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SCIENCE 100
We have learnt about the motion of objects and
force as the cause of motion. We have learnt
that a force is needed to change the speed or
the direction of motion of an object. We always
observe that an object dropped from a height
falls towards the earth. We know that all the
planets go around the Sun. The moon goes
around the earth. In all these cases, there must
be some force acting on the objects, the planets
and on the moon. Isaac Newton could grasp
that the same force is responsible for all these.
This force is called the gravitational force.
In this chapter we shall learn about
gravitation and the universal law of
gravitation. We shall discuss the motion of
objects under the influence of gravitational
force on the earth. We shall study how the
weight of a body varies from place to place.
We shall also discuss the conditions for
objects to float in liquids.
9.1 Gravitation
We know that the moon goes around the earth.
An object when thrown upwards, reaches a
certain height and then falls downwards. It is
said that when Newton was sitting under a tree,
an apple fell on him. The fall of the apple made
Newton start thinking. He thought that: if the
earth can attract an apple, can it not attract
the moon? Is the force the same in both cases?
He conjectured that the same type of force is
responsible in both the cases. He argued that
at each point of its orbit, the moon falls
towards the earth, instead of going off in a
straight line. So, it must be attracted by the
earth. But we do not really see the moon falling
towards the earth.
Let us try to understand the motion of the
moon by recalling activity 7.11.
Activity ______________9.1
• Take a piece of thread.
• Tie a small stone at one end. Hold the
other end of the thread and whirl it
round, as shown in Fig. 9.1.
• Note the motion of the stone.
• Release the thread.
• Again, note the direction of motion of
the stone.
Fig. 9.1: A stone describing a circular path with a
velocity of constant magnitude.
Before the thread is released, the stone
moves in a circular path with a certain speed
and changes direction at every point.
The change in direction involves change in
velocity or acceleration. The force that causes
this acceleration and keeps the body moving
along the circular path is acting towards
the centre. This force is called the
centripetal (meaning ‘centre-seeking’) force.
9
G G G G GRAVITATION RAVITATION RAVITATION RAVITATION RAVITATION
Chapter
2024-25
GRAVITATION 101
9.1.1 UNIVERSAL LAW OF GRAVITATION
Every object in the universe attracts every
other object with a force which is proportional
to the product of their masses and inversely
proportional to the square of the distance
between them. The force is along the line
joining the centres of two objects.
In the absence of this force, the stone flies off
along a straight line. This straight line will be
a tangent to the circular path.
More to know
Tangent to a circle
A straight line that meets the circle at
one and only one point is called a
tangent to the circle. Straight line
ABC is a tangent to the circle at
point B.
The motion of the moon around the earth
is due to the centripetal force. The centripetal
force is provided by the force of attraction of
the earth. If there were no such force, the
moon would pursue a uniform straight line
motion.
It is seen that a falling apple is attracted
towards the earth. Does the apple attract the
earth? If so, we do not see the earth moving
towards an apple. Why?
According to the third law of motion, the
apple does attract the earth. But according
to the second law of motion, for a given force,
acceleration is inversely proportional to the
mass of an object [Eq. (8.4)]. The mass of an
apple is negligibly small compared to that of
the earth. So, we do not see the earth moving
towards the apple. Extend the same argument
for why the earth does not move towards the
moon.
In our solar system, all the planets go
around the Sun. By arguing the same way,
we can say that there exists a force between
the Sun and the planets. From the above facts
Newton concluded that not only does the
earth attract an apple and the moon, but all
objects in the universe attract each other. This
force of attraction between objects is called
the gravitational force.
G
2
Mm
F =
d
Fig. 9.2: The gravitational force between two
uniform objects is directed along the line
joining their centres.
Let two objects A and B of masses M and
m lie at a distance d from each other as shown
in Fig. 9.2. Let the force of attraction between
two objects be F. According to the universal
law of gravitation, the force between two
objects is directly proportional to the product
of their masses. That is,
F
?
M × m (9.1)
And the force between two objects is inversely
proportional to the square of the distance
between them, that is,
?
2
1
F
d
(9.2)
Combining Eqs. (10.1) and (10.2), we get
F
?

2
× M m
d
(9.3)
or,
G
2
M × m
F =
d
(9.4)
where G is the constant of proportionality and
is called the universal gravitation constant.
By multiplying crosswise, Eq. (9.4) gives
F × d
2
= G M × m
2024-25
SCIENCE 102
From Eq. (9.4), the force exerted by the
earth on the moon is
G
2
M × m
F =
d
11 2 -2 24 22
8 2
6.7 10 N m kg 6 10 kg 7.4 10 kg
(3.84 10 m)
-
× × × × ×
=
×
= 2.02 × 10
20
N.
Thus, the force exerted by the earth on
the moon is 2.02 × 10
20
N.
uestions
1. State the universal law of
gravitation.
2. Write the formula to find the
magnitude of the gravitational
force between the earth and an
object on the surface of the earth.
9.1.2 IMPORTANCE OF THE UNIVERSAL
LAW OF GRAVITATION
The universal law of gravitation successfully
explained several phenomena which were
believed to be unconnected:
(i) the force that binds us to the earth;
(ii) the motion of the moon around the
earth;
(iii) the motion of planets around the Sun;
and
(iv) the tides due to the moon and the Sun.
9.2 Free Fall
Let us try to understand the meaning of free
fall by performing this activity.
Activity Activity Activity Activity Activity 9. 9. 9. 9. 9.2 2 2 2 2
• Take a stone.
• Throw it upwards.
• It reaches a certain height and then it
starts falling down.
We have learnt that the earth attracts
objects towards it. This is due to the
gravitational force. Whenever objects fall
towards the earth under this force alone, we
say that the objects are in free fall. Is there any
or
2
G=
×
F d
M m
(9.5)
The SI unit of G can be obtained by
substituting the units of force, distance and
mass in Eq. (9.5) as N m
2
kg
–2
.
The value of G was found out by
Henry Cavendish (1731 – 1810) by using a
sensitive balance. The accepted value of G is
6.673 × 10
–11
N m
2
kg
–2
.
We know that there exists a force of
attraction between any two objects. Compute
the value of this force between you and your
friend sitting closeby. Conclude how you do
not experience this force!
The law is universal in the sense that
it is applicable to all bodies, whether
the bodies are big or small, whether
they are celestial or terrestrial.
Inverse-square
Saying that F is inversely
proportional to the square of d
means, for example, that if d gets
bigger by a factor of 6, F becomes
1
36
times smaller.
Example 9.1 The mass of the earth is
6 × 10
24
kg and that of the moon is
7.4 × 10
22
kg. If the distance between the
earth and the moon is 3.84×10
5
km,
calculate the force exerted by the earth on
the moon. (Take G = 6.7 × 10
–11
N m
2
kg
-2
)
Solution:
The mass of the earth, M = 6 × 10
24
kg
The mass of the moon,
m = 7.4 × 10
22
kg
The distance between the earth and the
moon,
d = 3.84 × 10
5
km
= 3.84 × 10
5
× 1000 m
= 3.84 × 10
8
m
G = 6.7 × 10
–11
N m
2
kg
–2
More to know
Q
2024-25
GRAVITATION 103
calculations, we can take g to be more or less
constant on or near the earth. But for objects
far from the earth, the acceleration due to
gravitational force of earth is given by
Eq. (9.7).
9.2.1 TO CALCULATE THE VALUE OF g
To calculate the value of g, we should put the
values of G, M and R in Eq. (9.9), namely,
universal gravitational constant, G = 6.7 × 10
–11
N m
2
kg
-2
, mass of the earth, M = 6 × 10
24
kg,
and radius of the earth, R = 6.4 × 10
6
m.
G
2
M
g =
R
-11 2 -2 24
6 2
6.7 10 N m kg 6 10 kg
=
(6.4 10 m)
× × ×
×
= 9.8 m s
–2
.
Thus, the value of acceleration due to gravity
of the earth, g = 9.8 m s
–2
.
9.2.2 MOTION OF OBJECTS UNDER THE
INFLUENCE OF GRAVITATIONAL
FORCE OF THE EARTH
Let us do an activity to understand whether
all objects hollow or solid, big or small, will
fall from a height at the same rate.
Activity 9.3 Activity 9.3 Activity 9.3 Activity 9.3 Activity 9.3
• Take a sheet of paper and a stone.
Drop them simultaneously from the
first floor of a building. Observe
whether both of them reach the
ground simultaneously.
• We see that paper reaches the ground
little later than the stone. This happens
because of air resistance. The air offers
resistance due to friction to the motion
of the falling objects. The resistance
offered by air to the paper is more than
the resistance offered to the stone. If
we do the experiment in a glass jar
from which air has been sucked out,
the paper and the stone would fall at
the same rate.
change in the velocity of falling objects? While
falling, there is no change in the direction of
motion of the objects. But due to the earth’s
attraction, there will be a change in the
magnitude of the velocity. Any change in
velocity involves acceleration. Whenever an
object falls towards the earth, an acceleration
is involved. This acceleration is due to the
earth’s gravitational force. Therefore, this
acceleration is called the acceleration due to
the gravitational force of the earth (or
acceleration due to gravity). It is denoted by
g. The unit of g is the same as that of
acceleration, that is, m s
–2
.
We know from the second law of motion
that force is the product of mass and
acceleration. Let the mass of the stone in
activity 9.2 be m. We already know that there
is acceleration involved in falling objects due
to the gravitational force and is denoted by g.
Therefore the magnitude of the gravitational
force F will be equal to the product of mass
and acceleration due to the gravitational
force, that is,
F = m g (9.6)
From Eqs. (9.4) and (9.6) we have
2
= G
× M m
m g
d
or
G
2
M
g =
d
(9.7)
where M is the mass of the earth, and d is the
distance between the object and the earth.
Let an object be on or near the surface of
the earth. The distance d in Eq. (9.7) will be
equal to R, the radius of the earth. Thus, for
objects on or near the surface of the earth,
G
2
M × m
mg =
R
(9.8)

G
2
M
g =
R
(9.9)
The earth is not a perfect sphere. As the
radius of the earth increases from the poles to
the equator, the value of g becomes greater at
the poles than at the equator. For most
2024-25
SCIENCE 104
We know that an object experiences
acceleration during free fall. From Eq. (9.9),
this acceleration experienced by an object is
independent of its mass. This means that all
objects hollow or solid, big or small, should
fall at the same rate. According to a story,
Galileo dropped different objects from the top
of the Leaning Tower of Pisa in Italy to prove
the same.
As g is constant near the earth, all the
equations for the uniformly accelerated
motion of objects become valid with
acceleration a replaced by g.
The equations are:
v = u + at (9.10)
s = ut +
1
2
at
2
(9.11)
v
2
= u
2
+ 2as (9.12)
where u and v are the initial and final velocities
and s is the distance covered in time, t.
In applying these equations, we will take
acceleration, a to be positive when it is in the
direction of the velocity, that is, in the
direction of motion. The acceleration, a will
be taken as negative when it opposes the
motion.
Example 9.2 A car falls off a ledge and
drops to the ground in 0.5 s. Let
g = 10 m s
–2
(for simplifying the
calculations).
(i) What is its speed on striking the
ground?
(ii) What is its average speed during the
0.5 s?
(iii) How high is the ledge from the
ground?
Solution:
Time, t = ½ second
Initial velocity, u = 0 m s
–1
Acceleration due to gravity, g = 10 m s
–2
Acceleration of the car, a = + 10 m s
–2
(downward)
(i) speed v = a t
v = 10 m s
–2
× 0.5 s
= 5 m s
–1
(ii) average speed
=
2
u +v
= (0 m s
–1
+ 5 m s
–1
)/2
= 2.5 m s
–1
(iii) distance travelled, s = ½ a t
2
= ½ × 10 m s
–2
× (0.5 s)
2
= ½ × 10 m s
–2
× 0.25 s
2
= 1.25 m
Thus,
(i) its speed on striking the ground
= 5 m s
–1
(ii) its average speed during the 0.5 s
= 2.5 m s
–1
(iii) height of the ledge from the ground
= 1.25 m.
Example 9.3 An object is thrown vertically
upwards and rises to a height of 10 m.
Calculate (i) the velocity with which the
object was thrown upwards and (ii) the
time taken by the object to reach the
highest point.
Solution:
Distance travelled, s = 10 m
Final velocity, v = 0 m s
–1
Acceleration due to gravity, g = 9.8 m s
–2
Acceleration of the object, a = –9.8 m s
–2
(upward motion)
(i) v
2
= u
2
+ 2a s
0 = u
2
+ 2 × (–9.8 m s
–2
)

× 10 m
–u
2
= –2 × 9.8 × 10 m
2
s
–2
u =
196
m s
-1
u = 14 m s
-1
(ii) v = u + a t
0 = 14 m s
–1
– 9.8 m s
–2
× t
t = 1.43 s.
Thus,
(i) Initial velocity, u = 14 m s
–1
, and
(ii) Time taken, t = 1.43 s.
uestions
1. What do you mean by free fall?
2. What do you mean by acceleration
due to gravity?
Q
2024-25

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FAQs on NCERT Textbook - Gravitation - Science Class 9

1. What is gravitation and how does it affect objects on Earth?

Ans. Gravitation is a force of attraction between objects with mass. On Earth, the force of gravity pulls objects towards the center of the Earth, giving them weight and keeping them grounded.

2. How does the mass of an object affect the force of gravity acting on it?

Ans. The force of gravity between two objects is directly proportional to the product of their masses. Therefore, as the mass of an object increases, the force of gravity acting on it also increases.

3. What is the difference between weight and mass?

Ans. Mass is the amount of matter in an object and is constant, while weight is the force exerted on an object due to gravity and varies depending on the strength of the gravitational field. Mass is measured in kilograms, while weight is measured in newtons.

4. How does the distance between two objects affect the force of gravity between them?

Ans. According to the law of gravitation, the force of gravity between two objects is inversely proportional to the square of the distance between their centers. As the distance between two objects increases, the force of gravity between them decreases.

5. Can you explain the concept of escape velocity?

Ans. Escape velocity is the minimum velocity an object needs to escape the gravitational pull of a celestial body, such as Earth. It is the velocity required for the object to overcome the gravitational force and move away indefinitely. The escape velocity depends on the mass and radius of the celestial body.

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