Lesson 1: Lesson 1-Task 1:

Use Newton’s law of gravitation to infer Score: ______________

gravitational force, weight, and acceleration Done? [ ] Yes [ ] No
due to gravity (STEM_GP12G-IIb-16) Lesson 1-Task 2:
Task 1: Crossword Puzzle Score: ______________
Done? [ ] Yes [ ] No
Task 2: Gravity of the Lesson 1-Task 3:
situation Task 3: Evaluation Score: ______________
Lesson 2: Done? [ ] Yes [ ] No
Discuss the physical significance of Lesson 2-Task 1:
gravitational field (STEM_GP12G-IIb-17) Score: ______________
Task 1: Word Find Done? [ ] Yes [ ] No
Task 2: Exploring the Lesson 2-Task 2:
field Task 3: Evaluation Score: ______________
Lesson 3: Done? [ ] Yes [ ] No
Apply the concept of gravitational potential Lesson 2-Task 3:
energy in physics problems (STEM_GP12G- Score: ______________
Done? [ ] Yes [ ] No
IIb-18)
Lesson 3-Task 1:
Task 1: Potential around us
Score: ______________
Task 2: Calculating potential Done? [ ] Yes [ ] No
Task 3: Evaluation Lesson 3-Task 2:
Lesson 4: Score: ______________
Calculate quantities related to planetary Done? [ ] Yes [ ] No
or satellite motion (STEM_GP12G-IIb-19) Lesson 3-Task 3:
Task 1: Check your understanding Score: ______________
Task 2: Debunking Satellite Motion Done? [ ] Yes [ ] No
Task 3: Evaluation Lesson 4-Task 1:
Feedback (Learner/Parent/Guardian): Score: ______________
Done? [ ] Yes [ ] No
_____________________________________________
Lesson 4-Task 2:
_____________________________________________
Score: ______________
_____________________________________________ Done? [ ] Yes [ ] No
_____________________________________________ Lesson 4-Task 3:
_____________________________________________ Score: ______________
_____________________________________________ Done? [ ] Yes [ ] No
_____________________________________________ Parallel Test:
_____________________________________________ Score: ______________
_____________________________________________ Done? [ ] Yes [ ] No
_____________________________________________ Note of Guidance (To be filled out by the
_____________________________________________ subject teacher for feedbacking of learner’s
_____________________________________________ performance):
_____________________________________________ __________________________________________
__________________________________________
__________________________________________
__________________________________________
__________________________________________
__________________________________________
Lesson 1

If you are asked why objects fall, a very quick answer would be
given: Objects fall because of gravity. Gravity is probably the most
familiar force in our everyday lives. But what is gravity?

CROSSWORD PUZZLE

Directions: Identify the word that

is being described and fill-out the
corresponding number in the
puzzle.

The idea of gravity started way back in the ancient times when people tried to explain why things
fell on Earth. Many ideas followed and, now, our common view of gravity is attributed to the
idea developed by Newton.
Through Newton’s analysis of the motion of the planets around the Sun, Newton found out that
there are two factors that affect the force of gravity: the distance between two bodies and the
masses of both bodies. With this, he was able to arrive at the form of the law of universal
gravitation which states that:
Every object in the universe attracts every other object with a force that is directly proportional
to the product of the masses of the two objects and inversely proportional to the square of the
distance between the center of the two objects, and in a line connecting their centers.
In an equation, the force of gravity Fg between two bodies of masses m1 and m2, separated by a
𝒎 𝒎
distance R is given by: 𝑭𝒈 = 𝑮 𝟏 𝟐 𝟐
𝑹
𝑵∙𝒎𝟐
where G is a universal constant and has a value of 𝑮 = 𝟔. 𝟔𝟕𝒙𝟏𝟎−𝟏𝟏 .
𝒌𝒈𝟐

2
 GRAVITY OF THE SITUATION

Direction: Predict the outcome of the following situation using the Law of Universal Gravitation.
1. Suppose that two objects attract each other with a gravitational force of 16 units.
a. If the distance between the two objects is doubled, what is the new force of
attraction between the two objects?
b. If the distance between the two objects is reduced in half, then what is the new
force of attraction between the two objects?
c. If the mass of both objects was doubled, and if the distance between the objects
remained the same, then what would be the new force of attraction between the
two objects?
d. If the mass of both objects was doubled, and if the distance between the objects
was doubled, then what would be the new force of attraction between the two
objects?
e. If the mass of both objects was tripled, and if the distance between the objects
was doubled, then what would be the new force of attraction between the two
objects?

Direction: Complete the table below using the formula for the Law of Universal
Gravitation.
Fg m1 m2 d
1. 2m1 m2 d
𝑚1 𝑚2 2.

3𝐹𝑔 𝑚1 3. d
4. 𝑚1 𝑚2 2d
5. 𝑚2 d

Lesson 2

All the objects with mass have a gravitational field around them.
All matter has a gravitational field that attracts other objects. The
more mass an object has, the greater its gravitational field will
be. Let’s try to explore the field.
WORD FIND

Lesson 2

3
Directions: Find the following words below and give the

GRAVITY FIELD
MASS GRAVITY
FORCE
WEIGHT

Any object with mass

m experiences a force
exerted by the earth’s
gravitational field. A
gravitational force of
9.8 N is exerted on each kilogram of
mass and so the earth’s gravitational field strength is 9.8 N/kg which is equivalent to 9.8 m/s,
the acceleration due to gravity, which is represented by the symbol g. For convenience, both the
gravitational field strength and the acceleration due to gravity can be rounded up to 10.
Therefore,
Weight= mass x g where g = 10 N/kg.
In symbols,
𝐹𝑊 = mg
where:
 gravitational force is measured in newtons (N)
 mass is measured in kilograms (kg)
 gravitational field strength (g) is measured in newtons per kilogram (N/kg) The Earth’s
gravitational field strength (g) is 9.8 or 10 N/kg. This means that for each kg of mass, an
object will experience 10 N of force.

What is weight?

Weight refers to the gravity force of planets and other bodies in the Universe, and the
effect that this has on objects. It is important to remember that weight is not the same as
mass - the weight of an object and its mass are directly proportional. This means that for a
given gravitational field strength, the greater the mass of the object, the greater its weight.
Weight is a non-contact force because gravity exerts its force through a field. An
object does not need to be touching the Earth to have a weight. The weight of an object can
be measured using a calibrated spring-balance, often called a Newton meter. Weight is also
referred to as ‘gravitational force’.

The equation can be rearranged to make mass the subject:

4
 EXPLORING THE FIELD

Directions: Answer the following questions and discuss your answer using the concept of
gravitational field strength.

1. What would happen to objects on the Earth’s surface if:

a. the Earth’s gravitational field gradually disappeared?
b. the Earth’s gravitational field was fine, but the Earth slowly stopped rotating?
2. What effect, if any, would removing the Earth’s core have on the gravitational field at
its surface? (Assume the size and shape of the Earth does not change; that is,
assume the Earth was partially hollowed out.)
3. The Earth has a radius about twice as great and a mass ten times greater than the
planet Mars. How does the acceleration due to gravity on Mars compare to that on
Earth?

Direction: Choose the correct answer to the following questions below.

1. The gravitational field is directed_______________.

A. towards the earth

B. away from earth
C. has no direction
D. in a specific direction making angle with earth
2. The value of gravitation on moon's surface is 1.6 m/s. The weight of a 100 kg body
on the moon's surface will be___________.
A. 80 N
B. 100 N C. 120 N
D. 160 N
3. The gravitational pull of the Earth acting on the body whether the body is in contact
with the Earth or not, such force is known as
A. force of inertia
B. force of limiting friction
C. field of inertia
D. field of force
4. What will happen to the value of gravitational field strength if the distance increases
between two bodies?
A. increases C. remains constant
B. decreases D. may increase or decrease
5. How much force does each 1 kg mass on Earth experiences?
A. 7 N B. 9.8 N C. 20 N D. 100 N

5
Lesson 3

POTENTIAL AROUND
When you an objectUS
to a certain height you are using your
energy. Being able to calculate the quantities of energy are useful
for gathering data.

Directions: Read about four types of potential energy in the chart

below. Then write down as many examples of each that you can think
of. One is already don for you.

GRAVITATIONAL POTENTIAL
ENERGY
Consider a block with a mass, and is
tied to the end of a rope and goes over a
pulley while the other end is being by a man. If a man
lets go of the rope, the rope will be pulled downward
with a force of gravity of the block. The work performed
by the block depends on the weight and height, ∆h. The
work done will be:
W= Fd
Since F= mg,
Then: W=(mg)(∆h)
Where d=∆h
The more work to perform and energy stored in the block, the higher the block is from the ground.

6
 Gravitational potential energy, PEgrav, is the energy stored of an object because of
its distance above the surface of the Earth. The change in gravitational potential energy of
an object is expressed as:

∆PEgrav = mg∆h

Where m is the mass of the object in kilograms

g is the acceleration due to gravity at 9.8 m/s2
∆h is the vertical displacement of the object in meters
∆PEgrav is the object’s change in gravitational potential energy in Joules

Sample Problem
1. How much gravitational potential energy a 4.0 kg block has if it is lifted 25m?
Given: Solution: ΔPEgrav = mgΔh
m=4.0 kg = (4.0 kg)(9.80m/s2)(25 m)
g= 9.80m/s2 = 39.2 kg•m/s2 (25m)
Δh= 25m = 39.2 N (25 m)
= 980 N•m = 980 Joules (J)

2. A 61.2 kg boy fell 0.500 m out of the bed. How much potential energy is lost?
Given: Solution: ∆PEgrav = mg∆h
m=61.2 kg
= (61.2 kg)(9.80m/s2)(-0.500 m)
g= 9.80m/s2
= 599.76 kg•m/s2 (-0.500m)
∆h= -0.500 m = 599.76 N (-0.500m)
= -299.8 N•m = -300 Joules (J)

CALCULATING POTENTIAL

Directions: Complete the table below by calculating the gravitational potential energy.
Object Mass,kg Height from the Gravitational
ground,m Potential Energy, J

A 50 kg 50 m
B 25 kg 100 m
C 67 kg 7m
D 100 kg 25 m
E 110 kg 12 m

1. A bird flying over the sea has a weight of 15 N and has a constant 765 J of gravitational
potential energy. How far above the sea does the bird fly?
2. A woman walks along a bridge, holding in her hand a small cell phone with a weight of
0.75N. The phone is held at a height of 1.2 m above surface of the bridge. The bridge is
4.3m above the surface of a river. How much greater is the gravitational potential energy
of the phone held in the woman’s hand on the bridge than the gravitational potential
energy of the phone at the surface of the river?
3. A speck of dust with a mass of 0.22g is at a distance of 1,250m form a tiny shard of ice
that has a mass of 1.5g. Both of the objects are located in deep space, extremely far
7
 away from any stars or other objects of any kind. The gravitational force that the objects
exert on each other is negligible, as is the gravitational force on either the speck of dust
or on the shard of ice from any other objects.
a. What is the gravitational potential energy of the speck of dust?
b. What is the gravitational potential energy of the shard of ice?

Lesson 4

The motion of objects is governed by Newton's laws. The same

simple
CHECK YOURlawsUNDERSTANDING
that govern the motion of objects on earth also
extend to the heavens to govern the motion of planets, moons,
and other satellites. The mathematics that describes a satellite's
motion is the same mathematics presented for circular motion.

Directions: Answer the following questions and explain your answer.

1. The fact that satellites can maintain their motion and their distance
above the Earth is fascinating to many. How can it be? What keeps a
satellite up?
2. If there is an inward force acting upon an earth orbiting satellite, then why doesn't the
satellite collide into the Earth?

Orbital Speed Equation

Consider a satellite with mass Msat orbiting a central body with a mass of
mass MCentral. The central body could be a planet, the sun or some other
large mass capable of causing sufficient acceleration on a less massive
nearby object. If the satellite moves in circular motion, then the net centripetal force acting upon
this orbiting satellite is given by the relationship
Fnet = ( Msat • v2 ) / R
This net centripetal force is the result of the gravitational force that attracts the satellite towards
the central body and can be represented as
Fgrav = ( G • Msat • MCentral ) / R2
Since Fgrav = Fnet, the above expressions for centripetal force and gravitational force can be set
equal to each other. Thus,
(Msat • v2) / R = (G • Msat • MCentral ) / R2
Observe that the mass of the satellite is present on both sides of the equation; thus it can be
canceled by dividing through by Msat. Then both sides of the equation can be multiplied by R,
leaving the following equation.
v2 = (G • MCentral ) / R
Taking the square root of each side, leaves the following equation for the velocity of a satellite
moving about a central body in circular motion.

G*Mcentral
v= √
R
where G is 6.673 x 10-11 N•m2/kg2, Mcentral is the mass of the central body about which the
satellite orbits, and R is the radius of orbit for the satellite.

The Acceleration Equation

8
Similar reasoning can be used to determine an equation for the acceleration of our satellite that
is expressed in terms of masses and radius of orbit. The acceleration value of a satellite is equal
to the acceleration of gravity of the satellite at whatever location that it is orbiting. In Lesson 3,
the equation for the acceleration of gravity was given as
g = (G • Mcentral)/R2
Thus, the acceleration of a satellite in circular motion about some central body is given by the
following equation

G*Mcentral
a= √
R2
where G is 6.673 x 10-11 N•m2/kg2, Mcentral is the mass of the central body about which the
satellite orbits, and R is the average radius of orbit for the satellite.

Orbital Period Equation

The final equation that is useful in describing the motion of satellites is Newton's form of Kepler's
third law. Since the logic behind the development of the equation has been presented elsewhere,
only the equation will be presented here. The period of a satellite (T) and the mean distance from
the central body (R) are related by the following equation:
T2 4*𝜋 2
= √
R3 G*Mcentral
where T is the period of the satellite, R is the average radius of orbit for the satellite
(distance from center of central planet), and G is 6.673 x 10-11 N•m2/kg2.

There is an important concept evident in all three of these equations - the period, speed and
the acceleration of an orbiting satellite are not dependent upon the mass of the satellite.
G*Mcentral G*Mcentral T2 4*𝜋 2
v= √ a= √ = √
R R2 R3 G*Mcentral
None of these three equations has the variable Msatellite in them. The period, speed and
acceleration of a satellite are only dependent upon the radius of orbit and the mass of the central
body that the satellite is orbiting. Just as in the case of the motion of projectiles on earth, the
mass of the projectile has no effect upon the acceleration towards the earth and the speed at any
instant. When air resistance is negligible and only gravity is present, the mass of the moving
object becomes a non-factor. Such is the case of orbiting satellites.

Example Problems
To illustrate the usefulness of the above equations, consider the following practice problems.

Practice Problem #1
A satellite wishes to orbit the earth at a height of 100 km (approximately 60 miles) above the
surface of the earth. Determine the speed, acceleration and orbital period of the satellite. (Given:
Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m).

Like most problems in physics, this problem begins by identifying known and unknown
information and selecting the appropriate equation capable of solving for the unknown. For this
problem, the knowns and unknowns are listed below.
Unknown:
Given/Known:
R = Rearth + height = 6.47 x 106 m V=?
Mearth = 5.98x1024 kg a=?
G = 6.673 x 10-11 N m2/kg2 T=?

9
Note that the radius of a satellite's orbit can be found from
the knowledge of the earth's radius and the height of the
satellite above the earth. As shown in the diagram at the
right, the radius of orbit for a satellite is equal to the sum of
the earth's radius and the height above the earth. These two
quantities can be added to yield the orbital radius. In this
problem, the 100 km must first be converted to 100 000 m
before being added to the radius of the earth. The equations
needed to determine the unknown are listed above. We will
begin by determining the orbital speed of the satellite using the following equation:
v = SQRT [ (G•MCentral ) / R ]
The substitution and solution are as follows:

v = SQRT [ (6.673 x 10-11 N m2/kg2) • (5.98 x 1024 kg) / (6.47 x 106 m) ]

v = 7.85 x 103 m/s
The acceleration can be found from either one of the following
equations:

(1) a = (G • Mcentral)/R2
(2) a = v2/R
Equation (1) was derived above. Equation (2) is a general equation for circular motion. Either
equation can be used to calculate the acceleration. The use of equation (1) will be demonstrated
here.
a = (G •Mcentral)/R2
a = (6.673 x 10-11 N m2/kg2) • (5.98 x 1024 kg) / (6.47 x 106 m)2
a = 9.53 m/s2
Observe that this acceleration is slightly less than the 9.8 m/s2 value expected on earth's
surface. As discussed in Lesson 3, the increased distance from the center of the earth lowers the
value of g.
Finally, the period can be calculated using the following equation:

T2 4*𝜋 2
= √
R3 G*Mcentral
The equation can be rearranged to the following form

T = SQRT [(4 • pi2 • R3) / (G*Mcentral)]

The substitution and solution are as follows:

T = SQRT [(4 • (3.1415)2 • (6.47 x 106 m)3) / (6.673 x 10-11 N m2/kg2) • (5.98x1024 kg) ]
T = 5176 s = 1.44 hrs

Practice Problem #2
The period of the moon is approximately 27.2 days (2.35 x 106 s). Determine the radius of the
moon's orbit and the orbital speed of the moon. (Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x
106 m)
Given/Known:
Unknown:
T = 2.35 x 106 s
Mearth = 5.98 x 1024 kg R=?
G = 6.673 x 10-11 N m2/kg2 v=?
The radius of orbit can be calculated using the following equation:
T2 4*𝜋 2
= √
R3 G*Mcentral

10
The equation can be rearranged to the following form
R3 = [ (T2 • G • Mcentral) / (4 • pi2) ]

The substitution and solution are as follows:

R3 = [ ((2.35x106 s)2 • (6.673 x 10-11 N m2/kg2) • (5.98x1024 kg) ) / (4 • (3.1415)2) ]

R3 = 5.58 x 1025 m3

By taking the cube root of 5.58 x 1025 m3, the radius can be determined as follows:
R = 3.82 x 108 m
The orbital speed of the satellite can be computed from
either of the following equations:

(1) v = SQRT [ (G • MCentral ) / R ] (2) v = (2 • pi • R)/T

Equation (1) was derived above. Equation (2) is a general equation for circular motion. Either
equation can be used to calculate the orbital speed; the use of equation (1) will be demonstrated
here. The substitution of values into this equation and solution are as follows:
v = SQRT [ (6.673 x 10-11 N.m2/kg2)*(5.98x1024 kg) / (3.82 x 108 m) ]
v = 1.02 x 103 m/s

DEBUNKING SATELLITE MOTION

Directions: Use the information below and the relationship above to calculate the T2/R3 ratio
for the planets about the Sun, the moon about the Earth, and the moons of Saturn about the
planet Saturn.

The value of G is 6.673 x 10-11 N•m2/kg2.

Sun M = 2.0 x 1030 kg
Earth M = 6.0 x 1024 kg
Saturn M = 5.7 x 1026 kg

a. T2/R3 for planets about sun

b. T2/R3 for the moon about Earth
c. T2/R3 for moons about Saturn

Directions: Solve the following problems below.

1. One of Saturn's moons is named Mimas. The mean orbital distance of Mimas is 1.87 x 108 m.
The mean orbital period of Mimas is approximately 23 hours (8.28x104 s). Use this information
to estimate a mass for the planet Saturn.
2. Consider a satellite which is in a low orbit about the Earth at an altitude of 220 km above
Earth's surface. Determine the orbital speed of this satellite. Use the information given below.
G = 6.673 x 10-11
Nm2/kg2
Mearth = 5.98 x 1024 kg
Rearth = 6.37 x 106 m
3. Suppose the Space Shuttle is in orbit about the earth at 400 km above its surface. Use the
information given in the previous question to determine the orbital speed and the orbital period
of the Space Shuttle.

11
 7. The radius of the earth is 6.38 × 106 m.
Find the orbital velocity of a satellite if it
orbited 8.50 × 104 m above the surface of
the earth.
Direction: Choose the correct answer to the
following questions below.
1. If the mass of Earth increased, with no a. 6.17 × 1015 m/s
change in radius, your weight would _____. b. 3.49 × 102 m/s
a. stay the same c. 7.85 × 103 m/s
b. increase also 8. The orbital period is determined by
c. decrease what two variables?
d. cannot be determined a. Mass of the orbiting object and the
distance between the two objects.
2. Two objects move toward each other because
b. The universal constant and the mass of
of gravitational attraction. As the objects get
the orbiting object.
closer and closer, the force between them _____.
c. Mass of the orbited object and distance
a. remains constant between the two objects.
b. decreases d. Masses of the orbited and orbiting
c. increases objects.
d. none of the above 9. The radius of the earth is 6.38 × 106 m.
3. The mass of an object is doubled, yet the Find the orbital period (in hours) of a
mass and the distance between the two objects satellite if it orbits 8.50 × 104 m above the
remains the same. Determine the impact on the surface of the earth.
gravitational force. a. 0.931 h
a. The gravitational force remains the same. b. 23.3 h
b. The gravitational force doubles. c. 5180 h
c. The gravitational force is quadrupled. d. 1.43 h
d. The gravitational force is cut in half.
10. What happens to the orbital velocity as
4. What is the purpose of "G" in the Universal a comet, in an elliptical orbit, comes closer
Gravitation formula? to the sun?
a. "G" is a universal gravitational constant. It a. The orbital velocity remains the same.
does not change.
b. The orbital velocity decreases.
b. "G" is a universal gravitational constant, but
fluctuates depending upon the planet an object c. The orbital velocity increases.
is on or is orbiting. d. The orbital velocity becomes
c. "G" is used to emphasize that the force is a random.
gravitational force and not a strong or weak
force.
d. "G" represents whether the gravitational force
would be positive or negative.
5. Two bowling balls, each with a mass of 7.27
kg, are suspended from a ceiling. Find the
gravitational force between the two balls if they
are suspended 0.250 m from each other.
a. 846 N
b. 5640 N
c. 5.64 × 10–8 N
d. 8.46 × 10–8 N
6. What is the force responsible for maintaining
a satellite's orbital motion around another body?
a. Normal force
b. Frictional force
c. Centripetal force
12
d. Perpendicular force
d. 7.28 × 107 m/s
 http://samohiphysics.weebly.com/uploads/2/2/8/8/22
884236/universal_gravitation_practice_quiz.pdf
https://physics.info/gravitation/problems.shtml
https://mcqlearn.com/grade9/physics/gravitation-
multiple-choice-questions-answers.php?page=3
https://mcqslearn.com/a-
level/physics/mcq/gravitational-field-multiple-choice-
questions-answers.php
https://www.physicsclassroom.com/class/circles/Lesso
n-4/Mathematics-of-Satellite-Motion

10. c
9. d
8. c
7. c
6. c
5. c
4. a
3. b
2. b
1. a
Test
Parallel

What new information have I learned from these Lessons and how will I apply
this knowledge in the real world?
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