SCIENCE100

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.

RAVITATIONRAVITATION

RAVITATION

hapter

Rationalised 2023-24

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 for

ce 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.

F =

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,

∝

m (9.1)

And the force between two objects is inversely

proportional to the square of the distance

between them, that is,

∝

(9.2)

Combining Eqs. (10.1) and (10.2), we get

∝

M m

(9.3)

or,

M × m

F =

(9.4)

where G is the constant of proportionality and

is called the universal gravitation constant.

By multiplying crosswise, Eq. (9.4) gives

= G M

Rationalised 2023-24

SCIENCE102

From Eq. (9.4), the force exerted by the

earth on the moon is

M × m

F =

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

Thus, the force exerted by the earth on

the moon is 2.02 × 10

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 ______________ 9.2

• 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

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

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

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

times smaller.

Example 9.1 The mass of the earth is

6 × 10

kg and that of the moon is

7.4 × 10

kg. If the distance between the

earth and the moon is 3.84×10

km,

calculate the force exerted by the earth on

the moon. (Take G = 6.7 × 10

–11

N m

-2

)

Solution:

The mass of the earth, M = 6 × 10

The mass of the moon,

m = 7.4 × 10

The distance between the earth and the

moon,

d = 3.84 × 10

= 3.84 × 10

× 1000 m

= 3.84 × 10

G = 6.7 × 10

–11

N m

–2

More to know

Rationalised 2023-24

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

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

–

N m

-2

, mass of the earth, M = 6 × 10

kg,

and radius of the earth, R = 6.4 × 10

g =

-11 2 -2

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

• 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

= G

M m

m g

g =

(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,

M × m

mg =

(9.8)

g =

(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

Rationalised 2023-24

SCIENCE104

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 +

(9.11)

= u

+ 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

dir

ection 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

u + v

= (0 m s

–1

+ 5 m s

–1

)/2

= 2.5 m s

–1

(iii) distance travelled, s = ½

a t

= ½ × 10 m s

–2

× (0.5 s)

= ½ × 10 m s

–2

× 0.25 s

= 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

= u

+ 2a s

0 = u

+ 2 × (–9.8 m s

–2

)

× 10 m

–u

= –2 × 9.8 × 10 m

–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?

Rationalised 2023-24

GRAVITATION 105

9.3 Mass

We have learnt in the previous chapter that the

mass of an object is the measure of its inertia .

We have also learnt that greater the mass, the

greater is the inertia. It remains the same

whether the object is on the earth, the moon

or even in outer space. Thus, the mass of an

object is constant and does not change from

place to place.

9.4 Weight

We know that the earth attracts every object

with a certain force and this force depends on

the mass (m) of the object and the acceleration

due to the gravity (g). The weight of an object

is the force with which it is attracted towards

the earth.

We know that

F = m

a, (9.13)

that is,

F = m

g. (9.14)

The force of attraction of the earth on an

object is known as the weight of the object. It

is denoted by W. Substituting the same in Eq.

(9.14), we have

W = m

g (9.15)

As the weight of an object is the force with

which it is attracted towards the earth, the SI

unit of weight is the same as that of force, that

is, newton (N). The weight is a force acting

vertically downwards; it has both magnitude

and direction.

We have learnt that the value of g is

constant at a given place. Therefore at a given

place, the weight of an object is directly

proportional to the mass, say m, of the object,

that is, W

∝

m. It is due to this reason that at

a given place, we can use the weight of an

object as a measure of its mass. The mass of

an object remains the same everywhere, that

is, on the earth and on any planet whereas its

weight depends on its location because g

depends on location.

9.4.1 WEIGHT OF AN OBJECT ON

THE MOON

We have learnt that the weight of an object on

the earth is the force with which the earth

attracts the object. In the same way, the weight

of an object on the moon is the force with

which the moon attracts that object. The mass

of the moon is less than that of the earth. Due

to this the moon exerts lesser force of attraction

on objects.

Let the mass of an object be m. Let its

weight on the moon be W

. Let the mass of

the moon be M

and its radius be

By applying the universal law of

gravitation, the weight of the object on the

moon will be

M m

(9.16)

Let the weight of the same object on the

earth be W

. The mass of the earth is M and its

radius is

Table 9.1

Celestial Mass (kg) Radius (m)

body

Earth 5.98 × 10

6.37 × ×10

Moon 7.36 ××10

1.74 × ×10

From Eqs. (9.9) and (9.15) we have,

M m

(9.17)

Substituting the values from Table 10.1 in

Eqs. (9.16) and (9.17), we get

(

)

7.36 10 kg

1.74

10 m

× ×

2.431 10 G ×

= ×

(9.18a)

and

1.474 10 G ×

= ×

(9.18b)

Dividing Eq. (9.18a) by Eq. (9.18b), we get

2.431 10

1.474 10

0.165

= ≈

(9.19)

Weight of the object on the moon

Weight of the object on theearth 6

Weight of the object on the moon

= (1/6) × its weight on the earth.

Rationalised 2023-24

SCIENCE106

Example 9.4 Mass of an object is 10 kg.

What is its weight on the earth?

Solution:

Mass, m = 10 kg

Acceleration due to gravity, g = 9.8 m s

–2

W = m × g

W = 10 kg × 9.8 m s

-2

= 98 N

Thus, the weight of the object is 98 N.

Example 9.5 An object weighs 10 N when

measur

ed on the surface of the earth.

What would be its weight when

measured on the surface of the moon?

Solution:

We know,

Weight of object on the moon

= (1/6) × its weight on the earth.

That is,

6 6

W = =

= 1.67 N.

Thus, the weight of object on the

surface of the moon would be 1.67 N.

uestions

1. What are the differences between

the mass of an object and its

weight?

2. Why is the weight of an object on

the moon

its weight on the

earth?

9.5 Thrust and Pressure

Have you ever wondered why a camel can run

in a desert easily? Why an army tank weighing

more than a thousand tonne rests upon a

continuous chain? Why a truck or a motorbus

has much wider tyres? Why cutting tools have

sharp edges? In order to address these

questions and understand the phenomena

involved, it helps to introduce the concepts

of the net force in a particular direction (thrust)

and the force per unit area (pressure) acting

on the object concerned.

Let us try to understand the meanings of

thrust and pressure by considering the

following situations:

Situation 1:

You wish to fix a poster on a

bulletin board, as shown in Fig 9.3. To do this

task you will have to press drawing pins with

your thumb. You apply a force on the surface

area of the head of the pin. This force is directed

perpendicular to the surface area of the board.

This force acts on a smaller area at the tip of

the pin.

Fig. 9.3: To fix a poster, drawing pins are pressed

with the thumb perpendicular to the board.

Situation 2: You stand on loose sand. Your

feet go deep into the sand. Now, lie down on

the sand. You will find that your body will not

go that deep in the sand. In both cases the

force exerted on the sand is the weight of your

body.

Rationalised 2023-24

GRAVITATION 107

You have learnt that weight is the force

acting vertically downwards. Here the force is

acting perpendicular to the surface of the sand.

The force acting on an object perpendicular to

the surface is called thrust.

When you stand on loose sand, the force,

that is, the weight of your body is acting on

an area equal to area of your feet. When you

lie down, the same force acts on an area equal

to the contact area of your whole body, which

is larger than the area of your feet. Thus, the

effects of forces of the same magnitude on

different areas are different. In the above

cases, thrust is the same. But effects are

different. Therefore the effect of thrust

depends on the area on which it acts.

The effect of thrust on sand is larger while

standing than while lying. The thrust on unit

area is called pressure. Thus,

thrust

Pressure =

area

(9.20)

Substituting the SI unit of thrust and area in

Eq. (9.20), we get the SI unit of pressure as N/

or N m

–2

In honour of scientist Blaise Pascal, the

SI unit of pressure is called pascal, denoted

as Pa.

Let us consider a numerical example to

understand the effects of thrust acting on

different ar

eas.

Example 9.6 A block of wood is kept on a

tabletop. The mass of wooden block is 5

kg and its dimensions are 40 cm × 20

cm × 10 cm. Find the pressure exerted

by the wooden block on the table top if

it is made to lie on the table top with its

sides of dimensions (a) 20 cm

× 10 cm

and (b) 40 cm × 20 cm.

Solution:

The mass of the wooden block = 5 kg

The dimensions

= 40 cm × 20 cm × 10 cm

Here, the weight of the wooden block

applies a thrust on the table top.

That is,

Thrust = F = m

= 5 kg × 9.8 m s

–2

= 49 N

Area of a side = length × breadth

= 20 cm × 10 cm

= 200 cm

= 0.02 m

From Eq. (9.20),

Pressure =

49 N

0.02 m

= 2450 N m

-2

When the block lies on its side of

dimensions 40 cm ×

20 cm, it exerts

the same thrust.

Area= length × breadth

= 40 cm × 20 cm

= 800 cm

= 0.08 m

From Eq. (9.20),

Pressure =

49 N

0.08 m

= 612.5 N m

–2

The pressure exerted by the side 20 cm

× 10 cm is 2450 N m

–2

and by the side

40 cm × 20 cm is 612.5 N m

–2

Thus, the same force acting on a smaller

area exerts a larger pressure, and a smaller

pressure on a larger area. This is the reason

why a nail has a pointed tip, knives have sharp

edges and buildings have wide foundations.

9.5.1 PRESSURE IN FLUIDS

All liquids and gases are fluids. A solid exerts

pressure on a surface due to its weight.

Similarly, fluids have weight, and they alsoFig. 9.4

Rationalised 2023-24

SCIENCE108

water on the bottle is greater than its weight.

Therefore it rises up when released.

To keep the bottle completely immersed,

the upward force on the bottle due to water

must be balanced. This can be achieved by

an externally applied force acting downwards.

This force must at least be equal to the

differ

ence between the upward force and the

weight of the bottle.

The upward force exerted by the water on

the bottle is known as upthrust or buoyant

force. In fact, all objects experience a force of

buoyancy when they are immersed in a fluid.

The magnitude of this buoyant force depends

on the density of the fluid.

9.5.3 WHY OBJECTS FLOAT OR SINK

WHEN PLACED ON THE SURFACE OF

WATER?

Let us do the following activities to arrive at

an answer for the above question.

Activity ______________ 9.5

• Take a beaker filled with water.

• Take an iron nail and place it on the

surface of the water.

• Observe what happens.

The nail sinks. The force due to the

gravitational attraction of the earth on the

iron nail pulls it downwards. There is an

upthrust of water on the nail, which pushes

it upwards. But the downward force acting

on the nail is greater than the upthrust of

water on the nail. So it sinks (Fig. 9.5).

exert pressure on the base and walls of the

container in which they are enclosed. Pressure

exerted in any confined mass of fluid is

transmitted undiminished in all directions.

9.5.2 BUOYANCY

Have you ever had a swim in a pool and felt

lighter? Have you ever drawn water from a

well and felt that the bucket of water is heavier

when it is out of the water? Have you ever

wondered why a ship made of iron and steel

does not sink in sea water, but while the same

amount of iron and steel in the form of a sheet

would sink? These questions can be answered

by taking buoyancy in consideration. Let us

understand the meaning of buoyancy by

doing an activity.

Activity ______________ 9.4

• Take an empty plastic bottle. Close

the mouth of the bottle with an

airtight stopper. Put it in a bucket

filled with water. You see that the

bottle floats.

• Push the bottle into the water. You feel

an upward push. Try to push it further

down. You will find it difficult to push

deeper and deeper. This indicates that

water exerts a force on the bottle in the

upward direction. The upward force

exerted by the water goes on increasing

as the bottle is pushed deeper till it is

completely immersed.

• Now, release the bottle. It bounces

back to the surface.

• Does the force due to the gravitational

attraction of the earth act on this

bottle? If so, why doesn’t the bottle stay

immersed in water after it is released?

How can you immerse the bottle in

water?

The force due to the gravitational

attraction of the earth acts on the bottle in

the downward direction. So the bottle is

pulled downwards. But the water exerts an

upward force on the bottle. Thus, the bottle is

pushed upwards. We have learnt that weight

of an object is the force due to gravitational

attraction of the earth. When the bottle is

immersed, the upward force exerted by the

Fig. 9.5: An iron nail sinks and a cork floats when

placed on the surface of water.

Rationalised 2023-24

GRAVITATION 109

• Observe what happens to elongation

of the string or the reading on the

balance.

You will find that the elongation of the string

or the reading of the balance decreases as the

stone is gradually lowered in the water. However,

no further change is observed once the stone

gets fully immersed in the water. What do you

infer from the decrease in the extension of the

string or the reading of the spring balance?

We know that the elongation produced in

the string or the spring balance is due to the

weight of the stone. Since the extension

decreases once the stone is lowered in water,

it means that some force acts on the stone in

upward direction. As a result, the net force on

the string decreases and hence the elongation

also decreases. As discussed earlier, this

upward force exerted by water is known as

the force of buoyancy.

What is the magnitude of the buoyant

force experienced by a body? Is it the same

in all fluids for a given body? Do all bodies

in a given fluid experience the same buoyant

force? The answer to these questions is

contained in Archimedes’ principle, stated as

follows:

Activity ______________ 9.6

• Take a beaker filled with water.

• Take a piece of cork and an iron nail of

equal mass.

• Place them on the surface of water.

• Observe what happens.

The cork floats while the nail sinks. This

happens because of the difference in their

densities. The density of a substance is

defined as the mass per unit volume. The

density of cork is less than the density of

water. This means that the upthrust of water

on the cork is greater than the weight of the

cork. So it floats (Fig. 9.5).

The density of an iron nail is more than

the density of water. This means that the

upthrust of water on the iron nail is less than

the weight of the nail. So it sinks.

Therefore objects of density less than that

of a liquid float on the liquid. The objects of

density greater than that of a liquid sink in

the liquid.

uestions

1. Why is it difficult to hold a school

bag having a strap made of a thin

and strong string?

2. What do you mean by buoyancy?

3. Why does an object float or sink

when placed on the surface of

water?

9.6 Archimedes’ Principle

Activity ______________ 9.7

• Take a piece of stone and tie it to one

end of a rubber string or a spring

balance.

• Suspend the stone by holding the

balance or the string as shown in

Fig. 9.6 (a).

• Note the elongation of the string or

the reading on the spring balance due

to the weight of the stone.

• Now, slowly dip the stone in the water

in a container as shown in

Fig. 9.6 (b).

Fig. 9.6: (a) Observe the elongation of the rubber

string due to the weight of a piece of stone

suspended from it in air. (b) The elongation

decreases as the stone is immersed

in water.

(a)

(b)

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SCIENCE110

What

you have

learnt

• The law of gravitation states that the force of attraction

between any two objects is proportional to the product of

their masses and inversely proportional to the square of

the distance between them. The law applies to objects

anywhere in the universe. Such a law is said to be universal.

• Gravitation is a weak force unless large masses are involved.

• The force of gravity decreases with altitude. It also varies

on the surface of the earth, decreasing from poles to the

equator.

• The weight of a body is the force with which the earth

attracts it.

• The weight is equal to the product of mass and acceleration

due to gravity.

• The weight may vary from place to place but the mass stays

constant.

When a body is immersed fully or partially

in a fluid, it experiences an upward force that

is equal to the weight of the fluid displaced

by it.

Now, can you explain why a further

decrease in the elongation of the string was

not observed in activity 9.7, as the stone was

fully immersed in water?

Archimedes was a Greek scientist. He

discovered the principle, subsequently named

after him, after noticing that

the water in a bathtub

overflowed when he stepped

into it. He ran through the

streets shouting “Eureka!”,

which means “I have got it”.

This knowledge helped him to

determine the purity of the

gold in the crown made for

the king.

His work in the field of Geometry and

Mechanics made him famous. His

understanding of levers, pulleys, wheels-

and-axle helped the Greek army in its war

with Roman army.

Archimedes

Archimedes’ principle has many

applications. It is used in designing ships and

submarines. Lactometers, which are used to

determine the purity of a sample of milk and

hydrometers used for determining density of

liquids, are based on this principle.

uestions

1. You find your mass to be 42 kg

on a weighing machine. Is your

mass more or less than 42 kg?

2. You have a bag of cotton and an

iron bar, each indicating a mass

of 100 kg when measured on a

weighing machine. In reality,

one is heavier than other. Can

you say which one is heavier

and why?

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GRAVITATION 111

• All objects experience a force of buoyancy when they are

immersed in a fluid.

• Objects having density less than that of the liquid in which

they are immersed, float on the surface of the liquid. If the

density of the object is more than the density of the liquid in

which it is immersed then it sinks in the liquid.

Exercises

1. How does the force of gravitation between two objects change

when the distance between them is reduced to half ?

2. Gravitational force acts on all objects in proportion to their

masses. Why then, a heavy object does not fall faster than

a light object?

3. What is the magnitude of the gravitational force between the

earth and a 1 kg object on its surface? (Mass of the earth is

6 × 10

kg and radius of the earth is 6.4 × 10

m.)

4. The earth and the moon are attracted to each other by

gravitational force. Does the earth attract the moon with a

force that is greater or smaller or the same as the force

with which the moon attracts the earth? Why?

5. If the moon attracts the earth, why does the earth not move

towards the moon?

6. What happens to the force between two objects, if

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?

7. What is the importance of universal law of gravitation?

8. What is the acceleration of free fall?

9. What do we call the gravitational force between the earth and

an object?

10. Amit buys few grams of gold at the poles as per the instruction

of one of his friends. He hands over the same when he meets

him at the equator. Will the friend agree with the weight of gold

bought? If not, why? [Hint: The value of g is greater at the

poles than at the equator.]

11. Why will a sheet of paper fall slower than one that is crumpled

into a ball?

12. Gravitational force on the surface of the moon is only

strong as gravitational force on the earth. What is the weight

in newtons of a 10 kg object on the moon and on the earth?

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SCIENCE112

13. A ball is thrown vertically upwards with a velocity of 49 m/s.

Calculate

(i) the maximum height to which it rises,

(ii) the total time it takes to return to the surface of the

earth.

14. A stone is released from the top of a tower of height 19.6 m.

Calculate its final velocity just before touching the ground.

15. A stone is thrown vertically upward with an initial velocity

of 40 m/s. Taking g = 10 m/s

, find the maximum height

reached by the stone. What is the net displacement and the

total distance covered by the stone?

16. Calculate the force of gravitation between the earth and the

Sun, given that the mass of the earth = 6 × 10

kg and of the

Sun = 2 × 10

kg. The average distance between the two is

1.5 × 10

17. A stone is allowed to fall from the top of a tower 100 m high

and at the same time another stone is projected vertically

upwards from the ground with a velocity of 25 m/s. Calculate

when and where the two stones will meet.

18. A ball thrown up vertically returns to the thrower after 6 s.

Find

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

19. In what direction does the buoyant force on an object

immersed in a liquid act?

20. Why does a block of plastic released under water come up

to the surface of water?

21. The volume of 50 g of a substance is 20 cm

. If the density of

water is 1 g cm

–3

, will the substance float or sink?

22. The volume of a 500 g sealed packet is 350 cm

. Will the

packet float or sink in water if the density of water is 1 g

–3

? What will be the mass of the water displaced by this

packet?

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