Buku Teks Fizik Tingkatan 4/physics-form-4-chapter-3.pdf

Let’s StudyLet’s Study

3.13.1 Newton’s Universal  Newton’s Universal
Law  of GravitationLaw  of Gravitation

3.23.2 Kepler’s Laws Kepler’s Laws
3.33.3 Man-made Satellites Man-made Satellites

What are Newton’s Universal Law of What are Newton’s Universal Law of
Gravitation and Kepler’s Laws?Gravitation and Kepler’s Laws?

Why is it important to know the values Why is it important to know the values
of the gravitational acceleration of of the gravitational acceleration of
planets in the Solar System?planets in the Solar System?

How do man-made satellites help to How do man-made satellites help to
improve human life?improve human life?

CHA
PTE

R
GRAVITATIONGRAVITATION3

76767676
There are various types of man-made satellites revolving
in their respective orbits in outer space. Satellites are invented
for communication purposes, weather forecasts and Earth
observations. Why are these satellites able to revolve in their
respective orbits?

Several hundred years ago, Isaac Newton, a scientist,
conceptualised a universal law connecting all heavenly bodies
and also man-made satellites. Curiosity about the universe has
encouraged scientists to launch spaceships and satellites which
can overcome the Earth’s gravity. At present, there are spaceships
which are moving far away from Earth. Such spaceships enable
photographs of planets to be taken to benefi t scientifi c inventions
and technology.

PageInformation

Animated launching of satellite

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Learning Standards and Learning Standards and
List of FormulaeList of Formulae

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Activity 3.1

Newton’s Universal Law of Gravitation Newton’s Universal Law of Gravitation 3.1

HISTORY
INTEGRATION OF

In 1667, the scientist,
Isaac Newton,
observed an apple
which fell vertically
to the ground and
the movement of the
Moon around the Earth. He
subsequently concluded that
 a force of attraction not only
exists between the Earth and
the apple but also between the
Earth and the Moon.

Figure 3.1 Situation at a durian orchard

Dad, why does
the Moon revolve
around the Earth?

Hahaha! My children
think like Isaac
Newton too.

Aim: To discuss gravitational force between two bodies in the universe

Instructions:
1. Carry out a Hot Seat activity in groups.
2. Study Figure 3.2 and discuss the three situations below.

Figure 3.2 Situations involving gravitational force between two bodies

3. Surf the web to gather relevant information.
4. Choose a group representative to answer questions from other groups.

A person who jumps up will return to the ground. What force causes the
person to return to the ground?

Th e Moon revolves around the Earth without drift ing away from its
orbit. Th e Earth exerts a pulling force on the Moon. Does the Moon
also exert a force on the Earth?

Dad, why do
durians fall directly
to the ground?

Air molecules remain in the atmosphere without escaping to outer
space. What force acts between the molecules in the atmosphere and
the Earth?

Pattern Recognition ICS ISS

78787878 3.1.13.1.1
In the year 1687, Isaac Newton presented two relationships that
involve gravitational force between two bodies:
• gravitational force is directly proportional to the product of

the masses of the two bodies, that is F ∝ m1m2
• gravitational force is inversely proportional to the square of the

distance between the centres of the two bodies, that is F ∝ 1
r 2

Gravitational force is known as universal force because it acts between any two bodies in
the universe. Figure 3.3 shows the gravitational force between the Sun, the Earth and the Moon.
How can the gravitational force between two bodies be explained?

Body 1

F F
m1

Distance, r

m2

Body 2

Figure 3.4 Gravitational force between two bodies

Th e two relationships above are formulated in Figure 3.5 to
obtain Newton’s Universal Law of Gravitation.

F ∝ m1m2

F ∝ 1
r 2

F ∝
m1m2

r 2

Figure 3.5 Formulation of Newton’s Universal Law of Gravitation

The Earth

The Moon

The Sun

Gravitational force
between the Earth
and the Moon

Gravitational force between
the Sun and the Earth

Gravitational force between
the Sun and the Moon

Figure 3.3 Gravitational force as a universal force

Gravitational force

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F ile
� Gravitational force exists

between two bodies.
� Both bodies experience

gravitational force of the
same magnitude.

F ile
Why does a fallen leaf move
towards the ground?

Both the leaf and the Earth
experience the same
gravitational force. This causes
the leaf and the Earth to move
towards one another. As the
mass of the Earth is very much
larger than the mass of the leaf,
gravitational force does not
have an apparent effect on the
Earth’s movement. As such,
we only observe the leaf falling
to the ground.

79797979

Chapter 3 Gravitation

3.1.13.1.1
F =
Gm1m2

r 2

F = gravitational force between two bodies
m1 = mass of fi rst body
m2 = mass of second body
r =  distance between the centre of the fi rst body and the centre

of  the second body
G = gravitational constant  (G = 6.67 × 10–11 N m2 kg–2)

INFOSMARTSMART

F =
Gm1m2

r 2

Two bodies of masses m1
and m2 that are separated at
a distance of r, experience
a gravitational force of F
respectively.

Newton’s Universal Law of Gravitation
can be expressed as follows:

Newton’s Universal Law of Gravitation states that the
gravitational force between two bodies is directly proportional to
the product of the masses of both bodies and inversely proportional
to the square of the distance between the centres of the two bodies.

If you know the masses of two bodies and the distance between the centres of the two bodies,
you can calculate the magnitude of the gravitational force between the two bodies. Study the
examples given.

Calculate the gravitational force between a durian and the Earth.
Mass of durian = 2.0 kg
Mass of the Earth = 5.97 × 1024 kg
Distance between the centre of the durian and the centre of the Earth
= 6.37 × 106 m

Solution:

F = (6.67 × 10–11) × 2.0 × (5.97 × 1024)
(6.37 × 106)2

= 19.63 N1
2

3Step �
Substitute numerical values into the formula
and perform the calculations.

Gravitational force,

F = Gm1m2
r 21

2
3Step �

Identify and write down the formula used.

m1  = 2.0 kg
m2  = 5.97 × 1024 kg
r  = 6.37 × 106 m
G  = 6.67 × 10–11 N m2 kg–21

4
2
4
3

Step �
List the given information in symbols.

Figure 3.6

Example 1

F ile
The value of gravitational
constant, G can be determined
through experiment.

80808080 3.1.13.1.1
Activity 3.2

Aim: To solve problems involving Newton’s Universal Law of Gravitation for two bodies at
rest on the Earth

Instructions:
1. Work in pairs.
2. Imagine you and your partner are bodies at rest on the Earth.
3. Record your mass, m1 and the mass of your partner, m2 in Table 3.1.

Table 3.1

Pair
Mass, m / kg

r / m F / N
m1 m2

1
2.0
4.0

2
2.0
4.0

4. Calculate the gravitational force, F using both your masses and the distances given in
the table.

5. Change partners and repeat steps 3 and 4.

Discussion:
1. How do the masses of two bodies infl uence the gravitational force between them?
2. What is the effect of distance between two bodies on gravitational force between them?
3. Why is the magnitude of gravitational force between you and your partner small?

Logical Reasoning

A rocket at a launching pad experiences a gravitational force of 4.98 × 105 N. What is the
mass of the rocket?
[Mass of the Earth = 5.97 × 1024 kg, distance between the centre of the Earth and the centre of
the rocket = 6.37 × 106 m]

Solution:

Gravitational force, F = 4.98 × 105 N
Mass of the Earth, m1 = 5.97 × 1024 kg
Mass of rocket = m2
Distance between the centre of the
Earth and the centre of the rocket,
r = 6.37 × 106 m
G = 6.67 × 10–11 N m2 kg–2

Gravitational force,

F =
Gm1m2

r 2

4.98 × 105 =
(6.67 × 10–11) × (5.97 × 1024) × m2

(6.37 × 106)2

m2 =
(4.98 × 105) × (6.37 × 106)2

(6.67 × 10–11) × (5.97 × 1024)
  = 5.07 × 104 kg

Solving Problems Involving Newton’s Universal Law of Gravitation
Gravitational force acts between any two bodies, including those on the Earth, planets, the Moon
or the Sun. What are the eff ects of mass and distance between two bodies on gravitational force?

Example 2

CPS

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Chapter 3 Gravitation

3.1.13.1.1 3.1.23.1.2
Th e eff ects of mass and distance between two objects on
gravitational force are shown in Figure 3.7.

F1
F1

F2F2

The larger the mass of the body,
the larger the gravitational force.

F1 � F2

Gravitational force decreases when the
distance between the two bodies increases.

F1 � F2

Figure 3.7 Eff ects of mass and distance between two bodies on gravitational force

Gravitational force between two bodies depends on the
mass of the bodies, as well as the distance between them.

Activity 3.3

Aim: To solve problems involving Newton’s Universal Law of Gravitation for
 (i) objects on the Earth’s surface
 (ii) the Earth and satellite
 (iii) the Earth and the Sun

Instructions:
1. Work in pairs.
2. Study Figure 3.8 and answer the questions.

The Sun
Mass = 1.99 × 1030 kg
Distance between the Earth
and the Sun = 1.50 × 1011 m

The Moon
Mass = 7.35 × 1022 kg

The Earth
Mass = 5.97 × 1024 kg
Radius = 6.37 × 106 m

Man-made satellite
Mass = 1.20 × 103 kg
Distance between the Earth
and the satellite = 4.22 × 107 m

Figure 3.8 Th e Sun, the Earth, the Moon and a man-made satellite
Discussion:
1. What is the gravitational force on the man-made satellite before it is launched?
2. Compare:

(a) the mass of the Earth, the mass of the man-made satellite and the mass of the Sun
(b) between the Earth-satellite distance and the Sun-Earth distance.

3. Predict the difference in the magnitude of the gravitational force between the Earth and
the man-made satellite and the gravitational force between the Sun and the Earth.

4. Calculate:
(a) the gravitational force between the Earth and the man-made satellite
(b) the gravitational force between the Earth and the Sun.

Does your answer match your prediction in question 3?
5. The gravitational force between the Earth and the Moon is 2.00 × 1020 N. What is the

distance between the centre of the Earth and the centre of the Moon?

Abstraction

Even though gravitational force
is a universal force, two persons
on the Earth’s surface will not
feel the effect of the gravitational
force. This is because the
gravitational force between two
bodies of small mass has a very
small magnitude. For example,
two persons of masses 50 kg
and 70 kg respectively, only
experience a gravitational force
of 2.3 × 10–7 N if they stand at a
distance of 1 m from each other.

F ile

CPS

82828282 3.1.23.1.2
Relating Gravitational Acceleration, g on the Surface of
the Earth with Universal Gravitational Constant, G

According to Newton’s Second Law of Motion, gravitational force
can be expressed as F = mg. From Newton’s Universal Law of

Gravitation, gravitational force is expressed as F =
Gm1m2

r 2 .

What is the relationship between g and G?

Activity 3.4

Aim:  To derive the formula for gravitational acceleration, g using the formulae

F = mg and F = GMm
r 2

Instructions:

1. Work in pairs.
2. Scan the QR code and download Figure 3.9 from the

website given.
3. Discuss and complete Figure 3.9 to derive the

relationship between g and G.

M = mass of the Earth
m = mass of the object
r = distance between the centre of the Earth and the centre of the object

Newton’s Second Law of Motion Newton’s Universal Law of Gravitation

F F=

=

=
1442443 1442443

1442443

Equate two equations

Cancel common factor, m

Relationship between g and G g =

Gravitational force causes object to
fall with acceleration, g

Gravitational force pulls object to
the centre of the Earth

Figure 3.9 Relationship between g and G

Discussion:

1. What is the relationship between gravitational acceleration, g and gravitational constant, G?
2. What are the factors that infl uence the value of gravitational acceleration?

Algorithms CPS

From Newton’s Second Law of
Motion, F = ma.
When involving gravitational
acceleration, g,
F = mg

F ile

Download Figure 3.9

http://bt.sasbadi.com/p4083

83838383

Chapter 3 Gravitation

3.1.33.1.3
Activity 3.5

Value g is inversely
proportional to the square
of the distance from centre
of the Earth for r � R.

Figure 3.11 Variation of g with r

Aim: To discuss the variation in the values of g with r
Instructions:
1. Work in pairs.
2. Calculate the value of gravitational acceleration for the

fi ve distances given in Figure 3.10.

• Mass of the Earth, M
 = 5.97 × 1024 kg
• Radius of the Earth, R
 = 6.37 × 106 m
• Gravitational constant, G
 = 6.67 × 10–11 N m2 kg–2

3. Complete Table 3.2.

Table 3.2

Distance from centre of the Earth, r R 2R 3R 4R 5R

Gravitational acceleration, g / m s–2

Discussion:
1. What is the value of gravitational acceleration on the Earth’s surface?
2. Plot a graph of g against r.
3. How does the value of gravitational acceleration change when the distance from the

centre of the Earth increases?
4. Discuss the condition where the value of gravitational acceleration is almost zero.

Figure 3.11 shows a sketch of a graph with various values of gravitational acceleration, g
and distance, r from the centre of the Earth.

Value g is directly
proportional to the distance
from the centre of the Earth
for r � R.

R 2R 3R 4R 5R

Earth

Gravitational acceleration
below the surface of the Earth

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R
0 2R 3R

r

g

g = GM–––R 2

Logical Reasoning CPS

Figure 3.10

84848484 3.1.33.1.3
Figure 3.12 A satellite at height, h from the surface of the Earth

Figure 3.12 shows a satellite at height, h from the surface of the Earth. R is the radius of
the Earth and r is the distance of the satellite from the centre of the Earth, which is the radius
of the orbit.

At height, h from the surface of the
Earth, distance from the centre of
the Earth is r = (R + h).

With this, gravitational acceleration,

g = GM
(R + h)2

On the surface of the Earth,
height, h = 0.
Th erefore,
r = radius of the Earth, R.

Gravitational acceleration
on the surface of the Earth,

g = GM
R2 , where

M is the mass of the Earth.

Mass of the Earth is 5.97 × 1024 kg and radius of the Earth is 6.37 × 106 m. Calculate
gravitational acceleration on the surface of the Earth. [G = 6.67 × 10–11 N m2 kg–2]

Solution:

Step �
List the given information with symbols.

M = 5.97 × 1024 kg
G = 6.67 × 10–11 N m2 kg–2

r = 6.37 × 106 m

Step �
Identify and write down the formula used. g = GM

r 2

Step �
Make numerical substitution into the formula
and perform the calculations.

g = (6.67 × 10–11) × (5.97 × 1024)
(6.37 × 106)2

= 9.81 m s–2

A radar imaging satellite orbits around the Earth at a height of 480 km. What is the value of
gravitational acceleration at the position of the satellite?
[G = 6.67 × 10–11 N m2 kg–2, M = 5.97 × 1024 kg, R = 6.37 × 106 m]

Solution:

Height of orbit, h = 480 km
 = 480 000 m

g = GM
(R + h)2

=
(6.67 × 10–11) × (5.97 × 1024)

(6.37 × 106 + 480 000)2

= 8.49 m s–2

Satellite

r
h

RR

Sa

Earth

1
2

3
1

2
3

1
4

2
4

3

Example 1

Example 2

85858585

Chapter 3 Gravitation

3.1.33.1.3
When the value of the gravitational acceleration on the surface of a planet is known, the
magnitude of the gravitational force acting on an object on the surface of the planet can be
calculated. Knowledge on the value of gravitational acceleration plays an important role in space
exploration and continuity of life.

Activity 3.7

Aim: To discuss the importance of knowledge on gravitational acceleration of planets in
space exploration and continuity of life

Instructions:

1. Work in groups.

2. Gather information on the importance of knowledge on
gravitational acceleration of planets in space exploration
and continuity of life.

3. Present the results of your discussion in the form of
i-Think map.

Activity 3.6

Importance of Knowing the Value of Gravitational Acceleration

Aim: To compare different gravitational accelerations for the Moon, the Sun and the
planets in the Solar System

Instructions:
1. Carry out the Think-Pair-Share activity.

2. Gather information on the mass, M and radius, R of the Sun, the Moon as well as the
other remaining planets in the Solar System.

3. Present the gathered information in a table.

4. Calculate the gravitational acceleration, g for each of the bodies.

Discussion:

1. Which planet has the largest gravitational acceleration?

2. Which planet has gravitational acceleration closest to the gravitational acceleration
of the Earth?

3. What factors determine the value of the gravitational acceleration of a planet?

Space debris

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Logical Reasoning CPS ISS

Logical Reasoning ICS ISS

Gravitational force is a universal force. Th erefore, the formula g = GM
R2

can be used to calculate gravitational acceleration on the surface of
other bodies such as planets, the Moon and the Sun. Which planet
has the largest gravitational acceleration? Which planet has the
smallest gravitational acceleration?

Gravitational acceleration

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86868686 3.1.43.1.4
Activity 3.8

Aim: To gather information on the effects of gravity on the growth of humans

Instructions:

1. Carry out the Round Table activity.

Table 3.3

Factor Eff ect of low gravity Eff ect of high gravity

Diff erence
in density

Bone
fragility

Size of
lungs

Blood
circulatory
system

Blood
pressure

2. Based on Table 3.3, gather information on the effects of
gravity on the growth of humans by visiting websites or
from other suitable reading materials.

3. Complete Table 3.3.

4. Present a multimedia presentation entitled Effects of
Gravity on the Growth of Humans.

Effects of gravity

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We live on Earth where gravitational acceleration is 9.81 m s–2. While exploring space
whether far from Earth or near other planets, the body of astronauts can be exposed to low or high
gravity conditions. What are the eff ects of gravity on the growth of humans?

Logical Reasoning ISS ICS

87878787

Chapter 3 Gravitation

3.1.43.1.4
Centripetal Force in the Motion of Satellites and Planets

Figure 3.13 shows three positions of a satellite which orbits around the Earth at a uniform speed.
Observe the direction of the velocity of the satellite at each position.

Figure 3.13 Satellite in circular motion

An object in circular motion always experiences changes
in the direction of its motion even though its speed is fi xed.
In Chapter 2, you learnt that a force is required to change
the direction of motion of a body. What force acts on a body
which is in a circular motion?

Direction of velocity

Direction of
velocity

Direction of
velocity

Earth

INFOSMARTSMART

When a body moves in a circle
at uniform speed, the body is
said to be in uniform circular
motion.

88888888
Activity 3.9

Aim: To understand centripetal force using Centripetal Force Kit
Apparatus:  Centripetal Force Kit (a plastic tube, rubber stopper, slotted weight holder,

three 50 g slotted weights, crocodile clip, thick string) and ruler

Instructions:
1.  Set up the apparatus as shown in Figure 3.14 for circular motion of radius,

r = 50 cm. The total mass of the slotted weights and its holder is 100 g.

Crocodile clip

r = 50 cm Plastic tube

Slotted
weights

Thick string

Rubber stopper

Figure 3.14

2. Hold the plastic tube in your right hand and the slotted weights in your left hand. Rotate
the rubber stopper at a constant speed in a horizontal circle above your head as shown in
Figure 3.15. Make sure that the crocodile clip stays at a distance of about 1 cm from the
lower end of the plastic tube so that the radius is fi xed.

Rubber stopper

Slotted weights

Crocodile clip

Thick stringRadius of circle, r
Plastic tube

Figure 3.15

3. Release the slotted weights and continue rotating the rubber stopper. Observe the
speed of motion of the rubber stopper.

4. Repeat steps 1 to 3 with a total mass of 200 g of slotted weights. Compare the speed
of motion of the rubber stopper with the speed of motion in step 3.

5. Repeat step 4. When the rubber stopper is rotating, pull the lower end of the string
downwards so that the rubber stopper rotates with a decreasing radius. Feel how the
tension in the string acting on your left hand changes.

Discussion:
1. When the rubber stopper makes a circular motion, the stretched string exerts a force on

the rubber stopper. What is the direction of the force?
2. What is the relationship between the speed of the rubber stopper and the centripetal force?
3. How does the centripetal force change when the rubber stopper makes a circular

motion with a smaller radius?

INFOSMARTSMART

The slotted weights cause a
tension in the string that acts
as centripetal force during
the circular motion of the
rubber stopper.

Video demonstrating the use
of Centripetal Force Kit

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Chapter 3 Gravitation

3.1.53.1.5
Example 1

Centripetal force, F = 7.2 × 202

1.8

= 1 600 N

Figure 3.17

For a body in circular motion, a force acts on the body in a direction towards the centre of
the circle. Th is force is called centripetal force.
 Figure 3.16 shows the tension in the rope that acts as the centripetal force for the motion
of the rubber stopper. Th e magnitude of the centripetal force depends on the mass of the body,
the linear speed and the radius of the circle. Centripetal force can be calculated using the formula:

Step �
List the given information with symbols.

Step �
Identify and write down the formula used.

Step �
Make numerical substitution into the formula
and perform the calculations.

Figure 3.17 shows a hammer throw athlete swinging an iron
ball in a horizontal circle before releasing it. What is the
centripetal force that acts on the iron ball when the iron ball is
moving at a speed of 20 m s–1?

Solution:

1
2

3
1

2
3

1
4

2
4

3

m = 7.2 kg
v  = 20 m s–1

r  = 1.8 m

F = mv 2

r

TT

T T

v

v

v

v

Centre of circle

T = Tension in rope
v = Linear speed

Slotted weights

r = 1.8 m

m = 7.2 kg

Figure 3.16 Tension in the string acting as centripetal force

INFOSMARTSMART

Linear speed shows how fast a
body moves in a circular motion.

F = mv 2

r ,  where F = centripetal force
   m = mass
   v = linear speed
   r = radius of circle

INFOSMARTSMART

When a body is rotated at
a certain uniform speed with
the string almost horizontal,
the effect of gravitational force
on the circular motion of the
body can be ignored. Though
the speed is uniform, the
direction of motion of the body
keeps changing.

3.1.53.1.590909090
Object launched with low linear
speed will follow trajectory ➊ and
reaches the Earth at Q.

Object launched with suffi  ciently
high linear speed will follow
trajectory ➋ circulating the Earth.
Th e object will not return to
the Earth.

Figure 3.18 Prediction of Isaac Newton

Can a satellite orbit the Earth without being driven by a rocket engine? Th e possibility of
such movement was predicted by Isaac Newton in the 17th century as shown in Figure 3.18.

Q

P

QQQ

Object launched from P with linear speed, v

Trajectory for free
falling object
circulating the Earth

Earth

Th e prediction of Newton is now a reality with so
many man-made satellites orbiting around the Earth without
being driven by any thrust. Satellites always experience
gravitational force acting towards the centre of the Earth.
Th e gravitational force on satellites acts as centripetal force.

HISTORY
INTEGRATION OF

Although Isaac Newton did
not have the facilities to carry
out simulation or experiment,
he was able to visualise the
experiment on the movement
of bodies around the Earth.
His original sketch is
shown below.

➊

➋

Simulation of prediction of
Isaac Newton

http://bt.sasbadi.com/p4091b

By comparing the formula for force, F = ma

and formula for centripetal force, F = mv 2

r
, we obtain:

Centripetal acceleration, a = v 2

r
,

where v = linear speed of satellite
 r = radius of the orbit of satellite

9191

Chapter 3 Gravitation

3.1.53.1.5
Example 1

Figure 3.19 shows a weather satellite orbiting the Earth at a height, h = 480 km. Linear speed
of the satellite is 7.62 × 103 m s–1. Th e radius of the Earth, R is R = 6.37 × 106 m. What is the
centripetal acceleration of the satellite?

Weather
satellite

Orbit of weather
satellite

v

h m

M

Earth

Weather
m

Figure 3.19

Solution:

Step �
List the given information in symbols.

1
4

2
4

3

Height of satellite, h = 480 km
 = 480 000 m
Linear speed of satellite, v = 7.62 × 103 m s–1

Radius of the Earth, R = 6.37 × 106 m

Step �
Identify and write down the formula used.

1
2

3 a = v 2

r

Step �
Substitute numerical values into the formula
and perform the calculations.

1
4

2
4

3

a = v 2

(R + h)

=  (7.62 × 103)2

(6.37 × 106 + 480 000)
 = 8.48 m s–2

9292 3.1.53.1.5
Activity 3.10

Mass of the Earth and the Sun
Formula for the mass of the Earth and the Sun can be derived by using the formula of Newton’s
Universal Law of Gravitation and the formula for centripetal force.

Algorithms CPS

Aim: To determine the mass of the Earth and the Sun

Instructions:

1. Observe Figure 3.20.
2. Figure 3.20 shows the orbit of the Moon around the Earth.

v

r

M

m

Earth
Moon

Orbit of the Moon

Figure 3.20

M = mass of the Earth
m = mass of the Moon
r = radius of the Moon’s orbit
T = period of revolution of the

Moon around the Earth
v = linear speed of the Moon

3. Discuss and complete the boxes below.

Distance travelled by the Moon when making
one complete orbit around the Earth

Linear speed of the Moon, v

v

Distance
Time

=

=

=

INFOSMARTSMART

Circumference of a circle
with radius r, is 2πr.

9393

Chapter 3 Gravitation

3.1.63.1.6
4. Scan the QR code, download and print Figure 3.21 from
the website given. Complete it to determine the formula
for the mass of the Earth.

Newton’s Universal Law
of Gravitation Centripetal force

F F=

=

=

=

=

1442443 1442443

1442443

1442443

1442443

Equate two equations

Cancel common factor, m

Substitute v = 2πr
T

Rearrange so that M becomes
the subject of the formula

Mass of the Earth M =

Figure 3.21 Determining formula for the mass of the Earth

Discussion:
1. What is the formula to determine the mass of the Earth?
2. Period of revolution of the Moon around the Earth, T is 2.36 × 106 s and radius of the

Moon’s orbit, r is 3.83 × 108 m. Calculate the mass of the Earth, M.
3. The Earth revolves around the Sun in a period of one year and the radius of the orbit is

1.50 × 1011 m. Calculate the mass of the Sun.

Download Figure 3.21

http://bt.sasbadi.com/p4094

9494 3.1.63.1.6
M = 4π 2r 3

GT 2

• radius of the orbit of any satellite or the
Moon

• period of revolution

• radius of the orbit of any planet
• period of revolution of planet concerned

Formula used to determine the mass of the
Earth or the Sun

Data needed to calculate the mass of
the Earth

Data needed to calculate the mass of
the Sun

Figure 3.22 Formula and data used to calculate the mass of the Earth and the Sun

Formative Practice 3.13.1

1. State Newton’s Universal Law of Gravitation.

2. State two factors which infl uence the magnitude of the gravitational force between
two bodies.

3. A piece of space junk of mass 24 kg is at a distance of 7.00 × 106 m from the centre of the
Earth. What is the gravitational force between the space junk and the Earth?

[G = 6.67 × 10–11 N m2 kg–2, mass of the Earth = 5.97 × 1024 kg]

4. A weather satellite orbits the Earth at a height of 560 km. What is the value of
gravitational acceleration at the position of the satellite?

[G = 6.67 × 10–11 N m2 kg–2, mass of the Earth = 5.97 × 1024 kg, radius of the Earth
= 6.37 × 106 m]

5. A man-made satellite of mass 400 kg orbits the Earth with a radius of 8.2 × 106 m.
  Linear speed of the satellite is 6.96 × 103 m s–1. What is the centripetal force acting on the

satellite?
p

6. Figure 3.23 shows Mercury orbiting the Sun with a radius of 5.79 × 1010 m and a period of
revolution of 7.57 × 106 s. Calculate the mass of the Sun.

Orbit of Mercury

Sun

Mercury

r = 5.79 × 1010 m

Figure 3.23

9595

Chapter 3 Gravitation

3.1.63.1.6
Kepler’s First, Second and Third Laws

When you were in Form 3, you knew about Kepler, a German
astronomist, mathematician and astrologist who modifi ed the
heliocentric model according to Kepler’s Law. Do you know
that there are three Kepler’s Laws? Let us get to know these
three laws.

All planets move in elliptical orbits with the Sun
at one focus (Law of Orbits)

Kepler’s
First Law

Carry out Activity 3.11 to get a clear picture regarding
Kepler’s fi rst law.

Activity 3.11

Kepler’s Laws3.2

HISTORY
INTEGRATION OF

Johannes Kepler worked as
an assistant to astronomist
Tycho Brahe. His strong
determination motivated him
to study Brahe’s astronomical
data. Finally, Kepler succeeded
in formulating three laws that
describe the movement of
planets around the Sun.

Aim: To sketch the shape of an ellipse based on the concept of dual foci of ellipse
Materials: Pencil, 20 cm thread, two thumbtacks, A4 paper, softboard and cellophane tape

Instructions:

1. Scan the QR code and print the template from the website
given on a piece of A4 paper. Place it on a softboard.

2. Stick the thumbtacks at points F1 and F2 on the softboard.

3. Tie two ends of the thread to the two thumbtacks
respectively.

4. Tighten the thread with the tip of a pencil as
shown in Figure 3.24.

5. Move the pencil from the major axis to the
left of F1 to the major axis to the right of F2
to sketch half an ellipse.

6. Repeat step 5 below major axis to obtain the
shape of a complete ellipse.

7. Remove the thumbtacks and thread.

8. Draw a small circle to represent the Sun at F1.

Draw a small circle to represent the Earth on the circumference of the ellipse.

Discussion:

1. Describe how the distance between the Earth and the Sun changes when the Earth
makes a complete orbit around the Sun.

2. Discuss how the shape of the Earth’s orbit would be if the major axis is almost as long
as the minor axis.

Minor axis

Major axis

Thumbtack

Thumbtack

Pencil

Thread

F1 F2

Template Activity 3.11

http://bt.sasbadi.com/p4096

Figure 3.24

9696 3.2.13.2.1
Th e planets in the Solar System have elliptical shaped orbits. Figure 3.25 shows the Sun
always stays on a focus of the ellipse. Th e major axis is longer than the minor axis. Most orbits
of the planets in the Solar System have major axis and minor axis of almost the same length. As
such, the shape of the elliptical orbit of the planets in the Solar System is almost round. Planets
can be assumed to make circular motion around the Sun. Th e radius of orbit is the average
value of the distance between the planet and the Sun.

Minor axis

Planet

Sun
Major
axis

Figure 3.25 Orbit of planet around the Sun

A line that connects a planet to the Sun sweeps out equal areas in equal times
(Law of Areas).

Kepler’s
Second Law

Observe Figure 3.26. If a planet takes the same amount of time to move from A to B and
from C to D, the area AFB is the same as the area CFD. Distance AB is longer than distance CD.
Th is means the planet is moving at a higher linear speed from A to B than from C to D.

Elliptical orbit

B

F

A

D

C

Planet

Sun

F

Figure 3.26 Motion of planet in its orbit

9797

Chapter 3 Gravitation

3.2.13.2.1
Th e square of the orbital period of any planet is
directly proportional to the cube of the radius
of its orbit (Law of Periods).

Kepler’s
Th ird Law

A planet which orbits with a larger radius has a longer
orbital period. As such, planets which are further from the Sun
take a longer time to complete one orbit around the Sun.
 For example, the Earth takes 1 year to make one complete
orbit while Saturn takes 29.5 years. Figure 3.27 shows the orbits
and orbital periods of planets.

Mathematically,
T 2 ∝ r 3

T = orbital period of
a planet

r = radius of orbit

Neptune
(164.8 years)

Uranus
(84.0 years)

Saturn
(29.5 years)

Sun

Mercury
(0.2 years)

Venus
(0.6 years)

Earth
(1.0 year)

Mars
(1.9 year)

MMMMM

E tt

MMMMMMM

SS ttt

UUUU

Jupiter
(11.9 years)

Figure 3.27Figure 3.27 Orbits and orbital periods of planets Orbits and orbital periods of planets

9898 3.2.13.2.1
Kepler’s third law can be formulated using Newton’s Universal Law of Gravitation and
concept of circular motion. Planets make circular motions around the Sun. Th e centripetal
force is the same as the gravitational force between the Sun and the planet. Observe Figure 3.28
which shows the orbit of a planet around the Sun.
 Assuming that the orbit of the planet around the Sun is circular, we can derive the relationship
between the orbital period of the planet and the radius of the orbit as in Kepler’s third law.

Mass of the Sun = M
Mass of the planet = m
Radius of orbit = r
Gravitational force = F
Linear speed of planet = v
Orbital period = T

v

M

r

F

Sun

Planet
m

Figure 3.28 Orbit of a planet

Gravitational force acting on the planet, F = GMm
r 2

Th e gravitational force acts as a centripetal force for the planet to make circular motion around
the Sun.
Centripetal force, F = mv 2

r
Th erefore,
 Centripetal force = Gravitational force

mv 2

r
 = GMm

r 2

v 2 = GM
r

............... [1]

Linear speed of planet, v = Distance travelled in one complete orbit
Orbital period

= 2πr
T

............. [2]

Substitute [2] into [1]:

� 2πr
T �2 = GM

r

T 2 = � 4π 2

GM �r 3

As GM is constant, T 2 ∝ r 3

T 2 ∝ r 3 is Kepler’s third law.

INFOSMARTSMART

Perimeter of orbit = 2πr

9999

Chapter 3 Gravitation

3.2.23.2.2
Example 1

From Kepler’s third law,
relationship between orbital
period, T and radius of
orbit, r is

T 2 = � 4π 2

GM � r 3

Compare two planets.

For planet 1, T1
2 = � 4π 2

GM � r1
3 …………… (1)

For planet 2, T2
2 = � 4π 2

GM � r2
3 …………… (2)

(1) ÷ (2) gives T1
2

T2
2

= r1
3

r2
3

Figure 3.29 shows the formulation of Kepler’s third law. When Kepler’s third law is applied
in the system of planets and the Sun, M is referred to as the mass of the Sun. Kepler’s third law
can also be applied to the system of satellites and the Earth, with M referring to the mass of
the Earth.

the relationship between
orbital period of planet, T
and radius of its orbit, r is

T 2 = � 4π 2

GM � r 3

T 2 = kr 3

where the constant, k = 4π 2

GM

Kepler’s
third law,
that is
T 2 ∝ r 3

centripetal force,

F = mv 2

r

gravitational force,

F = GMm
r 2

v = 2πr
T

As
such

can be
formulated
by equating

Figure 3.29 Formulating Kepler’s Th ird Law

Solving Problems Using Kepler’s Third Law Formula

Th e equation
T1

2

T2
2  =

r1
3

r2
3  can be used to calculate the orbital period, T or radius of orbit, r.

3.2.23.2.2 3.2.33.2.3

Figure 3.30 shows the planets, Earth and Mars, orbiting the Sun.

r1

r2Sun

Sun at the centre

Orbital period
of Earth,T1

Orbital period
of Mars, T2

Figure 3.30

100100
Figure 3.31 shows that a research satellite needs to orbit at a height of 380 km to capture clear
images of the surface of the Earth. What is the orbital period of the satellite?

r1

h1

h2

r2

Earth

Earth at the centre

Moon

Satellite 1

Figure 3.31
[Radius of the orbit of the Moon = 3.83 × 108 m, orbital period of the Moon = 655.2 hours]

(a) Th e radius of the orbit of planet Mars can be determined by comparing the orbit of Mars
with the orbit of the Earth. What information is needed to determine the radius of the
orbit of Mars?

(b) Th e radius of the orbit of the Earth is 1.50 × 1011 m, orbital period of the Earth and Mars
is 1.00 year and 1.88 years respectively. Calculate the radius of the orbit of Mars.

Solution:

(a) Radius of the orbit of the Earth, orbital period of the Earth and orbital period of Mars.
(b)

1.002

1.882
 = (1.50 × 1011)3

r2
3

r2
3 = (1.50 × 1011)3 × 1.882

1.002

r2 = 3 (1.50 × 1011)3 × 1.882

1.002

= 2.28 × 1011 m

���������

Step �
List the given information
in symbols.

Step �
Identify and write down the
formula used.

Step �
Substitute numerical values
into the formula and
perform the calculations.

1
4

2
4

3
1

4
2

4
3

1
4

4
4

4
2

4
4

4
4

3

Radius of orbit of the Earth, r1 = 1.50 × 1011 m
Radius of orbit of Mars = r2
Orbital period of the Earth, T1 = 1.00 years
Orbital period of Mars, T2 = 1.88 years

T1
2

T2
2
 = r1

3

r2
3

INFOSMARTSMART

The equation
T1

2

T2
2 =

r1
3

r2
3  involves

the orbital period of a planet
divided by the orbital period
of another planet. The same
unit needs to be used for
both periods.

Example 2

101101

Chapter 3 Gravitation

3.2.33.2.3
Formative Practice 3.23.2

1. State Kepler’s fi rst law.

2. (a) State Kepler’s second law.
(b) Figure 3.32 shows the orbit of a planet around the Sun. Compare the linear speed of

the planet at positions X, Y and Z.

Sun

Planet

X

Z

Y

Figure 3.32

3. (a) State Kepler’s third law.
  (b) At what height should a satellite be if the satellite is required to orbit the Earth in a

period of 24 hours?
   [Orbital period of the Moon = 27.3 days, radius of orbit of the Moon = 3.83 × 108 m]

Solution:

Radius of orbit of the satellite, r1 = (6.37 × 106) + (380 × 103)
  = 6.75 × 106 m
Radius of orbit of the Moon, r2 = 3.83 × 108 m
Orbital period of the satellite = T1
Orbital period of the Moon, T2 = 655.2 hours
T1

2

T2
2  =

r1
3

r2
3

T1

2

655.22
 =

(6.75 × 106)3

(3.83 × 108)3

T1
2 =

(6.75 × 106)3 × 655.22

(3.83 × 108)3

T1 =
(6.75 × 106)3 × 655.22

(3.83 × 108)3

= 1.53 hours
����������

102102 3.2.33.2.3
Man-made SatellitesMan-made Satellites3.3

Orbit of Satellite
Figure 3.33 shows the International Space
Station, ISS, and MEASAT satellite. ISS can be
seen from the Earth because of its large size and
orbits at a height of 408 km. MEASAT satellite
is diffi  cult to be seen because of its small size
and orbits at a height of 35 786 km. Satellites
move in orbits at specifi c heights and suitable
linear speeds.
 Formulae for centripetal force and
Newton’s Universal Law of Gravitation are
used to establish and determine the linear
speed of satellites. Figure 3.34 shows the orbit
of a satellite around the Earth.

r

M F

v

m

Earth

SatelitSatelliteSS

Mass of the Earth = M
Mass of satellite = m
Radius of orbit of satellite = r
Linear speed of satellite = v
Orbital period = T

Figure 3.34 Orbit of a satellite

A satellite moving in a circular orbit around the Earth experiences
centripetal force, which is gravitational force.

Position and path of ISS

http://bt.sasbadi.com/p4103

MEASATInternational Space
Station (ISS)

Figure 3.33 Man-made satellite orbiting the Earth

Gravitational force between satellite and the Earth, F = GMm
r 2

Centripetal force on satellite, F = mv 2

r
Centripetal force = Gravitational force

mv 2

r  = GMm
r 2

v 2 = GM

r

v = ���GM
r

Astronautical engineering
involves the fi eld of
Physics on orbital
mechanics, outer space
environment, guidance
and control of height,
telecommunications,
aerospace structure and
rocket propulsion.

CAREER
iNFO

103103

Chapter 3 Gravitation

3.3.13.3.1
As GM is constant, linear speed of the satellite only depends on the radius of its orbit. If a
satellite is at a height, h above the surface of the Earth,

Radius of its orbit, r = R + h
that is R = radius of the Earth

Th erefore, linear speed of the satellite, v = ���GM
R + h

Man-made satellites can be launched to keep orbiting at specifi c heights around the Earth at

radius of orbit, r if the satellite is given linear speed, v =  GM
r

. Figure 3.35 shows a Global
Positioning System (GPS) satellite at an altitude of 20 200 km from the Earth.

= 20 200 kmh

r

Earth

Radius of orbit = (Radius of Earth) + (Height of orbit)
r = R + h

Global Positioning System
(GPS) satellite

RR

Figure 3.35 GPS satellite orbits the Earth

Height, h = 20 200 × 1000 m
 = 2.02 × 107 m

Radius of orbit, r = (6.37 × 106) + (2.02 × 107)
 = 2.657 × 107 m

Linear speed
of satellite, v =   GM

r

=   (6.67 × 10–11) × (5.97 × 1024)
2.657 × 107

= 3.87 × 103 m s–1

���

�����������

In a stable orbit, linear speed of satellite is

v =
GM

r���. Th is linear speed is high enough for

the satellite to move in a circular orbit around
the Earth. Centripetal acceleration is the same as
gravitational acceleration.

If the linear speed of the satellite becomes
less than the required linear speed, the satellite
will fall to a lower orbit and continue to revolve
towards the Earth until it enters the atmosphere.
Th e movement of the satellite at a high linear
speed against air resistance will generate heat
and eventually causes the satellite to burn.

How does GPS work?

http://bt.sasbadi.com/p4104

���

104104 3.3.13.3.1
Earth

Satellite B B

Satellite A

Satellite A always
above x

Geostationary
Earth Orbit

Non-geostationary satellite

• Normally in a lower or higher orbit than the Geostationary
Earth Orbit

• Orbital period is shorter or longer than 24 hours
• Above diff erent geographical locations at diff erent times

Figure 3.36 Geostationary and non-geostationary satellites

Geostationary satellite

• In a special orbit named the Geostationary Earth Orbit
• Moves around the Earth in the same direction as the

direction of the Earth’s rotation on its axis
• Orbital period T = 24 hours, that is the same as the period

of rotation of the Earth
• Always above the same geographical location

Geostationary and Non-Geostationary Satellites
Figure 3.36 shows two types of satellites orbiting the Earth, which is geostationary satellite and
non-geostationary satellite. Study the features of the satellites.

Geostationary satellite

http://bt.sasbadi.com/p4105a

Non-geostationary satellite

http://bt.sasbadi.com/p4105b

105105

Chapter 3 Gravitation

3.3.23.3.2
Activity 3.12

Aim: To gather information on geostationary and non-geostationary satellites in terms of
functions and lifespans

Instructions:

1. Work in groups.

2. Surf the internet to gather information on functions and lifespans for one geostationary
satellite and one non-geostationary satellite.

3. Present your fi ndings in the form of a folio and display it at the Resource Centre of your
school.

Discussion:

1. What are the advantages of a non-geostationary satellite?

2. Why do communication satellites need to be in geostationary orbits?

Figure 3.37 shows the comparison between geostationary and non-geostationary satellites
as well as examples of the satellites.

Figure 3.37 Comparison between geostationary and non-geostationary satellites

Geostationary
satellite

Non-
geostationary

satellite

Direction
of motion

same as the
direction
of Earth
rotation

Direction of
motion need not
be the same as
the direction of
Earth rotation

MEASAT

• TiungSAT
• RazakSAT
• Pipit
• ISS

T = 24 hours
T is shorter

or longer than
24 hours

• Earth
imaging

• GPS
• Weather

forecast

Communication
satellite

Above
the same

geographical
location

Above
diff erent

geographical
location

Orbits the
Earth

v =   GM
r���

T =  4π 2r 3

GM
   ����

ICS

106106 3.3.23.3.2
Escape Velocity
Escape velocity, v is the minimum velocity needed by an object
on the surface of the Earth to overcome the gravitational force
and escape to outer space.

If the distance of an object from the centre of the Earth is r,
the mass of the object is m, and the mass of the Earth is M, then

the object possesses gravitational potential energy, U = – GMm
r

.

Figure 3.38 shows an object launched at escape velocity, v.
Th is object can overcome gravitational force and move an infi nite
distance from the Earth.

M

m
Object moving an infinite

distance from the Earth

r

v
Kinetic energy =     mv 2

Earth

1–2

GMm–––––rGravitational potential energy, U = –

Figure 3.38 Object launched at escape velocity

Escape velocity is achieved when the minimum kinetic
energy of an object is able to overcome its gravitational potential
energy. As such:

Escape velocity, v of an object depends on the mass of the Earth, M and distance, r of the
object from the centre of the Earth. Escape velocity does not depend on mass of the object, m.

3.3.33.3.3

F ile
For an object on the surface of
the Earth, its distance from the
centre is the same as the radius
of the Earth, R.

Escape velocity of the object is

v  =    2GM
R����

F ile
As the Earth has a large mass,
escape velocity from the Earth
has a high value, 11 200 m s–1
or 40 300 km h–1.

Minimum kinetic energy + Gravitational potential energy = 0

that is, 1
2mv 2 + �–

GMm
r � = 0

v 2 = 2GM
r

Escape velocity, v = 2GM
r ����

Mass of the Earth, M = 5.97 × 1024 kg
Radius of the Earth, R = 6.37 × 106 m

Escape velocity from the Earth, v =  2GM
R

=  2 × (6.67 × 10–11) × (5.97 × 1024)
(6.37 × 106)

= 1.12 × 104 m s–1

= 11.2 × 103 m s–1

= 11.2 km s–1

����
�������������

Escape velocity

http://bt.sasbadi.com/p4107a

http://bt.sasbadi.com/p4107b

107107

Chapter 3 Gravitation
Benefi ts and Implications of Escape Velocity

Photograph 3.2Photograph 3.2 Rocket launching Rocket launching

High escape velocity of the Earth also enables
commercial aircraft s or fi ghter jets to fl y to high
levels in the atmosphere without the possibility of
escaping into outer space. Commercial aircraft s can
fl y at linear speed of 250 m s–1 while fi ghter jets can
achieve supersonic linear speed of up to 2 200 m s–1.
Both their linear speeds are lower than the escape
velocity from the Earth.

Figure 3.39Figure 3.39 Earth’s atmosphere Earth’s atmosphere

Photograph 3.1Photograph 3.1 Commercial aircraft  Commercial aircraft

Th e launching of rockets requires large quantities of
fuel to produce high thrust that enables the rocket
to achieve escape velocity of the Earth. Hence, it can
send the spacecraft  into outer space.

High escape velocity of the Earth, that is 11 200 m s–1
has benefi ts and implications on humans. One of the
benefi ts is the Earth is able to maintain a layer of
atmosphere around it. Molecules in the atmosphere
move at average linear speed of 500 m s–1, that is lower
than the escape velocity from the Earth. As such, air
molecules that are moving randomly will not be able
to escape from the Earth into outer space.

108108 3.3.33.3.3
Activity 3.13

Aim: To discuss escape velocity from planets

Instructions:

1. Work in pairs.

2. Copy and complete Table 3.4 by calculating the value of escape velocity.

Table 3.4

Planet Mass, M / kg Radius, R / m Escape velocity, v / m s–1

Venus 4.87 × 1024 6.05 × 106

Mars 6.42 × 1023 3.40 × 106

Jupiter 1.90 × 1027 6.99 × 107

Escape velocity of Mars is low, therefore the gases
are easier to escape to outer space. Th is causes the density of
the atmosphere on Mars to be low (100 times less dense than
Earth’s atmosphere). Jupiter on the other hand has such a high
escape velocity that hot gases on its surface cannot escape into
outer space. Knowledge on escape velocity is important to
determine how spacecraft s can land and take off  safely from
a planet.

Recall

Characteristics of
planets in the
Solar System
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Th e Moon and the Sun are two bodies in the Solar System. Table 3.5 shows the values of the
mass and radius of the Moon and the Sun. Compare:
(i) gravitational acceleration on the Moon and the Sun
(ii) escape velocity from the Moon and from the Sun
based on data provided in Table 3.5.

Table 3.5

Body Mass, M / kg Radius, R / m

Moon 7.35 × 1022 1.74 × 106

Sun 1.99 × 1030 6.96 × 108

Solving Problems Involving Escape Velocity

You have calculated escape velocity from the Earth using the formula v =  2GM
R

. In fact, this
formula can also be used to calculate escape velocity from other bodies such as the Moon, Mars and
the Sun.

����

Logical Reasoning CPS

Example 1

109109

Chapter 3 Gravitation

3.3.43.3.4
Solution:

(i) Gravitational acceleration is calculated using the formula g = GM
R2

Th e Moon

g =  (6.67 × 10–11) × (7.35 × 1022)
(1.74 × 106)2

= 1.62 m s–2

Th e Sun

g =  (6.67 × 10–11) × (1.99 × 1030)
(6.96 × 108)2

= 274.0 m s–2

(ii) Escape velocity is calculated using the formula v =  2GM
R

Th e Moon

v =   2 × (6.67 × 10–11) × (7.35 × 1022)
1.74 × 106

= 2.37 × 103 m s–1

Th e Sun

v =   2 × (6.67 × 10–11) × (1.99 × 1030)
(6.96 × 108)

= 6.18 × 105 m s–1

• Th e Moon has low gravitational acceleration and escape velocity because the mass of the
Moon is smaller than that of the Sun.

• Th e Sun is the largest body in the Solar System. Gravitational acceleration on the Sun and
escape velocity from the Sun have the highest values compared with those of the Moon as
well as other planets.

����

����������������

�����������������

Formative Practice 3.33.3

1. Compare and contrast geostationary and non-geostationary satellites.

2. What factors determine the linear speed of satellites orbiting the Earth?

3. State two factors which infl uence the value of escape velocity from a planet.

4. Discuss whether escape velocity from the Earth for spacecraft  X of mass 1 500 kg is
diff erent from spacecraft  Y of mass 2 000 kg.

5. Proba-1 satellite orbits the Earth at a height of 700 km. What is the linear speed of this
satellite?

[G = 6.67 × 10–11 N m2 kg–2, mass of the Earth = 5.97 × 1024 kg, radius of the Earth =
6.37 × 106 m]

110110 3.3.43.3.4
Su
n

–
Pl

an
et

S
ys

te
m

G
eo

st
at

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na

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sa
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lli
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 –

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at

ell
ite

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te
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pl

er
’s

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co

nd
 L

aw

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ne

ar
 sp

ee
d

of
 sa

te
lli

te
, v

=

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pl

er
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ird

L
aw

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2  ∝

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N
ew

to
n’s

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rs

al

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ta
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111111

Chapter 3 Gravitation
SELF REFLECTIONSELF-REFLECTION

1. New things I learnt in this chapter on gravitation are ✎ .

2. Th e most interesting thing I learnt in this chapter on gravitation is ✎ .

3. Th ings I still do not fully understand or comprehend are ✎ .

4. My performance in this chapter ,

Poor  1 2 3 4 5   Excellent

5. I need to ✎  to improve my performance in
this chapter.

Download and print
Self-refl ection Chapter 3

http://bt.sasbadi.com/p4112

1. Figure 1 shows planet Mars orbits the Sun in a circular motion with orbital period, T.

M
r

v

m = mass of Mars
M = mass of the Sun
r  = radius of orbit of Mars
v = linear speed of Mars

Figure 1

(a) For planet Mars, write the formula for:
(i) gravitational force in terms of m, M and r
(ii) centripetal force in terms of m, v and r
(iii) linear speed in terms of r and T

(b) Derive an expression for the mass of the Sun in terms of r and T by using the three
formulae in (a).

(c) Radius of orbit of Mars is r = 2.28 × 1011 m and its orbital period is T = 687 days.
Calculate the mass of the Sun.

Performance EvaluationPerformance Evaluation

112112
2. A satellite orbits the Earth with radius, r and orbital period, T.
(a) Write down the linear speed of the satellite in terms of r and T.
(b) Use other suitable formulae to establish the formula for linear speed of the satellite in

terms of r and M. M is the mass of the Earth.
(c) Why does the linear speed of a satellite orbiting the Earth not depend on the mass of

the satellite?

3. Figure 2 shows the orbit of planet Uranus.

Sun

Uranus

B

A

Figure 2

Describe the change in linear speed of planet Uranus when it moves from point A to
point B.

4. Figure 3 shows the Earth, the Moon and a satellite.

Earth

Moon

Satellite

Figure 3
(a) Which pair of bodies experience the smallest gravitational force?

Give a reason for your answer.
(b) Calculate the gravitational force between the Earth and the satellite.
 [Mass of the Earth = 5.97 × 1024 kg, mass of satellite = 1.2 × 103 kg, distance between

centre of the Earth and centre of the satellite = 7.87 × 106 m]

5. (a) What are the factors that determine the value of the gravitational acceleration?
(b) A satellite is at a distance of 4.20 × 107 m from the centre of the Earth.
 What is the value of the gravitational acceleration at this position?
 [Mass of the Earth = 5.97 × 1024 kg]

113113

Chapter 3 Gravitation
6. Figure 4 shows the Earth and planet Neptune.

r2

r1r1Orbital period
of Neptune = T2 T1

Sun

Orbit of Neptune

Orbit of Earth

Orbital period
of Earth =

Figure 4

(a) Write down the relationship between orbital period and radius of orbit for the Earth
and Neptune.

(b) Orbital period of the Earth is 365 days and radius of orbit of the Earth is 1.50 × 1011 m.
 Calculate the radius of orbit of Neptune if its orbital period is 5.98 × 104 days.

7. Th e Earth orbits the Sun with radius of orbit of 1.50 × 1011 m and orbital period of 1 year.
Radius of orbit of planet Saturn is 1.43 × 1012 m. What is the orbital period of Saturn?

8. A spacecraft  orbits the Earth at a height of 1 600 km. Calculate the escape velocity for
the spacecraft .

[G = 6.67 × 10–11 N m2 kg–2, mass of the Earth = 5.97 × 1024 kg, radius of the Earth
= 6.37 × 106 m]

9. Figure 5 shows planet Saturn with rings made up of small particles around it. Planet
Saturn has a mass of 5.68 × 1026 kg and radius of 6.03 × 107 m.

Rings of planet
Saturn

Figure 5

(a) Calculate the escape velocity of planet Saturn.

(b) Discuss the possibility of the small particles in the rings of planet Saturn escaping into
the outer space.

114114
10. Figure 6 shows three bodies A, B and C. It is given that the gravitational force between A
and B is P.

100 kg

5 m 5 m

A
100 kg

B
200 kg

C

Figure 6

State in terms of P, the gravitational force between
(i) B and C
(ii) A and C

11. Table 1 shows information on three types of orbits X, Y and Z for a satellite orbiting
the Earth.

Table 1

Orbit Shape of orbit Height of orbit / m Orbital period / hours

X Ellipse 6.70 × 103 1.41

Y Circle 3.59 × 107 24.04

Z Circle 5.43 × 107 41.33

A space agency wishes to launch two satellites, P and Q to orbit the Earth. Satellite
P is an Earth imaging satellite that can capture images of various locations on the
surface of the Earth while satellite Q is a communication satellite.

Using the information in Table 1, determine which orbit is suitable for satellite P and
satellite Q. Explain your choice.

12. Assume you are a scientist. Your group has found a new system of bodies. Th is system is
made up of a star at the centre and fi ve planets in a circular orbit around the star. Table 2
shows information on this system of bodies.

Table 2

Body Mass / kg Radius of body / m Radius of orbit / m

Star 5.90 × 1029 6.96 × 108 –

Planet A 2.80 × 1022 1.07 × 106 2.86 × 1010

Planet B 6.30 × 1023 2.30 × 106 9.85 × 1010

Planet C 7.40 × 1022 3.41 × 106 1.15 × 1011

Planet D 4.60 × 1025 1.32 × 107 5.32 × 1011

Planet E 1.90 × 1021 2.42 × 105 2.13 × 1012

115115

Chapter 3 Gravitation
(a) Calculate the gravitational acceleration, escape velocity and orbital period of
each planet.

ggggggggggggg

(b) How do the values of gravitational acceleration, escape velocity and orbital period
infl uence the suitability of a new planet to be inhabited by humans?

(c) Choose the most suitable planet to be inhabited by humans. Give a reason for your
choice.

13. Assuming humans have successfully inhabited planet Mars. You and a group of scientists
are required to invent a system of man-made satellites around Mars. Th ese man-made
satellites consist of weather satellites, planet surface mapping satellites and communication
satellites. Table 3 shows information on planet Mars.

Table 3

Mass / kg 6.42 × 1023

Radius of planet / m 3.40 × 106

Period of revolution / hours 24.6

Based on the information in Table 3, propose the characteristics of the satellite
orbit in terms of orbital height, orbital period, linear speed of satellite, launch base
as well as other suitable factors.

Enrichment Corner
Enrichment Corner

116116
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