CHAPTER 7

Gravitation
Gravity is an attractive field force that acts between objects with mass.

SECTIONS
1 Planetary Motion and
Gravitation
2 Using the Law of Universal
Gravitation

LaunchLAB iLab Station

MODEL MERCURY’S MOTION

How can measurements of angles and distances
be used to draw a model of an orbit?

WATCH THIS! Video

PLANETARY MOTION
What can looking through a telescope tell you about
gravitation? Explore the physics of amateur astronomy
and find out!
(l)PhotoLink/Getty Images, (r)NASA

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 Go online!
connec
connectED.mcgraw-hill.com

Chapter 7 • Gravitation 177

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 SECTION 1 Planetary Motion and Gravitation
PHYSICS Our solar system includes the Sun, Earth and seven other
major planets, dwarf planets, and interplanetary dust and

4 YOU gas. Various moons orbit the planets. What holds all this
together?

Early Observations
In ancient times, the Sun, the Moon, the planets, and the stars were
assumed to revolve around Earth. Nicholas Copernicus, a Polish astrono-
MAIN IDEA mer, noticed that the best available observations of the movements
The gravitational force between two of planets did not fully agree with the Earth-centered model.
objects is proportional to the product of The results of his many years of work were published in 1543, when
their masses divided by the square of Copernicus was on his deathbed. His book showed that the motion of
the distance between them.
planets is much more easily understood by assuming that Earth and
other planets revolve around the Sun. His model helped explain phenomena
Essential Questions such as the inner planets Mercury and Venus always appearing near the
• What is the relationship between a Sun. Copernicus's view advanced our understanding of planetary
planet’s orbital radius and period? motion. He incorrectly assumed, however, that planetary orbits are cir-
• What is Newton’s law of universal cular. This assumption did not fit well with observations, and modifica-
gravitation, and how does it relate to tion of Copernicus's model was necessary to make it accurate.
Kepler’s laws? Tycho Brahe was born a few years after Copernicus died. As a boy
• Why was Cavendish’s investigation of 14 in Denmark, Tycho observed an eclipse of the Sun on August 21,
important? 1560. The fact that it had been predicted inspired him toward a career
in astronomy.
Review Vocabulary As Tycho studied astronomy, he realized that the charts of the time
Newton's third law states all forces did not accurately predict astronomical events. Tycho recognized that
come in pairs and that the two forces in measurements were required from one location over a long period of
a pair act on different objects, are equal time. He was granted an estate on the Danish island of Hven and the
in strength, and are opposite in direction
funding to build an early research institute. Telescopes had not been
New Vocabulary invented, so to make measurements, Tycho used huge instruments that
he designed and built in his own shop, such as those shown in Figure 1.
Kepler’s first law
Tycho is credited with the most accurate measurements of the time.
Kepler’s second law
Kepler’s third law Figure 1 Instruments such as these were used by Tycho to measure the positions of planets.
gravitational force
law of universal gravitation

(t)StockTrek/Getty Images, (bl br)Newberry Library/SuperStock

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 Figure 2 The orbit of each planet is an
Foci ellipse, with the Sun at one focus.

Planet
Sun

(Not to scale)

Ellipse

Kepler’s Laws View an animation of Kepler's first law.

In 1600 Tycho moved to Prague where Johannes Kepler, a 29-year-old Concepts In Motion
German, became one of his assistants. Kepler analyzed Tycho’s observa-
tions. After Tycho’s death in 1601, Kepler continued to study Tycho’s data
and used geometry and mathematics to explain the motion of the planets.
After seven years of careful analysis of Tycho’s data on Mars, Kepler discov-
ered the laws that describe the motion of every planet and satellite, natural
or artificial. Here, the laws are presented in terms of planets.
Kepler’s first law states that the paths of the planets are ellipses,
with the Sun at one focus. An ellipse has two foci, as shown in Figure 2.
Although exaggerated ellipses are used in the diagrams, Earth’s actual
orbit is very nearly circular. You would not be able to distinguish it from
a circle visually.
Kepler found that the planets move faster when they are closer to the
Sun and slower when they are farther away from the Sun. Kepler’s second
law states that an imaginary line from the Sun to a planet sweeps out
equal areas in equal time intervals, as illustrated in Figure 3.
READING CHECK Compare the distances traveled from point 1 to point 2 and
from point 6 to point 7 in Figure 3. Through which distance would Earth be
traveling fastest?

A period is the time it takes for one revolution of an orbiting body.

Kepler also discovered a mathematical relationship between periods of
planets and their mean distances away from the Sun.
11
12 10

9
Planet

Figure 3 Kepler found that elliptical orbits

Sun
sweep out equal areas in equal time periods.
1 7
Explain why the equal time areas are
shaped differently.
6
View an animation and a simulation of
5
Kepler’s second law.
2
(Not to scale) Concepts In Motion
3 4

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 Table 1 Solar System Data
Name Average Radius (m) Mass (kg) Average Distance from the Sun (m)
Sun 6.96×10 8 1.99×10 30 —

Mercury 2.44×10 6 3.30×10 23 5.79×10 10

Venus 6.05×10 6 4.87×10 24 1.08×10 11

Earth 6.38×10 6 5.97×10 24 1.50×10 11

Mars 3.40×10 6 6.42×10 23 2.28×10 11

Jupiter 7.15×10 7 1.90×10 27 7.78×10 11

Saturn 6.03×10 7 5.69×10 26 1.43×10 12

Uranus 2.56×10 7 8.68×10 25 2.87×10 12

Neptune 2.48×10 7 1.02×10 26 4.50×10 12

Kepler’s third law states that the square of the ratio of the periods of
any two planets revolving about the Sun is equal to the cube of the ratio
of their average distances from the Sun. Thus, if the periods of the
planets are T A and T B and their average distances from the Sun are r A and
r B, Kepler’s third law can be expressed as follows.
View an animation and a simulation of
Kepler’s third law. KEPLER’S THIRD LAW
Concepts In Motion The square of the ratio of the period of planet A to the period of planet B is
equal to the cube of the ratio of the distance between the centers of planet
A and the Sun to the distance between the centers of planet B and the Sun.

2
() ()
TA
_
TB
rA
= _
r B
3

Note that Kepler's first two laws apply to each planet, moon, and sat-
ellite individually. The third law, however, relates the motion of two
objects around a single body. For example, it can be used to compare the
planets’ distances from the Sun, shown in Table 1, to their periods around
the Sun. It also can be used to compare distances and periods of the
PhysicsLAB Moon and artificial satellites orbiting Earth.
MODELING ORBITS Comet periods Comets are classified as long-period comets or short-
What is the shape of the orbits of period comets based on orbital periods. Long-period comets have orbital
planets and satellites in the solar periods longer than 200 years and short-period comets have orbital peri-
system? ods shorter than 200 years. Comet Hale-Bopp, shown in Figure 4, with a
iLab Station
period of approximately 2400 years, is an example of a long-period
comet. Comet Halley, with a period of 76 years, is an example of a short-
period comet. Comets also obey Kepler's laws. Unlike planets, however,
comets have highly elliptical orbits.
Jamie Cooper/SSPL/Getty Images

Figure 4 Hale-Bopp is a long-period comet,

with a period of 2400 years. This photo
was taken in 1997, when Hale-Bopp was
highly visible.

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 EXAMPLE PROBLEM 1 Find help with isolating a variable. Math Handbook

EXAMPLE PR
CALLISTO’S DISTANCE FROM JUPITER Galileo measured the orbital radii of Jupiter’s moons
using the diameter of Jupiter as a unit of measure. He found that Io, the closest moon to
Jupiter, has a period of 1.8 days and is 4.2 units from the center of Jupiter. Callisto, the
fourth moon from Jupiter, has a period of 16.7 days. Using the same units that Galileo
used, predict Callisto’s distance from Jupiter.

1 ANALYZE AND SKETCH THE PROBLEM

• Sketch the orbits of Io and Callisto.

ROBLEM
• Label the radii.
Jupiter
KNOWN UNKNOWN o
T C = 16.7 days rC = ? r
T I = 1.8 days
r I = 4.2 units Callisto
rC
2 SOLVE FOR CALLISTO’S DISTANCE FROM JUPITER
Solve Kepler’s third law for r C.
2
(_) = (_)
TC
TI
rC
rI
3

2
r = r (_)
3 3 TC
C I
TI
 2
r = r (_)
√ 3 TC
▼

3
Substitute r = 4.2 units, T = 16.7 days, T = 1.8 days
I C I
C I
TI

 2
= (4.2 units) (_)
√ 3
3 16.7 days
1.8 days
3
= √
6.4×10 3 units 3
= 19 units

3 EVALUATE THE ANSWER

• Are the units correct? r C should be in Galileo’s units, like r I.
• Is the magnitude realistic? The period is larger, so the radius should be larger.
PRACTICE PROBLEMS Do additional problems. Online Practice

PRACTICE PROBLEMS
1. If Ganymede, one of Jupiter’s moons, has a period of 32 days, how many units is its
orbital radius? Use the information given in Example Problem 1.
2. An asteroid revolves around the Sun with a mean orbital radius twice that of Earth’s.
Predict the period of the asteroid in Earth years.
3. Venus has a period of revolution of 225 Earth days. Find the distance between the
Sun and Venus as a multiple of Earth’s average distance from the Sun.
4. Uranus requires 84 years to circle the Sun. Find Uranus’s average distance from the
Sun as a multiple of Earth’s average distance from the Sun.
5. From Table 1 you can find that, on average, Mars is 1.52 times as far from the Sun as
Earth is. Predict the time required for Mars to orbit the Sun in Earth days.
6. The Moon has a period of 27.3 days and a mean distance of 3.9×10 5 km from its
center to the center of Earth.
a. Use Kepler’s laws to find the period of a satellite in orbit 6.70×10 3 km from the
center of Earth.
b. How far above Earth’s surface is this satellite?
7. CHALLENGE Using the data in the previous problem for the period and radius of
revolution of the Moon, predict what the mean distance from Earth's center would be
for an artificial satellite that has a period of exactly 1.00 day.

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 Newton’s Law of Universal Gravitation
In 1666, Isaac Newton began his studies of planetary motion. It has
been said that seeing an apple fall made Newton wonder if the force that
caused the apple to fall might extend to the Moon, or even beyond. He
found that the magnitude of the force (F g) on a planet due to the Sun var-
ies inversely with the square of the distance (r) between the centers of
the planet and the Sun. That is, F g is proportional to _
1 . The force (F )
g
r2
acts in the direction of the line connecting the centers of the two objects,
Gravitation between objects depends as shown in Figure 5.
on the product of the masses of the Newton found that both the apple's and the Moon’s accelerations
objects. agree with the _1 relationship. According to his own third law, the force
r2
M627-02C-MSS02_A Earth exerts on the apple is exactly the same as the force the apple exerts
on Earth. Even though these forces are exactly the same, you can easily
observe the effect of the force on the apple because it has much lower
mass than Earth. The force of attraction between two objects must be
proportional to the objects’ masses and is known as the gravitational force.
Newton was confident that the same force of attraction would act
between any two objects anywhere in the universe. He proposed the
law of universal gravitation, which states that objects attract other
objects with a force that is proportional to the product of their masses
The gravitational force between objects
and inversely proportional to the square of the distance between them
is inversely proportional to the square
of the distance between the objects. as shown below.

Figure 5 Mass and distance affect the LAW OF UNIVERSAL GRAVITATION

magnitude of the gravitational force The gravitational force is equal to the universal gravitational constant,
between objects. times the mass of object 1, times the mass of object 2, divided by the
distance between the centers of the objects, squared.
View an animation of the law of universal
Gm m
gravitation.
Fg = _
1 2
2
Concepts In Motion r
According to Newton’s equation, F is directly proportional to m 1 and
m 2. If the mass of a planet near the Sun doubles, the force of attraction
doubles. Use the Connecting Math to Physics feature below to examine
how changing one variable affects another. Figure 6 illustrates the inverse
square relationship graphically. The term G is the universal gravitational
constant and will be discussed in the next sections.
Figure 6 This is a graphical representation
of the inverse square relationship.

Inverse Square Law

CONNECTING MATH TO PHYSICS
Direct and Inverse Relationships Newton’s law of universal gravitation has
80 both direct and inverse relationships.

F g ∝ m1m2 Fg ∝
_1
60
r2
Force (N)

Change Result Change Result

40
(2m1)m2 2F g 2r _1 F g
4
(3m1)m2 3F g 3r _1 F g
20 9
(2m1)(3m2) 6F g _1 r 4F g
2
0
0.5
Distance (m)
(_12 )m m
1 2
_1 F g
2
_1 r
3
9F g

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 Universal Gravitation and Kepler’s Third Law
Newton stated the law of universal gravitation in terms that applied
to the motion of planets about the Sun. This agreed with Kepler’s third Sun
law and confirmed that Newton’s law fit the best observations of the day.
Consider a planet orbiting the Sun, as shown in Figure 7. Newton’s
second law of motion, F net = ma, can be written as F net = m pa c, where
F net is the magnitude of the gravitational force, m p is the mass of the r
planet, and a c is the centripetal acceleration of the planet. For simplicity,
assume circular orbits. Recall from your study of uniform circular
4π 2r mp
motion that for a circular orbit a c2 = _ . This means that F net = m pa c Planet
m p4π r T2 (Not to scale)
may now be written F net = _ . In this equation, T is the time in sec-
T2
onds required for the planet to make one complete revolution about the
Sun. If you set the right side of this equation equal to the right side of Figure 7 A planet with mass m p and average
the law of universal gravitation, you arrive at the following result: distance from the Sun r orbits the Sun. The
mass of the Sun is m S.
G m Sm p
_ m p4π 2r
=_
r2 T2

( )
T2 = _
4π 2 3
Gm S
r


T = (_)r
√ 4π 2
Gm S
3

The period of a planet orbiting the Sun can be expressed as follows.

PERIOD OF A PLANET ORBITING THE SUN

The period of a planet orbiting the Sun is equal to 2π times the square
root of the average distance from the Sun cubed, divided by the product
of the universal gravitational constant and the mass of the Sun.


r3
T = 2π _ √Gm S

Squaring both sides makes it apparent that this equation is Kepler’s

third law of planetary motion: the square of the period is proportional
to the cube of the distance that separates the masses. The factor _
4π 2

Gm S
depends on the mass of the Sun and the universal gravitational constant.
Newton found that this factor applied to elliptical orbits as well.

PHYSICS CHALLENGE
Astronomers have detected three planets that orbit the star Upsilon D
B
Andromedae. Planet B has an average orbital radius of 0.0595 AU and
a period of 4.6171 days. Planet C has an average orbital radius of rD
0.832 AU and a period of 241.33 days. Planet D has an average orbital rB
C rC
radius of 2.53 AU and a period of 1278.1 days. (Distances are given in
astronomical units (AU)—Earth’s average distance from the Sun. The Upsilon
distance from Earth to the Sun is 1.00 AU.) Andromedae
1. Do these planets obey Kepler’s third law?
2. Find the mass of the star Upsilon Andromedae in units of the Sun’s
mass. Hint: compare _
r 3
for these planets with that of Earth in the
2 T (Not to scale)
same units (AU and days).

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 Measuring the Universal Gravitational Constant
How large is the constant G? As you know, the force of gravitational
attraction between two objects on Earth is relatively small. The slightest
attraction, even between two massive bowling balls, is almost impossible
to detect. In fact, it took 100 years from the time of Newton’s work for
scientists to develop an apparatus that was sensitive enough to measure
the force of gravitational attraction.
Cavendish’s apparatus In 1798 English scientist Henry Cavendish
used equipment similar to the apparatus shown in Figure 8 to measure
the gravitational force between two objects. The apparatus has a horizontal
rod with small lead spheres attached to each end. The rod is suspended
at its midpoint so that it can rotate. Because the rod is suspended by a
thin wire, the rod and spheres are very sensitive to horizontal forces.
To measure G, two large spheres are placed in a fixed position close
to each of the two small spheres, as shown in Figure 8. The force of attrac-
tion between the large and small spheres causes the rod to rotate. When
Investigate universal gravitation.
the force required to twist the wire equals the gravitational force
Virtual Investigation between the spheres, the rod stops rotating. By measuring the angle
through which the rod turns, the attractive force between the objects can
be calculated.
READING CHECK Explain why the rod and sphere in Cavendish’s apparatus
must be sensitive to horizontal forces.

The angle through which the rod turns is measured by using a beam
of light that is reflected from the mirror. The distances between the
sphere’s centers and the force can both be measured. The masses of the
spheres are known. By substituting the values for force, mass, and distance
into Newton’s law of universal gravitation, an experimental value for G
is found: when m 1 and m 2 are measured in kilograms, r in meters, and F
in newtons, G = 6.67×10 −11 N·m 2/kg 2.

n Cavendish Balance

Large sphere
Thin wire

Small sphere

Mirror

Light source
Pivot

Figure 8 A Cavendish balance uses a light Direction

source and a mirror to measure the movement of rotation
of the spheres.

View an animation of Cavendish’s Scale

investigation.
Concepts In Motion

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 The importance of G Cavendish’s investigation often is called
“weighing Earth” because it helped determine Earth’s mass. Once the
value of G is known, not only the mass of Earth, but also the mass of the
Sun can be determined. In addition, the gravitational force between any
two objects can be calculated by using Newton’s law of universal gravita-
tion. For example, the attractive gravitational force (F g) between two
bowling balls of mass 7.26 kg, with their centers separated by 0.30 m,
can be calculated as follows:
(6.67×10 -11 N·m 2/kg 2)(7.26 kg)(7.26 kg)
F g = ____
2
= 3.9×10 -8 N
(0.30 m)

You know that on Earth’s surface, the weight of an object of mass m is

a measure of Earth’s gravitational attraction: F g = mg. If Earth’s mass is
represented by m E and Earth’s radius is represented by r E, the following
is true:
Gm m Gm
Fg = _E
2
= mg, and so g = _2
E
rE rE

This equation can be rearranged to solve for m E.

gr 2
mE = _E
G

Using g = 9.8 N/kg, r E = 6.38×10 6 m, and G = 6.67×10 -11 N·m2/kg2,

the following result is obtained for Earth’s mass:
(9.8 N/kg)(6.38×106 m) 2
m E = ___
-11 2 2
= 5.98×10 24 kg
6.67×10 N·m /kg
Figure 9 Cavendish’s
When you compare the mass of Earth to that of a bowling ball, you investigations helped calculate
can see why the gravitational attraction between everyday objects is not the mass of Earth.
easily observed. Cavendish’s investigation determined the value of G,
confirmed Newton’s prediction that a gravitational force exists between
any two objects, and helped calculate the mass of Earth (Figure 9).

SECTION 1 REVIEW Section Self-Check Check your understanding.

8. MAI
MAINN IDEA What is the gravitational force between 10. Gravity If Earth began to shrink, but its mass
two 15-kg balls whose centers are 35 m apart? What remained the same, what would happen to the value
fraction is this of the weight of one ball? of g on Earth’s surface?
9. Neptune’s Orbital Period Neptune orbits the Sun at an
11. Universal Gravitational Constant Cavendish did his
average distance given in Figure 10, which allows
investigation using lead spheres. Suppose he had
gases, such as methane, to condense and form an
replaced the lead spheres with copper spheres of
atmosphere. If the mass of the Sun is 1.99×10 30 kg,
equal mass. Would his value of G be the same or
calculate the period of Neptune’s orbit.
NASA/Goddard Space Flight Center Scientific Visualization Studio

different? Explain.

12. Kepler’s three statements and Newton’s equation for

gravitational attraction are called laws. Were they ever
theories? Will they ever become theories?
Sun
13. Critical Thinking Picking up a rock requires less effort
on the Moon than on Earth.
a. How will the Moon’s gravitational force affect the
(Not to r = 4.496×1012 m path of the rock if it is thrown horizontally?
scale)
b. If the thrower accidentally drops the rock on her
toe, will it hurt more or less than it would on Earth?
Figure 10 Neptune Explain.

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 SECTION 2 Using the Law of Universal Gravitation
PHYSICS Have you ever used a device to locate your position or to
map where you want to go? Where does that device get

4 YOU information? The Global Positioning System (GPS) consists of

many satellites circling Earth. GPS satellites give accurate
position data anywhere on or near Earth.

Orbits of Planets and Satellites

The planet Uranus was discovered in 1781. By 1830 it was clear that
the law of gravitation didn’t correctly predict its orbit. Two astronomers
proposed that Uranus was being attracted by the Sun and by an undis-
covered planet. They calculated the orbit of such a planet in 1845, and,
MAIN IDEA one year later, astronomers at the Berlin Observatory found the planet
All objects are surrounded by a gravitational now called Neptune. How is it possible for planets, such as Neptune and
field that affects the motions of other Uranus, to remain in orbit around the Sun?
objects. Newton used a drawing similar to the one shown in Figure 11 to
illustrate a thought experiment on the motion of satellites. Imagine a
Essential Questions cannon, perched high atop a mountain, firing a cannonball horizontally
• How can you describe orbital motion? with a given horizontal speed. The cannonball is a projectile, and its
• How are gravitational mass and inertial motion has both vertical and horizontal components. Like all projectiles
mass alike, and how are they on Earth, it would follow a parabolic trajectory and fall back to the ground.
different? If the cannonball’s horizontal speed were increased, it would travel
• How is gravitational force explained, farther across the surface of Earth and still fall back to the ground. If an
and what did Einstein propose about extremely powerful cannon were used, however, the cannonball would
gravitational force? travel all the way around Earth and keep going. It would fall toward Earth
at the same rate that Earth’s surface curves away. In other words, the
Review Vocabulary curvature of the projectile would continue to just match the curvature of
centripetal acceleration the center- Earth so that the cannonball would never get any closer to or farther away
seeking acceleration of an object moving from Earth’s curved surface. The cannonball would, therefore, be in orbit.
in a circle at a constant speed
Figure 11 Newton imagined a projectile launched parallel to Earth. If it has enough speed
New Vocabulary it will fall toward Earth with a curvature that matches the curvature of Earth's surface.
inertial mass Identify the factor that is not considered in this example.
gravitational mass

Investigate Newton’s cannon.

Virtual Investigation

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 Newton’s thought experiment ignored air resistance. For the
cannonball to be free of air resistance, the mountain on which the
cannon is perched would have to be more than 150 km above Earth’s
surface. By way of comparison, the mountain would have to be much
taller than the peak of Mount Everest, the world’s tallest mountain,
which is only 8.85 km in height. A cannonball launched from a mountain
that is 150 km above Earth’s surface would encounter little or no air
resistance at an altitude of 150 km because the mountain would be
above most of the atmosphere. Thus, a cannonball or any object or
satellite at or above this altitude could orbit Earth.
A satellite’s speed A satellite in an orbit that is always the same
height above Earth moves in uniform circular motion. Recall that its
v2
centripetal acceleration is given by a c = _r . Newton’s second law,
2
F net = ma c, can thus be rewritten F net = _
mv
r . If Earth’s mass is m E, then
this expression combined with Newton’s law of universal gravitation
produces the following equation:
Gm Em
_ mv 2
=_ r
r2
Solving for the speed of a satellite in circular orbit about Earth (v)
yields the following.

SPEED OF A SATELLITE ORBITING EARTH

The speed of a satellite orbiting Earth is equal to the square root of the universal gravitational
constant times the mass of Earth, divided by the radius of the orbit.

v= √
Gm
_
r
E
REAL-WORLD
A satellite’s orbital period A satellite’s orbit around Earth is similar
to a planet’s orbit about the Sun. Recall that the period of a planet orbiting
PHYSICS
the Sun is expressed by the following equation: GEOSYNCHRONOUS ORBIT The
 GOES weather satellites orbit Earth
r3
T = 2π _ √ Gm once a day at an altitude of 35,785 km.
S
The orbital speed of the satellite
Thus, the period for a satellite orbiting Earth is given by the following matches Earth’s rate of rotation. Thus,
equation. to an observer on Earth, the satellite
appears to remain above one spot.
PERIOD OF A SATELLITE ORBITING EARTH Dish antennas on Earth can be direct-
The period for a satellite orbiting Earth is equal to 2π times the square ed to one point in the sky and remain
root of the radius of the orbit cubed, divided by the product of the in a fixed position as the satellite
universal gravitational constant and the mass of Earth. orbits.


r3
T = 2π _ √Gm E

The equations for the speed and period of a satellite can be used for
any object in orbit about another. The mass of the central body will
replace m E in the equations, and r will be the distance between the
centers of the orbiting body and the central body. Orbital speed (v) and
StockTrek/Getty Images

period (T) are independent of the mass of the satellite.

READING CHECK Describe how the mass of a satellite affects that satellite's
orbital speed and period.

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 EXAMPLE PROBLEM 2 Get help with orbital speed. Personal Tutor
ROBLEM
ORBITAL SPEED AND PERIOD Assume that a satellite orbits Earth 225 km above
its surface. Given that the mass of Earth is 5.97×10 24 kg and the radius of
Earth is 6.38×10 6 m, what are the satellite’s orbital speed and period?

1 ANALYZE AND SKETCH THE PROBLEM

EXAMPLE PR

Sketch the situation showing the height of the satellite’s orbit. Satellite
KNOWN UNKNOWN rE
h = 2.25×10 5 m v=?
h
r E = 6.38×10 6 m T=?
m E = 5.97×10 24 kg
G = 6.67×10 −11 N·m 2/kg 2 Earth r

2 SOLVE FOR ORBITAL SPEED AND PERIOD

Determine the orbital radius by adding the height of the satellite’s orbit to Earth’s radius.
r = h + rE
= 2.25×10 5 m + 6.38×10 6 m = 6.60×10 6 m

▼
Substitute h = 2.25×10 5 m and r E = 6.38×10 6 m.

Solve for the speed.

√
Gm
_ E
Substitute G = 6.67×10 −11 N•m 2/kg 2,
▼
v= r
m E = 5.97×10 24 kg, and r = 6.60×10 6 m.

(6.67×10 -11 N·m 2/kg 2)(5.97×10 24 kg)
= √____ 6.60×10 6 m

= 7.77×10 3 m/s

Solve for the period.

r3
T = 2π _
√Gm Substitute r = 6.60×10 6 m, G = 6.67×10 -11 N•m 2/kg 2, and
▼

E
m E = 5.97×10 24 kg.

(6.60×10 6 m) 3
√
= 2π ____
-11 2
(6.67×10 2 24
N·m /kg )(5.97×10 kg)

= 5.34×10 3 s

This is approximately 89 min, or 1.5 h.

3 EVALUATE THE ANSWER

Are the units correct? The unit for speed is meters per second, and the unit for period is seconds.

PRACTICE PROBLEMS Do additional problems. Online Practice

PRACTICE PROBLEMS
S

For the following problems, assume a circular orbit for all 16. Use Newton’s thought experiment on the motion of
calculations. satellites to solve the following.
14. Suppose that the satellite in Example Problem 2 is a. Calculate the speed that a satellite shot from a
moved to an orbit that is 24 km larger in radius than cannon must have to orbit Earth 150 km above its
its previous orbit. surface.
a. What is its speed? b. How long, in seconds and minutes, would it take
b. Is this faster or slower than its previous speed? for the satellite to complete one orbit and return to
the cannon?
c. Why do you think this is so?
17. CHALLENGE Use the data for Mercury in Table 1 to
15. Uranus has 27 known moons. One of these moons is find the following.
Miranda, which orbits at a radius of 1.29×10 8 m.
Uranus has a mass of 8.68×10 25 kg. Find the orbital a. the speed of a satellite that is in orbit 260 km
speed of Miranda. How many Earth days does it take above Mercury's surface
Miranda to complete one orbit? b. the period of the satellite

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 CONNECTION TO EARTH SCIENCE Landsat 7, shown in Figure 12, is
an artificial satellite that provides images of Earth’s continental surfaces.
Landsat 7 images have been used to create maps, study land use, and
monitor resources and global changes. The Landsat 7 system enables
researchers to monitor small-scale processes, such as deforestation, on a
global scale. Satellites, such as Landsat 7, are accelerated to the speeds
necessary for them to achieve orbit by large rockets, such as shuttle-
booster rockets. Because the acceleration of any mass must follow
Newton’s second law of motion, F net = ma, more force is required to
launch a more massive satellite into orbit. Thus, the mass of a satellite is
limited by the capability of the rocket used to launch it.

Free-Fall Acceleration
The acceleration of objects due to Earth’s gravity can be found by
using Newton’s law of universal gravitation and his second law of
motion. For a free-falling object of mass m, the following is true:
Gm m Gm
F=_E
2
= ma, so a = _2
E
Figure 12 Landsat 7 is capable of providing
r r
up to 532 images of Earth per day.
If you set a = gE and r = r E on Earth’s surface, the following equation
can be written:
Gm gr E 2
g=_2
E
, thus, m E = _
rE G
Gm
You saw above that a = _2
E
for a free-falling object. Substitution
r
of the above expression for m E yields the following:

( )
a=_
gr E
G _
G
2

r2 Figure 13 Astronauts in orbit around Earth

are in free fall because their spacecraft and
rE 2
a=g _( )
r
everything in it is accelerating toward Earth
along with the astronauts. That is, the floor is
constantly falling from beneath their feet.
On the surface of Earth, r = r E and so a = g. But, as you move farther
from Earth’s center, r becomes larger than r E, and the free-fall accelera-
tion is reduced according to this inverse square relationship. What
happens to your mass as you move farther and farther from Earth’s
center?
Weight and weightlessness You may have seen photos similar to
Figure 13 in which astronauts are on a spacecraft in an environment often
called zero-g or weightlessness. The spacecraft orbits about 400 km above
(t)Science & Society Picture Library/SSPL/Getty Images, (b)NASA

Earth’s surface. At that distance, g = 8.7 N/kg, only slightly less than that
on Earth’s surface. Earth’s gravitational force is certainly not zero in the
shuttle. In fact, gravity causes the shuttle to orbit Earth. Why, then, do
the astronauts appear to have no weight?
Remember that you sense weight when something, such as the floor
or your chair, exerts a contact force on you. But if you, your chair, and
the floor all are accelerating toward Earth together, then no contact
forces are exerted on you. Thus, your apparent weight is zero and you
experience apparent weightlessness. Similarly, the astronauts experience
apparent weightlessness as the shuttle and everything in it falls freely
toward Earth.

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 MiniLABs The Gravitational Field
Recall from studying motion that many common forces are contact
WEIGHTLESS WATER
What are the effects of weightless- forces. Friction is exerted where two objects touch; for example, the floor
ness in free fall? and your chair or desk push on you when you are in contact with them.
Gravity, however, is different. It acts on an apple falling from a tree and
WEIGHT IN FREE FALL on the Moon in orbit. In other words, gravity acts over a distance. It acts
What is the effect of free fall between objects that are not touching or that are not close together.
on mass? Newton was puzzled by this concept. He wondered how the Sun could
exert a force on planet Earth, which is hundreds of millions of kilome-
iLab Station
ters away.
Field concept The answer to the puzzle arose from a study of magne-
tism. In the nineteenth century, Michael Faraday developed the concept
of a field to explain how a magnet attracts objects. Later, the field
concept was applied to gravity.
Any object with mass is surrounded by a gravitational field, which
exerts a force that is directly proportional to the mass of the object and
inversely proportional to the square of the distance from the object’s
center. Another object experiences a force due to the interaction between
its mass and the gravitational field (g) at its location. The direction of g
and the gravitational force is toward the center of the object producing
Figure 14 Earth’s gravitational field can be the field. Gravitational field strength is expressed by the following equation.
represented by vectors pointing toward
Earth’s center. The decrease in g’s magni- GRAVITATIONAL FIELD
tude follows an inverse-square relationship The gravitational field strength produced by an object is equal to the universal gravitational
as the distance from Earth’s center constant times the object’s mass, divided by the square of the distance from the object’s center.
increases.
Explain why the value of g never reaches g=_
Gm
2
zero. r
Suppose the gravitational field is created by the Sun. Then a planet of
mass m in the Sun's gravitational field has a force exerted on it that
depends on its mass and the magnitude of the gravitational field at its
location. That is, F g = mg, toward the Sun. The force is caused by the
interaction of the planet’s mass with the gravitational field at its loca-
tion, not with the Sun millions of kilometers away. To find the gravita-
tional field caused by more than one object, calculate all gravitational
fields and add them as vectors.
The gravitational field is measured by placing an object with a small
mass (m) in the gravitational field and measuring the force (F g) on it.
Fg
The gravitational field is calculated using g = _m . The gravitational field
is measured in units of newtons per kilogram (N/kg).
N N
g = 9.8 kg g = 7.4 kg On Earth’s surface, the strength of the gravitational field is 9.8 N/kg,
N and its direction is toward Earth’s center. The field can be represented by
g = 4.9 kg
a vector of length g pointing toward the center of the object producing
the field. You can picture the gravitational field produced by Earth as a
collection of vectors surrounding Earth and pointing toward it, as shown
in Figure 14. The strength of Earth's gravitational field varies inversely
with the square of the distance from Earth's center. Earth's gravitational
field depends on Earth’s mass but not on the mass of the object experi-
N encing it.
N g = 2.5 kg
g = 0.98 kg

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 Two Kinds of Mass
You read that mass can be defined as the slope of a graph of force
versus acceleration. That is, mass is equal to the net force exerted on an
object divided by its acceleration. This kind of mass, related to the
inertia of an object, is called inertial mass and is represented by the
following equation. PhysicsLABs
HOW CAN YOU MEASURE
INERTIAL MASS MASS?
Inertial mass is equal to the net force exerted on the object divided How is an inertial balance used to
by the acceleration of the object. measure mass?
F
m inertial = _
net
a INERTIAL MASS AND
GRAVITATIONAL MASS
Inertial mass You know that it is much easier to push an empty How can you determine the
cardboard box across the floor than it is to push one that is full of books. relationship between inertial
The full box has greater inertial mass than the empty one. The inertial mass and gravitational mass?
mass of an object is a measure of the object’s resistance to any type of
iLab Station
force. Inertial mass of an object is measured by exerting a force on the
object and measuring the object’s acceleration. The more inertial mass
an object has, the less acceleration it undergoes as a result of a net force
exerted on it.
Gravitational mass Newton’s law of universal gravitation,
Gm 1m 2
Fg = _ 2
, also involves mass—but a different kind of mass. Mass as
r
used in the law of universal gravitation is a quantity that measures an
object’s response to gravitational force and is called gravitational mass.
It can be measured by using a simple balance, such as the one shown in
Figure 15. If you measure the magnitude of the attractive force exerted on
an object by another object of mass m, at a distance r, then you can
define the gravitational mass in the following way.

GRAVITATIONAL MASS
The gravitational mass of an object is equal to the distance between the centers
of the objects squared, times the gravitational force, divided by the product of the
universal gravitational constant, times the mass of the other object.
r 2F g
mg = _
Gm
How different are these two kinds of mass? Suppose you have a
watermelon in the trunk of your car. If you accelerate the car forward,
the watermelon will roll backward relative to the trunk. This is a result
Figure 15 A simple balance is used to
of its inertial mass—its resistance to acceleration. Now, suppose your car determine the gravitational mass of an object.
climbs a steep hill at a constant speed. The watermelon will again roll
Stephen Frisch/The McGraw-Hill Companies, Inc.

backward. But this time, it moves as a result of its gravitational mass.

The watermelon is pulled downward toward Earth.
Newton made the claim that inertial mass and gravitational mass
are equal in magnitude. This hypothesis is called the principle of
equivalence. All investigations conducted so far have yielded data
that support this principle. Most of the time we refer simply to the
mass of an object. Albert Einstein also was intrigued by the principle
of equivalence and made it a central point in his theory of gravity.

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 Curved Space Einstein’s Theory of Gravity
Newton’s law of universal gravitation allows us to
calculate the gravitational force that exists between
two objects because of their masses. Newton was
puzzled, however, as to how two objects could exert
forces on each other if those two objects were millions
of kilometers away from each other. Albert Einstein
proposed that gravity is not a force but rather an effect
of space itself. According to Einstein’s explanation of
gravity, mass changes the space around it. Mass causes
space to be curved, and other bodies are accelerated
because of the way they follow this curved space.
Ships Traveling South Curved space One way to picture how mass
N affects space is to model three-dimensional space as a
large, two-dimensional sheet, as shown in the top
part of Figure 16. The yellow ball on the sheet repre-
Equator sents a massive object. The ball forms an indentation
on the sheet. A red ball rolling across the sheet simu-
lates the motion of an object in space. If the red ball
moves near the sagging region of the sheet, it will be
accelerated. In a similar way, Earth and the Sun are
attracted to each other because of the way space is
S distorted by the two objects.
Ships traveling south The following is another
Apples Dropped to Earth analogy that might help you understand the curvature
of space. Suppose you watch from space as two ships
travel due south from the equator. At the equator, the
ships are separated by 4000 km. As they approach the
South Pole, the distance decreases to 1 km. To the
sailors, their paths are straight lines, but because of
Earth’s curvature, they travel in a curve, as viewed far
Center from Earth’s surface, as in Figure 16.
Apples dropped to Earth Consider a similar
motion. Two apples are dropped to Earth, initially
traveling in parallel paths, as in Figure 16. As they
approach Earth, they are pulled toward Earth’s center.
Their paths converge.
Converging Lines
Converging lines This convergence can be attrib-
Curved space uted to the curvature of space near Earth. Far from
near Earth
Flat space Parallel paths any massive object, such as a planet or star, space is
far from flat, and parallel lines remain parallel. Then they
Earth begin to converge. In flat space, the parallel lines
would remain parallel. In curved space, the lines
converge.
Ted Kinsman/Photo Researchers

Converging
paths Einstein’s theory or explanation, called the general
Earth theory of relativity, makes many predictions about
how massive objects affect one another. In every test
Figure 16 Visualizing how space is curved is difficult. Analogies can conducted to date, Einstein’s theory has been shown
help you understand difficult concepts. to give the correct results.

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 Figure 17 Light is bent around massive
Star Actual position objects in space, altering the apparent
position of the light’s source.
Sun Describe how this effect contradicts your
Moon experience of light’s behavior.
Earth
Apparent position

(Not to scale)
Reference star

Deflection of light Einstein’s theory predicts that massive objects

deflect and bend light. Light follows the curvature of space around the
massive object and is deflected, as shown in Figure 17. In 1919, during an
eclipse of the Sun, astronomers found that light from distant stars that
passed near the Sun was deflected an amount that agreed with Einstein’s
predictions.
Another result of general relativity is the effect of gravity on light
from extremely massive objects. If an object is massive and dense
enough, the light leaving it is totally bent back to the object. No light
ever escapes the object. Objects such as these, called black holes, have
been identified as a result of their effects on nearby stars. Black holes
have been detected through the radiation produced when matter is
pulled into them.
While Einstein’s theory provides very accurate predictions of gravity’s
effects, it is still incomplete. It does not explain the origin of mass or
how mass curves space. Physicists are working to understand the deeper
meaning of gravity and the origin of mass itself.

SECTION 2 REVIEW Section Self-Check Check your understanding.

18. MAI
MAINN IDEA The Moon is 3.9×10 5 km from Earth’s 20. Gravitational Field The mass of the Moon is 7.3×1022 kg
center and Earth is 14.96×10 7
km from the Sun’s and its radius is 1785 km. What is the strength of the
center. The masses of Earth and the Sun are gravitational field on the surface of the Moon?
5.97×10 24 kg and 1.99×10 30 kg, respectively. During a
21. Orbital Period and Speed Two satellites are in circular
full moon, the Sun, Earth, and the Moon are in line with
orbits about Earth. One is 150 km above the surface,
each other, as shown in Figure 18.
the other is 160 km.
a. Find the ratio of the gravitational fields due to Earth
a. Which satellite has the larger orbital period?
and the Sun at the center of the Moon.
b. Which has the greater speed?
b. What is the net gravitational field due to the Sun and
Earth at the center of the Moon? 22. Theories and Laws Why is Einstein’s description of
gravity called a theory, while Newton’s is a law?
23. Astronaut What would be the strength of Earth’s
gravitational field at a point where an 80.0-kg astronaut
Sun would experience a 25.0 percent reduction in weight?
Earth Moon 24. A Satellite’s Mass When the first artificial satellite was
(Not to scale) launched into orbit by the former Soviet Union in
Figure 18 1957, U.S. president Dwight D. Eisenhower asked
scientists to calculate the mass of the satellite. Would
they have been able to make this calculation? Explain.
19. Apparent Weightlessness Chairs in an orbiting spacecraft
are weightless. If you were on board such a spacecraft 25. Critical Thinking It is easier to launch a satellite from
and you were barefoot, would you stub your toe if you Earth into an orbit that circles eastward than it is to
kicked a chair? Explain. launch one that circles westward. Explain.

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NO ESCAPE
Is a BLACK HOLE really a hole?
You may have wondered about black holes.
What are they and where do they come from?

A star explodes What happens when a giant star

runs low on fuel? The star cannot maintain its
temperature, causing it to collapse un u der its own
weigi ht. The resulting explosion, callled a supernova,
is usually brighter than the entire galaxy it is in.
If what’s left of the star is massive enough (more
than three times the mass of the Sun), the remnant
becomes one of the strangest objects in the
universe: a black hole.
Escape from Earth Imagine standing on the FIGURE 2 This artist’s
depiction shows a star like our
surface of Earth and throwing a ball straight up. As
Sun being consumed by a
the ball moves up, gravitational force robs the ball
nearby black hole.
of its upward velocity. Finally, when the upward
velocity reaches zero, the ball begins to fall back
down. Figure 1 illustrates this.
Of course, this happens due to the limitations of
your throwing arm. If you could throw the ball fast
enough (about 11,000 m/s), it would not fall back
to Earth but instead would escape Earth entirely.
This speed is the escape velocity from the surface
of our planet.
Escape from a black hole? A black hole is a very
comppact, massive object with an escape velocity
so high that nothing, not even light, can escape.
The image in Figure 2 shows a star being
swallowed by a black hole. Anything that passes FIGURE 1 Escape
speed greater than vesc
through the boundary of the influence of a black velocity from the
hole is truly lost forever. surface of any
object depends on
Are black holes really holes? NO! They are speed smaller than vesc
the mass and the
extremely dense objects in space. radius of the object
in question.

Research Light traveling through a vacuum

cannot slow down. So how, then, does a black
hole prevent a light beam from escaping? To find
out, investigate and write about an effect known
NASA

as the gravitational red shift.

194 Chapter 7 • Gravitation

 CHAPTER 7 STUDY GUIDE
Gravity is an attractive field force that acts between objects with mass.

VOCABULARY SECTION 1 Planetary Motion and Gravitation

Gravit
• Kepler’s first law (p. 179)
MAIN IDEA The gravitational force between two objects is proportional to the
• Kepler’s second law (p. 179) product of their masses divided by the square of the distance
• Kepler’s third law (p. 180) between them.
• gravitational force (p. 182) • Kepler’s first law states that planets move in elliptical orbits, with the Sun at one focus, and
• law of universal gravitation Kepler’s second law states that an imaginary line from the Sun to a planet sweeps out equal
(p. 182)
areas in equal times. Kepler’s third law states that the square of the ratio of the periods of any
two planets is equal to the cube of the ratio of the distances between the centers of the
planets and the center of the Sun.
2
(_) = (_)
TA
TB
rA
rB
3

• Newton’s law of universal gravitation can be used to rewrite Kepler’s third law to relate the
radius and period of a planet to the mass of the Sun. Newton’s law of universal gravitation
states that the gravitational force between any two objects is directly proportional to the
product of their masses and inversely proportional to the square of the distance between their
centers. The force is attractive and along a line connecting the centers of the masses.
Gm m
F=_
1 2
2 r

• Cavendish’s investigation determined the value of G, confirmed Newton’s prediction that a

gravitational force exists between two objects, and helped calculate the mass of Earth.

VOCABULARY SECTION 2 Using the Law of Universal Gravitation

G
• inertial mass (p. 191)
MAIN IDEA All objects are surrounded by a gravitational field that affects the
• gravitational mass (p. 191) motions of other objects.
• The speed and period of a satellite in circular orbit describe orbital motion. Orbital speed and
period for any object in orbit around another are calculated with Newton’s second law.
• Gravitational mass and inertial mass are two essentially different concepts. The gravitational
and inertial masses of an object, however, are numerically equal.
• All objects have gravitational fields surrounding them. Any object within a gravitational field
experiences a gravitational force exerted on it by the gravitational field. Einstein’s general
theory of relativity explains gravitational force as a property of space itself.

Games and Multilingual eGlossary

Vocabulary Practice

Chapter 7 • Study Guide 195

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 CHAPTER 7 ASSESSMENT Chapter Self-Check

SECTION 1 39. Figure 19 shows a Cavendish apparatus like the one

used to find G. It has a large lead sphere that is
Planetary Motion and Gravitation 5.9 kg in mass and a small one with a mass of
Mastering Concepts 0.047 kg. Their centers are separated by 0.055 m.
26. Problem Posing Complete this problem so that it Find the force of attraction between them.
can be solved using Kepler’s third law: “Suppose a
new planet was found orbiting the Sun in the region 5.9 kg
between Jupiter and Saturn . . .” 0.047 kg

27. In 1609 Galileo looked through his telescope at

Jupiter and saw four moons. The name of one of the 0.055 m
0.055 m
moons that he saw is Io. Restate Kepler’s first law for
Io and Jupiter.
0.047 kg
28. Earth moves more slowly in its orbit during summer
5.9 kg
in the northern hemisphere than it does during Figure 19
winter. Is it closer to the Sun in summer or in winter?
40. A ssume that your mass is 50.0 kg. Earth's mass is
29. Is the area swept out per unit of time by Earth 5.97×10 24 kg, and its radius is 6.38×10 6 m.
moving around the Sun equal to the area swept out
per unit of time by Mars moving around the Sun? a. What is the force of gravitational attraction
Explain your answer. between you and Earth?
b. What is your weight?
30. Why did Newton think that a force must act on
the Moon? 41. A 1.0-kg mass weighs 9.8 N on Earth’s surface, and
the radius of Earth is roughly 6.4×10 6 m.
31. How did Cavendish demonstrate that a gravitational
a. Calculate the mass of Earth.
force of attraction exists between two small objects?
b. Calculate the average density of Earth.
32. What happens to the gravitational force between
two masses when the distance between the masses 42. Tom's mass is 70.0 kg, and Sally's mass is 50.0 kg.
is doubled? Tom and Sally are standing 20.0 m apart on the
dance floor. Sally looks up and sees Tom. She feels
33. According to Newton’s version of Kepler’s third law, an attraction. Supposing that the attraction is
how would the ratio _
T 2 change if the mass of the
3
gravitational, find its size. Assume that both Tom
r and Sally can be replaced by spherical masses.
Sun were doubled?
43. BIGIDEA
BI The centers of two balls are 2.0 m apart,
Mastering Problems as shown in Figure 20. One ball has a mass of 8.0 kg.
The other has a mass of 6.0 kg. What is the gravita-
34. Jupiter is 5.2 times farther from the Sun than Earth tional force between them?
is. Find the length of Jupiter’s orbital period in Earth
years.
35. The dwarf planet Pluto has a mean distance from
the Sun of 5.87×10 12 m. What is its orbital period of 2.0 m
Pluto around the Sun in years? 8.0 kg 6.0 kg
Figure 20
36. Use Table 1 to compute the gravitational force that
the Sun exerts on Jupiter.
44. The star HD102272 has two planets. Planet A has a
37. The gravitational force between two electrons that period of 127.5 days and a mean orbital radius of
are 1.00 m apart is 5.54×10 -71 N. Find the mass of 0.615 AU. Planet B has a period of 520 days and a
an electron. mean orbital radius of 1.57 AU. What is the mass of
the star in units of the Sun’s mass?
38. Two bowling balls each have a mass of 6.8 kg.
The centers of the bowling balls are located 21.8 cm 45. If a small planet, D, were located 8.0 times as far
apart. What gravitational force do the two bowling from the Sun as Earth is, how many years would it
balls exert on each other? take the planet to orbit the Sun?

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 Chapter Self-Check

46. The Moon’s center is 3.9×10 8 m from Earth’s center. SECTION 2

The Moon is 1.5×10 8 km from the Sun’s center.
If the mass of the Moon is 7.3×10 22 kg, find the
Using the Law of Universal Gravitation
ratio of the gravitational forces exerted by Earth and Mastering Concepts
the Sun on the Moon. 54. How do you answer the question, “What keeps a
47. Ranking Task Using the solar system data in the satellite up?”
reference tables at the end of the book, rank the 55. A satellite is orbiting Earth. On which of the follow-
following pairs of planets according to the gravita- ing does its speed depend?
tional force they exert on each other, from least to
greatest. Specifically indicate any ties. a. mass of the satellite
b. distance from Earth
A. Mercury and Venus, when 5.0×10 7 km apart
c. mass of Earth
B. Jupiter and Saturn, when 6.6×10 8 km apart
C. Jupiter and Earth, when 6.3×10 8 km apart 56. What provides the force that causes the centripetal
acceleration of a satellite in orbit?
D. Mercury and Earth, when 9.2×10 7 km apart
E. Jupiter and Mercury, when 7.2×10 8 km apart 57. During space flight, astronauts often refer to forces
as multiples of the force of gravity on Earth’s surface.
48. Two spheres are placed so that their centers are What does a force of 5g mean to an astronaut?
2.6 m apart. The gravitational force between the two
spheres is 2.75×10 −12 N. What is the mass of each 58. Newton assumed that a gravitational force acts
sphere if one of the spheres is twice the mass of the directly between Earth and the Moon. How does
other sphere? Einstein’s view of the attractive force between the
two bodies differ from Newton’s view?
49. Toy Boat A force of 40.0 N is required to pull a
10.0-kg wooden toy boat at a constant velocity 59. Show that the units of g, previously given as N/kg,
across a smooth glass surface on Earth. What is the are also m/s 2.
force that would be required to pull the same 60. If Earth were twice as massive but remained the
wooden toy boat across the same glass surface on same size, what would happen to the value of g?
the planet Jupiter?

50. Mimas, one of Saturn’s moons, has an orbital radius Mastering Problems
of 1.87×10 8 m and an orbital period of 23.0 h. Use 61. Satellite A geosynchronous satellite is one that
Newton’s version of Kepler’s third law to find appears to remain over one spot on Earth, as shown
Saturn’s mass. in Figure 21. Assume that a geosynchronous satellite
51. Halley’s Comet Every 76 years, comet Halley is has an orbital radius of 4.23×10 7 m.
visible from Earth. Find the average distance of the a. Calculate its speed in orbit.
comet from the Sun in astronomical units. (AU is b. Calculate its period.
equal to the Earth’s average distance from the Sun.
The distance from Earth to the Sun is defined as
1.00 AU.) Satellite
52. Area is measured in m 2,so the rate at which area is r
swept out by a planet or satellite is measured in m 2/s. (Not to scale)
a. How quickly is an area swept out by Earth in its Earth
orbit about the Sun?
b. How quickly is an area swept out by the Moon in
its orbit about Earth? Use 3.9×10 8 m as the Figure 21
average distance between Earth and the Moon
and 27.33 days as the period of the Moon. 62. Asteroid The dwarf planet Ceres has a mass of
53. The orbital radius of Earth's Moon is 3.9×10 8 m. 7×10 20 kg and a radius of 500 km.
Use Newton’s version of Kepler’s third law to a. What is g on the surface of Ceres?
calculate the period of Earth’s Moon if the orbital b. How much would a 90-kg astronaut weigh
radius were doubled. on Ceres?

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 ASSESSMENT
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63. The Moon’s mass is 7.34×10 22 kg, and it has an orbital 70. On the surface of the Moon, a 91.0-kg physics
radius of 3.9×10 8 m from Earth. Earth’s mass is teacher weighs only 145.6 N. What is the value of
5.97×10 24 kg. the Moon’s gravitational field at its surface?
a. Calculate the gravitational force of attraction 71. The mass of an electron is 9.1×10 −31 kg. The mass of
between Earth and the Moon. a proton is 1.7×10 −27 kg. An electron and a proton
b. Find the magnitudes of Earth’s gravitational field are about 0.59×10 −10 m apart in a hydrogen atom.
at the Moon. What gravitational force exists between the proton
and the electron of a hydrogen atom?
64. Reverse Problem Write a physics problem with
real-life objects for which the following equation 72. Consider two spherical 8.0-kg objects that are
would be part of the solution: 5.0 m apart.

√
(6.67×1 0 -11 2 2
N·m /kg )(5.97×10 kg)24
___ a. What is the gravitational force between the two
8.3×10 3 m/s = r objects?
65. The radius of Earth is about 6.38×10 3 km. A space- b. What is the gravitational force between them
craft with a weight of 7.20×10 3 N travels away from when they are 5.0×10 1 m apart?
Earth. What is the weight of the spacecraft at each of 73. If you weigh 637 N on Earth’s surface, how much
the following distances from the surface of Earth? would you weigh on the planet Mars? Mars has a
a. 6.38×10 3 km mass of 6.42×10 23 kg and a radius of 3.40×10 6 m.
b. 1.28×10 4 km 74. Find the value of g, the gravitational field at Earth’s
c. 2.55×10 4 km surface, in the following situations.
66. Rocket How high does a rocket have to go above a. Earth’s mass is triple its actual value, but its
Earth’s surface before its weight is half of what it is radius remains the same.
on Earth? b. Earth’s radius is tripled, but its mass remains
the same.
67. Two satellites of equal mass are put into orbit
c. Both the mass and the radius of Earth are doubled.
30.0 m apart. The gravitational force between them
is 2.0×10 -7 N.
a. What is the mass of each satellite?
Applying Concepts
b. What is the initial acceleration given to each 75. Acceleration The force of gravity acting on an object
satellite by gravitational force? near Earth’s surface is proportional to the mass of
the object. Figure 23 shows a table-tennis ball and a
68. Two large spheres are suspended close to each other. golf ball in free fall. Why does a golf ball not fall
Their centers are 4.0 m apart, as shown in Figure 22. faster than a table-tennis ball?
One sphere weighs 9.8×10 2 N. The other sphere
weighs 1.96×10 2 N. What is the gravitational force
between them?

Richard Megna/Fundamental Photographs, NYC

4.0 m

9.8×102 N 1.96×102 N Figure 23

Figure 22
76. What information do you need to find the mass of
69. Suppose the centers of Earth and the Moon are
Jupiter using Newton’s version of Kepler’s third law?
3.9×10 8 m apart, and the gravitational force
between them is about 1.9×10 20 N. What is the 77. Why was the mass of the dwarf planet Pluto not
approximate mass of the Moon? known until a satellite of Pluto was discovered?

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 Chapter Self-Check

78. A satellite is one Earth radius above Earth's surface. 86. Mars has about one-ninth the mass of Earth.
How does the acceleration due to gravity at that Figure 26 shows satellite M, which orbits Mars with
location compare to acceleration at the surface of the same orbital radius as satellite E, which orbits
Earth? Earth. Which satellite has a smaller period?
79. What would happen to the value of G if Earth were
twice as massive, but remained the same size?
80. An object in Earth’s gravitational field doubles in
Mars Earth
mass. How does the force exerted by the field on the
object change? rM rE

81. Decide whether each of the orbits shown in Figure 24 E

is a possible orbit for a planet. M
Figure 26 (Not to scale)

87. Weight Suppose that yesterday your body had a

Sun mass of 50.0 kg. This morning you stepped on a
Sun scale and found that you had gained weight. Assume
your location has not changed.
a. What happened, if anything, to your mass?
Sun
b. What happened, if anything, to the ratio of your
Sun weight to your mass?

Figure 24 88. As an astronaut in an orbiting space shuttle, how would

you go about “dropping” an object down to Earth?
82. The Moon and Earth are attracted to each other by
gravitational force. Does the more massive Earth 89. Weather Satellites The weather pictures you see on
attract the Moon with a greater force than the Moon television come from a spacecraft that is in a station-
attracts Earth? Explain. ary position relative to Earth, 35,700 km above the
83. Figure 25 shows a satellite orbiting Earth. Examine equator. Explain how the satellite can stay in exactly
the same position. What would happen if it were

Gm E
the equation v = _ √
r , relating the speed of an closer? Farther out? Hint: Draw a pictorial model.
orbiting satellite and its distance from the center of
Earth. Does a satellite with a large or small orbital Mixed Review
radius have the greater velocity?
90. Use the information for Earth to calculate the mass
of the Sun, using Newton’s version of Kepler’s third law.
Satellite
91. The Moon’s mass is 7.3×10 22 kg and its radius is
1738 km. Suppose you perform Newton’s thought
experiment in which a cannonball is fired horizon-
r
tally from a very high mountain on the Moon.
a. How fast would the cannonball have to be fired
Earth to remain in orbit?
(Not to scale) b. How long would it take the cannonball to return
to the cannon?
Figure 25

84. Space Shuttle If a space shuttle goes into a higher 92. Car Races Suppose that a Martian base has been
orbit, what happens to the shuttle’s period? established. The inhabitants want to hold car races
for entertainment. They want to construct a flat,
85. Jupiter has about 300 times the mass of Earth and circular race track. If a car can reach speeds of
about ten times Earth’s radius. Estimate the size of g 12 m/s, what is the smallest radius of a track for
on the surface of Jupiter. which the coefficient of friction is 0.50?

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 ASSESSMENT
Chapter Self-Check

93. Apollo 11 On July 19, 1969, Apollo 11 was adjusted to 99. Analyze and Conclude The tides on Earth are caused
orbit the Moon at a height of 111 km. The Moon's by the pull of the Moon. Is this statement true?
radius is 1738 km, and its mass is 7.3×10 22 kg. a. Determine the forces (in newtons) that the Moon
a. What was the period of Apollo 11 in minutes? and the Sun exert on a mass (m) of water on
b. At what velocity did Apollo 11 orbit the Moon? Earth. Answer in terms of m.
b. Which celestial body, the Sun or the Moon, has a
94. The Moon's period is one month. Answer the greater pull on the waters of Earth?
following assuming the mass of Earth is doubled.
c. What is the difference in force exerted by the
a. What would the Moon's period be in months? Moon on water at the near surface and water at
b. Where would a satellite with an orbital period of the far surface (on the opposite side) of Earth, as
one month be located? illustrated in Figure 28. Answer in terms of m.
c. How would the length of a year on Earth change?
Far tidal bulge
95. Satellite A satellite is in orbit, as in Figure 27, with an
orbital radius that is half that of the Moon’s. Find
(Not to scale)
the satellite's period in units of the Moon's period.

Moon
rM
Earth Near tidal bulge
1r
2 M Figure 28 Earth Moon

Satellite d. Determine the difference in force exerted by the

(Not to scale) Sun on water at the near surface and on water at
Figure 27 the far surface (on the opposite side) of Earth.
96. How fast would a planet of Earth’s mass and size e. Which celestial body has a greater difference in
pull from one side of Earth to the other?
have to spin so that an object at the equator would
be weightless? Give the period in minutes. f. Why is it misleading to say the tides result from
the pull of the Moon? Make a correct statement to
explain how the Moon causes tides on Earth.
Thinking Critically
97. Make and Use Graphs Use Newton’s law of universal Writing in Physics
gravitation to find an equation where x is equal to
100. Research and report the history of how the distance
an object’s distance from Earth’s center and y is its
between the Sun and Earth was determined.
acceleration due to gravity. Use a graphing calculator
to graph this equation, using 6400–6600 km 101. Explore the discovery of planets around other stars.
as the range for x and 9–10 m/s 2 as the range for y. What methods did the astronomers use? What
measurements did they take? How did they use
The equation should be of the form y = c _
1 . Use
(x )
2 Kepler’s third law?
this graph and find y for these locations: sea level,
6400 km; the top of Mt. Everest, 6410 km; a satellite in
typical orbit, 6500 km; a satellite in higher orbit,
Cumulative Review
6600 km. 102. Airplanes A jet left Pittsburgh at 2:20 P.M. and
landed in Washington, DC, at 3:15 P.M. on the same
98. Suppose the Sun were to disappear—its mass
day. If the jet’s average speed was 441.0 km/h, what
destroyed. If the gravitational force were action at a
is the distance between the cities?
distance, Earth would experience the loss of the
gravitational force of the Sun immediately. But, if 103. Carolyn wants to weigh her brother Jared. He
the force were caused by a field or Einstein’s curva- agrees to stand on a scale, but only if they ride in
ture of space, the information that the Sun was gone an elevator. He steps on the scale while the elevator
would travel at the speed of light. How long would it accelerates upward at 1.75 m/s 2. The scale reads
take this information to reach Earth? 716 N. What is Jared’s usual weight?

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 CHAPTER 7 STANDARDIZED TEST PRACTICE
MULTIPLE CHOICE 4. A moon revolves around a planet with a speed of
9.0×10 3 m/s. The distance from the moon to the
1. Two satellites are in orbit around a planet. One center of the planet is 5.43×10 6 m. What is the orbital
satellite has an orbital radius of 8.0×10 6 m. The period of the moon?
period of revolution for this satellite is 1.0×10 6 s. The A. 1.2π×10 2 s C. 1.2π×10 3 s
other satellite has an orbital radius of 2.0×10 7 m.
B. 6.0π×10 2 s D. 1.2π×10 9 s
What is this satellite’s period of revolution?
A. 5.0×10 5 s C. 4.0×10 6 s 5. A moon in orbit around a planet experiences a
B. 2.5×10 6 s D. 1.3×10 7 s gravitational force not only from the planet, but also
from the Sun. The illustration below shows a moon
2. The illustration below shows a satellite in orbit during a solar eclipse, when the planet, the moon,
around a small planet. The satellite’s orbital radius is and the Sun are aligned. The moon has a mass of
6.7×10 4 km and its speed is 2.0×10 5 m/s. What is the about 3.9×10 21 kg. The mass of the planet is
mass of the planet around which the satellite orbits? 2.4×10 26 kg, and the mass of the Sun is
(G = 6.67×10 −11 N·m 2/kg 2) 1.99×10 30 kg. The distance from the moon to the
A. 2.5×10 18 kg C. 2.5×10 23 kg center of the planet is 6.0×10 8 m, and the distance
from the moon to the Sun is 1.5×10 11 m. What is the
B. 4.0×10 20 kg D. 4.0×10 28 kg
ratio of the gravitational force on the moon due to the
planet, compared to its gravitational force due to the
Satellite Sun during the solar eclipse?
A. 0.5 C. 5.0
6.7×10 4 km B. 2.5 D. 7.5

Sun
Planet 6.0×10 8 m

Planet
(Not to scale)
1.5×10 11 m
(Not to scale)

3. Two satellites are in orbit around the same planet.

Satellite A has a mass of 1.5×10 2 kg, and satellite B FREE RESPONSE
has a mass of 4.5×10 3 kg. The mass of the planet is
6.6×10 24 kg. Both satellites have the same orbital 6. Two satellites are in orbit around a planet. Satellite
radius of 6.8×10 6 m. What is the difference in the S 1 takes 20 days to orbit the planet at a distance of
orbital periods of the satellites? 2×10 5 km from the center of the planet. Satellite S 2
A. no difference C. 2.2×10 2 s takes 160 days to orbit the planet. What is the dis-
tance of satellite S 2 from the center of the planet?
B. 1.5×10 2 s D. 3.0×10 2 s

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