation
V = 4 πR 3
3
where π (the Greek letter pi) has a value of approximately 3.14. Although planets are not perfect spheres, this
equation works well enough. The masses and diameters of the planets are given in Table 8.2. For data on selected
moons, see Appendix D. Let’s use Saturn’s moon Mimas as our example, with a mass of 4 × 1019 kg and a
diameter of approximately 400 km (radius, 200 km = 2 × 105 m).
Solution
The volume of Mimas is
3
4 × 3.14 × ⎛2 × 10 5 m⎞ = 3.3 × 10 16 m 3.
3 ⎝ ⎠
Answer:
For a sphere,
density = ⎛mass3⎞ kg/m 3.
4
⎝ 3 πR ⎠
Figure 8.7 This 2007 Cassini image shows Saturn and its complex system of rings, taken from a distance of about 1.2 million
kilometers. This natural-color image is a composite of 36 images taken over the course of 2.5 hours. (credit: modification of work
by NASA/JPL/Space Science Institute)
The solar system has many other less-conspicuous members. Another group is the asteroids, rocky bodies that orbit the
Sun like miniature planets, mostly in the space between Mars and Jupiter (although some do cross the orbits of planets like
Earth—see Figure 8.8). Most asteroids are remnants of the initial population of the solar system that existed before the
planets themselves formed. Some of the smallest moons of the planets, such as the moons of Mars, are very likely captured
asteroids.
270 Chapter 8 | Overview of the Solar System
Figure 8.8 This small Earth-crossing asteroid image was taken by the NEAR-Shoemaker spacecraft from an altitude of about
100 kilometers. This view of the heavily cratered surface is about 10 kilometers wide. The spacecraft orbited Eros for a year
before landing gently on its surface. (credit: modification of work by NASA/JHUAPL)
Another class of small bodies is composed mostly of ice, made of frozen gases such as water, carbon dioxide, and carbon
monoxide; these objects are called comets (see Figure 8.9). Comets also are remnants from the formation of the solar
system, but they were formed and continue (with rare exceptions) to orbit the Sun in distant, cooler regions—stored in a
sort of cosmic deep freeze. This is also the realm of the larger icy worlds, called dwarf planets.
Figure 8.9 This image shows Comet Churyumov-Gerasimenko, also known as 67P, near its closest approach to the Sun in
2015, as seen from the Rosetta spacecraft. Note the jets of gas escaping from the solid surface. (credit: modification of work by
ESA/Rosetta/NAVACAM, CC BY-SA IGO 3.0 (http://creativecommons.org/licenses/by-sa/3.0/igo/) )
Finally, there are countless grains of broken rock, which we call cosmic dust, scattered throughout the solar system. When
these particles enter Earth’s atmosphere (as millions do each day) they burn up, producing a brief flash of light in the
night sky known as a meteor (meteors are often referred to as shooting stars). Occasionally, some larger chunk of rocky or
metallic material survives its passage through the atmosphere and lands on Earth. Any piece that strikes the ground is known
as a meteorite. (You can see meteorites on display in many natural history museums and can sometimes even purchase
pieces of them from gem and mineral dealers.)
Figure 8.10 Sagan was Tyson’s inspiration to become a scientist. (credit “Sagan”:
modification of work by NASA, JPL; credit “Tyson”: modification of work by Bruce F. Press)
In the early 1960s, when many scientists still thought Venus might turn out to be a hospitable place, Sagan calculated
that the thick atmosphere of Venus could act like a giant greenhouse, keeping the heat in and raising the temperature
enormously. He showed that the seasonal changes astronomers had seen on Mars were caused, not by vegetation, but
by wind-blown dust. He was a member of the scientific teams for many of the robotic missions that explored the solar
system and was instrumental in getting NASA to put a message-bearing plaque aboard the Pioneer spacecraft, as well
as audio-video records on the Voyager spacecraft—all of them destined to leave our solar system entirely and send
these little bits of Earth technology out among the stars.
To encourage public interest and public support of planetary exploration, Sagan helped found The Planetary Society,
now the largest space-interest organization in the world. He was a tireless and eloquent advocate of the need to study
the solar system close-up and the value of learning about other worlds in order to take better care of our own.
Sagan simulated conditions on early Earth to demonstrate how some of life’s fundamental building blocks might have
formed from the “primordial soup” of natural compounds on our planet. In addition, he and his colleagues developed
computer models showing the consequences of nuclear war for Earth would be even more devastating than anyone
had thought (this is now called the nuclear winter hypothesis) and demonstrating some of the serious consequences of
continued pollution of our atmosphere.
Sagan was perhaps best known, however, as a brilliant popularizer of astronomy and the author of many books on
science, including the best-selling Cosmos, and several evocative tributes to solar system exploration such as The
Cosmic Connection and Pale Blue Dot. His book The Demon Haunted World, completed just before his death in
1996, is perhaps the best antidote to fuzzy thinking about pseudo-science and irrationality in print today. An intriguing
science fiction novel he wrote, titled Contact, which became a successful film as well, is still recommended by many
science instructors as a scenario for making contact with life elsewhere that is much more reasonable than most science
fiction.
Sagan was a master, too, of the television medium. His 13-part public television series, Cosmos, was seen by an
estimated 500 million people in 60 countries and has become one of the most-watched series in the history of public
broadcasting. A few astronomers scoffed at a scientist who spent so much time in the public eye, but it is probably fair
to say that Sagan’s enthusiasm and skill as an explainer won more friends for the science of astronomy than anyone or
anything else in the second half of the twentieth century.
In the two decades since Sagan’s death, no other scientist has achieved the same level of public recognition. Perhaps
closest is the director of the Hayden Planetarium, Neil deGrasse Tyson, who followed in Sagan’s footsteps by making
an updated version of the Cosmos program in 2014. Tyson is quick to point out that Sagan was his inspiration to
become a scientist, telling how Sagan invited him to visit for a day at Cornell when he was a high school student
looking for a career. However, the media environment has fragmented a great deal since Sagan’s time. It is interesting
to speculate whether Sagan could have adapted his communication style to the world of cable television, Twitter,
Facebook, and podcasts.
272 Chapter 8 | Overview of the Solar System
Two imaginative videos provide a tour of the solar system objects we have been discussing. Shane Gellert’s I
Need Some Space (https://openstaxcollege.org/l/30needsomespace) uses NASA photography and models
to show the various worlds with which we share our system. In the more science fiction-oriented Wanderers
(https://openstaxcollege.org/l/30wanderers) video, we see some of the planets and moons as tourist
destinations for future explorers, with commentary taken from recordings by Carl Sagan.
are named for women who have made significant contributions to human knowledge and welfare. Volcanic features
on Jupiter’s moon Io, which is in a constant state of volcanic activity, are named after gods of fire and thunder
from the mythologies of many cultures. Craters on Mercury commemorate famous novelists, playwrights, artists, and
composers. On Saturn’s moon Tethys, all the features are named after characters and places in Homer’s great epic
poem, The Odyssey. As we explore further, it may well turn out that more places in the solar system need names than
Earth history can provide. Perhaps by then, explorers and settlers on these worlds will be ready to develop their own
names for the places they may (if but for a while) call home.
You may be surprised to know that the meaning of the word planet has recently become controversial because we
have discovered many other planetary systems that don’t look very much like our own. Even within our solar system,
the planets differ greatly in size and chemical properties. The biggest dispute concerns Pluto, which is much smaller
than the other eight major planets. The category of dwarf planet was invented to include Pluto and similar icy objects
beyond Neptune. But is a dwarf planet also a planet? Logically, it should be, but even this simple issue of grammar has
been the subject of heated debate among both astronomers and the general public.
The fact that there are two distinct kinds of planets—the rocky terrestrial planets and the gas-rich jovian planets—leads us
to believe that they formed under different conditions. Certainly their compositions are dominated by different elements.
Let us look at each type in more detail.
Figure 8.11 This true-color image of Jupiter was taken from the Cassini spacecraft in 2000. (credit: modification of work by
NASA/JPL/University of Arizona)
Uranus and Neptune are much smaller than Jupiter and Saturn, but each also has a core of rock, metal, and ice. Uranus and
Neptune were less efficient at attracting hydrogen and helium gas, so they have much smaller atmospheres in proportion to
their cores.
Chemically, each giant planet is dominated by hydrogen and its many compounds. Nearly all the oxygen present is
combined chemically with hydrogen to form water (H2O). Chemists call such a hydrogen-dominated composition reduced.
Throughout the outer solar system, we find abundant water (mostly in the form of ice) and reducing chemistry.
Figure 8.12 This view of Jupiter’s moon Ganymede was taken in June 1996 by the Galileo spacecraft. The brownish gray
color of the surface indicates a dusty mixture of rocky material and ice. The bright spots are places where recent impacts have
uncovered fresh ice from underneath. (credit: modification of work by NASA/JPL)
Most of the asteroids and comets, as well as the smallest moons, were probably never heated to the melting point. However,
some of the largest asteroids, such as Vesta, appear to be differentiated; others are fragments from differentiated bodies.
Because most asteroids and comets retain their original composition, they represent relatively unmodified material dating
back to the time of the formation of the solar system. In a sense, they act as chemical fossils, helping us to learn about a
time long ago whose traces have been erased on larger worlds.
we build in the systems of the giant planets may well have to be in space or one of their moons—none of which is
particularly hospitable to a luxury hotel with a swimming pool and palm trees. Perhaps we will find warmer havens
deep inside the clouds of Jupiter or in the ocean under the frozen ice of its moon Europa.
All of this suggests that we had better take good care of Earth because it is the only site where life as we know it could
survive. Recent human activity may be reducing the habitability of our planet by adding pollutants to the atmosphere,
especially the potent greenhouse gas carbon dioxide. Human civilization is changing our planet dramatically, and
these changes are not necessarily for the better. In a solar system that seems unready to receive us, making Earth less
hospitable to life may be a grave mistake.
Geological Activity
The crusts of all of the terrestrial planets, as well as of the larger moons, have been modified over their histories by
both internal and external forces. Externally, each has been battered by a slow rain of projectiles from space, leaving
their surfaces pockmarked by impact craters of all sizes (see m59816 (https://legacy.cnx.org/content/m59816/
latest/#OSC_Astro_07_01_Mercury) ). We have good evidence that this bombardment was far greater in the early
history of the solar system, but it certainly continues to this day, even if at a lower rate. The collision of more than 20 large
pieces of Comet Shoemaker–Levy 9 with Jupiter in the summer of 1994 (see Figure 8.13) is one dramatic example of this
process.
Figure 8.13 In this image of Comet Shoemaker–Levy 9 taken on May 17, 1994, by NASA’s Hubble Space Telescope, you can
see about 20 icy fragments into which the comet broke. The comet was approximately 660 million kilometers from Earth,
heading on a collision course with Jupiter. (credit: modification of work by NASA, ESA, H. Weaver (STScl), E. Smith (STScl))
Figure 8.14 shows the aftermath of these collisions, when debris clouds larger than Earth could be seen in Jupiter’s
atmosphere.
Figure 8.14 The Hubble Space Telescope took this sequence of images of Jupiter in summer 1994, when fragments of Comet
Shoemaker–Levy 9 collided with the giant planet. Here we see the site hit by fragment G, from five minutes to five days after
impact. Several of the dust clouds generated by the collisions became larger than Earth. (credit: modification of work by H.
Hammel, NASA)
During the time all the planets have been subject to such impacts, internal forces on the terrestrial planets have buckled and
twisted their crusts, built up mountain ranges, erupted as volcanoes, and generally reshaped the surfaces in what we call
geological activity. (The prefix geo means “Earth,” so this is a bit of an “Earth-chauvinist” term, but it is so widely used
that we bow to tradition.) Among the terrestrial planets, Earth and Venus have experienced the most geological activity over
their histories, although some of the moons in the outer solar system are also surprisingly active. In contrast, our own Moon
is a dead world where geological activity ceased billions of years ago.
Geological activity on a planet is the result of a hot interior. The forces of volcanism and mountain building are driven by
heat escaping from the interiors of planets. As we will see, each of the planets was heated at the time of its birth, and this
primordial heat initially powered extensive volcanic activity, even on our Moon. But, small objects such as the Moon soon
cooled off. The larger the planet or moon, the longer it retains its internal heat, and therefore the more we expect to see
surface evidence of continuing geological activity. The effect is similar to our own experience with a hot baked potato: the
larger the potato, the more slowly it cools. If we want a potato to cool quickly, we cut it into small pieces.
For the most part, the history of volcanic activity on the terrestrial planets conforms to the predictions of this simple theory.
The Moon, the smallest of these objects, is a geologically dead world. Although we know less about Mercury, it seems
likely that this planet, too, ceased most volcanic activity about the same time the Moon did. Mars represents an intermediate
case. It has been much more active than the Moon, but less so than Earth. Earth and Venus, the largest terrestrial planets,
still have molten interiors even today, some 4.5 billion years after their birth.
Much of astronomy is motivated by a desire to understand the origin of things: to find at least partial answers to age-old
questions of where the universe, the Sun, Earth, and we ourselves came from. Each planet and moon is a fascinating place
that may stimulate our imagination as we try to picture what it would be like to visit. Taken together, the members of the
solar system preserve patterns that can tell us about the formation of the entire system. As we begin our exploration of the
planets, we want to introduce our modern picture of how the solar system formed.
The recent discovery of hundreds of planets in orbit around other stars has shown astronomers that many exoplanetary
systems can be quite different from our own solar system. For example, it is common for these systems to include planets
intermediate in size between our terrestrial and giant planets. These are often called superearths. Some exoplanet systems
even have giant planets close to the star, reversing the order we see in our system. In The Birth of Stars and the
Discovery of Planets outside the Solar System (https://legacy.cnx.org/content/m59915/latest/) , we will
look at these exoplanet systems. But for now, let us focus on theories of how our own particular system has formed and
evolved.
Figure 8.15 This artist’s conception of the solar nebula shows the flattened
cloud of gas and dust from which our planetary system formed. Icy and rocky
planetesimals (precursors of the planets) can be seen in the foreground. The bright
center is where the Sun is forming. (credit: William K. Hartmann, Planetary
Science Institute)
The composition of the planets gives another clue about origins. Spectroscopic analysis allows us to determine which
elements are present in the Sun and the planets. The Sun has the same hydrogen-dominated composition as Jupiter and
Saturn, and therefore appears to have been formed from the same reservoir of material. In comparison, the terrestrial planets
and our Moon are relatively deficient in the light gases and the various ices that form from the common elements oxygen,
carbon, and nitrogen. Instead, on Earth and its neighbors, we see mostly the rarer heavy elements such as iron and silicon.
This pattern suggests that the processes that led to planet formation in the inner solar system must somehow have excluded
much of the lighter materials that are common elsewhere. These lighter materials must have escaped, leaving a residue of
heavy stuff.
The reason for this is not hard to guess, bearing in mind the heat of the Sun. The inner planets and most of the asteroids are
made of rock and metal, which can survive heat, but they contain very little ice or gas, which evaporate when temperatures
are high. (To see what we mean, just compare how long a rock and an ice cube survive when they are placed in the sunlight.)
In the outer solar system, where it has always been cooler, the planets and their moons, as well as icy dwarf planets and
comets, are composed mostly of ice and gas.
Figure 8.16 These Hubble Space Telescope photos show sections of the Orion Nebula, a relatively close-by region where stars
are currently forming. Each image shows an embedded circumstellar disk orbiting a very young star. Seen from different angles,
some are energized to glow by the light of a nearby star while others are dark and seen in silhouette against the bright glowing
gas of the Orion Nebula. Each is a contemporary analog of our own solar nebula—a location where planets are probably being
formed today. (credit: modification of work by NASA/ESA, L. Ricci (ESO))
Building Planets
Circumstellar disks are a common occurrence around very young stars, suggesting that disks and stars form together.
Astronomers can use theoretical calculations to see how solid bodies might form from the gas and dust in these disks as they
cool. These models show that material begins to coalesce first by forming smaller objects, precursors of the planets, which
we call planetesimals.
Today’s fast computers can simulate the way millions of planetesimals, probably no larger than 100 kilometers in diameter,
might gather together under their mutual gravity to form the planets we see today. We are beginning to understand that this
process was a violent one, with planetesimals crashing into each other and sometimes even disrupting the growing planets
themselves. As a consequence of those violent impacts (and the heat from radioactive elements in them), all the planets
were heated until they were liquid and gas, and therefore differentiated, which helps explain their present internal structures.
The process of impacts and collisions in the early solar system was complex and, apparently, often random. The solar nebula
model can explain many of the regularities we find in the solar system, but the random collisions of massive collections
of planetesimals could be the reason for some exceptions to the “rules” of solar system behavior. For example, why do the
planets Uranus and Pluto spin on their sides? Why does Venus spin slowly and in the opposite direction from the other
planets? Why does the composition of the Moon resemble Earth in many ways and yet exhibit substantial differences? The
answers to such questions probably lie in enormous collisions that took place in the solar system long before life on Earth
began.
Today, some 4.5 billion years after its origin, the solar system is—thank goodness—a much less violent place. As we will
see, however, some planetesimals have continued to interact and collide, and their fragments move about the solar system
as roving “transients” that can make trouble for the established members of the Sun’s family, such as our own Earth. (We
discuss this “troublemaking” in Comets and Asteroids: Debris of the Solar System (https://legacy.cnx.org/
content/m59865/latest/) .)
Figure 8.17 This space habitat and laboratory orbits Earth once every 90 minutes. (credit: modification of work by NASA)
How would you find a new planet at the outskirts of our solar system that is too dim to be seen with the unaided eye and is
so far away that it moves very slowly among the stars? This was the problem confronting astronomers during the nineteenth
century as they tried to pin down a full inventory of our solar system.
If we could look down on the solar system from somewhere out in space, interpreting planetary motions would be much
simpler. But the fact is, we must observe the positions of all the other planets from our own moving planet. Scientists of
the Renaissance did not know the details of Earth’s motions any better than the motions of the other planets. Their problem
was that they had to deduce the nature of all planetary motion using only their earthbound observations of the other planets’
positions in the sky. To solve this complex problem more fully, better observations and better models of the planetary system
were needed.
At about the time that Galileo was beginning his experiments with falling bodies, the efforts of two other scientists
dramatically advanced our understanding of the motions of the planets. These two astronomers were the observer Tycho
Brahe and the mathematician Johannes Kepler. Together, they placed the speculations of Copernicus on a sound
mathematical basis and paved the way for the work of Isaac Newton in the next century.
Figure 8.18 (a) A stylized engraving shows Tycho Brahe using his instruments to measure the altitude of celestial objects
above the horizon. The large curved instrument in the foreground allowed him to measure precise angles in the sky. Note that the
scene includes hints of the grandeur of Brahe’s observatory at Hven. (b) Kepler was a German mathematician and astronomer.
His discovery of the basic laws that describe planetary motion placed the heliocentric cosmology of Copernicus on a firm
mathematical basis.
At Hven, Brahe made a continuous record of the positions of the Sun, Moon, and planets for almost 20 years. His extensive
and precise observations enabled him to note that the positions of the planets varied from those given in published tables,
which were based on the work of Ptolemy. These data were extremely valuable, but Brahe didn’t have the ability to analyze
them and develop a better model than what Ptolemy had published. He was further inhibited because he was an extravagant
and cantankerous fellow, and he accumulated enemies among government officials. When his patron, Frederick II, died
in 1597, Brahe lost his political base and decided to leave Denmark. He took up residence in Prague, where he became
court astronomer to Emperor Rudolf of Bohemia. There, in the year before his death, Brahe found a most able young
mathematician, Johannes Kepler, to assist him in analyzing his extensive planetary data.
Johannes Kepler
Johannes Kepler was born into a poor family in the German province of Württemberg and lived much of his life amid the
turmoil of the Thirty Years’ War (see Figure 8.18). He attended university at Tubingen and studied for a theological career.
There, he learned the principles of the Copernican system and became converted to the heliocentric hypothesis. Eventually,
Kepler went to Prague to serve as an assistant to Brahe, who set him to work trying to find a satisfactory theory of planetary
motion—one that was compatible with the long series of observations made at Hven. Brahe was reluctant to provide Kepler
with much material at any one time for fear that Kepler would discover the secrets of the universal motion by himself,
thereby robbing Brahe of some of the glory. Only after Brahe’s death in 1601 did Kepler get full possession of the priceless
records. Their study occupied most of Kepler’s time for more than 20 years.
Through his analysis of the motions of the planets, Kepler developed a series of principles, now known as Kepler’s three
laws, which described the behavior of planets based on their paths through space. The first two laws of planetary motion
were published in 1609 in The New Astronomy. Their discovery was a profound step in the development of modern science.
eventually discovered that the orbit of that planet had the shape of a somewhat flattened circle, or ellipse. Next to the circle,
the ellipse is the simplest kind of closed curve, belonging to a family of curves known as conic sections (Figure 8.19).
Using the precise data collected by Tycho Brahe, Johannes Kepler carefully analyzed the positions in the sky of all the
known planets and the Moon, plotting their positions at regular intervals of time. From this analysis, he formulated three
laws, which we address in this section.
Figure 8.20 (a) An ellipse is a curve in which the sum of the distances from a point on the curve to two foci
( f 1 and f 2) is a constant. From this definition, you can see that an ellipse can be created in the following way. Place a
pin at each focus, then place a loop of string around a pencil and the pins. Keeping the string taught, move the pencil
around in a complete circuit. If the two foci occupy the same place, the result is a circle—a special case of an ellipse. (b)
For an elliptical orbit, if m ≪ M , then m follows an elliptical path with M at one focus. More exactly, both m and M
move in their own ellipse about the common center of mass.
For elliptical orbits, the point of closest approach of a planet to the Sun is called the perihelion. It is labeled point A in
Figure 8.20. The farthest point is the aphelion and is labeled point B in the figure. For the Moon’s orbit about Earth, those
points are called the perigee and apogee, respectively.
An ellipse has several mathematical forms, but all are a specific case of the more general equation for conic sections. There
are four different conic sections, all given by the equation
α = 1 + ecosθ. (8.1)
r
The variables r and θ are shown in Figure 8.21 in the case of an ellipse. The values of α and e determine which of the
four conic sections (hyperbola, parabola, ellipse or circle) represents the path of the satellite. For an ellipse, 0 ≤ e < 1,
with an eccentricity of zero meaning a circular orbit.
You can see an animation of two interacting objects at the My Solar System page at Phet
(https://openstaxcollege.org/l/21mysolarsys) . Choose the Sun and Planet preset option. You can also view
the more complicated multiple body problems as well. You may find the actual path of the Moon quite surprising,
yet is obeying Newton’s simple laws of motion.
Figure 8.22 The shaded regions shown have equal areas and
represent the same time interval.
Comparing the areas in the figure and the distance traveled along the ellipse in each case, we can see that in order for the
areas to be equal, the planet must speed up as it gets closer to the Sun and slow down as it moves away.
Now consider Figure 8.23. A small triangular area ΔA is swept out in time Δt . The velocity is along the path and it
makes an angle θ with the radial direction. Hence, the perpendicular velocity is given by v perp = vsinθ . The planet moves
a distance Δs = vΔtsinθ projected along the direction perpendicular to r. Since the area of a triangle is one-half the base
(r) times the height (Δs) , for a small displacement, the area is given by ΔA = 1 rΔs .
2
Figure 8.23 The element of area ΔA swept out in time Δt as the planet moves
through angle Δϕ . The angle between the radial direction and
→
v is θ .
The areal velocity is simply the rate of change of area with time, so we have
1 r Δs (8.2)
areal velocity = ΔA = 2 = 1r v
Δt Δt 2
The fact that the areal velocity remains constant, then, implies that the product of the planet's distance from the Sun ( r ) and
its instantaneous speed ( v ) is a constant of the motion. Hence, the closer it is to the Sun, the faster it moves, and vice versa.
You can view an animated version (https://openstaxcollege.org/l/21animationgrav) of Figure 8.22,
and many other interesting animations as well, at the School of Physics (University of New South Wales) site.
Kepler’s third law applies to all objects orbiting the Sun, including Earth, and provides a means for calculating their relative
distances from the Sun from the time they take to orbit. Let’s look at a specific example to illustrate how useful Kepler’s
third law is.
For instance, suppose you time how long Mars takes to go around the Sun (in Earth years). Kepler’s third law can then
be used to calculate Mars’ average distance from the Sun. Mars’ orbital period (1.88 Earth years) squared, or T2, is 1.882
= 3.53, and according to the equation for Kepler’s third law, this equals the cube of its semimajor axis, or a3. So what
number must be cubed to give 3.53? The answer is 1.52 (since 1.52 × 1.52 × 1.52 = 3.53). Thus, Mars’ semimajor axis in
astronomical units must be 1.52 AU. In other words, to go around the Sun in a little less than two years, Mars must be about
50% (half again) as far from the Sun as Earth is.
Example 8.2
Calculating Periods
Imagine an object is traveling around the Sun. What would be the orbital period of the object if its orbit has a
semimajor axis of 50 AU?
Solution
From Kepler’s third law, we know that (when we use units of years and AU)
T 2 = a3
If the object’s orbit has a semimajor axis of 50 AU (a = 50), we can cube 50 and then take the square root of the
result to get T:
T = a3
T = 50 × 50 × 50 = 125,000 = 353.6 years
Therefore, the orbital period of the object is about 350 years. This would place our hypothetical object beyond
the orbit of Pluto.
Kepler’s third law states that the square of the period is proportional to the cube of the semi-major axis of the orbit. When
written in the form of Equation 8.3, it is simple and easy to use for objects orbiting our Sun. The distances are measured
in AU and the periods in years. These units came naturally out of Kepler's work on the solar system. But, in fact, Kepler's
Third Law is much more general, and can be applied to bodies orbiting any large object. To do so, it makes the most sense to
measure quantities in SI units. Before we do so, however, we must realize that Kepler's Third Law in the form of Equation
8.3 does in fact contain a missing constant of proportionality, which is the the inverse of the mass of the Sun. The reason
that it does not appear in Equation 8.3 is that it is measured in units of "solar masses", which for the Sun has a value of
exactly 1. Nevertheless, the full equation for Kepler's Third Law should read:
286 Chapter 8 | Overview of the Solar System
In this equation, the period T is measured in seconds, the distance a in meters, and the mass M in kg. The constant G is
Newton's universal gravitational constant, which we will encounter in Chapter 9.
Example 8.3
⎛ ⎞
1/3
a = GM2 T 2
⎝ 4π ⎠
⎛(6.67 × 10 −11 N · m 2 /kg 2)(2.00 × 10 30 kg) ⎞
1/3
=⎜ (75.3 yr × 365 days/yr × 24 hr/day × 3600 s/hr) 2⎟ .
⎝ 4π 2 ⎠
a = 1 (aphelion + perihelion)
2
aphelion = 2a − perihelion.
Substituting for the values, we found for the semi-major axis and the value given for the perihelion, we find the
value of the aphelion to be 35.0 AU.
Significance
Edmond Halley, a contemporary of Newton, first suspected that three comets, reported in 1531, 1607, and 1682,
were actually the same comet. Before Tycho Brahe made measurements of comets, it was believed that they were
one-time events, perhaps disturbances in the atmosphere, and that they were not affected by the Sun. Halley used
Newton’s new mechanics to predict his namesake comet’s return in 1758.
8.1 Check Your Understanding The nearly circular orbit of Saturn has an average radius of about 9.5 AU
and has a period of 30 years, whereas Uranus averages about 19 AU and has a period of 84 years. Is this
consistent with our results for Halley’s comet?
288 Chapter 8 | Overview of the Solar System
CHAPTER 8 REVIEW
KEY TERMS
aphelion farthest point from the Sun of an orbiting body; the corresponding term for the Moon’s farthest point from Earth
is the apogee
asteroid a stony or metallic object orbiting the Sun that is smaller than a major planet but that shows no evidence of an
atmosphere or of other types of activity associated with comets
astronomical unit (AU) the unit of length defined as the average distance between Earth and the Sun; this distance is
about 1.5 × 108 kilometers
comet a small body of icy and dusty matter that revolves about the Sun; when a comet comes near the Sun, some of its
material vaporizes, forming a large head of tenuous gas and often a tail
differentiation gravitational separation of materials of different density into layers in the interior of a planet or moon
eccentricity in an ellipse, the ratio of the distance between the foci to the major axis
ellipse a closed curve for which the sum of the distances from any point on the ellipse to two points inside (called the
foci) is always the same
focus (plural: foci) one of two fixed points inside an ellipse from which the sum of the distances to any point on the
ellipse is constant
giant planet any of the planets Jupiter, Saturn, Uranus, and Neptune in our solar system, or planets of roughly that mass
and composition in other planetary systems
Kepler’s first law each planet moves around the Sun in an orbit that is an ellipse, with the Sun at one focus of the ellipse
Kepler’s second law the straight line joining a planet and the Sun sweeps out equal areas in space in equal intervals of
time
Kepler’s third law the square of a planet’s orbital period is directly proportional to the cube of the semimajor axis of its
orbit
major axis the maximum diameter of an ellipse
meteor a small piece of solid matter that enters Earth’s atmosphere and burns up, popularly called a shooting star because
it is seen as a small flash of light
meteorite a portion of a meteor that survives passage through an atmosphere and strikes the ground
orbit the path of an object that is in revolution about another object or point
orbital period (P) the time it takes an object to travel once around the Sun
orbital speed the speed at which an object (usually a planet) orbits around the mass of another object; in the case of a
planet, the speed at which each planet moves along its ellipse
perihelion point of closest approach to the Sun of an orbiting body; the corresponding term for the Moon’s closest
approach to Earth is the perigee
planetesimals objects, from tens to hundreds of kilometers in diameter, that formed in the solar nebula as an
intermediate step between tiny grains and the larger planetary objects we see today; the comets and some asteroids
may be leftover planetesimals
semimajor axis half of the major axis of a conic section, such as an ellipse
solar nebula the cloud of gas and dust from which the solar system formed
terrestrial planet any of the planets Mercury, Venus, Earth, or Mars; sometimes the Moon is included in the list
SUMMARY
8.1 Overview of Our Planetary System
Our solar system currently consists of the Sun, eight planets, five dwarf planets, nearly 200 known moons, and a host of
smaller objects. The planets can be divided into two groups: the inner terrestrial planets and the outer giant planets. Pluto,
Eris, Haumea, and Makemake do not fit into either category; as icy dwarf planets, they exist in an ice realm on the fringes
of the main planetary system. The giant planets are composed mostly of liquids and gases. Smaller members of the solar
system include asteroids (including the dwarf planet Ceres), which are rocky and metallic objects found mostly between
Mars and Jupiter; comets, which are made mostly of frozen gases and generally orbit far from the Sun; and countless smaller
grains of cosmic dust. When a meteor survives its passage through our atmosphere and falls to Earth, we call it a meteorite.
The giant planets have dense cores roughly 10 times the mass of Earth, surrounded by layers of hydrogen and helium. The
terrestrial planets consist mostly of rocks and metals. They were once molten, which allowed their structures to differentiate
(that is, their denser materials sank to the center). The Moon resembles the terrestrial planets in composition, but most of
the other moons—which orbit the giant planets—have larger quantities of frozen ice within them. In general, worlds closer
to the Sun have higher surface temperatures. The surfaces of terrestrial planets have been modified by impacts from space
and by varying degrees of geological activity.
Regularities among the planets have led astronomers to hypothesize that the Sun and the planets formed together in a giant,
spinning cloud of gas and dust called the solar nebula. Astronomical observations show tantalizingly similar circumstellar
disks around other stars. Within the solar nebula, material first coalesced into planetesimals; many of these gathered together
to make the planets and moons. The remainder can still be seen as comets and asteroids. Probably all planetary systems have
formed in similar ways, but many exoplanet systems have evolved along quite different paths, as we will see in Cosmic
Samples and the Origin of the Solar System (https://legacy.cnx.org/content/m59870/latest/) .
CONCEPTUAL QUESTIONS
4. Which type of planets have the most moons? Where did
8.3 Origin of the Solar System these moons likely originate?
1. Venus rotates backward and Uranus and Pluto spin
about an axis tipped nearly on its side. Based on what you 5. What is the difference between a meteor and a
learned about the motion of small bodies in the solar system meteorite?
and the surfaces of the planets, what might be the cause of
these strange rotations? 6. Explain our ideas about why the terrestrial planets are
rocky and have less gas than the giant planets.
2. What is the difference between a differentiated body
and an undifferentiated body, and how might that influence 7. Do all planetary systems look the same as our own?
a body’s ability to retain heat for the age of the solar
system?
8. What is comparative planetology and why is it useful to
astronomers?
3. What does a planet need in order to retain an
atmosphere? How does an atmosphere affect the surface of
9. What changed in our understanding of the Moon and
a planet and the ability of life to exist?
290 Chapter 8 | Overview of the Solar System
Moon-Earth system as a result of humans landing on the with Earth at a distance of about one city block from the
Moon’s surface? Sun. If you were to make a model of the distances in the
solar system to match your height, with the Sun at the top
10. If Earth was to be hit by an extraterrestrial object, of your head and Pluto at your feet, which planet would
where in the solar system could it come from and how be near your waist? How far down would the zone of the
would we know its source region? terrestrial planets reach?
11. List some reasons that the study of the planets has 22. Seasons are a result of the inclination of a planet’s
progressed more in the past few decades than any other axial tilt being inclined from the normal of the planet’s
branch of astronomy. orbital plane. For example, Earth has an axis tilt of 23.4°
(Appendix F (https://legacy.cnx.org/content/
m59998/latest/) ). Using information about just the
12. Imagine you are a travel agent in the next century.
inclination alone, which planets might you expect to have
An eccentric billionaire asks you to arrange a “Guinness
seasonal cycles similar to Earth, although different in
Book of Solar System Records” kind of tour. Where would
duration because orbital periods around the Sun are
you direct him to find the following (use this chapter and
different?
Appendix D:
A. the least-dense planet
B. the densest planet 23. Again using Appendix D, which planet(s) might you
C. the largest moon in the solar system expect not to have significant seasonal activity? Why?
D. excluding the jovian planets, the planet where
you would weigh the most on its surface (Hint: 24. Again using Appendix D, which planets might you
Weight is directly proportional to surface gravity.) expect to have extreme seasons? Why?
E. the smallest planet
F. the planet that takes the longest time to rotate 25. Using some of the astronomical resources in your
G. the planet that takes the shortest time to rotate college library or the Internet, find five names of features
H. the planet with a diameter closest to Earth’s on each of three other worlds that are named after real
I. the moon with the thickest atmosphere people. In a sentence or two, describe each of these people
J. the densest moon and what contributions they made to the progress of science
K. the most massive moon or human thought.
13. What characteristics do the worlds in our solar system 26. Explain why the planet Venus is differentiated, but
have in common that lead astronomers to believe that they asteroid Fraknoi, a very boring and small member of the
all formed from the same “mother cloud” (solar nebula)? asteroid belt, is not.
14. How do terrestrial and giant planets differ? List as 27. Would you expect as many impact craters per unit area
many ways as you can think of. on the surface of Venus as on the surface of Mars? Why or
why not?
15. Why are there so many craters on the Moon and so few
on Earth? 28. Interview a sample of 20 people who are not taking
an astronomy class and ask them if they can name a living
16. How do asteroids and comets differ? astronomer. What percentage of those interviewed were
able to name one? Typically, the two living astronomers
17. How and why is Earth’s Moon different from the the public knows these days are Stephen Hawking and Neil
larger moons of the giant planets? deGrasse Tyson. Why are they better known than most
astronomers? How would your result have differed if you
had asked the same people to name a movie star or a
18. Where would you look for some “original”
professional basketball player?
planetesimals left over from the formation of our solar
system?
29. Using Appendix B, complete the following table that
describes the characteristics of the Galilean moons of
19. What was the solar nebula like? Why did the Sun form
Jupiter, starting from Jupiter and moving outward in
at its center?
distance.
Moon Semimajor Diameter Density about a much larger mass, indicate where its speed is the
Axis (km3) (g/cm3) greatest and where it is the least. What conservation law
dictates this behavior? Indicate the directions of the force,
Io acceleration, and velocity at these points. Draw vectors for
these same three quantities at the two points where the
Europa
y-axis intersects (along the semi-minor axis) and from this
Ganymede determine whether the speed is increasing decreasing, or at
a max/min.
Callisto
Table A
PROBLEMS
the Sun’s commonly listed value of 1.989 × 10 30 kg .
8.3 Origin of the Solar System
32. Calculate the density of Jupiter. Show your work. Is it
38. Io orbits Jupiter with an average radius of 421,700 km
more or less dense than Earth? Why?
and a period of 1.769 days. Based upon these data, what is
the mass of Jupiter?
33. Calculate the density of Saturn. Show your work. How
does it compare with the density of water? Explain how this
39. The “mean” orbital radius listed for astronomical
can be.
objects orbiting the Sun is typically not an integrated
average but is calculated such that it gives the correct
34. What is the density of Jupiter’s moon Europa (see period when applied to the equation for circular orbits.
Appendix D for data on moons)? Show your work. Given that, what is the mean orbital radius in terms of
aphelion and perihelion?
35. Look at Appendix F (https://legacy.cnx.org/
content/m59998/latest/) and Appendix G 40. The perihelion of Halley’s comet is 0.586 AU and
(https://legacy.cnx.org/content/m59999/latest/) and the aphelion is 17.8 AU. Given that its speed at perihelion
indicate the moon with a diameter that is the largest fraction is 55 km/s, what is the speed at aphelion (
of the diameter of the planet or dwarf planet it orbits.
1 AU = 1.496 × 10 11 m )? (Hint: You may use either
conservation of energy or angular momentum, but the latter
36. Barnard’s Star, the second closest star to us, is about
is much easier.)
56 trillion (5.6 × 1012) km away. Calculate how far it
would be using the scale model of the solar system given in
Overview of Our Planetary System. 41. The perihelion of the comet Lagerkvist is 2.61 AU and
it has a period of 7.36 years. Show that the aphelion for this
comet is 4.95 AU.
8.4 Kepler's Laws of Planetary Motion
42. What is the ratio of the speed at perihelion to that at
37. Calculate the mass of the Sun based on data for aphelion for the comet Lagerkvist in the previous problem?
average Earth’s orbit and compare the value obtained with
292 Chapter 8 | Overview of the Solar System
43. Eros has an elliptical orbit about the Sun, with a 1.78 AU. What is the period of its orbit?
perihelion distance of 1.13 AU and aphelion distance of
9 | NEWTON'S SYNTHESIS
294 Chapter 9 | Newton's Synthesis
12 | MOMENTUM IN ONE
DIMENSION
300 Chapter 12 | Momentum in One Dimension
13 | ANGULAR MOMENTUM
302 Chapter 13 | Angular Momentum
14 | REFLECTION AND
REFRACTION
304 Chapter 14 | Reflection and Refraction
15 | DISPERSION,
LUMINOSITY AND
APPARENT BRIGHTNESS
306 Chapter 15 | Dispersion, Luminosity and Apparent Brightness
16 | IMAGE FORMATION
308 Chapter 16 | Image Formation
17 | WAVE PROPERTIES OF
LIGHT
310 Chapter 17 | Wave Properties of Light
18 | INTERFERENCE AND
DIFFRACTION
312 Chapter 18 | Interference and Diffraction
19 | SPECTROSCOPY AND
THE DOPPLER EFFECT
314 Chapter 19 | Spectroscopy and the Doppler Effect
20 | QUANTUM OPTICS
316 Chapter 20 | Quantum Optics
21 | TEMPERATURE
318 Chapter 21 | Temperature
23 | KINETIC THEORY
322 Chapter 23 | Kinetic Theory
25 | NUCLEAR PHYSICS
AND COMPARATIVE
GEOLOGY
326 Chapter 25 | Nuclear Physics and Comparative Geology
26 | COMPARATIVE
ATMOSPHERES
328 Chapter 26 | Comparative Atmospheres
27 | EXOPLANETS
330 Chapter 27 | Exoplanets
28 | THE SUN
332 Chapter 28 | The Sun
29 | STELLAR PROPERTIES
334 Chapter 29 | Stellar Properties
32 | SURVEY OF GALAXIES
340 Chapter 32 | Survey of Galaxies
33 | GALACTIC EVOLUTION
342 Chapter 33 | Galactic Evolution
34 | BIG BANG
COSMOLOGY
344 Chapter 34 | Big Bang Cosmology
APPENDIX A | UNITS
Quantity Common Unit Unit in Terms of Base SI
Symbol Units
Acceleration →
a m/s2 m/s2
Angular acceleration →
α rad/s2 s−2
Angular velocity →
ω rad/s s−1
Area A m2 m2
Atomic number Z
Capacitance C farad (F) A 2 · s 4 /kg · m 2
Current I ampere A
Current density → A/m2 A/m2
J
Dielectric constant κ
Electric dipole moment →
p C·m A·s·m
→
Electric field E N/C kg · m/A · s 3
Electric flux Φ N · m 2 /C kg · m 3 /A · s 3
Entropy S J/K kg · m 2 /s 2 · K
Length: ℓ, L meter m
Displacement Δx, Δ →
r
Distance d, h
Position x, y, z, →
r
→ N · J/T
Magnetic dipole μ A · m2
moment
Mass m, M kilogram kg
Molar specific heat C J/mol · K kg · m 2 /s 2 · mol · K
Moment of inertia I kg · m 2 kg · m 2
Momentum →
p kg · m/s kg · m/s
Period T s s
Permeability of free μ0 2
N/A =(H/m) kg · m/A 2 · s 2
space
Permittivity of free ε0 C 2 /N · m 2 =(F/m) A 2 · s 4 /kg · m 3
space
Potential V volt(V) = (J/C) kg · m 2 /A · s 3
Temperature T kelvin K
Time t second s
→ N·m
Torque τ kg · m 2 /s 2
Volume V m3 m3
Wavelength λ m m
Work W joule(J) = (N · m) kg · m 2 /s 2
APPENDIX B | UNIT
CONVERSIONS
m cm km
1 meter 1 102 10−3
1 centimeter 10−2 1 10−5
1 kilometer 103 105 1
1 inch 2.540 × 10 −2 2.540 2.540 × 10 −5
1 foot 0.3048 30.48 3.048 × 10 −4
1 mile 1609 1.609 × 10 4 1.609
1 angstrom 10−10
1 fermi 10−15
1 light-year 9.461 × 10 15 9.461 × 10 12
1 parsec 3.084 × 10 16 3.084 × 10 13
in. ft mi
1 meter 39.37 3.281 6.214 × 10 −4
1 centimeter 0.3937 3.281 × 10 −2 6.214 × 10 −6
1 kilometer 3.937 × 10 4 3.281 × 10 3 0.6214
Table B1 Length
Area
1 cm 2 = 0.155 in. 2
1 m 2 = 10 4 cm 2 = 10.76 ft 2
1 in. 2 = 6.452 cm 2
s min h day yr
1 second 1 1.667 × 10 −2 2.778 × 10 −4 1.157 × 10 −5 3.169 × 10 −8
1 minute 60 1 1.667 × 10 −2 6.944 × 10 −4 1.901 × 10 −6
1 hour 3600 60 1 4.167 × 10 −2 1.141 × 10 −4
1 day 8.640 × 10 4 1440 24 1 2.738 × 10 −3
1 year 3.156 × 10 7 5.259 × 10 5 8.766 × 10 3 365.25 1
Table B2 Time
3.281 × 10 −2 2.237 × 10 −2
−2
1 centimeter/second 10 1
Table B3 Speed
Acceleration
1 m/s 2 = 100 cm/s 2 = 3.281 ft/s 2
kg g slug u
1 kilogram 1 103 6.852 × 10 −2 6.024 × 10 26
1 gram 10−3 1 6.852 × 10 −5 6.024 × 10 23
1 slug 14.59 1.459 × 10 4 1 8.789 × 10 27
1 atomic mass unit 1.661 × 10 −27 1.661 × 10 −24 1.138 × 10 −28 1
Table B4 Mass
N dyne lb
1 newton 1 105 0.2248
Table B5 Force
1 bar 105
1 torr 1 (mmHg)
*Where the acceleration due to gravity is 9.80665 m/s2 and the temperature is 0°C
Table B6 Pressure
J erg ft.lb
1 joule 1 107 0.7376
1 erg 10−7 1 7.376 × 10 −8
1 foot-pound 1.356 1.356 × 10 7 1
Power
1 W = 1 J/s
1 hp = 746 W = 550 ft · lb/s
1 Btu/h = 0.293 W
Angle
354 Appendix B
APPENDIX C |
FUNDAMENTAL PHYSICS
CONSTANTS
Quantity Symbol Value
Atomic mass unit u 1.660 538 782 (83) × 10 −27 kg
931.494 028 (23) MeV/c 2
Bohr radius
a0 = ℏ2 5.291 772 085 9 (36) × 10 −11 m
me e2 ke
Table C1 Fundamental Constants Note: These constants are the values recommended in 2006
by CODATA, based on a least-squares adjustment of data from different measurements. The
numbers in parentheses for the values represent the uncertainties of the last two digits.
356 Appendix C
Table C1 Fundamental Constants Note: These constants are the values recommended in 2006
by CODATA, based on a least-squares adjustment of data from different measurements. The
numbers in parentheses for the values represent the uncertainties of the last two digits.
APPENDIX D |
ASTRONOMICAL DATA
Astronomical Constants
Name Value
astronomical unit (AU) 1.496 × 1011 m
Light-year (ly) 9.461 × 1015 m
parsec (pc) 3.086 × 1016 m = 3.262 light-years
sidereal year (y) 3.156 × 107 s
mass of Earth (MEarth) 5.974 × 1024 kg
equatorial radius of Earth (REarth) 6.378 × 106 m
obliquity of ecliptic 23.4° 26’
surface gravity of Earth (g) 9.807 m/s2
escape velocity of Earth (vEarth) 1.119 × 104 m/s
mass of Sun (MSun) 1.989 × 1030 kg
equatorial radius of Sun (RSun) 6.960 × 108 m
luminosity of Sun (LSun) 3.85 × 1026 W
solar constant (flux of energy 1.368 × 103 W/m2
received at Earth) (S)
Hubble constant (H0) approximately 20 km/s per million light-years, or approximately
70 km/s per megaparsec
Table D1
Table D3
Table D4
Table D5
Table D6
discovered on a regular basis. Of the major planets, only Mercury and Venus do not have moons. In addition to moons
of the planets, there are many moons of asteroids. In this appendix, we list only the largest and most interesting objects
that orbit each planet (including dwarf planets). The number given for each planet is discoveries through 2015. For
further information see https://solarsystem.nasa.gov/planets/solarsystem/moons and https://en.wikipedia.org/wiki/
List_of_natural_satellites.
Table D7
Table D7
2. R stands for retrograde rotation (backward from the direction that most objects in the solar system revolve
and rotate).
3. R stands for retrograde rotation (backward from the direction that most objects in the solar system revolve
and rotate).
362 Appendix D
Table D7
Table D8
Table D8
Table D8
D4 Stellar Data
Note: These are the stars that appear the brightest visually, as seen from our vantage point on Earth. They are not the stars
that are intrinsically the most luminous.
8. Brown dwarf
9. Brown dwarf
10. Brown dwarf
Figure D1 The brightest stars typically have names from antiquity. Next to each star’s ancient name, we have added a column
with its name in the system originated by Bayer (see the Naming Stars (https://legacy.cnx.org/content/m59905/
latest/#fs-id1165721974313) feature box.) The distances of the more remote stars are estimated from their spectral types and
apparent brightnesses and are only approximate. The luminosities for those stars are approximate to the same degree. Right
ascension and declination is given for Epoch 2000.0.
366 Appendix D
APPENDIX E | USEFUL
MATHEMATICAL
FORMULAS
Quadratic formula
2
If ax 2 + bx + c = 0, then x = −b ± b − 4ac
2a
Table E1 Geometry
Trigonometry
Trigonometric Identities
1. sin θ = 1/csc θ
2. cos θ = 1/sec θ
3. tan θ = 1/cot θ
7. sin 2 θ + cos 2 θ = 1
8. sec 2 θ − tan 2 θ = 1
9. tan θ = sin θ/cos θ
10. sin⎛⎝α ± β⎞⎠ = sin α cos β ± cos α sin β
tan α ± tan β
12. tan⎛⎝α ± β⎞⎠ =
1 ∓ tan α tan β
1. Law of sines: a = b = c
sin α sin β sin γ
3. Pythagorean theorem: a 2 + b 2 = c 2
Series expansions
n(n − 1)a n − 2 b 2 n(n − 1)(n − 2)a n − 3 b 3
1. Binomial theorem: (a + b) n = a n + na n − 1 b + + + ···
2! 3!
n(n − 1)x 2
2. (1 ± x) n = 1 ± nx + ± ···⎛⎝x 2 < 1⎞⎠
1! 2!
n(n + 1)x 2
3. (1 ± x) −n = 1 ∓ nx + ∓ ···⎛⎝x 2 < 1⎞⎠
1! 2!
3 5
4. sin x = x − x + x − ···
3! 5!
2 4
5. cos x = 1 − x + x − ···
2! 4!
3 5
6. tan x = x + x + 2x + ···
3 15
2
7. e x = 1 + x + x + ···
2!
Derivatives
5. d x m = mx m − 1
dx
6. d sin x = cos x
dx
7. d cos x = −sin x
dx
8. d tan x = sec 2 x
dx
9. d cot x = −csc 2 x
dx
12. d ex = ex
dx
13. d ln x = 1
dx x
14. d sin −1 x = 1
dx 1 − x2
15. d cos −1 x = − 1
dx 1 − x2
16. d tan −1 x = − 1
dx 1 + x2
370 Appendix E
APPENDIX F | PERIODIC
TABLE OF THE ELEMENTS
372 Appendix F
Gamma Γ γ Omicron O ο
Delta Δ δ Pi Π π
Epsilon E ε Rho P ρ
Zeta Z ζ Sigma Σ σ
Eta H η Tau T τ
Theta Θ θ Upsilon ϒ υ
lota I ι Phi Φ ϕ
Kappa K κ Chi X χ
Lambda Λ λ Psi ψ ψ
Mu M μ Omega Ω ω
ANSWER KEY
CHAPTER 1
CHECK YOUR UNDERSTANDING
1.1. 4.79 × 10 2 Mg or 479 Mg
1.2. 3 × 10 8 m/s
1.3. 10 8 km 2
1.4. The numbers were too small, by a factor of 4.45.
1.5. 4πr 3 /3
1.6. yes
1.7. 3 × 10 4 m or 30 km. It is probably an underestimate because the density of the atmosphere decreases with altitude. (In fact,
30 km does not even get us out of the stratosphere.)
1.8. No, the coach’s new stopwatch will not be helpful. The uncertainty in the stopwatch is too great to differentiate between the
sprint times effectively.
CONCEPTUAL QUESTIONS
1. Physics is the science concerned with describing the interactions of energy, matter, space, and time to uncover the fundamental
mechanisms that underlie every phenomenon.
3. No, neither of these two theories is more valid than the other. Experimentation is the ultimate decider. If experimental evidence
does not suggest one theory over the other, then both are equally valid. A given physicist might prefer one theory over another
on the grounds that one seems more simple, more natural, or more beautiful than the other, but that physicist would quickly
acknowledge that he or she cannot say the other theory is invalid. Rather, he or she would be honest about the fact that more
experimental evidence is needed to determine which theory is a better description of nature.
5. Probably not. As the saying goes, “Extraordinary claims require extraordinary evidence.”
7. Conversions between units require factors of 10 only, which simplifies calculations. Also, the same basic units can be scaled up
or down using metric prefixes to sizes appropriate for the problem at hand.
9. a. Base units are defined by a particular process of measuring a base quantity whereas derived units are defined as algebraic
combinations of base units. b. A base quantity is chosen by convention and practical considerations. Derived quantities are
expressed as algebraic combinations of base quantities. c. A base unit is a standard for expressing the measurement of a base
quantity within a particular system of units. So, a measurement of a base quantity could be expressed in terms of a base unit in any
system of units using the same base quantities. For example, length is a base quantity in both SI and the English system, but the
meter is a base unit in the SI system only.
11. a. Uncertainty is a quantitative measure of precision. b. Discrepancy is a quantitative measure of accuracy.
13. Check to make sure it makes sense and assess its significance.
PROBLEMS
15. a. 103; b. 105; c. 102; d. 1015; e. 102; f. 1057
17. 102 generations
19. 1011 atoms
21. 103 nerve impulses/s
23. 1026 floating-point operations per human lifetime
25. a. 957 ks; b. 4.5 cs or 45 ms; c. 550 ns; d. 31.6 Ms
27. a. 75.9 Mm; b. 7.4 mm; c. 88 pm; d. 16.3 Tm
29. a. 3.8 cg or 38 mg; b. 230 Eg; c. 24 ng; d. 8 Eg e. 4.2 g
31. a. 27.8 m/s; b. 62 mi/h
33. a. 3.6 km/h; b. 2.2 mi/h
35. 1.05 × 10 5 ft 2
37. 8.847 km
39. a. 1.3 × 10 −9 m; b. 40 km/My
41. 10 6 Mg/μL
43. 62.4 lbm/ft3
45. 0.017 rad
47. 1 light-nanosecond
49. 3.6 × 10 −4 m 3
51. a. Yes, both terms have dimension L2T-2 b. No. c. Yes, both terms have dimension LT-1 d. Yes, both terms have dimension
376 Answer Key
LT-2
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
53. a. [v] = LT–1; b. [a] = LT–2; c. ⎣∫ vdt⎦ = L; d. ⎣∫ adt⎦ = LT ; e. ⎣ da ⎦ = LT
–1 –3
dt
55. a. L; b. L; c. L0 = 1 (that is, it is dimensionless)
57. 1028 atoms
59. 1051 molecules
61. 1016 solar systems
63. a. Volume = 1027 m3, diameter is 109 m.; b. 1011 m
65. a. A reasonable estimate might be one operation per second for a total of 109 in a lifetime.; b. about (109)(10–17 s) = 10–8 s, or
about 10 ns
67. 2 kg
69. 4%
71. 67 mL
73. a. The number 99 has 2 significant figures; 100. has 3 significant figures. b. 1.00%; c. percent uncertainties
75. a. 2%; b. 1 mm Hg
77. 7.557 cm2
79. a. 37.2 lb; because the number of bags is an exact value, it is not considered in the significant figures; b. 1.4 N; because the
value 55 kg has only two significant figures, the final value must also contain two significant figures
ADDITIONAL PROBLEMS
81. a. [s 0] = L and units are meters (m); b. [v 0] = LT −1 and units are meters per second (m/s); c. [a 0] = LT −2 and units
are meters per second squared (m/s2); d. [ j 0] = LT −3 and units are meters per second cubed (m/s3); e. [S 0] = LT −4 and units
CHAPTER 3
CHECK YOUR UNDERSTANDING
3.1. (a) The rider’s displacement is Δx = x f − x 0 = −1 km . (The displacement is negative because we take east to be positive
and west to be negative.) (b) The distance traveled is 3 km + 2 km = 5 km. (c) The magnitude of the displacement is 1 km.
3.2. Inserting the knowns, we have
–a = Δv = 2.0 × 10 7 m/s − 0 = 2.0 × 10 11 m/s 2.
Δt 10 −4 s − 0
3.3. If we take east to be positive, then the airplane has negative acceleration because it is accelerating toward the west. It is also
decelerating; its acceleration is opposite in direction to its velocity.
3.4. To answer this, choose an equation that allows us to solve for time t, given only a , v0 , and v:
v = v 0 + at.
Rearrange to solve for t:
v − v 0 400 m/s − 0 m/s
t= a = = 20 s.
20 m/s 2
2
3.5. a = m/s .
2
3
3.6. It takes 2.47 s to hit the water. The quantity distance traveled increases faster.
CONCEPTUAL QUESTIONS
1. You drive your car into town and return to drive past your house to a friend’s house.
3. If the bacteria are moving back and forth, then the displacements are canceling each other and the final displacement is small.
5. Distance traveled
7. Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your
home, your average speed is a positive number. Since Average velocity = Displacement/Elapsed time, your average velocity is
zero.
9. Average speed. They are the same if the car doesn’t reverse direction.
11. No, in one dimension constant speed requires zero acceleration.
13. A ball is thrown into the air and its velocity is zero at the apex of the throw, but acceleration is not zero.
15. Plus, minus
17. If the acceleration, time, and displacement are the knowns, and the initial and final velocities are the unknowns, then two
kinematic equations must be solved simultaneously. Also if the final velocity, time, and displacement are the knowns then two
kinematic equations must be solved for the initial velocity and acceleration.
19. a. at the top of its trajectory; b. yes, at the top of its trajectory; c. yes
g g
21. Earth v = v 0 − gt = −gt ; Moon v′ = t′ v = v′ − gt = − t′ t′ = 6t ; Earth y = − 1 gt 2 Moon
6 6 2
g ⎛ ⎞
y′ = − 1 (6t) 2 = − 1 g6t 2 = −6⎝1 gt 2⎠ = −6y
26 2 2
PROBLEMS
25. a. →
^ ^
x 1 = (−2.0 m) i , →
x 2 = (5.0 m) i ; b. 7.0 m east
27. a. t = 2.0 s; b. x(6.0) − x(3.0) = −8.0 − (−2.0) = −6.0 m
–
29. a. 150.0 s, v = 156.7 m/s ; b. 45.7% the speed of sound at sea level
31.
33.
34. a = 4.29m/s 2
378 Answer Key
36.
38. a = 11.1g
40. 150 m
42. a. 525 m;
b. v = 180 m/s
44. a.
c. x = x 0 + v 0 t + 1 at = 1 at = 2.40 m/s (12.0 s) = 172.80 m , the answer seems reasonable at about 172.8 m; d.
2 2 2 2
2 2
v = 28.8 m/s
52. a.
a = −40.0 m/s 2
c.
a = 4.08 g
60. Knowns: x = 3 m, v = 0 m/s, v 0 = 54 m/s . We want a, so we can use this equation: a = −486 m/s 2 .
62. a. a = 32.58 m/s 2 ;
b. v = 161.85 m/s ;
c. v > v max , because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears and would have
a greater acceleration in first gear than second gear than third gear, and so on. The acceleration would be greatest at the beginning,
so it would not be accelerating at 32.6 m/s 2 during the last few meters, but substantially less, and the final velocity would be less
than 162 m/s .
y = −8.23 m
64. a. ;
v 1 = −18.9 m/s
y = −18.9 m
b. ;
v 2 = 23.8 m/s
y = −32.0 m
c. ;
v 3 = −28.7 m/s
y = −47.6 m
d. ;
v 4 = −33.6 m/s
y = −65.6 m
e.
v 5 = −38.5 m/s
b. y = y 0 + v 0 t − 1 gt y = v 0 t − 1 gt = −1.4 m/s(1.8 sec) − 1 (9.8)(1.8 s) = −18.4 m and the origin is at the rescuers,
2 2
2 2 2
who are 18.4 m above the water.
2 2
v
68. a. v 2 = v 20 − 2g(y − y 0) y 0 = 0 v = 0 y = 0 = (4.0 m/s) = 0.82 m ; b. to the apex v = 0.41 s times 2 to the
2g 2(9.80)
380 Answer Key
board = 0.82 s from the board to the water y = y 0 + v 0 t − 1 gt 2 y = −1.80 m y 0 = 0 v 0 = 4.0 m/s
2
−1.8 = 4.0t − 4.9t 2 4.9t 2 − 4.0t − 1.80 = 0 , solution to quadratic equation gives 1.13 s; c.
2
v = v 20 − 2g(y − y 0) y0 = 0 v 0 = 4.0 m/s y = −1.80 m
v = 7.16 m/s
70. Time to the apex: t = 1.12 s times 2 equals 2.24 s to a height of 2.20 m. To 1.80 m in height is an additional 0.40 m.
Take the positive root, so the time to go the additional 0.4 m is 0.04 s. Total time is 2.24 s + 0.04 s = 2.28 s .
2 2
72. a. v = v 0 − 2g(y − y 0) y 0 = 0 v = 0 y = 2.50 m ; b. t = 0.72 s times 2 gives 1.44 s in the air
v 20 = 2gy ⇒ v 0 = 2(9.80)(2.50) = 7.0 m/s
74. a. v = 70.0 m/s ; b. time heard after rock begins to fall: 0.75 s, time to reach the ground: 6.09 s
ADDITIONAL PROBLEMS
76. Take west to be the positive direction.
–
1st plane: ν = 600 km/h
2nd plane –
ν = 667.0 km/h
v−v −3.4 cm/s − v 0
78. a= t−t0, t = 0, a = = 1.2 cm/s 2 ⇒ v 0 = − 8.2 cm/s v = v 0 + at = − 8.2 + 1.2 t ;
0 4s
v = −7.0 cm/s v = −1.0 cm/s
79. a = −3 m/s 2
81. a.
v = 8.7 × 10 5 m/s ;
b. t = 7.8 × 10 −8 s
equation for displacement with x 0 = 0 and v 0 = 0 , since the police car starts from rest: x = x 0 + v 0 t + 1 at = 1 at ; Now
2 2
2 2
we have an equation of motion for each car with a common parameter, which can be eliminated to find the solution. In this case,
– –
we solve for t . Step 1, eliminating x : x = v t = 1 at ; Step 2, solving for t : t = 2v
2
2 a . The speeding car has a constant velocity
of 40 m/s, which is its average velocity. The acceleration of the police car is 4 m/s2. Evaluating t, the time for the police car to
– 2(40)
reach the speeding car, we have t = 2v
a = = 20 s .
4
−v
89. At this acceleration she comes to a full stop in t = a 0 = 8 = 16 s , but the distance covered is
0.5
x = 8 m/s(16 s) − 1 (0.5)(16 s) 2 = 64 m , which is less than the distance she is away from the finish line, so she never finishes
2
the race.
3
91. x 1 = v 0 t
2
x2 = 5 x1
3
93. v 0 = 7.9 m/s velocity at the bottom of the window.
v = 7.9 m/s
v 0 = 14.1 m/s
95. a. v = 5.42 m/s ;
b. v = 4.64 m/s ;
c. a = 2874.28 m/s 2 ;
d. (x − x 0) = 5.11 × 10 −3 m
97. Consider the players fall from rest at the height 1.0 m and 0.3 m.
0.9 s
0.5 s
99. a. t = 6.37 s taking the positive root;
b. v = 59.5 m/s
101. a. y = 4.9 m ;
b. v = 38.3 m/s ;
c. −33.3 m
103. h = 1 gt , h = total height and time to drop to ground
2
2
2 h = 1 g(t − 1) 2 in t – 1 seconds it drops 2/3h
3 2
2 1 gt 2⎞ = 1 g(t − 1) 2 or t 2 = 1 (t − 1) 2
⎛
3 ⎝2 ⎠ 2 3 2
2
0 = t 2 − 6t + 3 t = 6 ± 6 − 4 · 3 = 3 ± 24
2 2
2
t = 5.45 s and h = 145.5 m. Other root is less than 1 s. Check for t = 4.45 s h = 1 gt = 97.0 m = (145.5)
2
2 3
CHALLENGE PROBLEMS
CHAPTER 4
CHECK YOUR UNDERSTANDING
4.1. a. 40.0 rev/s = 2π(40.0) rad/s , – = Δω = 2π(40.0) − 0 rad/s = 2π(2.0) = 4.0π rad/s 2 ; b. Since the angular
α
Δt 20.0 s
velocity increases linearly, there has to be a constant acceleration throughout the indicated time. Therefore, the instantaneous
angular acceleration at any time is the solution to 4.0π rad/s 2 .
7000.0(2π rad)
4.2. a. Using Equation 4.22, we have 7000 rpm = = 733.0 rad/s,
60.0 s
ω − ω 0 733.0 rad/s
α= t = = 73.3 rad/s 2 ;
10.0 s
b. Using Equation 4.25, we have
ω 2 − ω 20 0 − (733.0 rad/s) 2
ω 2 = ω 20 + 2αΔθ ⇒ Δθ = = = 3665.2 rad
2α 2(73.3 rad/s 2)
(5.0 − 0)rad/s
4.3. The angular acceleration is α = = 0.25 rad/s 2 . Therefore, the total angle that the boy passes through is
20.0 s
ω 2 − ω 20 (5.0) 2 − 0
Δθ = = = 50 rad .
2α 2(0.25)
Thus, we calculate
382 Answer Key
CONCEPTUAL QUESTIONS
1. The second hand rotates clockwise, so by the right-hand rule, the angular velocity vector is into the wall.
3. They have the same angular velocity. Points further out on the bat have greater tangential speeds.
4. straight line, linear in time variable
6. constant
PROBLEMS
9. ω =
2π rad = 0.14 rad/s
45.0 s
s 3.0 m = 2.0 rad ; b. ω = 2.0 rad = 2.0 rad/s
11. a. θ = r =
1.5 m 1.0 s
Δω 0 rad/s − 10.0(2π) rad/s = 31.4 s
13. The propeller takes only Δt = α = to come to rest, when the propeller is at 0 rad/s,
−2.0 rad/s 2
it would start rotating in the opposite direction. This would be impossible due to the magnitude of forces involved in getting the
propeller to stop and start rotating in the opposite direction.
15. a. ω = 25.0(2.0 s) = 50.0 rad/s ; b. α =
dω = 25.0 rad/s 2
dt
16. a. ω = 54.8 rad/s ;
b. t = 11.0 s
18. a. 0.87 rad/s 2 ;
b. θ = 66,264 rad
20. a. ω = 42.0 rad/s ;
v t = 42 m/s
b. θ = 200 rad ; c.
a t = 4.0 m/s 2
CHAPTER 5
CHECK YOUR UNDERSTANDING
5.1. Special relativity applies only to objects moving at constant velocity, whereas general relativity applies to objects that undergo
acceleration.
γ= 1 = 1 = 1.32
5.2. 1− v2 (0.650c) 2
1−
c2 c2
Δt = Δτ = 2.10 × 10 −8 s = 2.71 × 10 −8 s.
5.3. a. 1− v2 8 m/s)
2
c2 1 − (1.90 × 10 2
(3.00 × 10 8 m/s)
5.3. b. Only the relative speed of the two spacecraft matters because there is no absolute motion through space. The signal is
emitted from a fixed location in the frame of reference of A, so the proper time interval of its emission is τ = 1.00 s. The duration
of the signal measured from frame of reference B is then
Δt = Δτ = 1.00 s = 1.01 s.
v2 2
1− (4.00 × 10 7 m/s)
c2 1− 2
(3.00 × 10 8 m/s)
2
5.4. L = L 0 1 − v = (2.50 km) 1 −
(0.750c) 2
2 2
= 1.65 km
c c
5.5. Start with the definition of the proper time increment:
dτ = −(ds) 2 /c 2 = dt 2 − ⎛⎝dx 2 + dx 2 + dx 2⎞⎠/c 2.
where (dx, dy, dx, cdt) are measured in the inertial frame of an observer who does not necessarily see that particle at rest. This
therefore becomes
dτ = −(ds) 2 /c 2 = dt 2 − ⎡⎣(dx) 2 + ⎛⎝dy⎞⎠ 2 + (dz) 2⎤⎦/c 2
⎡ ⎛dy ⎞ 2⎤ 2
⎛ ⎞ ⎛ ⎞
2
= dt 1 − ⎢⎝dx ⎠ + ⎝ ⎠ + ⎝dz ⎠ ⎥/c 2
⎣ dt dt dt ⎦
= dt 1 − v 2 /c 2
dt = γdτ.
5.6. Although displacements perpendicular to the relative motion are the same in both frames of reference, the time interval
between events differ, and differences in dt and dt′ lead to different velocities seen from the two frames.
CONCEPTUAL QUESTIONS
1. the second postulate, involving the speed of light; classical physics already included the idea that the laws of mechanics, at least,
were the same in all inertial frames, but the velocity of a light pulse was different in different frames moving with respect to each
other
3. yes, provided the plane is flying at constant velocity relative to the Earth; in that case, an object with no force acting on it within
the plane has no change in velocity relative to the plane and no change in velocity relative to the Earth; both the plane and the
ground are inertial frames for describing the motion of the object
5. The observer moving with the process sees its interval of proper time, which is the shortest seen by any observer.
7. The length of an object is greatest to an observer who is moving with the object, and therefore measures its proper length.
9. a. No, not within the astronaut’s own frame of reference. b. He sees Earth clocks to be in their rest frame moving by him, and
therefore sees them slowed. c. No, not within the astronaut’s own frame of reference. d. Yes, he measures the distance between the
two stars to be shorter. e. The two observers agree on their relative speed.
PROBLEMS
10. a. 1.0328; b. 1.15
12. 5.96 × 10 −8 s
14. 0.800c
16. 0.140c
18. 48.6 m
20. Using the values given in m58563 (https://legacy.cnx.org/content/m58563/latest/#fs-id1167793912924) : a. 1.39
km; b. 0.433 km; c. 0.433 km
22. a. 10.0c; b. The resulting speed of the canister is greater than c, an impossibility. c. It is unreasonable to assume that the canister
will move toward the earth at 1.20c.
24. The angle α approaches 45°, and the t′- and x′-axes rotate toward the edge of the light cone.
26. 15 m/s east
28. 32 m/s
30. a. The second ball approaches with velocity −v and comes to rest while the other ball continues with velocity −v; b. This
conserves momentum.
t 1′ = 0; x 1′ = 0; t 1′ = 0; x 1′ = 0;
; τ −vτ
32. a.
t 2′ = τ; x 2′ = 0 b. t 2′ = 2 2
; x 2′ =
1 − v /c 1 − v 2 /c 2
34. 0.615c
36. 0.696c
38. (Proof)
ADDITIONAL PROBLEMS
40. a. 0.866c; b. 0.995c
42. a. 4.303 y to four digits to show any effect; b. 0.1434 y; c. 1/ ⎛⎝1 − v 2 /c 2⎞⎠ = 29.88.
44. a. 4.00; b. v = 0.867c
46. a. A sends a radio pulse at each heartbeat to B, who knows their relative velocity and uses the time dilation formula to calculate
384 Answer Key
the proper time interval between heartbeats from the observed signal. b. (66 beats/min) 1 − v 2/c 2 = 57.1 beats/min
48. a. first photon: (0, 0, 0) at t = t′; second photon:
2
t′ = −vx/c 2 = −(c/2)(1.00 m)/c = − 0.577 m = 1.93 × 10 −9 s
0.75 c
1 − v 2 /c 2
x′ = x = 1.00 m = 1.15 m
2 2
1 − v /c 0.75
b. simultaneous in A, not simultaneous in B
⎛ ⎛ 3 m⎞
2 ⎝4.5 × 10 −4 s⎞⎠ − (0.6c)⎝150 × 10
2 ⎠
t′ = t − vx/c = c
1 − v 2 /c 2 1 − (0.6) 2
= 1.88 × 10 −4 s
3 ⎛ 8 ⎞⎛ −4 ⎞
50.
x′ = x − vt = 150 × 10 m − (0.60)⎝3.00 × 10 m/s⎠⎝4.5 × 10 s⎠
1 − v 2 /c 2 1 − (0.6) 2
= −1.01 × 10 5 m = −101 km
y = y′ = 15 km
z = z′ = 1 km
2
Δt = Δt′ + vΔx′/c
2 2
1 − v /c
52.
Δt′ + v(500 m)/c 2
0 = ;
1 − v 2 /c 2
since v ≪ c, we can ignore the term v 2 /c 2 and find
(50 m/s)(500 m)
Δt′ = − 2
= −2.78 × 10 −13 s
⎛ 8 ⎞
⎝3.00 × 10 m/s⎠
The breakdown of Newtonian simultaneity is negligibly small, but not exactly zero, at realistic train speeds of 50 m/s.
2
Δt′ = Δt − vΔx/c
2 2
1 − v /c
⎛ ⎞
(v)⎝2.0 × 10 9 m⎠
(0.30 s) − 2
⎛ 8 ⎞
⎝3.00 × 10 m/s⎠
54. 0 =
1 − v 2 /c 2
(0.30 s) ⎛ 2
v = ⎛ 9 ⎞⎝
3.00 × 10 8 m/s⎞⎠
⎝2.0 × 10 m⎠
v = 1.35 × 10 7 m/s
56. Note that all answers to this problem are reported to five significant figures, to distinguish the results. a. 0.99947c; b.
1.2064 × 10 11 y; c. 1.2058 × 10 11 y
58. a. –0.400c; b. –0.909c
60. a. 1.65 km/s; b. Yes, if the speed of light were this small, speeds that we can achieve in everyday life would be larger than 1%
of the speed of light and we could observe relativistic effects much more often.
CHAPTER 6
CHECK YOUR UNDERSTANDING
6.1. a. not equal because they are orthogonal; b. not equal because they have different magnitudes; c. not equal because they have
different magnitudes and directions; d. not equal because they are antiparallel; e. equal.
6.2. 16 m; → D = −16 m ^ u
6.3. G = 28.2 cm, θ G = 291°
6.4. → ^ ^
D = (−5.0 i − 3.0 j )cm ; the fly moved 5.0 cm to the left and 3.0 cm down from its landing site.
CONCEPTUAL QUESTIONS
1. scalar
3. answers may vary
5. parallel, sum of magnitudes, antiparallel, zero
7. no, yes
9. zero, yes
11. no
13. equal, equal, the same
15. a unit vector of the x-axis
17. They are equal.
19. yes
PROBLEMS
21. → h = −49 m ^ u , 49 m
23. 30.8 m, 35.7° west of north
25. 134 km, 80°
27. 7.34 km, 63.5° south of east
29. 3.8 km east, 3.2 km north, 7.0 km
31. 14.3 km, 65°
33. a. → ^ ^ → ^ ^ → ^ ^
A = + 8.66 i + 5.00 j , b. B = + 30.09 i + 39.93 j , c. C = + 6.00 i − 10.39 j , d.
→ ^ ^ → ^ ^
D = −15.97 i + 12.04 j , f. F = −17.32 i − 10.00 j
43. a. → → ^ ^
A + B = −4 i − 6 j , | → →
|
A + B = 7.211, θ = 213.7° ; b. → → ^ ^
A − B =2i −2j ,
| → →
|
A − B = 2 2, θ = −45°
45. a. → ^ ^ ^
C = (5.0 i − 1.0 j − 3.0 k )m, C = 5.92 m ;
b. → ^ ^ ^
D = (4.0 i − 11.0 j + 15.0 k )m, D = 19.03 m
47. → ^ ^ ^
D = (3.3 i − 6.6 j )km , i is to the east, 7.34 km, −63.5°
49. a. → ^ ^ → ^ ^
R = −1.35 i − 22.04 j , b. R = −17.98 i + 0.89 j
51. → ^ ^
D = (200 i + 300 j )yd , D = 360.5 yd, 56.3° north of east; The numerical answers would stay the same but the physical
unit would be meters. The physical meaning and distances would be about the same because 1 yd is comparable with 1 m.
53. → ^ ^
R = −3 i − 16 j
55. → ^ E = −357.8V/m , E z = 0.0V/m , θ E = −tan −1(2)
E = EE , E x = + 178.9V/m , y
→ ^ ^ ^ → ^ ^
57. a. R B = (12.278 i + 7.089 j + 2.500 k )km , R D = (−0.262 i + 3.000 k )km ; b.
| →
R B−
→
R D| = 14.414 km
ADDITIONAL PROBLEMS
58. a. 18.4 km and 26.2 km, b. 31.5 km and 5.56 km
60. a. (r, φ + π/2) , b. (2r, φ + 2π) , (c) (3r, −φ)
62. d PM = 33.12 nmi = 61.34 km, d NP = 35.47 nmi = 65.69 km
64. proof
66. a. 10.00 m, b. 5π m , c. 0
68. 22.2 km/h, 35.8° south of west
70. 240.2 m, 2.2° south of west
CHAPTER 7
CHECK YOUR UNDERSTANDING
7.1. (a) Taking the derivative with respect to time of the position function, we have
→ ^ ^
v (t) = 9.0t 2 i and →
v (3.0s) = 81.0 i m/s. (b) Since the velocity function is nonlinear, we suspect the average velocity is
not equal to the instantaneous velocity. We check this and find
→ ^ ^
→ r (t 2) − → r (t 1) →
r (4.0 s) − → r (2.0 s) (144.0 i − 36.0 i ) m ^
v avg = t2 − t1 = = = 54.0 i m/s,
4.0 s − 2.0 s 2.0 s
which is different from → ^
v (3.0s) = 81.0 i m/s.
7.2. The acceleration vector is constant and doesn’t change with time. If a, b, and c are not zero, then the velocity function must be
→ ^ ^ ^ ^ ^ ^
linear in time. We have →
v (t) = ∫ a dt = ∫ (a i + b j + c k )dt = (a i + b j + c k )t m/s, since taking the derivative of
y = 1 (v 0y + v y)t = 1 v y t, v y = −gt, y = − 1 gt 2, and v 2y = −2gy. (d) We use the kinematic equations to find the
2 2 2
x and y components of the velocity at the point of impact. Using v 2y = −2gy and noting the point of impact is −100.0 m, we
find the y component of the velocity at impact is v y = 44.3 m/s. We are given the x component, v x = 15.0 m/s, so we can
calculate the total velocity at impact: v = 46.8 m/s and θ = 71.3° below the horizontal.
7.4. The golf shot at 30°.
7.5. 134.0 cm/s
7.6. Labeling subscripts for the vector equation, we have B = boat, R = river, and E = Earth. The vector equation becomes
→v BE = → v BR + → v RE. We have right triangle geometry shown in Figure 04_05_BoatRiv_img. Solving for → v BE , we
have
v BE = v 2BR + v 2RE = 4.5 2 + 3.0 2
⎛ ⎞
v BE = 5.4 m/s, θ = tan −1 ⎝3.0 ⎠ = 33.7°.
4.5
CONCEPTUAL QUESTIONS
1. straight line
3. The slope must be zero because the velocity vector is tangent to the graph of the position function.
5. No, motions in perpendicular directions are independent.
7. a. no; b. minimum at apex of trajectory and maximum at launch and impact; c. no, velocity is a vector; d. yes, where it lands
9. They both hit the ground at the same time.
11. yes
13. If he is going to pass the ball to another player, he needs to keep his eyes on the reference frame in which the other players on
the team are located.
388 Answer Key
15.
PROBLEMS
17. → ^ ^ ^
r = 1.0 i − 4.0 j + 6.0 k
19. Δ →
^ ^
r Total = 472.0 m i + 80.3 m j
23. a. → ^ ^ → → ^ ^
v (t) = 8.0t i + 6.0t 2 k , v (0) = 0, v (1.0) = 8.0 i + 6.0 k m/s ,
^ ^
b. →
v avg = 4.0 i + 2.0 k m/s
25. Δ →
^ ^ ^
r 1 = 20.00 m j , Δ →
r 2 = (2.000 × 10 4 m) (cos30° i + sin 30° j )
^ ^
Δ→
r = 1.700 × 10 4 m i + 1.002 × 10 4 m j
^ ^ → ^ ^
27. a. → r (t) = (2.0t 2 i + 3 t 2 j ) m ,
v (t) = (4.0t i + 3.0t j )m/s, 2
2 3 2 2 x
b. x(t) = 2.0t m, y(t) = t m, t = ⇒ y = x
3
2 2 4
29. a. → ^ ^ ^
v (t) = (6.0t i − 21.0t 2 j + 10.0t −3 k )m/s ,
b. → ^ ^ ^
a (t) = (6.0 i − 42.0t j − 30t −4 k )m/s 2 ,
c. → ^ ^ ^
v (2.0s) = (12.0 i − 84.0 j + 1.25 k )m/s ,
^ ^ ^
d. →
v (1.0 s) = 6.0 i − 21.0 j + 10.0 k m/s, | →v (1.0 s)| = 24.0 m/s
v (3.0 s) = 18.0 i − 189.0 j + 0.37 k m/s, | v (3.0 s)| = 199.0 m/s ,
→ ^ ^ ^ →
e. → ^ ^ ^
r (t) = (3.0t 2 i − 7.0t 3 j − 5.0t −2 k )cm
→ ^ ^ ^
v avg = 9.0 i − 49.0 j − 6.3 k m/s
^ ^ ^ ^ ^
v (t) = −sin(1.0t) i + cos(1.0t) j + k , b. →
31. a. → a (t) = −cos(1.0t) i − sin(1.0t) j
33. a. t = 0.55 s , b. x = 110 m
35. a. t = 0.24s, d = 0.28 m , b. They aim high.
37. a., t = 12.8 s, x = 5619 m b. v y = 125.0 m/s, v x = 439.0 m/s, | →v | = 456.0 m/s
39. a. v y = v 0y − gt, t = 10s, v y = 0, v 0y = 98.0 m/s, v 0 = 196.0 m/s , b. h = 490.0 m,
c. v 0x = 169.7 m/s, x = 3394.0 m,
x = 2545.5 m
d. = 465.5 m
y
→ ^ ^
s = 2545.5 m i + 465.5 m j
b. t = 3 s v 0x = 18 m/s x = 54 m ,
c. y = −100 m y 0 = 0 y − y 0 = v 0y t − 1 gt
2
− 100 = 24t − 4.9t 2 ⇒ t = 7.58 s ,
2
d. x = 136.44 m ,
e. t = 2.0 s y = 28.4 m x = 36 m
t = 4.0 s y = 17.6 m x = 22.4 m
t = 6.0 s y = −32.4 m x = 108 m
47. v 0y = 12.9 m/s y − y 0 = v 0y t − 1 gt
2
− 20.0 = 12.9t − 4.9t 2
2
t = 3.7 s v 0x = 15.3 m/s ⇒ x = 56.7 m
So the golfer’s shot lands 13.3 m short of the green.
49. a. R = 60.8 m ,
390 Answer Key
b. R = 137.8 m
2 2
51. a. v y = v 0y − 2gy ⇒ y = 2.9 m/s
y = 3.3 m/s
v 20y (v 0 sinθ) 2
y= = ⇒ sinθ = 0.91 ⇒ θ = 65.5°
2g 2g
53. R = 18.5 m
⎡ g ⎤
55. y = (tanθ 0)x − ⎢ ⎥x 2 ⇒ v 0 = 16.4 m/s
⎣2(v 0 cosθ 0) ⎦
2
2
v sin 2θ 0
57. R = 0
g ⇒ θ 0 = 15.0°
59. It takes the wide receiver 1.1 s to cover the last 10 m of his run.
2(v 0 sinθ)
T tof = g ⇒ sinθ = 0.27 ⇒ θ = 15.6°
61. a C = 40 m/s 2
2
63. a C = v ⇒ v 2 = r a C = 78.4, v = 8.85 m/s
r
T = 5.68 s, which is 0.176 rev/s = 10.6 rev/min
65. Venus is 108.2 million km from the Sun and has an orbital period of 0.6152 y.
r = 1.082 × 10 11 m T = 1.94 × 10 7 s
v = 3.5 × 10 4 m/s, a C = 1.135 × 10 −2 m/s 2
67. 360 rev/min = 6 rev/s
v = 3.8 m/s a C = 144. m/s 2
t = 6.0 min
Downstream = 0.3 km
→
v = →
→
v
77. AG v CGAC +
|→
|
v AC = 25 km/h → |
v CG = 15 km/h → | |
v AG = 29.15 km/h → | v AG = → v AC + → v CG
→ →
The angle between v AC and v AG is 31°, so the direction of the wind is 14° north of east.
ADDITIONAL PROBLEMS
79. a C = 39.6 m/s 2
81. 90.0 km/h = 25.0 m/s, 9.0 km/h = 2.5 m/s, 60.0 km/h = 16.7 m/s
2 2 2
a T = −2.5 m/s , a C = 1.86 m/s , a = 3.1 m/s
4π 2 R E cos λ
83. The radius of the circle of revolution at latitude λ is R E cos λ. The velocity of the body is 2πr . a C = for
T 2
T
392 Answer Key
λ = 40°, a C = 0.26% g
85. a T = 3.00 m/s 2
v(5 s) = 15.00 m/s a C = 150.00 m/s 2 θ = 88.8° with respect to the tangent to the circle of revolution directed inward.
| |
→a = 150.03 m/s 2
87. → ^ ^
a (t) = −Aω 2 cos ωt i − Aω 2 sin ωt j
a C = 5.0 mω 2 ω = 0.89 rad/s
→ ^ ^
v (t) = −2.24 m/s i − 3.87 m/s j
89. →
^ ^ ^ ^ ^
r 1 = 1.5 j + 4.0 k →
r 2 =Δ→
r + →
r 1 = 2.5 i + 4.7 j + 2.8 k
91. v x(t) = 265.0 m/s
v y(t) = 20.0 m/s
→ ^ ^
v (5.0 s) = (265.0 i + 20.0 j )m/s
93. R = 1.07 m
95. v 0 = 20.1 m/s
97. v = 3072.5 m/s
a C = 0.223 m/s 2
CHALLENGE PROBLEMS
v 2
99. a. −400.0 m = v 0y t − 4.9t 2 359.0 m = v 0x t t = 359.0 − 400.0 = 359.0 0y − 4.9( 359.0 )
v 0x v 0x v 0x
CHAPTER 8
CHECK YOUR UNDERSTANDING
8.1. The semi-major axis for the highly elliptical orbit of Halley’s comet is 17.8 AU and is the average of the perihelion and
aphelion. This lies between the 9.5 AU and 19 AU orbital radii for Saturn and Uranus, respectively. The radius for a circular orbit
is the same as the semi-major axis, and since the period increases with an increase of the semi-major axis, the fact that Halley’s
period is between the periods of Saturn and Uranus is expected.
CONCEPTUAL QUESTIONS
31. The speed is greatest where the satellite is closest to the large mass and least where farther away—at the periapsis and apoapsis,
respectively. It is conservation of angular momentum that governs this relationship. But it can also be gleaned from conservation
of energy, the kinetic energy must be greatest where the gravitational potential energy is the least (most negative). The force, and
hence acceleration, is always directed towards M in the diagram, and the velocity is always tangent to the path at all points. The
acceleration vector has a tangential component along the direction of the velocity at the upper location on the y-axis; hence, the
satellite is speeding up. Just the opposite is true at the lower position.
PROBLEMS
37. 1.98 × 10 30 kg ; The values are the same within 0.05%.
39. Compare m58348 (https://legacy.cnx.org/content/m58348/latest/#fs-id1168327874347) and ??? to see that they
differ only in that the circular radius, r, is replaced by the semi-major axis, a. Therefore, the mean radius is one-half the sum of the
aphelion and perihelion, the same as the semi-major axis.
41. The semi-major axis, 3.78 AU is found from the equation for the period. This is one-half the sum of the aphelion and
perihelion, giving an aphelion distance of 4.95 AU.
43. 1.75 years
394 Answer Key
INDEX
A differentiation, 288 International Astronomical
acceleration due to gravity, 91, dimension, 25, 39 Union, 272
97 dimensionally consistent, 26, 39
dimensionless, 26, 39 J
acceleration vector, 222, 251
direction angle, 186, 204 Jupiter, 265, 267, 268, 273, 274,
accuracy, 32, 39
discrepancy, 33, 39 276, 276
air resistance, 91
Albert Einstein, 126 displacement, 64, 97, 171, 204 K
Andromeda galaxy, 56, 57 displacement vector, 215, 251 Kepler, 281, 285
angular acceleration, 111, 121 distance traveled, 66, 97 Kepler’s first law, 282, 288
angular frequency, 241, 251 distributive, 176, 204 Kepler’s second law, 283, 288
angular position, 109, 121 Kepler’s third law, 285, 288
E
angular velocity, 110, 121 kilogram, 20, 39
Earth, 51, 268, 274
antiparallel, 172 kinematics, 63, 97
eccentricity, 288
antiparallel vectors, 204 kinematics of rotational motion,
elapsed time, 66, 97
aphelion, 283, 288 114, 121
ellipse, 282, 288
associative, 176, 204
English units, 18, 39 L
asteroid, 288
equal vectors, 195, 204 Large Magellanic Cloud, 56
asteroids, 269
Eris, 265, 266 law, 17, 39
astronomical unit (AU), 285, 288
estimation, 29, 39 Length contraction, 143
average acceleration, 73, 97
event, 147, 161 length contraction, 161
average speed, 71, 97
average velocity, 66, 97 F light cone, 152
Fermi, 16 Local Group, 58
B Lorentz factor, 134
first postulate, 128
base quantities, 19 Lorentz transformation, 149,
first postulate of special
base quantity, 39 161
relativity, 161
base unit, 19, 39
focus, 288 M
Brahe, 280
frame of reference, 64 magnitude, 171, 204
Brownian motion, 217
free fall, 91, 97 major axis, 288
C Makemake, 266
G
centripetal acceleration, 239, Mars, 10, 263, 268, 274
Galilean relativity, 127, 161
251 Mars Climate Orbiter, 25
Galilean transformation, 148,
Ceres, 108, 265, 266 Mercury, 266, 267, 274, 275
161
classical (Galilean) velocity meteor, 270, 288
Galileo, 90, 280
addition, 158, 161 meteorite, 270, 288
Ganymede, 275
comet, 288 meter, 20, 39
giant planet, 288
Comet Shoemaker–Levy 9, 276, method of adding percents, 39
giant planets, 267
276 metric system, 21, 39
comets, 270 H Michelson-Morley experiment,
commutative, 176, 204 Halley, 286 128, 161
component form of a vector, 204 Halley’s comet, 286 Milky Way, 12
conversion factor, 23, 39 Haumea, 266 Milky Way Galaxy, 52, 54
conversion factors, 23 model, 16, 39
coordinate system, 183 I Moon, 51, 264, 266, 274
Crab Nebula, 11 inertial frame of reference, 127, muons, 135
Curie, 16 161
instantaneous acceleration, 75, N
D 97 Neptune, 267, 268, 274
density, 268 instantaneous angular null vector, 195, 204
derived quantity, 19, 39 acceleration, 111, 121
derived units, 19, 39 instantaneous angular velocity, O
difference of two vectors, 176, 110, 121 orbit, 281, 288
204 instantaneous speed, 72, 97 orbital period, 285
Differentiation, 274 instantaneous velocity, 69, 97 orbital period (P), 288
396 Index
ATTRIBUTIONS
Collection: The Cosmic Universe
Edited by: Steven Mellema
URL: https://legacy.cnx.org/content/col12169/1.46/
Copyright: Steven Mellema
License: http://creativecommons.org/licenses/by/4.0/
License: http://creativecommons.org/licenses/by/4.0/
Based on: Free Fall <http://legacy.cnx.org/content/m58286/1.4> by OpenStax.
Module: Introduction
By: OpenStax
URL: https://legacy.cnx.org/content/m58276/1.4/
Copyright: Rice University
License: http://creativecommons.org/licenses/by/4.0/
URL: https://legacy.cnx.org/content/m64017/1.1/
Copyright: Steven Mellema
License: http://creativecommons.org/licenses/by/4.0/
Based on: Mathematical Formulas <http://legacy.cnx.org/content/m58618/1.5> by OpenStax.
Module: Chemistry
Used here as: Periodic Table of the Elements
By: OpenStax
URL: https://legacy.cnx.org/content/m58619/1.5/
Copyright: Rice University
License: http://creativecommons.org/licenses/by/4.0/
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