First mechanics final

Circular Motion | 151

Where 'a' is the acceleration of the particle moving in the uniform circular motion and it is called
centripetal acceleration. The centripetal acceleration is the acceleration produced due to the centripetal
force.

Expression for Centripetal Acceleration

C
v

Or vQ vQ ∆v
θ vP
Q θ A
r θ B

P -vP

(a) (b)

[Fig. 6.4. Expression for centripetal force]

Let a particle having mass 'm' is moving along a circle of radius 'r' with constant speed 'v' as shown in

→→

figure 6.4 (a). Let the particle moves from point P to Q in small time interval ∆t. Let vP and vQ be the

velocities of the particle at points P and Q respectively such that

→→

|vP | = |vQ | = v
→ →
Here, and have equal magnitudes but different directions. Let 'θ' be the small angular displacement
vP vQ

in small time ∆t. That is, ∠POQ = θ. The angle between two tangents is equal to the angle between the
→→

two corresponding radii. So, angle between vP and vQ is also θ. Now, draw a triangle ABC in such a way

→→ →→
that sides AB and BC represent – vP and vQ respectively as shown in figure 6.4 (b).

→→ →
Here, AB = – vP ;
AB = |– vP | = v

→→ →
BC = vQ ;
BC = |vQ | = v

Using triangle law of vectors to ∆ABC, we get

→→ →
AB + BC = AC

→→ →
or, –vP + vQ = AC

∴ → →→
AC = vQ – vP

This shows that side → represents the small change in velocity → in small time ∆t. Therefore, let,
AC
∆v

A→C → →

= ∆v ; AC = |∆v | = ∆v

Obviously, ∠ABC = θ.

The two isosceles triangles ABC and POQ have equal vertex angle θ. So, these triangles must be similar.

The corresponding sides of similar triangles are proportional. Therefore,

AC BC
PQ = OQ

∆v v
or, PQ = r

Here, PQ is the small linear displacement in small time ∆t with constant speed v.

∴ PQ = v.∆t

152 | Essential Physics

∆v v
Therefore, v.∆t = r

∆v v2
or, ∆t = r

Taking limit as ∆t → 0, we get
lim ∆v v2

∆t → 0 ∆t = r
dv v2
or, dt = r

∴ v2 . . (2)
a= r .

dv lim 0 ∆v , is the centripetal acceleration. Equation (2) is the required expression for
Where, a = dt = ∆t → ∆t

magnitude of centripetal acceleration.

When ∆t → 0 then points P and Q are very close to each other. In such condition → and → will be almost

→ →→ vP vQ
→→
parallel to each other and hence ∆v will be perpendicular to both vP and vQ . Here, vP and vQ are along
→→

the tangents to the circle, so, ∆v must be along the radius (normal) of the circle. The acceleration 'a ' is in
→→

the direction of ∆v . Thus, the centripetal acceleration 'a ' is also directed towards the centre of the

circle along the radius.

Now, from equations (1) and (2), we get . . . (3)
mv2

F= r

Equation (3) gives the magnitude of centripetal force

Using v = rω in equation (3), we get

F = mv2 = mrω2 . . . (4)
r

Examples of Centripetal Force

i. The gravitational force of attraction between Sun and Earth provides the necessary centripetal
force for the Earth to keep on moving around he sun in circular orbit.

ii. For an electron revolving around the nucleus of an atom in circular orbit, the necessary centripetal
force is provided by the electrostatic force of attraction between the protons in the nucleus and
electrons at the orbit.

iii. When a stone tied to a string is whirled in a circle, then the tension in the string provides the
necessary centripetal force.

6.4 Centrifugal Force

An outward force experienced by a particle while moving on a circular path is called centrifugal
force. For example; the passengers of a bus jerk away when the bus takes a turn on a circular path due to
centrifugal force. The magnitudes of centripetal force and centrifugal force are equal but they are directed
in opposite directions. It is also called fictitious force or pseudo force because it comes in to play only
when there is a centripetal force. The centrifugal force is not a real force and it arises due to the inertial
property of the material body.

Centrifuge: The magnitude of centrifugal force is, F = mrω2. For constant r and ω, F ∝ m. The
centrifugal force is directly proportional to the mass of the particle. When a mixture containing light

Circular Motion | 153

particles and heavy particles is set into a circular motion, then the heavy particles move away from the
axis of rotation while the light particles stay close to the axis. This is the principle of centrifuge. The
principle of centrifuge is used to separate cream from milk, to separate honey from wax, to dry clothes in
a drying machine, etc.

6.5 Application of Centripetal Forces

I. Motion in a Horizontal Circle: [Conical Pendulum]

When a system consisting of a small heavy bob suspended by an inextensible string inclined to the
vertical is whirled round in a horizontal circle at a constant speed, then the system of bob and the
string describes a conical path. Such type of system is called a conical pendulum.

Let a particle having mass 'm' is whirled round in a horizontal circle of radius 'r' with constant
speed 'v' by the help of a string inclined to the vertical as shown in figure 6.5. Let 'θ' be the angle
of inclination of the string to the vertical and 'T' be the tension in the string at position A. The
weight 'mg' of the particle acts vertically downwards. The tension 'T' in the string can be resolved
into two rectangular components:

Tcosθ → opposite to weight 'mg' of the particle, and Tsinθ → along the centre B of the circle.

The 'Tcosθ' component balances the weight 'mg' of the particle and 'Tsinθ' component provides the

necessary centripetal force. That is, . . . (1) O
Tcosθ = mg

and, Tsinθ = mv2 . . . (2) θ
r

Dividing equation (2) by equation (1), we get l h

v2 Tcosθ
rg
Tanθ = . . . (3) θ T v

Time period (T) of a conical pendulum: Tsinθ = mv2
From equation (3), we have r
A r
v2 = rgTanθ B

or, v = rgTanθ mg

[Fig. 6.5, Motion in a horizontal
circle: conical pendulum]

or, rω = rgTanθ [ since,v = rω]

or, r2Tπ = rgTanθ [since, ω= 2π ]
T

2πr
or, T =

rgTanθ

∴ T = 2π r . . . (4)
gTanθ

From ∆AOB, we have, Sinθ = AB = r
OA l

or, r = l sinθ . . . (5), where 'l' is the length of the string.

From equations (4) and (5), we get,

T = 2π lsinθ
gTanθ

154 | Essential Physics

∴ Τ = 2π lcosθ . . . (6)
g

This is the required expression for time period of a conical pendulum.

II. Motion in a Vertical Circle

The tension on the string provides the necessary centripetal force for a particle to keep on moving

on a vertical circle. Let a particle having mass 'm' is whirled round in a vertical circle of radius 'r'

by the help of a string as shown in figure 6.6. Let the particle is at position P at any instant. At this

position P, the tension T in string acts along PO and weight 'mg' of the particle acts vertically

downwards. The weight 'mg' can be resolved into 'mgcosθ ' and 'mgsinθ ' components. The

'mgcosθ ' component opposes the tension T on the string and mgsinθ component acts along the

tangent to the circular path. The 'mgsinθ' component continuously decreases the speed as the

particle rises up. At this position P, the force towards the centre O of the circle is, T–mgcosθ,

which provides the necessary centripetal force. Therefore,

T – mgcosθ = mv2 . . . (1)
r
C
Consider four points A, B, C and D of the particle along the vC

circle. Let TA, TB, TC, TD and vA, vB, vC, vD be the TC vB
respective values of tensions and velocities at these four mg

points respectively.

Now, at the lowest point A, θ = 0° D TB B
mvA2 TD
∴ TA – mg cos0° = r O T
vD P
or, TA = mvA2 + mg . . . (2) θ
r TA

This is the required expression for maximum tension in A vA θ

the string.

At highest point C, θ = 180°. So, from equation (1), we get, mg mg
mvC2
TC – mgcos180° = r [Fig. 6.6, Motion in a vertical circle]

∴ TC = mvC2 – mg . . . (3)
r

This is the required expression for minimum tension in the string.

At horizontal points B or D, θ = 90° or 270°. So, from equation (1), we get,
mvB2
∴ TB – mgcos 90° = r

∴ TB = mvB2 . . . (4)
r

Also, TD = mvD2 . . . (5)
r

At the highest point C, if tension is zero, then

TC = 0
mvC2
or, r – mg = 0

∴ vC = rg . . . (6)

This is the minimum velocity of the particle at the highest point C to complete the circle called

critical velocity. If the velocity at the highest point C is less than this value, then the particle will

fall down from the highest point without completing the circle.

Circular Motion | 155

Now, we calculate minimum velocity at the lowest point A. From conservation of energy,

Total mechanical energy at A = Total mechanical energy at C.

or, K.EA + P.EA = K.EC + P.EC

or, K.EA = K.EC + (P.EC – P.EA)

or, 1 mvA2 = 1 mvC2 + mg (2r)
2 2

or, vA2 = vC2 + 4rg [Using equation (6)]
or, vA2 = rg + 4rg

∴ vA = 5rg . . . (7)

This is the required expression for minimum velocity at the lowest point A to complete the

vertical circle.

III. Motion of a Cyclist

In order to take a safe turn at a circular path, a cyclist has to bend a Rcosθ
little from his vertical position as shown in figure 6.7. When the R
cyclist bends, the horizontal component of normal reaction is
θ

directed towards the centre of the circular path and provides the

necessary centripetal force.

Let a cyclist having mass 'm' is moving along a circular path of Rsinθ

radius 'r' with constant speed 'v'. Let 'θ' be the angle of inclination

of the cyclist to the vertical. If the friction between the road and

cycle is ignored, then following two forces are acting on the θ mg
system.

a. Weight 'mg' of the system acting vertically downwards through

centre of gravity of the system, and

b. Reaction 'R' of the road to the system, which acts at an angle 'θ' rO
with the vertical.

The reaction 'R' can be resolved into two rectangular components:

Rcosθ → opposite to 'mg', and

Rsinθ → along the centre O of the circular path. [Fig. 6.7, Banding of cyclist
The 'Rcosθ' component balances the weight 'mg' of the cyclist and at a circular path]
'Rsinθ' component provides the necessary centripetal force. That

is,

Rcosθ = mg . . . (1)
mv2 . . . (2)
and, Rsinθ = r

Dividing equation (2) by equation (1), we get,

Tanθ = v2
rg

or, θ = Tan–1 v2 . . . (3)
rg

It follows from this equation (3) that, the angle θ through which a cyclist should bend would be

small if radius of circular path 'r' is large and speed of cyclist 'v' is small.

IV. Motion of Car on a Circular Level Road

The friction between road and tyres provides the necessary centripetal force for a car (vehicle)
while taking a turn on a circular level road. Consider a car having weight 'mg' is moving on a
circular level road of radius 'r' with constant speed 'v' as shown in figure 6.8. While taking the turn,
the tyres of the car tend to leave the road and move away from the centre of the circular path. So,

156 | Essential Physics

the forces of friction F1 and F2 act inward to the two tyres. If normal reactions of ground on the
tyres be R1 and R2, then, F1 = µR1 and F2 = µR2,
Where 'µ' is the coefficient of friction between the road and the tyres.

The total frictional force F is F = F1 + F2 = µR1 + µR2 = µ(R1 + R2)

or, F = µR . . . (1)

Where, R = R1 + R2 is the total normal reaction of the ground on the car.
The total frictional force 'F' provides the

necessary centripetal force and total normal R1 + R2
reaction R balances the weight 'mg' of the

car. That is, . . . (2) R1 R1
mv2

F= r

and, R = mg . . . (3)

From equation (1), (2) & (3), we get

mv2 = µ.mg F1 F2
r

or, v2 = µrg r

∴ v = µrg . . . (4) mg

This is the maximum limit of speed of a

vehicle (car, bus etc) while taking a turn on a

circular level road of radius 'r' and having

coefficient of friction 'µ'. If the speed exceeds [Fig. 6.8, Motion of car on a circular level road.]

this limit, then the car will skid and go off the road in a circle of radius greater than 'r' because the

maximum available friction will be insufficient to provide the necessary centripetal force.

V. Motion of Car on a Banked Circular Track

If a cyclist is to take a turn, he can bend from his vertical position to provide necessary centripetal

force. However, this is not possible in case of vehicles like car or bus or train. The tilting of the

vehicle can be achieved by raising outer edge of the circular track slightly above the inner edge.

This process is called banking of curved track. The angle through which the outer edge is raised

above the inner edge is called angle of banking. The main purpose of banking is to tilt the normal

reaction towards the centre of the curved track so that its horizontal component provides the

necessary centripetal force.

Let a car having mass 'm' is going round a circular track of radius 'r' with constant speed 'v' as

shown in figure 6.9. Let 'θ' be angle of banking (R1 + R2)cosθ
and 'R' be the normal reaction of the road to the

car. The normal reaction R has two rectangular R1 + R2

components : horizontal component Rsinθ and R1 R2
vertical component Rcosθ. The vertical (R1 + R2)sinθ
component Rcosθ balances the weight 'mg' of θ
the car and the horizontal component Rsinθ

provides the necessary centripetal force. That is, F2

Rcosθ = mg . . . (1)
mv2 . . . (2)
And, Rsinθ = r F1 θ

Dividing equation (2) by equation (1), we get

Tanθ = v2 . . . (3) r
rg mg

or, v2 = rg Tanθ [Fig. 6.9, Motion of car on banked circular track]

Circular Motion | 157

∴ v = rgTanθ . . . (4)

Equation (4) gives the maximum safe speed of the vehicle. In actual practice, some frictional

forces are always present. Therefore, the maximum safe speed is always much greater than that

given by equation (4).

Boost for Objectives

• Motion of particle in circular path with constant speed but variable acceleration is called uniform circular
motion.

• Centrifugal force is a apparent force or pseudo force and directed radially outward.
• When a car driver takes a circular turn and his body is pressed against the side of the car, the force acting on

him is centrifugal force.

• Water doesn't fall from bucket when it is rotated because centrifugal force balances the force of gravity.
• When vehicle moves on horizontal circular track then reaction of inner wheel is less than that of outer wheel.
• A car sometimes overturns while taking a turn; the inner wheel leaves the ground first.
• When a body moves in uniform circular motion no work is done on it.
• A particle is tied to an end of a string, which is whirled in a horizontal circle. If the string is cut, the particle

flies off tangentially.

• The minimum rotational speed so that a coin placed on a rotating table doesn't slip (i.e. move with the table)

is ωmin = µg . [Hint: µmg > mω2r]
r

• In uniform circular motion, velocity and acceleration act always at right angle.
mv2 mv2
• In vertical circle, maximum tension is Tmax = r + mg, and minimum tension is, Tmin = r – mg.

• The ratio of angular speeds of minnute hand and hour hand clock is 12:1. That is, ωhour = 2: 1
ωearth

Short Questions with Answers

1. If there is a net force on a particle in uniform circular motion, why does the particle's speed not
Ans:
2. change? [HSEB sample question]
Ans:
The force acting on a particle in uniform circular motion is centripetal force along the centre of the
3.
Ans: circle, which is always perpendicular to the direction of velocity of particle. Such a perpendicular force
4.
can change direction of velocity but not its magnitude (speed of the particle). The speed of the particle

can change only when the force acts on the particle at any angle θ with its velocity except θ = 90°.

A curve in a road has the banking angle calculated for 80 km/hr. However, the road is covered

with ice and you plan to creep around the highest lane at 20 km/her. What may happen to your

car? Why? [HSEB sample question]

The friction is much reduced due to presence of ice on a road. To balance the car, the centripetal force
mv2

[F = r ] must be decreased. Since the speed decreases from 80 km/hr to 20 km/hr and at highest lane,

the radius increases. Thus the necessary centripetal force decreases and hence reduced friction in the ice

covered road can provide the centripetal force and the car will not skid during creep around the highest

lane.

Why is it more difficult to revolve a stone by tying it to a longer string than by tying it to a shorter

string? [HSEB 2072 C]

The necessary centripetal force F for a stone of mass m to revolve in a circle of radius r with constant
angular velocity ω is, F = mrω2. Here, for constant angular velocity ω, F ∝ r. The longer string revolves

in a circular path with larger radius, it requires more centripetal force F than that with a shorter string.

So, it is more difficult to revolve a stone by tying it to a longer string than by tying it to a shorter string.

In a washing machine, there is a spin dryer, which removes the water from clothes easily. What

could be its working principle? Justify. [HSEB 2071 S]

158 | Essential Physics

Ans: In a washing machine, there is a spin dryer which works on the principle of centrifuge. The magnitude

5. of centrifugal force is, F = mrω2. For constant r and ω, F ∝ m. When a mixture containing light
Ans:
particles and heavy particles is set into a circular motion, then the heavy particles move away from the
6.
Ans: axis of rotation while the light particles stay close to the axis. This is the principle of centrifuge, which

is used to remove the water from clothes easily in a spin dryer.

Explain the significance of banking of a curved road. [HSEB 2071, 2067, 2062, 2059]

If a cyclist has to take a turn, he can bend from his vertical position to provide the necessary centripetal

force. However, this is not possible in case of vehicles like car or bus or train. The tilting of vehicles

can be achieved by raising outer edge of the circular track slightly above the inner edge. This process is

called banking of curved track. The angle through which the outer edge is raised above the inner edge

is called angle of banking. The main purpose of banking is to tilt the normal reaction towards the

centre of the curved track so that its horizontal component provides the necessary centripetal force.

Explain the difference between a simple and a conical pendulum. [HSEB 2071]

The differences between simple and conical pendulum are:

Simple Pendulum Conical Pendulum

1. It is a heavy point mass suspended on a 1. It is simply a simple pendulum whired in a

rigid support by an inextensible, flexible horizontal circle.

and weightless string.

2. Motion is to and fro. i.e. SHM. 2. Motion is circular.

3. Acceleration is always directed towards 3. Acceleration is always directed towards

mean position. centre of the circle.

4. Time period is, T = 2π l 4. lcos θ
g g
Time period is, T = 2π

7. Why does a cyclist lean from vertical while turning on curved track? [HSEB 2070, 2055, 2054]
Ans:
In order to take a safe turn at a circular path, a cyclist has to bend a little from his vertical position.
8.
Ans: When the cyclist bends, the normal reaction of the road makes an angle with the vertical. The horizontal

9. component of normal reaction is directed towards the centre of the circular path, which provides the
Ans:
necessary centripetal force to move the cyclist in the curved track.
10.
Ans: The positively charged nucleus of an atom attracts the electrons in the orbit. Why do electrons not

collapse into the nucleus? [HSEB 2063, 2056]

The necessary centripetal force for electrons to keep on moving around the charged nucleus of an atom

is provided by electric force of attraction between the protons at the nucleus and electrons at the orbit.

The direction of velocity (displacement) and centripetal force are perpendicular to each other (θ = 90°)

and the work done by centripetal force is zero. [W =→F .→S = F cos 90° = 0]. Hence, the electrons do not

collapse into the nucleus.

What is the source of centripetal force to a satellite revolving round the earth? [HSEB 2060]

The gravitational force of attraction between the earth and satellite is the source of centripetal force to a

satellite revolving around the earth. That is,

Centripetal force = Gravitational force
Mv2 GMm

or, r = r2

Where, M = mass of earth, m = mass of satellite.

v = orbital velocity of satellite, r = radius of orbit of satellite.

What is meant by angular velocity? [HSEB 2053]

The rate of change of angular displacement is called angular velocity and it is denoted by ω.

Mathematically,

ω = lim 0 ∆θ = dθ
∆t → ∆t dt

The S.I. Unit of angular velocity is rad s–1 and it is a vector quantity. It is represented by an arrow

drawn along the axis of rotation. It's dimensional formula is [M0L0T–1]

Circular Motion | 159

11. When a bus takes a turn, passengers are thrown away from the centre of the curved path. Why?
Ans:
12. An outward force experienced by a particle while moving on a circular path, is called centrifugal force.
Ans:
It is not a real force and it arises due to inertial property of the material body. When a bus takes a turn
13.
Ans: passengers are thrown away from the centre of the curved path because of centrifugal force.

14. A solid tied at the end of a string is revolved in vertical circle. At what point the tension in the
Ans:
string will be greatest? [HSEB 2052]
15.
Ans: When a solid having mass m is revolved in a vertical circle of radius r with constant speed v, then at the

16. highest point A, Vm
Ans: mv2 A
mg
TA + mg = r

∴ mv2 . . . (1) O TA
TA = r – mg TB

At the lowest point B, B V
Mv2

TB – mg = r

∴ Mv2 . . . (2) mg
TB = r + mg

From equations (1) and (2), we see that tension is maximum (greatest) at the lowest end B and

minimum (lowest) at highest end A of the vertical circle.

In uniform circular motion, the acceleration is perpendicular to the velocity at every instant. Is

this still true when the motion is not uniform, that is, when the speed is not constant? [HSEB 2068]

In uniform circular motion, the particle will possess centripetal acceleration

only which is always perpendicular to the velocity of particle at any instant. aT

But, in case of non–uniform circular motion, the particle will possess both θ

centripetal acceleration (→aC) and transverse acceleration (→aT) as shown in a
the figure. Then the magnitude of resultant (total) acceleration →a is given by,
a = ac2 + aT2, according to parallelogram law of vector addition. Obviously, ac
O

in this case of non–uniform circular motion, acceleration (→a ) is not

perpendicular to the velocity at every instant.

Does the acceleration of a body moving in a circular path remain constant? [HSEB Model]
v2
Acceleration is a vector quantity. The magnitude of acceleration in a uniform circular motion is a = r ,

which remains constant, But, its direction is always towards the centre of circular path which is
different at different points of the circular path. This centripetal acceleration is due to change in
direction of velocity of the particle in uniform circular motion. Thus, the acceleration of a body moving
in a circular path doesn't remain constant, only the magnitude of acceleration is constant.

A bucket containing water is rotated in a vertical circle with a velocity such that water does not

spill out even at the highest point of the circle. Is it possible? How? [HSEB 2062]

Yes. For the revolution of the bucket containing water in a vertical circle, a centripetal force is required.

The weight of water due to which water can fall is utilized in providing the necessary centripetal force

and the water doesn't spill out even at the highest point of the circle. For this, the minimum velocity at

the highest point should be rg and that at the lowest point should be 5rg, where r is the radius of the
circular path and g is acceleration due to gravity.

What is the direction of the angular velocity of the minute hand of a wall–clock?
The angular velocity (ω) is a vector quantity. It is represented by an arrow drawn along the axis of
rotation. The direction of arrow (i.e. ω) depends on the sense of rotation and is given by right hand
thumb rule. When the fingers of right hand are curled in the sense of revolution, then the thumb held
perpendicular to the curled fingers points in the direction of angular velocity. So, the direction of
angular velocity of minute hand of a wall–clock is normal to the wall and is directed inwards.

160 | Essential Physics

17. A motor cyclist is going to a vertical circle. What is the necessary condition so that he may not fall
Ans:
down?
18. mv2
Ans:
For the motor cyclist to loop safely in a vertical circle, the necessary centripetal force [ r ] should be at
19.
Ans: least equal to the weight (mg) of the motor cyclist. That is,
20. mv2
Ans: r ≥ mg

21. ∴ v2 ≥ g
Ans: r

v2
The speed of the motor cyclist should be high enough to keep centripetal acceleration [ r ] more than or

equal to acceleration due to gravity [g].

What is the angular velocity of minute hand of a watch?

For minute hand of a watch, the time taken to complete one revolution is,

T = 1 hour = 60 minute = 60 × 60 sec.

So, its angular velocity is, ω = 2π = 2π = π rad/sec.
T 60 × 60 1800

The magnitude and direction of acceleration of a body are constant. Will the path of the body

necessarily be a straight line?

If the magnitude and direction of acceleration of a body are constant, it is not necessary that the path of

the body be a straight line. For example, in projectile motion, the acceleration is equal to the
acceleration due to gravity g which is constant in magnitude (9.8 ms2) and direction (vertically

downward) but its path is not a straight line. The path of projectile is a parabola.

A plane has to tilt slightly if it has to change its direction, why?

A plane has to tilt slightly if it has to change its direction for the generation of centripetal force. When

the plane is tilted with angle θ with vertical, the reaction force R of air on the plane will generate

horizontal component Rsinθ and vertical component Rcosθ. The horizontal component Rsinθ provides
mv2

the necessary centripetal force r which helps the plane in achieving a circular path. The vertical

component Rcosθ helps in balancing the weight mg of the plane.

Is it correct to say that the banking of road reduces the wear and tear of the tyres of automobiles?

If yes, explain.

yes, when the road is not banked, then the frictional force between the road and the tyres provides the

necessary centripetal force for the automobiles. For a banked road, horizontal component Rsinθ of

normal reaction R provides the necessary centripetal force. Thus, banking of road reduces the friction

and which reduces the wear and tear of the tyres of the automobiles.

Numerical Examples

1. A stone with mass 0.80kg is attached to one end of a string 0.9m long. The string will break if its

tension exceeds 600N. The stone is whirled in a horizontal circle, the other end of the string remains

fixed. Find the maximum speed, the stone can attain without breaking the string. [HSEB 2072 D]

Solution:

Given, mass of stone (m) = 0.80 kg ; length of string (l) = 0.9 m

Radius of horizontal circle (r) = l = 0.9 m; maximum Tension (T) = 600N.

Maximum speed (v) = ?

The tension on the string provides the necessary centripetal force. Thus,

centripetal force = Tension on string.
mv2
or, r = T

Circular Motion | 161

∴ rT 0.5 × 600
v= m = 0.80 = 25.98 m/s.

The angular speed (ω) is,

ω = v 25.98 = 28.87 rad/s.
r = 0.9

Also, ω = 2πf

∴ Again frequency (f) = ω = 28.87 = 4.6 revs/s.
2π 2π

2. An object of mass 4 kg is whirled round a vertical circle of radius 1m with a constant speed of 3 ms–1.

Calculate the maximum tension in the string. [HSEB 2072 E]

Solution:

Given, mass of object (m) = 4kg constant speed (v) = 3 ms–1
Radius of vertical circle (r) = 1m;

Maximum tension (Tmax) = ?

The maximum tension is at the bottom of the vertical circle and it is given by
mv2 4 × 32

Tmax = r + mg = 1 + 4 × 10 = 36 + 40 = 76N

3. A mass of 0.2 kg is whirled in a horizontal circle of radius 0.5m by a string inclined at 30° to the
vertical. Calculate, (i) the tension in the string, and (ii) the speed of the mass in the horizontal circle.

Solution: [HSEB Sample question]

Given, Mass (m) = 0.2 kg B
θ T Tcosθ
Radius of circle (r) = 0.5 m
vθ
Angle of inclination (θ) = 30° rA

(i) Tension in the string (T) = ? O Tsinθ
mg
(ii) Speed of mass (v) = ?

From figure,

Tcosθ = mg . . . (1)
mv2 . . . (2)
and, Tsinθ = r

mg 0.2 × 10
From (1), T = cosθ = cos30° = 2.31N.

From (2), v2 = Tsinθ × r 2.31 × sin30° ×0.5 = 2.89
m = 0.2

∴ v = 2.89 = 1.70 ms–1

4. A mass of 1 kg is attached to the lower end of a string 1m long whose upper end is fixed. The mass is

made to rotate in a horizontal circle of radius 60cm. If the circular speed of the mass is constant, find

the tension in the string and the period of motion. [HSEB 2070 C]

Solution:

Given, mass (m) = 1 kg

From figure of question no. (3), length of string (l) = AB = 1m

radius of horizontal circle (r) = OA = 60cm = 0.60m, tension in the string (T) = ?

Period of motion (T1) = ?

We have, Tcosθ = mg …..(1)
mv2 …..(2)
and, Tsinθ = r

From ∆AOB, sinθ = OA r 0.60 = 0.60.
AB =l =1

162 | Essential Physics

∴ θ = sin–1 (0.60) = 36.87°
mg 1×10

Now, from equation (1), T = cosθ =cos36.87° = 12.50N

The period of motion is, T1 = 2π lcosθ = 2π 1 × cos36.87°
g 10 = 1.78 secs.

5. A certain string breaks when a weight of 25N acts on it. A mass of 500gm is attached to one end of

the string of 1m long and is rotated in a horizontal circle. Find the greatest number of revolutions

per minute, which can be made without breaking the string. [HSEB 2070 D]

Solution:

Maximum tension on string (T) = 25N, mass of object (m) = 500 gm = 0.500 kg

Radius of circle (r) = l = 1m, frequency (f) = ?

The tension on string provides the necessary centripetal force. That is,

Centripetal force = Tension on the string.

or, mrω2 = T

or, mr(2πf)2 = T [since,ω = 2πf]

or, f2 = T 25 = 1.27
4π2mr = 4π2 × 0.500×1

∴ f = 1.27 = 1.13 rev/sec = 1.13 × 60 rev/min = 67.52 rpm

6. A mass of 0.2kg is rotated by a string at a constant speed in a vertical circle of radius 1m. If the

minimum tension in the string is 3N, calculate the magnitude of the speed and the maximum tension

in the string. [HSEB 2071 C]

Solution:

Given, Mass (m) = 0.2 kg; radius of vertical circle (r) = 1m

Minimum tension (Tmin) = 3N, magnitude of speed (v) = ?

Maximum tension (Tmax) = ?

The tension is minimum at the top and maximum at bottom of the vertical circle.

mv2
Since, Tmin = r – mg

0.2 × v2
or, 3 = 1 – 0.2 × 10

or, v2 = 3+2 5 = 25
0.2 = 0.2

∴ v = 5 ms–1
mv2 0.2×52

And, Tmax = r + mg = 1 + 0.2 × 10 = 5 + 2 = 7N

7. An object of mass 8.0 kg is whirled round in a vertical circle of radius 2m with a constant speed of 6

ms–1. Calculate the maximum and the minimum tensions in the string. [HSEB 2069]

Solution:

Given, mass of object (m) = 8.0 kg, radius of vertical circle (r) = 2m
Constant speed (v) = 6 ms–1, maximum Tension (Tmax) = ?

Minimum Tension (Tmin) = ?

The tension is maximum at the bottom and minimum at the top of a vertical circle. We have,
mv2 8×62

Tmax = r + mg = 2 + 8×10 = 144 + 80 = 224 N

mv2 8×62
And, Tmin = r – mg = 2 – 8×10 = 144 – 80 = 64N.

Circular Motion | 163

8. At what angle should a circular road be banked so that a car running at 50 km/hr be safe to go round

the circular turn of 200m radius? [HSEB 2071 D]

Solution:

50 × 1000 125
Given, Speed of car (v) = 50km/hr = 60 × 60 m/s = 9 m/s.

Radius of circle (r) = 200m; angle of banking of road (θ) =?

We have, Tanθ = v2 (125/9)2 = 0.0965
rg = 200×10

∴ θ = Tan–1 (0.0965) = 5.51°

9. 1 rev. min–1 provided that the coin is not more than
A coin placed on a disc rotates with speed of 333

10 cm. from the axis. Calculate the coefficient of static friction between the coin and the disc.

[HSEB 2052]

Solution: R

1 rev.min–1 = 100 5 rev/s.
Given, Frequency (f) = 33 3 3×60 revls = 9

Radius (r) = 10 cm = 0.10m; coefficient of static friction (µs) = ? v

The necessary centripetal force for coin is provided by frictional force between Ff
mv2
coin and disc. That is, Ff = r . . . (1)

Also, Reaction (R) is, R = mg . . . (2) mg

We have,

µs = Fs mv2/r v2 (rω)2 rω2 r(2πf)2 4π2rf2 4π2 × 0.10 × (5/9)2 = 0.122
R = mg = rg = rg =g =g =g = 10

Important Numerical Problems

1. A train has to take a circular turn of radius 500m with a speed of 36 km/hr. By how much should the

outer rail be raised above the inner rail so that there is no side pressure on the rails? The distance

between the rails is 1m. [T.U. 2061]

Solution:

Given, radius of circular path (r) = 500m

36 × 1000 = 10ms–1 h=?
Speed (v) = 36 km/hr = 60 × 60

Let, outer rail should be raised above the inner rail by 'h'. θ
l = 1m
Distance between the rails (l) = 1m.

We have, tanθ = v2
rg

h v2
or, l = rg

∴ v2l 102 × 1 100
h = rg = 500 × 10 = 50000 = 0.02m.

2. A stone of mass 5 kg is attached to a string 8m long and is whirled in a horizontal circle. The string

can stand maximum tension 150N. What is the greatest number of revolutions per second of the

stone? [T.U. 2052]

Solution:

Given, mass of stone (m) = 5 kg; radius of horizontal circle (r) = l = 8m

Maximum tension (T) = 150N, greatest no. of revolutions per second (f) = ?

164 | Essential Physics

We have, Centripetal force = Tension in the string. [since,v = rω]
mv2 [since,ω = 2πf]

or, r = T

m(rω)2
or, r = T

or, mr (2πf)2 = T

∴ T 150
f = 4π2 mr = 4π2 × 5 × 8 = 0.31 rev/sec.

3. A spaceman in training is rotated in a seat at the end of a horizontal rotating arm of length 6m. If he

can withstand accelerations upto 8g, what is the maximum number of revolutions per second

permissible? [T.U. 2050]

Solution:

Given, Radius (r) = 6m; maximum acceleration (a) = 8g

Greatest number of revolutions per second (f) = ?

Let 'm' be the mass of spaceman. Then, the required centripetal force F is,
F = mrω2

Also, F = ma

∴ mrω2 = ma
or, r(2πf)2 = 8g [since,ω = 2πf, and a = 8g]

∴ 8g 8 × 10
f = uπ2r = 4π2×6 = 0.58 rev/sec

4. A stone of mass 500g is attached to a string of length 50cm which will break if the tension in it

exceeds 20N. The stone is whirled in a vertical circle, the axis of rotation being at height of 100cm

above the ground. The angular speed is very slowly increased until the string breaks. In what

position is this break most likely to occur, and at what angular speed? Where will the stone hit the

ground? [T.U. 2057]

Solution:

Given, mass of stone (m) = 500g = 0.500kg; radius of vertical circle (r) = l = 50cm = 0.50 m

Maximum tension (Tmax) = 20N.
Height of centre of circle 0 above the ground (h') = 100cm = 1m

The break is most likely to occur at the bottom of the vertical circle. Let, ω be the angular speed when

the string breaks. O
mv2

We have, Tmax = r + mg.

or, Tmax = mrω2 + mg u
or, 20 = 0.500 × 0.50 × ω2 + 0.500 ×10 h

or, ω2 = 20 – 5 = 60
0.25

∴ ω = 60 = 7.75 rads–1 R

Second case

When the string breaks, height above ground (h) = h' – r = 1 – 0.5 = 0.5m

Let, u be the horizontal velocity of the stone when the string breaks. The stone follows a parabolic path.

So, its horizontal range (R) is,

R =u 2h = rω 2h
g g

= 0.50 × 7 .75 × 2 × 0.5
10 = 1.225m

Circular Motion | 165

5. Find the maximum speed of vehicle moving on a circular track without skidding, if the radius of
circular path is 80m and the coefficient of friction between the road and the tyres of vehicles is 0.8

[T.U. 2064]

Solution:

Given, radius of circular path (r) = 80m; coefficient of friction (µ) = 0.8

Maximum speed of vehicle (v) = ?

Here, centripetal force = Frictional force

mv2
or, r = Ff

or, mv2 = µR [since,Ff = µR]
r

or, mv2 = µmg [since, R = mg]
r

∴ v = µrg = 0.8 × 80 × 10 = 25.30 ms–1

6. In a test of "g–suit", a volunteer is rotated in a horizontal circle of radius 7.0m. What is the period of
rotation at which the centripetal acceleration has a magnitude of (a) 3.0g ? (b) 10g?

Solutions:

Given, radius of horizontal circle (r) = 7.0m; period of rotation (T) = ?

(a) Centripetal acceleration (a) = 3.0 g
We have, a = rω2

or, ω= a
r

or, 2π = a
T r

∴ T = 2π r = 2π 7.0 7.0
a 3.0g = 3.0 × 10 = 3.04 sec.

(b) Centripetal acceleration (a) = 10g.

T = 2π r
a

= 2π 7.0 = 2π 7.0
10g 10×10 = 1.66 sec.

Practice Short Questions

1. Roads are banked at the turning, why? [HSEB 2067]

2. Explain why a cyclist inclines himself to the vertical while moving round a circular path? [HSEB 2054]

3. Can the direction of velocity of a body be changed when its acceleration is constant? [HSEB 2069]

4. Why are roads banked on curved path? [HSEB 2059]

5. A stone tied to the end of a string is whirled in a horizontal circle. When the string breaks, the stone flies
away tangentially, why?

6. If both the speed of a body and radius of its circular path are doubled, what will happen to centripetal
acceleration?

7. What is the angular velocity of earth about its own axis?

8. What is the angular velocity of hour hand of a watch?

9. For uniform circular motion, does the direction of centripetal force depend on the sense of rotation (i.e.
clockwise or anticlockwise)? Explain

10. Cream separates out when milk is churned. Why?

166 | Essential Physics

11. Why does a pilot not fall down, when the airplane loops a vertical loop?
12. Why is the earth bulged at the equator and flattened at the poles?
13. What are the differences between centripetal force and centrifugal force?
14. What physical quantities remain constant for a particle moving along a circular path in a horizontal plane?
15. The speed of driving a car safely depends upon the range of headlight. Explain.

Practice Long Questions

1. Define centripetal and centrifugal forces. Derive an expression for the force acting on a body moving with

uniform speed along a circular path. [HSEB 2072 C]

2. What do you mean by the banking of a curved path? Derive an expression for the banking angle. [HSEB 2071 D]

3. Define centripetal force. Show that the acceleration of a body moving in a circular path of radius 'r' with

uniform speed v is v2/r and is directed towards the centre of circular path. [HSEB 2070 A]

4. Define centripetal force. Show that the acceleration of a body moving in a circular path of radius 'r' with

uniform speed v is v2/r and draw a diagram to show the direction of the acceleration. [HSEB 2069 S]

5. Define centripetal force. Calculate the force acting on a body moving with a uniform speed along circular

path. [HSEB 2069 Old B]

6. Define centripetal force. Discuss the motion of a car moving round in a circular banked track.

[HSEB 2068 2056 ]

7. Define the centripetal force. Derive an expression for the force acting on a body moving with uniform speed

along a circular path. [HSEB 2066 Old]

8. What is conical pendulum? Show that the period of oscillation of this pendulum is give by T = 2π lcosθ
g

where symbols have their usual meanings. [NEB 2074, HSEB 2066, 2068]

9. What is centripetal force? Derive it in the case of motion of a bi–cycle on a curved road. [HSEB 2062]

10. Explain what is meant by angular velocity. Show that the acceleration of a body moving in a circular path of

radius r with uniform speed v is v2/r. [HSEB 2061]

11. Define centripetal acceleration. Derive a expression for it. [HSEB 2068]

12. Derive an expression for the force required to make a particle of mass m move in a circle of radius r with

uniform angular velocity ω. [ HSEB 2053]

13. Why a force is necessary to keep a body moving with uniform speed in a circular motion? Deduce its

expression. [HSEB 2051]

Practice Numerical Questions

1. A bucket of water having total mass 3 kg is whirled around the vertical circle of radius 1.5 m with speed of 6

ms–1. Find the maximum and minimum tensions. [TU. 2038] Ans: 102 N, 42 N]

2. What force is necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius 0.7 m with a period

of 0.5 s? What is the direction of this force? [An: 90 N]

3. A spacemen in training is rotated in a seat at the end of a horizontal rotating arm of length 5m. If he can
withstand acceleration upto 9g, what is the maximum number of revolutions per second permissible?
[Ans: 0.675 rev s–1 ]

4. The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) What is the radial
acceleration of an object at the earth's equator? Give your answer in m/s2 and as a fraction of g.

[Ans: 0.034ms–2, 3.4 × 10–3 g]

5. A model of a helicopter rotor has four blades, each 3.40 m in length from the central shaft to the blade tip.
The model is rotated in a wind tunnel at 550 rev/min. (a) What is the linear speed of the blade tip, in m/s? (b)
What is the radial acceleration of the blade tip expressed as a multiple of the acceleration due to gravity, g?
[Ans: 195.82 ms–1, 1151 g]

Circular Motion | 167

6. The radius of the earth's orbit around the sun (assumed to be circular) is 1.50 × 108 km, and the earth travels

around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth in m/s? (b) What is
the radical acceleration of the earth towards the sun in m/s2 ? Repeat parts (a) and (b) for the motion of the
planet Mercury (orbit radius = 5.79 × 107 km, orbital period = 88.0 days.)

[Ans: 2.99 × 104ms–1, 5.96 ×10–3ms–2, 4.78 × 104 ms–1, 0.04 ms–2]

7. An object of mass 4 kg moves round a circle 6 m with a constant speed of 12 ms–1. Calculate (i) the angular

velocity, (ii) the force towards the centre. [Ans: (i) 2 rads–1, (ii) 96N]

8.. An object of mass 10 kg is whirled round a horizontal circle of radius 4m by a revolving string inclined to the
vertical. If the uniform speed of the object is 5 ms–1, calculate (i) the tension in the string (ii) the angle of

inclination of the string to the vertical. [Ans: (i) 118 N (ii) 32°]

9. A racing car of 1000 kg moves round a banked track at a constant speed of 108 km s–1. Assuming the total

reaction at the wheels is normal to the track and the horizontal radius of the track is 100m; calculate the angle

of inclination of the track to the horizontal and reaction at the wheels. [Ans: 42°, 13456 N]

10. Calculate the mean angular velocity of the earth assuming it takes 24 hours to rotate about its axis. An object
of mass 2 kg is (i) at the poles, (ii) at the equator. Assuming the earth is a perfect sphere of radius 6.4 × 106

m. Calculate the change in weight of the mass when taken from poles to the equator.

[Ans: 73×10–4 rad sec–1, 0.0682 N]

11. An object of mass 0.5 kg is rotated in a horizontal circle by a string 1 m long The maximum tension in the
string before it breaks is 50N. What is the greatest number of revolutions per sec. of the object?
[Ans: 1.6 rev s–1 ]

12. A mass of 0.4 kg is rotated by a string at a constant speed v in a vertical circle of radius 1m. If the minimum

tension of the string is 3N, calculate (i) v, (ii) the maximum tension and (iii) the tension when the string is just

horizontal. [Ans: (i) 4.18 ms–1 (ii) 11.00 N (iii) 7.00 N]

MBBS, B.E., B.Sc. Nursing Entrance Preparation Questions

1. A particle moving in a circle with a uniform speed has constant

a. velocity b. acceleration c. kinetic energy d. displacement

2. A body in circular motion with constant speed v along circular path of radius r, has tangential acceleration

v2 v2 v2 d. zero
a. 2πr b. πr c. r

3. A cyclist turns around a curve at 20km/hr. If he turns at the double of this speed, the tendency to overturn
is

a. double b. quadruped c. halved d. unchanged

4. A boy swings a bucket of water in a vertical circle of diameter of 2m. The minimum speed with which he

must swing the bucket so that the water does not spill is

a. 1.56 ms–1 b. 9.8 ms–1 c. 3.13 ms–1 d. 4.4 ms–1

5. A body slides down from a frictionless track, which ends in a circular loop of radius r. What should be the
minimum height h of the body so that the body is just able to complete the vertical circular motion?

5r 5 2 r
a. 2 b. 4 r c. 5 r d. 2

6. In a circular motion, tangential velocity and tangential acceleration are

a. parallel b. anti–parallel c. both (a) and (b) d. perpendicular

7. Two particles of equal masses are revolving in circular paths of radii r1 and r2 respectively with same
speed. What will be the ratio of their forces

a. r2 b. r2 ( )c. r1 2 ( )d. r2 2
r1 r1 r2 r1

8. A particle of mass m is moving along a circular path of radius 'R' with a frequency 'n' and time period 'T'.

It's centripetal acceleration can be expressed as

a. 4π2R2n b. 4π2RT c. 4π2Rn2 d. 4π2R2T2

168 | Essential Physics

9. A body of mass 0.1 kg tied by a string is rotating around a vertical circle with a speed of 10ms–1 at radius
1m. What is the tension experienced by the string at the highest point.

a. 8 N b. 11 N c. 10 N d. 9 N

10. A body of mass 2 kg is attached to one end of string of length 1m and rotated in vertical circle such that
velocity at the highest point is 4ms–1. The minimum tension in the string is

a. 18 N b. 12 N c. 45 N d. 0

11. An object is tied to string and rotated in a vertical circle of radius 'r' and constant speed is maintained

( )along if Tmax = v2 is equal to
the trajectory. Tmin 2, then rg

a. 1 b. 2 c. 3 d. 4

12. A particle is making uniform circular motion with angular momentum L. If its kinetic energy is made half
and angular frequency be doubled, its new angular momentum will be

L L c. 2L d. 4L
a. 4 b. 2

13. A car of mass 1500 kg is moving with a speed of 12.5ms–1 on a circular path of radius 20m on a level road.
What should be the coefficient of friction between the car and the road so that the car does not slip?

a. 0.2 b. 0.4 c. 0.6 d. 0.8

14. A particle of mass m is rotating by means of string in a vertical circle. The difference in the tension at the
bottom and at top could be

a. 6mg b. 2mg c. 4mg d. zero

15. The normal (radial) components of a particle in a circular motion is due to

a. speed of the particle b. change in the direction of its velocity

c. change in magnitude of its velocity d. rate of change of acceleration.

16. A stone tied to one end of a light inextensible string is whirled in vertical circle. If the speed is gradually
increased, the string is more likely to break from

a. highest point b. lowest point

c. a point when the string is horizontal d. it never breaks

17. A particle of mass m is tied to one end of a long inextensible string whose other end is fixed to a rigid
support. The particle is taken one side so that the angular displacement from vertical is α. The particle is
released so that the angular displacement at any instant θ(<α). Then the tension in the string must be

a. mg cosθ b. mg cosα c. greater than mg cosθ d. less than mg cosθ

Answer Key

1. c 2. d 3. b 4. c 5. a 6. c 7. a 8. c 9. d
10. b 11. c 12. a 13. d 14. a 15. b 16. d 17. c

Ch apter Gravity and
Gravitation
7

Teaching Manual Physics Grade – XI, Higher Secondary Education Board
Curriculum and Training Division Sanothimi, Bhaktapur

Syllabus:

Gravitation–Newton's law of gravitation; acceleration due to gravity, g; Mass and weight; gravitational field
strength, variation in value of 'g' due to altitude, depth and rotation of earth; Weightlessness; Motion of a
satellite: Orbital velocity, and time period of the satellite, geostationary satellite, potential and kinetic
energy of the satellite; Gravitational potential: Gravitational potential energy; Escape velocity; Black holes.

Objectives:

The objective of this sub unit is to make the students able to:
– understand the universal law of gravitation,
– understand planetary motion,
– understand the motion of the natural and artificial satellite around the earth.
– understand the variation of g, and
– understand gravitational potential, escape velocity and black holes.

Activities (Micro syllabus) :

1. State and explain Newton's law of gravitation.
2. Define universal gravitational constant, G and acceleration due to gravity, g.
3. Discuss quantitatively the variation of g due to radius, altitude, depth and rotation of the earth.
4. Differentiate between mass and weight; inertial mass and gravitational mass.
5. Explain gravitational field and field strength.
6. Define gravitational potential and gravitational potential energy. Derive expressions for them.
7. Define and derive an expression for escape velocity. Explain black holes.
8. Explain weightlessness.
9. Define orbital velocity. Derive an expression for the orbital velocity and the time period of a satellite.
10. Determine the total energy of a satellite.
11. Numerical problems: Focused numerical problems given:

(a) As exercise in the University Physics (Ref.1) and
(b) In advanced level Physics (Ref.2) and also in Physics for XI (Ref.3).

170 | Essential Physics

7.1 Introduction

The theory of gravitation was discovered by Sir Issac Newton in 1687 A.D. The gravitational force arises
due to the force of attraction between two masses. This is the same interaction that makes an apple fall
out of a tree also keeps the planets in their orbits around the sun. The gravitational interaction is universal;
gravity acts in the same fundamental way between all bodies in the universe. The gravitational force is the
weakest force in nature. Due to gravitation, everybody in the universe has a tendency to move towards
one another.

7.2 Newton's Law of Gravitation

It states that, "Every particle of matter in this universe attracts every other particle with a force
which is directly proportional to the product of their masses and inversely proportional to the
square of distance between them." The force of attraction between any two particles is called
gravitational force. This force is mutual which acts along the line joining their centers.

Let m1 and m2 be the masses of two bodies in the universe and r be the distance between their centres. If F
be the magnitude of gravitational force between
them, then

F ∝ m1m2 . . . (1) m1 FF m2
. . . (2)
and F∝ 1
r2

Combining equations (1) and (2), we get r

F∝ m1m2 [Fig. 7.1, Gravitational force between two bodies ]
r2

or, F = Gm1m2 . . . (3)
r2

Where G is a constant of proportionality known as gravitational constant. It has same value everywhere
in the universe. So, G is called a universal constant of gravitation.

If m1 = m2 = 1 unit and r = 1 unit, then, F = G.

So, the universal gravitational constant G is numerically equal to the gravitational force between two unit
masses, separated by unit distance. S.I. unit of G is Nm2kg–2 and its value is 6.67 × 10–11 Nm2kg–2. The
dimensional formula of G is [M–1L3T–2]. The value of G is same for all pairs of masses and it is

independent with medium in which the masses are placed, chemical composition of masses, nature of

masses, directions of gravitational force between the masses, temperature, pressure etc.

Weight, Gravitation and Gravity

The force with which a body is attracted towards the centre of earth is called weight of the body. It has
same units and dimensions as those of force.
The force of gravitation is defined as the force of attraction between any two material bodies in the
universe. However, the force of attraction is called force of gravity if one of the attracting bodies is earth
or some other planet or natural satellite.

7.3 Inertial Mass and Gravitational Mass

The total quantity of matter contained in a body is called its mass. The inertial mass of a body is the mass
that measures inertia. The inertial mass is measured by using Newton's second law of motion. If 'a' be

Gravity and Gravitation | 171

the acceleration produced on a body of mass m by applying an external force F, then from Newton's
second law of motion,

F = ma

∴ F . . . (1)
m=a

Equation (1) measures inertial mass of a body and it is defined as the ratio of magnitude of force applied
on it to the magnitude of acceleration produced in it. The inertial mass is a measure of ability of a body to
resist the production of acceleration in it.

The mass of a body measured by using Newton's law of gravitation is called its gravitational mass.
If M and R be mass and radius of earth respectively, then from Newton's law of gravitation, the
gravitational force F on a body of mass 'm' on its surface is,

GMm
F = R2

∴ FR2 . . . (2)
m = GM

Equation (2) measures the gravitational mass of the body. Different experiments show that the inertial
mass and gravitational mass of a body are equivalent and share the same properties.

7.4 Acceleration Due to Gravity

The acceleration produced on a freely falling body on account of force of gravity is called

acceleration due to gravity. It is denoted by 'g'. At a given place, value of acceleration due to gravity is
same for all bodies irrespective of their masses. However the value of g differs from place to place on the

surface of earth. It's value also varies with altitude and depth. The value of g at sea – level and at latitude
45° is 9.81ms–2 which is taken as the standard for reference.

To find an expression for acceleration due to gravity at the surface of earth, assume that earth is a perfect
sphere having mass M and radius R. Consider a body of mass 'm' lying on the surface of earth as shown in

figure 7.2.

According to Newton's law of gravitation , the gravitational force of attraction between the earth and the

body is

GMm . . . (1) m
F = R2 mg

If 'g' be the acceleration due to gravity, then from Newton's second law R
of motion,

F = mg . . . (2)
From equations (1) and (2), we get
O

GMm M
mg = R2

∴ GM . . . (3) [Fig. 7.2, Acceleration due to gravity at
g = R2 surface of earth]

Equation (3) shows acceleration due to gravity is independent with the

mass of falling body and depends only on the mass and radius of earth.

If ρ be the mean density of earth, then mass of earth M is,

M = volume × density

or, M = 4 πR3ρ . . . (4)
3

172 | Essential Physics

From equations (3) and (4), we get

g = G × 4 πR3ρ
R2 3

∴ g = 4 π Rρ G . . . (5)
3

This is the expression for acceleration due to gravity on the surface of earth in terms of mean density ρ of
the earth.

Mass and Mean Density of Earth

From equation (3) mass of earth M is
gR2 9.8 × (6.4 × 106)2

M= G = 6.67 × 10–11

∴ M = 6.018 × 1024 kg ≈ 6 × 1024 kg

From equation (5), mean density ρ of the earth is

ρ = 3g 3 × 9.8
4πRG = 4π × 6.4 × 106 × 6.67 × 10–11

∴ ρ = 5.5 × 103 kg/m3

7.5 Variation of Acceleration Due to Gravity

GM
We have, g = R2 . Since, GM = constant

∴ g ∝ 1
R2

So, the acceleration due to gravity on the surface of earth is inversely proportional to square of radius of
earth.

The earth is not perfect spherical. It is flattened at the poles and bulged at the equator. So, value of g is
maximum at poles and minimum at equator on the surface of earth.

The variation in the value of g corresponds to the different factors such as altitude, depth and latitude as
discussed below:

I. Variation of 'g' Due to Altitude (Height)

Let earth be a perfect sphere having mass M and radius R as shown in figure 7.3. Let g be the

value of acceleration due to gravity on the surface of earth, then

GM . . . (1) m
g = R2

Let g' be the value of acceleration due to gravity at a height 'h' h
from the surface of earth, then

GM . . . (2) R
g' = (R + h)2

Dividing (2) by (1), we get O
g' GM R2 R2 M
g = (R + h)2 × GM = (R + h)2
Earth
g' 1 [Fig. 7.3, Variation of g with altitude]
or, =
g 1 h 2
+ R 

Gravity and Gravitation | 173

or, g' = 1 + Rh –2 . . . (3)
g

When h << R, then h << 1. So, expanding 1 + Rh–2 by using Binomial theorem and neglecting
R

h
the terms containing higher powers of R , we get

1 + Rh–2 2h . . . (4)
= 1 – R + ……

From equations (3) and (4)

g' 2h
g = 1 – R + ……

∴ g' = g1 – 2Rh . . . (5)

2h
Here, R < 1

∴ g' < g.

Also from equation (3), it shows that the value of g decreases with altitude h by following

inverse square law.

At height h from earth's surface,

Loss in Weight = mg – mg' = mg – m g1 – 2Rh  = mg – mg + 2mgh
R

∴ 2mgh
Loss in weight = R

II. Variation of 'g' Due to Depth

let earth be a perfect homogeneous sphere having mass M, radius R and mean density ρ as shown

in figure 7.4. Let g be the value of acceleration due to gravity on the surface of earth, then

GM G4 πR3ρ
g = R2 = R2 × 3

∴ g = 4 πRρG . . . (1)
3

Let g' be the value of acceleration due to gravity at a depth x from the surface of earth, then

GM' . . . (2) m
g' = (R – x)2

Here, acceleration due to gravity g' is only due to inner solid x

spherical section of earth of radius (R – x) and M' is the mass of

inner spherical section of earth of radius (R – x). R–x

So, M' = 4 π(R – x)3ρ . . . (3) MR
3 M'

From equations (2) and (3), we get

G × 4 π(R – x)3ρ
g' = (R – x)2 3

∴ g' = 4 π (R – x) ρG . . . (4) [Fig. 7.4, Variation of g with
3 depth]

Dividing equation (4) by equation (1), we get

174 | Essential Physics

g' 43π(R – x)ρG
g = 43πRρG

g' R – x
or, g = R

∴ g' = g1 – Rx . . . (5)

x
Since, R < 1

∴ g' < g

Equation (5) shows that the value of acceleration due to gravity decreases linearly with depth

below the surface of earth. g' At the surface
At the center of earth, x = R. g

∴ g' = g1 – RR = 0

And, weight (w) = mg' = 0

So, the value of acceleration

due to gravity and hence weight

of a body becomes zero at the

centre of earth. Figure 7.5

shows the variation of 'g' with

distance 'r' from the centre 'O'

of the earth. It shows that g O r
varies linearly with distance R
below the surface of earth,
attains a maximum value at the [Fig. 7.5, Variation of g above and below the surface of earth]

surface and decreases following inverse square law above the surface of earth.

III. Variation of 'g' Due to Latitude [Rotation of Earth]

ω

C rφ PA

Equatorial Axis of Oφ
Plane Rotation of Earth RB

North

P

Oφ

South (b)
(a)

[Fig. 7.6, Variation of g due to rotation of earth ]

Gravity and Gravitation | 175

The latitude φ of a point P on the surface of earth is defined as the angle made by a straight line

joining the point to the centre of earth with equatorial plane. Its value is 90° at poles and 0° at

equator.

Let φ be the latitude of a point P on the surface of earth. Let a body having mass 'm' is placed at

point P. Then OP = mg is the actual weight of the body which acts towards the centre O of the

earth.

Let 'ω' be the angular velocity of rotation of earth about its polar axis. Due to this rotation, the

body at point P moves along a circle of centre C and radius r = Rcos φ as shown in figure 7.6 (b).

The body at point P experiences centrifugal force given by
PA = mrω2 = m(Rcosφ) ω2 = mω2 Rcos φ

The body at point P is under the action of two forces – actual weight mg and centrifugal force
mω2 Rcosφ. The apparent weight mg' of the body at point P is given by diagonal PB in

parallelogram PABO according to parallelogram law of vector addition.

From parallelogram law of vector addition, in parallelogram PABO,

R2 = P2 + Q2 + 2PQ cos ∠OPA
or, PB2 = PO2 + PA2 + 2PO.PA. cos (180 – φ)
or, (mg')2 = (mg)2 + (mRω2cosφ)2 – 2.mg. mRω2 cosφ. cosφ

or, g' = g 1 + R2ω4cos2φ – 2Rω2 cos2φ1/2
g2 g

Since, ω is very small quantity, the term containing ω4 can be neglected. Therefore,

g' = g 1 – 2Rω2 cos2φ1/2
g

Here, 2Rω2cos2φ < 1.

g

So, expanding above equation by using Binomial theorem and neglecting the terms containing higher

powers of ω2, we get

g' = g 1 – 1 2R ω2 cos2φ + . . .
2 g

or, g' = g 1 – Rω2 cos2φ
g

∴ g' = g – Rω2cos2φ . . . (1)

This equation shows that the value of g decreases due to rotation of earth about its polar axis.

At the poles, φ = 90°and cos90° = 0

∴ g' = g

This shows that value of g at poles is unaffected by rotation of earth about its polar axis.

At the equator, φ = 0° and cos 0° = 1

∴ g' = g – Rω2 . . . (2)

This shows that value of g at equator is less than that at poles by Rω2.

If earth stops rotating, then ω = 0. So, from equation (2), g' = g.

This shows that if earth stops rotating, then the value of g at equator increases and becomes

equal to that at poles. In such condition, the value of g everywhere on the surface of earth

remains constant and the earth would be perfect spherical.

7.5 Gravitational Field

When a material particle is placed in space, it modifies the space around it. This modified space is called
its gravitational field. When another material particle is brought in this field, then it experiences a

176 | Essential Physics

gravitational force of attraction which varies inversely as the square of the distance between the centers of
the two material particles. Thus, the gravitational field may be defined as, "the space or region around a
mass over which it can exert gravitational forces on other masses." Theoretically, the gravitational
field extends up to infinity. But, in actual practice, it becomes too weak to be measured beyond a
particular distance.

7.6 Gravitational Field Intensity or Gravitational Field Strength

The gravitational field intensity at a point inside the gravitational field is defined as the gravitational
force experienced by a unit mass placed at that point. It is always directed towards the particle
producing the field. It is a vector quantity. It is denoted by E or I.

Let earth be a perfect sphere having mass M and radius R. Let a M
particle having mass m is placed at point P at a distance 'r' from
the centre of the earth. Then the gravitational force F experienced R r m∞
by the mass m is, O F
P
Earth
GMm . . . (1)
F = r2

From definition, the gravitational field intensity at point P is, [Fig.7.7, Gravitation field intensity.]

F . . . (2),
E=m

S.I. unit of E is Nkg–1 or ms–2. The dimensional formula of E is [M0L1T–2]

F GM . . . (3)
From equation (1), m = r2

From equations (2) and (3), we get,

GM . . . (4)
E = r2

Also, if g be the value of acceleration due to gravity at point P, then

GM . . . (5)
g = r2

Comparing equations (4) and (5), E = g. So, the gravitational field intensity at a point inside the
gravitational field is equal to the value of acceleration due to gravity at that point.

7.7 Gravitational Potential Energy

The gravitational potential energy of a body at a point inside the gravitational field is defined
as the amount of work done on bringing the body from infinity to that point without acceleration. It
is denoted by 'U'. It is a scalar quantity.

Let earth be a perfect sphere R dx ∞
having mass M and radius R. Let P be a M m
point at a distance 'r' from the centre O of
the earth where the gravitational potential O P B FA
energy has to be determined. Let a body Earth
of mass m which was initially at infinity,
is at point A at any instant of time, where r
OA = x. Then, the gravitational force F x

[Fig. 7.8, Gravitational Potential Energy at point P.]

Gravity and Gravitation | 177

experienced by mass 'm' at point A due to earth is,

GMm . . . (1)
F = x2

The small work done dW by this force on moving the body in small displacement dx from point A to B is,

dW = F . dx . . . (2)

From equations (1) and (2), we get

GMm . . . (3)
dW = x2 dx

The total work done W by this gravitational force on moving the body of mass m from infinity to point P
is obtained by integrating equation (3) from ∞ to r. Therefore,

∫W =r = ∫r GMm dx = GMm ∫∞r x–2 dx = GMm – x1∞r = – GMm 1 – 1
x2 r ∞
∞dW ∞

GMm
or, W = – r

This work done is equal to gravitational potential energy U of mass m at point P.

∴ GMm . . . (4)
U=– r

The negative sign suggests that the gravitational force is always attractive. This equation (4) also
shows that U has maximum value of 0 at r = ∞ and decreases gradually as the body moves from infinity
towards the surface of earth.

Gravitational Potential

The gravitational potential at a point inside the gravitational field is defined as the amount of work
done in bringing a unit mass from infinity to that point without acceleration. It is simply
gravitational potential energy per unit mass. It is denoted by V. So,

U . . . (5)
V=m

It is a scalar quantity. S.I. unit of V is J kg–1 and its dimensional formula is [M0L2T–2].

From equations (4) and (5), we get,

V = 1 – GMr m
m

∴ GM . . . (6)
V=– r

Gravitational Potential Difference

It is defined as the work done to move a unit mass from one point to another point inside the

gravitational field. The gravitational potential difference VAB between points A and B in the
gravitational field is,

VAB = VB – VA = GM – –GrMA 
– rB

∴ VAB = GM r1A – r1B . . . (7)

178 | Essential Physics

Where, rA and rB are respective distances of points A and B from the centre of earth.

7.8 Escape Velocity

The minimum velocity with which a body must be projected vertically upwards in order that it may
just cross the gravitational pull of Earth is called escape velocity. When a body is thrown vertically
upwards with a velocity less than escape velocity, then the body returns to the Earth's surface after some
time. However, if the body is thrown with a velocity equal to or more than the escape velocity, then the
body overcomes the gravitational pull of Earth and this body never returns to the Earth's surface again.

Let Earth be a perfect sphere having mass M and radius R. Let m be the ∞
mass of a body lying on the surface of earth. For the body to escape from
dx
the gravitational pull of the Earth, the K.E. of the body on the earth's surface m ve x
must be equal to the amount of work done on moving the body from the MR
Earth
surface of earth to infinity. That is, [Fig. 7.9, Escape velocity]

∫1mv2e= ∞ [ ve = escape velocity]

2 RdW

∞ ∞ GMm ∞
R x2 dx = GMm
R Fdx = R x–2 dx
∫ ∫ ∫=

= GMm – 1x∞R = – GMm 1 – 1
∞ R

or, 1 mv2e = GMm
2 R

∴ 2GM . . . (1)
ve = R . . . (2)

GM
But, g = R2

or, GM = gR2

From equations (1) and (2), we get

2gR2
ve = R

∴ ve = 2gR . . . (3)
For Earth, g = 9.8 ms2 and R = 6.4 × 106 m

∴ Ve = 2 × 9.8 × 6.4 × 106 ≈ 11.2 km/s

The escape velocity on the Earth's surface is 11.2 km/s. This is a very high velocity. Such a high velocity
is not easily acquired by gas molecules. This explains as to why gases have not escaped the surface of
earth. This explains the existence of atmosphere on earth. Similarly, the escape velocity is very large, in
the case of sun. So, the sun holds its gaseous envelope very firmly. On the other hand, the escape velocity
is only 2.4 km/s, in the case of moon. Such a small velocity is easily acquired by gas molecules and
escape out from the gravitational pull of moon. Therefore, the moon has practically no atmosphere.

7.9 Weightlessness

Reaction gives us feeling of weight. The reaction of weighing machine on a body gives the measurement
of weight of the body. If the weighing machine is at rest, the reaction measures the true weight (mg) of
the body. If the weighing machine is in motion, then the reaction may change and the weight measured by

Gravity and Gravitation | 179

the weighing machine is called apparent weight (mg') of the body. In some situations, the reaction may
become zero and then the body is said to be in the state of weightlessness. A body may be in the state

of weightlessness in the following different situations: R
a
i. A body inside a freely falling lift: Equation of motion for downward
motion is,

mg – R = ma

or, R = m (g – a) m F = ma
If the lift is falling freely, then a = g. Thus,
mg
R = m (g – g) [Fig. 7.10, Weightlessness]
∴ R=0

Here, reaction or apparent weight becomes zero and the body is in the state of weightlessness.

ii. A body inside an orbiting satellite: When a body is inside an orbiting satellite, then inward

gravitational pull on the body due to earth's gravitational field is balanced by outward centrifugal

force on the body and the resultant force on the body becomes zero. In such situation, the reaction

or apparent weight becomes zero and the body is in the state of weightlessness.

iii. At null points: Null points are the points in the outer space where the resultant gravitational force
due to different heavenly bodies becomes zero. At null points, the body feels truly

weightlessness. We lose our mental image of up and down at such null points.

iv. A body at the centre of Earth: At the centre of earth, value of acceleration due to gravity g = 0
and the weight of the body W = mg = 0. The body feels truly weightlessness at the centre of earth.

7.10 Satellites

A body which is continuously revolving around a bigger body is called a satellite. A satellite may be
natural or artificial. In our solar system, different planets are revolving around the sun. So, the planets can
be regarded as the natural satellites of the sun. Moon revolves around the earth in circular orbit. So, moon
is the natural satellite of earth. The man made satellites are called artificial satellites. Artificial satellites
are put into their orbits by the help of multi–stage rockets.

Principle of Launching of Artificial Satellites

Consider a very high launching pad on the surface of earth as shown in A1
figure 7.11. Suppose a body thrown horizontally from the top of A2
launching pad hits the earth at point A1 by following a parabolic path. A3
If the second body were projected with a greater velocity, then the Earth
body would have struck the earth at point A2. Now, let another third
body is projected with a still greater velocity, and the body hits the Orbit
earth at a point A3 (directly opposite to the launching point).
[Fig. 7.11, Launching of artificial
Let the fourth body be thrown from the top of the launching pad with a satellite]
velocity which is slightly greater than the velocity of projection of
third body, then the body will miss to hit the surface of earth. Then the
body will fall into an orbit and becomes a satellite of the earth. The
satellite will always be in a state of free fall under gravity in an
attempt to fall to the earth's surface but always misses to hit to the
surface of earth.

180 | Essential Physics

Orbital Velocity

The path followed by a satellite is called orbit of the satellite. The velocity required to keep a satellite

into its orbit is called the orbital velocity of the satellite. It is denoted by v0.
The necessary centripetal force for a satellite to keep on moving around a planet is provided by the

gravitational force of attraction between the planet and the satellite. That is,

Centripetal force on satellite = Gravitational force between planet and m
satellite.

or, mv20 = GMm . . . (1) v0 h
r r2 r
R
where, M = mass of planet (say, earth), M

m = mass of satellite Earth

r = radius of orbit of satellite,

v0 = orbital velocity of satellite [Fig. 7.12, Orbital Velocity]
From equation (1),

GM . . . (2)
v0 = r

GM . . . (3) [since, r = R + h]
or, v0 = R + h

gR2 GM
or, v0 = R + h . . . (4) [since, g = R2 ]

Where h is the height of orbit of satellite from Earth's surface.

When the satellite lies very close to earth's surface, then R + h → R, i.e. h → 0

From equation (4),

gR2
v0 = R + 0

or, v0 = gR . . . (5)
For the earth, v0 = 9.8 × 6.4 × 106 = 7.9 km/s
We know, escape velocity of Earth is,

ve = 2gR . . . (6)
or, ve = 2 ( gR)
or, ve = 2 (v0)
∴ Escape velocity = 2 (orbital velocity)

Time Period of a Satellite (T)

It is the time taken by the satellite to make one complete revolution around the Earth.

∴ circumference of orbit
T = orbital velocity

2πr . . . (7)
or, T = v0
From equations (2) and (7), we get

T = 2πr r
GM

Gravity and Gravitation | 181

or, T = 2π r3 . . . (8)
GM

or, T = 2π (R + h)3 . . . (9) [since, r = R + h and GM = gR2].

gR2

If the satellite is orbiting very close to the earth's surface, then R + h → R; h → 0

From equation (9), we get

T = 2π (R + 0)3 = 2π R3
gR2 gR2

∴ T = 2π R . . . (10)
g

or, T = 2 × 3.14 6.4 × 106 = 5077 sec ≈ 84 minutes.
9.8

Geostationary Satellite and Parking Orbit

The satellite which seems to be stationary when viewed from a point on the earth's surface, is called

geostationary satellite. For a satellite to be geostationary, following conditions should be satisfied:

i. The time period of revolution of the satellite should be same as the time period of rotation of earth

i.e. 24 hrs.

ii. The direction of revolution of the satellite should be same as the direction of rotation of earth i.e.

from west to east, and

iii. The satellite should be placed above the equatorial region of the earth.

The orbits of such geostationary satellites are called parking orbits.

From equation (9), we have,

T = 2π (R + h)3
gR2

Squaring, T2 = 4π2 (R + h)3
gR2

or, (R + h)3 = T2.gR2
4π2

or, (R + h) = T2gR21/3
 4π2 

∴ h = T42gπR2 21/3– R . . . (11)

Equation (11) is the expression for height of satellite above the surface of earth.

For a geostationary satellite,

T = 24 hrs = (24 × 60 × 60) s = 86400 s

R = 6.4 × 106 m, g = 9.8 ms–2

Substituting values in equation (11), we get

h = (86400)2 × 9.8 × (6.4 × 106)21/3– 6.4 × 106 ≈ 3.6 × 107 m = 36,000 km.
 4π2

The radius of parking orbit is,

r = R + h = 6400 km + 36000 km = 42,400 km

Also, the speed of geostationary satellite at the parking orbit is,

v0 = 2πr = 2π × 42400 × 1000 = 3083 m/s ≈ 3.1 km/s
T 24 × 60 × 60

182 | Essential Physics

Energy of Satellite

A satellite revolving around the earth has both kinetic energy and potential energy. Total energy of the

satellite is equal to sum as its kinetic energy and potential energy. That is,

Total energy = kinetic energy + potential energy

or, E = K.E. + P.E. . . . (12)

The kinetic energy is the energy due to motion of the satellite and is given by,

K.E. = 1 m v02 = 1 m  GM2 [using equation (2) ]
2 2  r ,

∴ GMm . . . (13)
K.E. = 2r

The gravitational potential energy of the satellite is because of gravitational force of attraction exerted by

earth on the satellite and it is given by

GMm . . . (14)
P.E. = – r

From equation (12), (13) and (14), we get

E = GMm + –GMr m
2r

∴ GMm . . . (15)
E = – 2r

The total energy of the satellite is negative. It suggests that the satellite is bound to the gravitational pull

of earth.

The energy required by a satellite to leave its orbit around the earth and escape to infinity is called

Binding Energy of the satellite. So,

Binding energy of satellite = E∞ – E = 0 – –GM2rm = GMm
2r

7.11 Black Holes

The expression for escape velocity on the surface of a spherical body having mass M, density ρ and
radius R is

2GM . . . (1)
v= R

2G × 4 π R3ρ
3
or, v =
R

or, v = R 8πGρ . . . (2)

3

For sun, M = 2 × 1030 kg, R = 6.98 × 108m and ρ = 1410 kg m–3

From equation (2), v = 6.98 × 108 8π × 6.67 × 10–11 × 1410 = 6.18 × 105 ms–1
3

So, the escape velocity at the surface of sun is about 5010 times the speed of light in vacuum i.e. 3 × 108

ms–1. For a heavenly body having average density same as that of sun and radius 500 times that of sun,

the escape velocity would be greater than or equal to that of speed of light in vacuum. In such condition,

all the light rays emitted from the surface of the heavenly body would return towards it. That means,

Gravity and Gravitation | 183

nothing (not even light) can escape from the surface of that heavenly body and such type of heavenly

body can't be seen from outer universe. Such heavenly body is called a black hole.

From equation (1), the radius R in terms of escape velocity is

2GM . . . (3)
R = v2

A heavenly body having mass M becomes a black hole if its radius R is less than or equal to certain

critical radius called Schwarzchild radius (Rs). In 1916, Karl Schwarzchild used Einstein's general
theory of relatively to derive an expression for Rs by setting v = c (speed of light in vacuum) in equation
(3) as,

2GM . . . (4)
Rs = c2

Thus, if a spherical, non–rotating heavenly body with mass M has a radius less than or equal to Rs, then
nothing (not even light) can escape from the surface of the body and the body transforms into a black

hole. Any other body within a distance Rs from the centre of the black hole is trapped by the immense
gravitational pull of the black hole and cannot escape from it. The surface of the sphere having radius Rs
surrounding a black hole is called an event horizon. Any event that occurs inside the event horizon

cannot be seen from outside universe as light cannot escape from it. All that can be known from outside

the event horizon about a black hole is its mass (from its gravitational pull on other bodies), its electric

charge (from electric force on other charged bodies) and its angular momentum (because a rotating black

hole tends to drag space and everything in that space–around it).

Astrophysical theory suggests that a burned out star may collapse to form a black hole if its mass is
at least three times the solar mass. That is,

M = 3(Solar mass) = 3 × 2 × 1030 kg

From equation (4), we can estimate the value of radius of event horizon (Rs).

From (4), Rs = 2 × 6.67 × 10–11 × (3 × 2 × 1030) = 8893.33 m ≈ 8900 m = 8.9 km
(3 × 108)2

Boost for Objectives

• At the same point, ve: v0 = 2 : 1
• If speed of satellite increases by 41.4% or its K.E. increases by 100% then it will escape to infinity.

• Since the satellite has total negative energy, it is bound to revolve in the orbit.

• When a satellite is orbiting in its orbit, zero energy is spent to keep it in its orbit.

• Mass of air bubble in material medium is negative.

• Spring balance measures apparent weight, physical balance measures gravitational mass.

• If polar ice caps melt, then moment of inertia (I) and time period of rotation of earth increase.

• Gravitational mass is proportional to gravitation force.

• Air friction increases the velocity of the satellite.

• An astronaut in a satellite cannot use a pendulum clock. However, he can use spring clock or digital clock.

• Escape velocity from moon = 2.4 km/sec where, gm = gc = 1.6m/s2
6

• Astronauts in space see the sky black due to lack of atmosphere above them.

• If a body falls freely from infinite height, it will reach the earth surface with the escape velocity i.e. 11.2

km/sec.

• A satellite will be seen Geostationary only when it is launched at a proper height from West to East (not east

to west) in equatorial plane since the earth rotates itself west to east.

• A bus weighs more while moving due west with sufficient speed than when it is at rest or is moving due east.
• The hydrogen balloon released on moon would fall with an acceleration 9.8/6 m/s2 (i.e. 1.6m/s2) since there is

no air to impart upthrust.

184 | Essential Physics

• If a piece of rock falls vertically on a satellite then the satellite will fall on the earth surface following a spiral
path.

• Two heavenly bodies not far off from each other are seen to revolve around their common centre of mass.
• The escape velocity of a body does not depend on its mass but depends on the mass of the planet.
• If the earth suddenly stops rotating about its axis, the value of g at the equator will increase by ω2R.

[Hint: ge = g – Rω2 (at equator) and at poles no rotational effect]
• The time period of geostationary satellite is 24 hours (that of earth).
• The height of communication satellite from earth's surface is 36000km.
• When velocity of satellite increases, K.E. increases and T.E. becomes less –ve. Thus, satellite begins to

revolve in orbit of greater radius.

• When KE = PE, satellite escapes away from gravitational pull of earth.
• If a body is released from an artificial satellite it does not fall to the earth but will continue orbiting along

with the satellite.
• For a freely falling body, apparent weight is zero.
• In a freely falling lift, time period of simple pendulum is infinity.
• When the total energy of a satellite is –ve, it will be moving in either a circular or an elliptical orbit.
• When the total energy of a satellite is zero, it will escape away from its orbit and the path becomes parabolic.
• Tidal waves in sea are principally due to gravitational effect of moon on earth.
• Moon has no atmosphere because r.m.s velocity of gas molecules there are greater than escape velocity of

moon.
• Rocket launched with escape velocity follows parabolic path.

Short Questions with Answers

1. Distinguish between gravitational potential and gravitational field strength. [NEB 2074]
Ans:

S.N. Gravitational Potential (V) S.N. Gravitation field Strength (E)

1. It is defined as the amount of work done in 1. It is defined as the gravitational force

bringing a unit mass from infinity to a point experienced by a unit mass placed at a

inside the gravitational field. point inside the gravitational field.

2. W Gm 2. F Gm
Formula : V = m = - r Formula: E = m = r2

3. It is a scalar quantity. 3. It is a vector quantity.

4. Unit is Jkg-1. 4. Unit is N kg-1.

5. o 2 -2 5. o -2

Dimensional formula is [M L T ] Dimensional formula is [M LT ]

2. An astronaut in a space capsule orbiting the earth experiences weightlessness, why?[HSEB 2072 C]
Ans:
The inward resultant force acting on the astronaut having mass m inside the satellite (space capsule) is,
3.
Ans: GMm
F = r2 – R

GMm
or, ma = r2 – R

or, m GrM2  = GMm – R
r2

∴ R=0

The reaction R of the surface to the astronaut becomes zero. The reaction is the measure of apparent

weight of the body. Hence, apparent weight of the astronaut becomes zero and it gives the feeling of

weightlessness to the astronaut. The astronaut sees all bodies floating weightlessly inside the space

capsule.

At what condition does a body becomes weightless at the equator? [HSEB 2072 D]

The apparent value of acceleration due to gravity at equator due to rotation of earth is

Gravity and Gravitation | 185

g' = g1 – Rgω2

For a body to become weightless at equator, the apparent weight (mg') should be zero. That is,

mg' = 0

or, g' = 0 [ g' = g1 – Rgω2 at equator]

or, g 1 – Rgω2 = 0

∴ g = Rω2
So, a body becomes weightless (W = mg' = 0) at equator if g = Rω2. Also, the body becomes weightless

if it is inside a freely falling lift.

4. Since the moon is constantly attracted towards the earth by the gravitational interaction, why
Ans:
doesn't it crash into the earth? [HSEB Model Question]
5.
Ans: The gravitational force of attraction between the earth and moon provides the necessary centripetal
6.
Ans: force for the moon to keep on moving around the earth in circular path. The gravitational force is

7. perpendicular to the direction of motion of the moon. So, the work done by the gravitational force is W
Ans:
= →F .→S = FS cos90° = 0, as θ = 90°. Therefore, the moon doesn't crash into the earth.
8.
Ans: Write the unit and dimension of gravitational field with its definition. [HSEB 2072, E]

The region or space around the earth up to which its gravitational force can be experienced is called its
gravitational field. This is just space, so, it is measured in area of m2 and dimension is [L2].

What is a black hole? Write down its characteristic features. [NEB 2074, HSEB 2071 C, 2071 D]

A burned out star may collapse to form a black hole if its mass is at least three times the solar mass. The

gravitational pull of the black hole is so high that nothing (not even light) can escape from its surface.

Some important characteristic features are:

i. Its radius is less than or equal to some critical radius called Schwarzchild's radius given by

2GM
Rs = c2 .

ii. Escape velocity on its surface is equal to or more than speed of light in vacuum.

iii. Light can go inward from its even horizon but cannot escape. Therefore, it can't be seen from

outer universe.

iv. The mass is so condensed that it has very high specific gravity (relative density).

If heavier bodies are attracted more strongly by the earth, why do they not fall faster than lighter

ones? (neglect air resistance), [HSEB 2070]

The gravitational force F on a body of mass m is,

GMm . . (1)
F = R2

The acceleration due to gravity of on the surface of earth is,

GM . . . (2)
g = R2

Equations (1) and (2) show that the gravitational force F depends on mass m of falling body but

acceleration due to gravity g is independent with the mass m of falling body. Therefore, the value of g is

same for all bodies, either heavier or lighter ones. Hence, heavier or lighter bodies fall at the same time

if air resistance is neglected.

What will happen to the value of acceleration due to gravity if the earth stops rotating about its

axis? [HSEB 2070 C, 2051]

The value of acceleration due to gravity due to rotation of earth is,

g' = g 1 – Rω2 cos2 θ
g

If earth stops rotating, then ω = 0

∴ g' = g

This shows that if earth stops rotating, then the value of g increases. In such condition value of g

everywhere on the surface of earth would remain constant and the earth would be perfect spherical.

186 | Essential Physics

9. Suppose the radius of earth were to shrink by 2%, its mass remaining the same, calculate the
Ans:
percent change in the acceleration due to gravity 'g'. [HSEB 2069 S]
10.
Ans: The acceleration due to gravity g on the surface of earth is,
GM
11.
Ans: g = R2
12.
Ans: If the radius of earth were to shrink by 2%, the new radius is, R' = R – 2% of R = 0.98 R

13. The new acceleration due to gravity g' is,
Ans:
g' = GM = GM R)2 = 1.0412 × GRM2  = 1.0412 × g
R'2 (0.98 ×

g'
or, g = 1.0412

Now, percent change in g = g' – g × 100% = gg' – 1× 100% = (1.0412 – 1) × 100% = 4.12%
g

So, the value of g increases by 4.12%

If the sun somehow collapsed to form a black hole, what effect would this event have on the orbit

of the earth? [HSEB 2066, 2067, 2069 A]

The gravitational force of attraction between sun and earth provides the necessary centripetal force for

earth to keep on moving around the sun in circular orbit. We have,
GMm

F = r2

Where, M is mass of sun, m is mass of earth and r is distance between centres of earth and sun. While

forming black hole, the mass of sun shrinks in a smaller region but its mass M and centre point would

not change. Hence, the gravitational force F remains constant and there would be no effect on the orbit

of earth.

What do you mean by geo–stationary satellite? Explain. [HSEB 2062]

A satellite which seems to be stationary when viewed from a point on the earth's surface is known as

geostationary satellite. The geostationary satellite revolves around the earth in its equatorial plane with

the same time period and in the same direction as that of rotation of earth about its polar axis. It has

time period of 24 hrs and it is about 36,000 km above from the surface of earth. Communication

satellites are geo–stationary satellites.

Explain why the moon has no atmosphere? [HSEB 2058]

The escape velocity (v) is,

2GM
v= R

The mass and radius of moon are less than that of earth. Due to this, the escape velocity on the surface

of moon is only about 2.4km/s. The r.m.s. speed of air molecules on the surface of moon is more

than the value of escape velocity on the surface of moon. The air molecules can easily cross the

gravitational field of moon and hence the moon has no atmosphere.

How does 'g' at a point vary with the distance from the centre of the earth? Where is the highest

value of g? Explain. [HSEB 2055]

Value of g at a depth x from the earth's surface g' At the surface
is, g

g' = g 1 – Rx

At the centre of earth, x = R, So, g' = 0

At the surface of earth, x = 0, So, g' = g

(maximum)

Again, the value of g above the surface of

earth, at height h from the surface is,

g' = g 1 + Rh–2

OR r

Gravity and Gravitation | 187

Adjoining figure shows that value of g is zero at the centre of earth, varies linearly with distance below

the surface of earth, attains a maximum value at the surface and decreases following inverse square law

above the surface of earth.

14. The weight of a body is less inside the earth than on the surface. Explain. [HSEB 2053]
Ans:
Let g and g' be the values of acceleration due to gravity on the surface of earth and at a depth x from the
15.
Ans: surface respectively. Then,

16. g' = g 1 – Rx
Ans:
Where R is the radius of earth. Multiplying both sides by m, we get
17.
Ans: mg' = mg 1 – Rx

18. ∴ mg' ∠ mg [ 0 ∠ x ∠ 1]
Ans: R

19. Thus, weight inside the earth (mg') ∠ weight on the surface (mg) of earth.
Ans:
Obtain an expression for gravitational potential energy and establish its dimension. [HSEB 2052]

The gravitational potential energy of a body at a point inside the gravitational field is the amount of

work done on bringing the body from infinity to that point without acceleration. It can be shown that

gravitational potential energy (U) is,

GMm
U=– r

Where r is the distance of mass m from the centre of earth and M is the mass of earth. Its dimensions is
same as that of energy or work i.e. [ML2 T–2]

Explain the concept of weightlessness giving an example. [HSEB 2062 S]

We feel our weight due to reaction of the surface on which we are standing. If the reaction becomes

zero, then this gives us feeling of zero weight which is the condition of weightlessness. As for

example, suppose a person of mass m is standing in a lift which is falling downward with acceleration

'a' as shown in figure. Then resultant downward force F is, R

F = mg – R

or, ma = mg – R a
or, R = m (g – a)

If the lift is falling freely, then a = g m F = ma
∴ R = m (g – g)

or, R = 0

So, the person feels weightlessness in a freely falling lift. mg

Which takes more fuel, a voyage from the earth to the moon or from the moon to the earth?

Explain. [HSEB 2068 S]

The gravitational pull on the surface of moon is about one–sixth that of gravitational pull on the surface

of earth. Also, the escape velocity on the surface of moon is very less than that on the surface of earth.

More fuel is required to escape from earth's gravitational field than to escape from moon's

gravitational field. Hence, a voyage from the earth to moon takes more fuel than from the moon to

earth.

Why are space rockets usually launched from west to east in the equatorial plane?

Every particle on the earth has a linear velocity directed from west to east due to rotation of earth about

its polar axis. The magnitude of the velocity is v = rω. At equator r = R (Radius of earth) and hence vmax
= Rω (Maximum velocity). When the rocket is launched from west to east at the equatorial plane, this

maximum velocity, vmax = Rω, is added to the launching velocity. Hence, the launching process
becomes easier.

An artificial satellite revolves around the earth without using any fuel. On the other hand, an

aeroplane requires fuel to fly. Why?

An artificial satellite revolves around the earth at a much higher height where air resistance is almost

negligible and so it requires no fuel to overcome air resistance. Also, the satellite is orbiting the earth

with much higher speed and the necessary centripetal force for it is provided by the gravitational force

188 | Essential Physics

20. of attraction of the earth. On the other hand, an aeroplane is comparatively at low height from earth's
Ans: surface and it requires fuel to overcome the air resistance. Moreover, its speed is low and do not
maintain the centripetal force provided by the gravitational attraction force. Hence aeroplane needs fuel
21. but satellite requires no fuel.
Ans: Explain why tidal waves (high tide and low tide) are formed on seas?
The gravitational pull of moon on sea water causes high tides. The high tides at one place causes the
22. low tides (ebbs) at another place. The gravitational pull by sun also causes tides but they have only of
Ans: half magnitudes. On full moon and no moon days, these both effects are added and hence very high
tides are formed.
23. What will be our weight at the centre of earth, if the earth were a hallow sphere?
Ans:
24. Below the surface of earth, the acceleration due to gravity is, g' = g1 – Rx. At the center of earth, depth
Ans:
x = R and so, g' = 0. Hence, our weight = mg' = 0 at the centre of earth.
When a train moves from west to east at high speed, does its weight increase or decrease?
The weight of the train decreases. This is because the apparent acceleration due to gravity due to
rotation of earth about its polar axis is, g' = g – Rω2cos2θ.
When the train moves west to east, the term (Rω2cos2θ), which is also directed west to east, increases
and therefore, g' decreases. Hence, the apparent weight mg' decreases.
What is the effect on our weight due to revolution of the earth about the sun?
For the expression for g, there is no any factor involved which is influenced by the revolution of earth
about the sun. Hence, there is no any effect in our weight due to the revolution of earth about the sun.
Is it possible to shield a body from gravitational effects? Explain.
No, it is not possible to shield a body from gravitational effects because value of G and hence
gravitational force is independent with the nature of the intervening medium.

Numerical Examples

1. The sun has mass 330,000 times that of the earth. For a person on the earth, the average distance to

the centre of the sun is 23,500 times the distance to the centre of earth. In magnitude, what is the

ratio of the sun's gravitational force on you to the earth's gravitational force on you? [HSEB model]

Solution: ME Ms

Given, Ms = 330,000 ME FE m Fs
Rs = 23,500 RE RE Rs
FFEs = ?

GMsm Earth Sun

Here, Fs = Rs2 = GMsm × RE2 = Ms × RE2 = 330000 ME × RE2 = 330000
FE GMEm Rs2 GMEm ME Rs2 ME (23500 RE)2 23500 × 23500
RE2

∴ Fs = 33
FE 55225

2. Taking the earth to be uniform sphere of radius 6400 km and the value of g at the surface to be 10
ms–2, calculate the total energy needed to raise a satellite of mass 2000 kg to a height of 800 km above

the ground and to set into circular orbit at that altitude. [HSEB 2072 C, 2067 HSEB]

Solution:

Given, radius of earth (R) = 6400 km = 6.4 × 106 m
g = 10 ms–2

Mass of satellite (m) = 2000 kg

Height of satellite (h) = 800 km = 8 × 105 m

Radius of orbit of satellite (r) = R + h = 6.4 × 106 + 8 × 105 = 7.2 × 106 m

Gravity and Gravitation | 189

Total energy needed to raise the satellite and to set into circular orbit at that altitude is,

E = Increase in P.E. + K.E.

= (P.E.2 – P.E.1) + K.E.

= – GMm – – GMR m + 1 mv2 m
r 2 Mh

R1 1r 1 Gm Rr
2 r
= GMm – + m.

R1 1 21r Earth
r
= GMm – +

= gR2m R1 – 21r [since, GM = gR2]

= mgR – R2r2

= 2000 × 10 6.4 × 106 – (6.4 × 106)2 
2 × 7.2 × 106

= 7.1 × 1010 Joules
3. An earth satellite moves in a circular orbit with a speed of 6.2 kms–1. Find the time of one revolution

and its centripetal acceleration. [HSEB 2071 D]

Solution:
Given, orbital velocity of satellite (V0) = 6.2 kms–1 = 6200 ms–1
Radius of earth (R) = 6.4 × 106 m
g = 10 ms–2

Time period (T) = ?

Centripetal acceleration (a) = ?

Since,

GM gR2
v0 = r = R + h

or, v02 = gR2
R+h

gR2
or, R + h = v02

or, h = gR2 – R = 10 × (6.4 × 106)2 – 6.4 × 106
v02 (6200)2

∴ h = 4255567 m = 4.26 × 106 m

Now,

T = 2π (R + h)3 = 2π (6.4 × 106 + 4.26 × 106)3
gR2 10 × (6.4 × 106)2 = 10805.23 sec = 180 min = 3 hrs.

And, centripetal acceleration (a) is,
v02 v02 (6200)2
a = r = R+h = 6.4 × 106 + 4.26 × 106 = 3.61 ms–2

4. What is the period of revolution of a satellite of mass m that orbits the earth in a circular path of

radius 7880 km about 1500 km above the surface of the earth? [HSEB 2066]

Solution:
Radius of orbit of satellite (r) = 7880 km = 7.88 × 106 m

Height of satellite from earth's surface (h) = 1500 km
Radius of earth (R) = r – h = (7880 – 1500) km = 6380 km = 6.38 × 106 m

Time period (T) = ?

Since, T = 2π r3 = 2π (7.88 × 106)3
gR2 10 × (6.38 × 106)2 sec = 6888.88 sec = 1.91 hrs.

190 | Essential Physics

5. Calculate the period of revolution of a satellite revolving at a distance of 20 km above the surface of

the earth. [Radius of the earth = 6400 km, acceleration due to gravity = 10 ms–2] [HSEB 2054]

Solution:

Period of revolution (T) = ?

Height from surface (h) = 20 km = 2.0 × 104 m

Radius of earth (R) = 6400 km = 6.4 × 106 m
Acceleration due to gravity (g) = 10 ms–2

Since,

T = 2π r3 = 2π (R + h)3 = 2π (6.4 × 106 + 2.0 × 104)3
gR2 gR2 10 × (6.4 × 106)2 = 5050.13 sec.

6. A 200 kg satellite is lifted to an orbit of 2.2 × 104 km radius. If the radius and mass of the earth are
6.37 × 106 m and 5.98 × 1024 kg respectively, how much additional potential energy is required to lift

the satellite? [HSEB 2051]

Solution:

Mass of satellite (m) = 200 kg
Radius of orbit of satellite (r) = 2.2 × 104 km = 2.2 × 107 m
Radius of earth (R) = 6.37 × 106 m
Mass of earth (M) = 5.98 × 1024 kg

Additional potential energy required = ?

Since,

Additional P.E. required = P.E. at orbit – P.E. on earth's surface

= – GMm – –GMR m
r

= GMm R1 – 1r = GMm r R–rR

= 6.67 × 10–11 × 5.98 × 1024 × 200 2.2 × 107 – 6.37 × 106
6.37 × 106 × 2.2 × 107

= 8.90 × 109 Joules

7. Calculate the points along a line joining the centers of earth and moon where there is no

gravitational force. [HSEB 2069 S]

Solution:

Given, Mass of earth (Me) = 6 × 1024 kg
Mass of moon (Mm) = 7.4 × 1022 kg

Distance between centers of earth and moon (r) = 3.8 × 108 m

Let P be a point at a distance x from the centre of earth where the gravitational force due to earth and

moon are equal and opposite. Then, for no net gravitational force,

Fe = Fm Me

or, GMem = GMmm Mm
x2 (r – x)2 r
Fe P Fm
or, Me = Mm
x2 (r – x)2 m
x r–x

6 × 1024 7.4 × 1022 Earth Moon
or, x2 = (r – x)2

600 7.4
or, x2 = (r – x)2

Taking square root on both sides,

600 7.4
x2 = (r – x)2

Gravity and Gravitation | 191

or, 24.5 = 2.72
x (r – x)

or, 24.5 r – 24.5 x = 2.72 x

∴ x = 24.5 r = 24.5 × 3.8 × 108 = 3.42 × 108 m
2.72 + 24.5 27.22

So, at a distance 3.42 × 108 m from the centre of earth at a line joining centres of earth and moon, there

is no gravitational force.

Important Numerical Problems

1. Thee period of moon revolving under the gravitational force of the earth is 27.3 days. Find the
distance of the moon from the centre of the earth if the mass of earth is 5.97 × 1024 kg. [NEB 2074]

Solution:
Period of revolution (T) = 27.3 days = 27.3 × 24 × 60 × 60 sec. = 2.36 × 106 secs.

Distance of moon from centre of earth (r) = ?
Mass of earth (M) = 5.97 × 1024 kg.
Gravitational constant (G) = 6.67 × 10-11 Nm2 kg-2

We know,

T = 2π 3

r
GM

3
r
or, T2 = 4π2 GM

or, r3 = GMT2
4π2

or, r3 = 6.67 × 10-11 × 5.97 × 1024 × (2.36× 106)2 = 5.62 × 1025
4 ×(3.14)2

or, r = [ 5.62 × 1025]1/3

∴ r = 3.83 × 108m
So, the distance of the moon from the centre of the earth is 3.83 × 108 m.

2. An Earth's satellite makes a circle around the earth in 90 minutes. Calculate the height of satellite
above the surface of earth. Given the radius of earth is 6400 km and g = 980 cms–2

Solution:

Given, time period (T) = 90 minutes = 5400 sec
Radius of earth (R) = 6400 km = 6.4 × 106 m
g = 980 cms–2 = 9.80 ms–2

Height of satellite (h) = ?

Since, (R + h)3
T = 2π gR2

Squaring both sides,
(R + h)3
or, T2 = 4π2 gR2

or, (R + h)3 = T2gR2
4π2

11
h = T42gπR2 2 3 (5400)2 106)2
∴ – R =  × 9.80 × (6.4 ×  3 – 6.4 × 106
4π2

= 6.668 × 106 – 6.4 × 106 = 2.68 × 105m = 268 × 103 m = 268 km

192 | Essential Physics

3. Calculate the escape velocity for a particle from a satellite of earth (neglecting height of satellite
compared to earth's radius) in the direction of the satellite. Take g = 9.8 ms–2 and R = 6.37 × 106 m.

Solution:

The orbital velocity v0 of the satellite is,

GM GM
v0 = r = R + h

If the height of the satellite compared to earth's radius is neglected, then h → 0.

∴ v0 = GM gR2 gR = 9.8 × 6.37 × 106 = 7.9 × 103 m/s =7.9 km/s
R= R=

Also, escape velocity ve is,
ve = 2gR = 2 × 9.8 × 6.37 × 106 = 11.2 km/s

Hence, Escape velocity from the satellite

= Escape velocity from earth's surface – orbital velocity of the satellite

= 11.2 – 7.9 = 3.3 km/s

4. What would be the size of earth if the escape velocity from it were to equal to the speed of light in free
space. [Mass of earth = 6 × 1024 kg, G = 6.67 × 10–11 Nm2 kg–2]

Solution: Given, Escape velocity (Ve) = C = 3 × 108 m/s
mass of earth (M) = 6 × 1024 kg

Size of earth (R) = ?

Since,

2GM
ve = R

or, ve2 = 2GM
R

∴ R = 2GM = 2 × 6.67 × 10–11 × 6 × 1024 = 8.89 × 10–3 m = 8.89 mm
ve2 (3 × 108)2

Therefore, radius (size) of earth must be 8.89 mm for its escape velocity to be equal to speed of light in

free space.
5. Assuming g = 9.8 ms–2 and radius of earth = 6.37 × 106 m, calculate the escape velocity. Neglecting

earth's motion and sun's effect, if a meteor starts from rest at infinite distance, what will be its

velocity when it strikes earth's surface?

Solution:
Given, g = 9.8 ms–2
Radius of earth (R) = 6.37 × 106 m

Escape velocity, ve = 2gR = 2 × 9.8 × 6.37 × 106 = 11.2 km/s
Let, m = mass of meteor

P.E. of the meteor at infinite distance = 0

GMm
P.E. of the meteor at earth's surface = – R

Decrease in P.E. = 0 – –GMR m = GMm
R

gR2m ∴ g = GRM2 
=R

= gRm

Let v be the velocity of the meteor when it strikes the ground. Then,

Gain in K.E. = Decrease in P.E.

or, 1 mv2 = gRm
2

or, v = 2gR = 2 × 9.8 × 6.37 × 106 = 11.2 km/s

Gravity and Gravitation | 193

6. Calculate the velocity with which a body be projected upwards so as to reach height 4RE from the
surface of earth where RE is radius of earth = 6.4 × 106 m.

Solution:

Let M = mass of earth

m = mass of body

GMm v=? 4RE
Gravitational P.E. at earth's surface = – RE m

Gravitational P.E. at height 4RE from earth's surface
GMm GMm

= – RE + 4RE = – 5RE

Let v be the initial velocity of the body. Then, RE
Loss in K.E. = Gain in P.E.

1 mv2 – 0 = – GMm – –GRMEm M
2 5RE

or, 1 mv2 = –GMm + 5GMm
2 5RE
Earth
1 4GMm
or, 2 mv2 = 5RE

or, v2 = 8gRE2 Q g = GM
5RE RE2

∴ v= 8gRE = 8 × 9.8 × 6.4 × 106 = 10018 ms–1 = 10.018 kms–1
5 5

7. Calculate the amount of work done to move 1 kg mass from the surface of the earth to a point 105

km from the centre of the earth. [mass of the earth = 6 × 1024 kg, radius of the earth = 6400 km, G

= 6.7 × 10–11 Nm2 kg–2]

Solution:
Distance of a point from earth's centre (r) = 105 km = 108m

Mass of body (m) = 1 kg
Mass of earth (M) = 6 × 1024 kg
Radius of earth (R) = 6400 km = 6.4 × 106 m

G = 6.7 × 10–11 Nm2 kg–2

Work done (W) = ?

P.E. of 1 kg mass on earth's surface,

GMm
W1 = – R
P.E. of 1 kg mass at a point 105 km from centre of earth,

GMm
W2 = – r

Required work done = Difference in P.E.

or, W2 – W1 = – GMm – –GMR m
r

= GMm R1 – 1r = 6.67 × 10–11 × 6 × 1024 × 1 6.4 1 106 – 1108 = 5.88 × 107 Joules.
×

8.. A 400 kg satellite is in a circular orbit of radius 2RE above the earth. How much energy is required to
transfer it to a circular orbit of radius 4RE? What are the changes in the kinetic energy and potential
energy?

Solution:
Total mechanical energy of a satellite is,
GMm
E = – 2r

194 | Essential Physics

For, r = 2RE, initial energy is,

GMm GMm . . . (1)
Ei = – 2(2RE) = – 4RE

For r = 4RE, final energy is,

GMm GMm . . . (2)
Ef = – 2(4RE) = – 8RE

Total Energy required, ∆E = Ef – Ei

or, ∆E = – GMm – – G4MREm = GMm
8RE 8RE

= gRE2m GM
8RE [since, g = RE2 ]

= gREm = 10 × 6.4 × 106 × 400 = 3.20 × 109 Joules
8 8

GMm
The kinetic energy is, Ek = 2r

So, numerically, change in K.E. is equal to the change in total energy but opposite in sign.
∴ ∆Ek = – 3.20 × 109 J

GMm
The potential energy is, Ep = – r

So, the change in P.E. is numerically double than the change in Kinetic energy.
So, ∆Ep = 2 ∆ Ek = 2 × 3.20 × 109 (Numerically)
∴ ∆Ep = 6.40 × 109 Joules

9. At what height from the surface of earth will the value of g be reduced by 36% from the value at the

surface? Radius of earth = 6400 km

Solution:

Let, h = height at which g is reduced by 36% = ?

g' = g – 36% of g = 0.64 g

g' . . . (1)
or, g = 0.64 . . . (2)

We know, g' = GM × R2 = (R R2
g (R + h)2 GM + h)2

∴ From (1) and (2), we get
R2

(R + h)2 = 0.64

R
or, R + h = 0.8

R
or, R + h = 0.8

∴ R
h = 0.8 – R

6400
= 0.8 – 6400 = 1600 km

10. An experiment using the Cavendish balance to measure the gravitational constant G found that a
uniform 0.400 kg sphere attracts another uniform 0.00300 kg sphere with a force of 8.00 × 10–10 N,

when the distance between the centres of the spheres is 0.0100 m. The acceleration due to gravity at
earth's surface is 9.80 m/s2, and the radius of earth is 6380 km. Compute the mass of the earth from

these data.

Solution:

Given, m1 = 0.400 kg; m2 = 0.00300 kg

Gravity and Gravitation | 195

Gravitational force between spheres (F) = 8.00 × 10–10N
Distance between centres of spheres (r) = 0.0100 m ; g = 9.80 ms–2

Radius of earth (R) = 6380 km = 6.380 ×106 m

mass of earth (M) = ?

Since, F = Gm1m2
r2

or, G = Fr2 = 8.00 × 10–10 × (0.0100)2 = 6.67 × 10–11 Nm2 kg–2
m1m2 0.400 × 0.00300

Also,

GM
g = R2

or, M = gR2 = 9.80 × (6.380 × 106)2 = 5.98 × 1024 kg
G 6.67 × 10–11

11. Astrophysical theory suggests that a burned out star will collapse under its own gravity to form a

black hole when its mass is at least three times solar mass. If it does, what is the radius of its event
horizon? (Solar mass = 1.99 × 1030 kg)

Solution:

Given: Solar mass = 1.99 × 1030 kg
Mass of star (M) = 3 × 1.99 × 1030 kg

Radius of event horizon (Rs) = ?

Since,

Rs = 2GM = 2 × 6.67 × 10–11 × (3 × 1.99 × 1030) = 8.9 × 103 m = 8.9 km
c2 (3 × 108)2

12. When in orbit, communication satellite attracts the earth with a force of 19.0 KN and earth–satellite
gravitational potential energy (relative to zero at infinite separation) is –1.39 × 1011 J. (a) Find the

satellite's altitude above the earth's surface. (b) Find the mass of the satellite.

Solution:

Given, gravitational force (F) = 19.0 KN = 19000 N
Gravitational potential energy (U) = –1.39 × 1011 J

(a) Satellite's altitude (h) = ?

(b) Mass of satellite (m) = ?

Let M be mass of earth and r be radius of orbit of satellite.

GMm GMm
We know, U = – r = – r2 . r = –F.r

or, – 1.39 × 1011 = – 19000.r

∴ r = 1.39 × 1011 = 7.32 × 106 m
19000

Also, r = R + h

or, 7.32 × 106 = 6.4 × 106 + h
∴ h = 7.32 × 106 – 6.4 × 106 = 9.20 × 105 m = 920 km

Again,

GMm
U=– r

∴ Ur (–1.39 × 1011) × 7.32 × 106
m = – GM = – 6.67 × 10–11 × 6 × 1024 = 2542 kg

13. Titania, the largest moon of the planet Uranus, has 18 the radius of the earth and 17100 the mass

of the earth. (a) What is the acceleration due to gravity at the surface of Titania? (b) What is the

average density of Titania?

196 | Essential Physics

Solution:
1

Given, Radius of Titania = 8 (Radius of earth)

or, RT = RE . . . (1)
8

1
Mass of Titania = 1700 (mass of earth)

or, MT = ME . . . (2)
1700

(a) g = ?
T

(b) ρ = ?
T

GMT G. ME 64 GRME2E 64 64
RT2 1700 1700 1700 1700
We know, g = = = × = × gE= × 9.80 = 0.37 ms–2
T R8E2

GMT G. 4 π RT3.ρT 4
RT2 3 RT2 3
Also, g = = = π RTGρT
or, T

3g 3 g
T T
ρ = =
T 4πRTG R8E
4π G

or, ρ = 6g = π × 6.4 × 6× 0.37 × 10–11
T T 106 × 6.67

πREG

∴ ρ = 1655 kg m–3
T

14. Suppose you want to place a 1000 kg weather satellite into a circular orbit 300 km above the earth

surface. (a) What speed, period and radial acceleration must it have? (b) How much work has to be
done to place this satellite in orbit? Assume mass of earth 5.98 × 1024 kg and radius 6380 km.

Solution:

Mass of satellite (m) = 1000 kg

height of satellite (h) = 300 km

Mass of earth (M) = 5.98 × 1024 kg
radius of earth (R) = 6380 km = 6.380 × 106 m
radius of orbit of satellite (r) = R + h = 6380 + 300 = 6680 km = 6.680 × 106 m

(a) Speed of satellite,

v0 = GM 6.67 × 10–11 × 5.98 × 1024 = 7727 ms–1
r= 6.680 × 106

2πr 2π × 6.680 × 106 = 5432 sec. = 90.53 min
Period, T = v0 =
7727

Radial acceleration, a = v02 = (7727)2 = 8.94 ms–2
r 6.680 × 106

(b) Work done to place satellite in orbit,

W = (P.E. at orbit – P.E. at earth) + K.E. at orbit

= – GMm – –GMR m + 1 mv02 = GMm – GMm + GMm GMm GMm
r 2 R r 2r = R – 2r

= GMm R1 – 21r = 6.67 × 10–11 × 5.98 × 1024 × 1000 6.3801× 106 – 2 × 1 × 106
6.68

= 3.27 × 1010 Joules

Gravity and Gravitation | 197

15. Obtain the value of g from the motion of moon, assuming that its period of rotation round the earth

is 27 days 8 hours and the radius of its orbit is 60.1 times the radius of the earth. (Radius of earth =
6.36 × 106m).

Solution:

Here, g = ?

Period (T) = 27 days 8 hrs = (27 × 24 + 8) hrs = 656 hrs = 656 × 60 × 60 secs = 2361600 secs
Radius of earth (R) = 6.36 × 106 m
Radius of orbit of moon (r) = 60.1 R = 60.1 × 6.36 × 106 = 3.82 × 108 m

We know,

T = 2π r3
gR2

or, T2 = 4π2 r3
gR2

∴ g = 4π2r3 = 4π2 × (3.82 × 108)3 = 9.77 ms–2 ≈ 9.8 ms–2
R2T2 (6.36 × 106)2 × (2361600)2

16. The maximum vertical distance through which a fully–dressed astronaut can jump on the earth is 0.5

m. Estimate the maximum vertical distance through which he can jump on the moon, which has a

mean density two–thirds that of the earth and a radius one–quarter that of the earth, stating any

assumptions made. Determine the ratio of time duration of his jump on the moon to that of his jump

on the Earth.

Solution:

Height jumped on earth, he = 0.5 m
The acceleration due to gravity on earth's surface is,

ge = GMe = G × 4 π Re3ρe
Re2 Re2 3

∴ ge = 4 πReρe G . . . (1)
3

Where, ρe is density of earth,
The work done against earth's pull i.e. potential energy gained is,

W = mgehe = mge × 0.5 = 0.5 mge J

Where m is the mass of astronaut.

Now, density of moon, ρm = 2 ρe
3

Radius, of moon, Rm = Re
4

The acceleration due to gravity on moon's surface is,

gm = 4 π Rm ρm G [From equation (1)]
3

or, gm = 4 π R4e. 23ρe. G = 1 34 π Re ρe G = 1 ge
3 6 6

∴ gm = 1 . . . (2)
ge 6

The work done against moon's pull i.e. potential energy gained is,

W = mgmhm
where, hm is height jumped on moon
But, the potential energy gained on earth and moon must be same.

i.e., mgehe = mgmhm

∴ hm = ggme . he = 6 × 0.5 = 3 m

So, the astronaut can jump 3m on moon.

198 | Essential Physics

Again, let te and tm be the durations of the astronaut's jump on earth and moon respectively. Then,

he = 1 ge te2
2

or, 0.5 = 1 ge te2
2

∴ te2 = 1 . . . (3)
And, ge

hm = 1 gm tm2
2

or, 3 = 1 gm tm2
2

∴ tm2 = 6 . . . (4)
gm

Dividing (4) and (3), we get

tm2 = 6 × ge = 6 ggme  = 6 × 6 = 62
te2 gm 1

or, tm = 6
te

∴ tm:te = 6:1

17. A preliminary stage of spacecraft Apollo 11's journey to the moon was to place it in an earth parking

orbit. This orbit was circular, maintaining an almost constant distance of 189 km from the earth's
surface. Assuming the gravitational field strength in this orbit is 9.4 NKg–1, calculate (i) the speed of

the spacecraft in this orbit and (ii) the time to complete one orbit. (Radius of the earth = 6370 km).

Solution:

Given, height of orbit of satellite (h) = 189 km
Gravitational field strength (E) = g = 9.4 Nkg–1

Radius of earth (R) = 6370 km = 6.370 × 106 m
Radius of orbit (r) = R + h = 6370 + 189 = 6559 km = 6.559 × 106m

(i) Speed of spacecraft in the orbit (v0) = ?
Since,

GM gr2 Q g = GM
v0 = r = r RE2

∴ v0 = rg = 6.559 × 106 × 9.4 = 7852 ms–1

(ii) Time to complete one orbit (T) = ?

2πr 2π × 6.559 × 106
Therefore, T = V0 = = 5249 sec.
7852

Practice Short Questions

1. Why an astronaut in a space capsule orbiting the earth experiences a feeling of weightlessness?

[HSEB 2054]

2. If the force of gravity acts on all bodies in proportion to their masses, why does not a heavy body fall faster

than a light body? [HSEB 2068]

3. If earth suddenly stops rotating about its axis, what would be the effect on g? [HSEB 2064]

4. The weight of a body is less in mine. Explain. [HSEB 2063 S]

5. Why is the weight of a body become zero at the centre of earth?

6. What is the difference between inertial mass and gravitational mass of a body.

7. A satellite does not need any fuel to circle around the earth, why?

8. An elephant and an ant are to be projected out of earth into space. Do you need different velocities to do so?

9. What do you mean by gravitational field strength?

Gravity and Gravitation | 199

10. What is escape velocity? Write down its minimum value on the surface of earth. [HSEB 2058, 2059]

11. What would happen if gravity suddenly disappear?

12. Why is G called universal gravitational constant.

13. Value of g at the equator is less than that of poles of earth. Explain.

14. Why do different planets have different escape velocities?

15. Why does the earth have atmosphere?

16. Lighter gases like H2, He etc are rare in earth's atmosphere, why?
17. Is it possible to describe condition of weightlessness as condition of masslessness? Explain

18. What do you mean by terms black hole and event horizon?

19. Earth is continuously pulling moon towards its centre. Why does the moon not fall on to the earth?

20. When a satellite is suddenly stopped in its orbit, what will happen to it?

Practice Long Questions

1. What is a geostationary satellite? Derive an expression for the time period of the satellite revolving around

the earth? [HSEB 2072 D]

2. Derive expressions for the time period and height of artificial satellite from the surface of the earth.[HSEB 2072 E]

3. Obtain an expression for variation of 'g' with rotation of earth. [HSEB 2071 S]

4. Explain the concept of geostationary satellite. Find an expression for the total energy of the moon revolving

around the earth, [HSEB 2071 C]

5. What is escape velocity? Show that the escape velocity of a body is 2 R.g . Where symbols have usual

meanings. [HSEB 2070 S]

6. Define gravitational potential energy and derive its expression. [HSEB 2070 S]

7. Define escape velocity. Find an expression for the escape velocity from the surface of the earth.

[HSEB 2070 C, 2068]

8. What is a gravitational potential energy? Obtain an expression for the gravitational potential energy of a body

at a distance r from the centre of the earth. [HSEB 2070 D]

9. What is geo stationary satellite? Obtain an expression for the total energy of satellite orbiting round the earth?

[HSEB 2069 A]

10. What do you mean by the terms black hole and event horizon? Obtain an expression for the Schwarzschild

radius. [HSEB 2069 B]

11. What do you mean by parking orbit? Derive an expression for the orbital velocity and hence find time period

of the satellite revolving around the earth. [HSEB 2069 B, 2065]

12. Derive an expression for the variation of acceleration due to gravity of the earth with the altitude and explain

its meaning. [HSEB 2068 Old]

13. What is escape velocity? Prove that escape velocity of a body from earth's surface is 2gR where g is the
acceleration due to gravity on the surface of the surface of the earth and R is the radius of the earth.
[HSEB 2066]

14. How does acceleration due to gravity vary with distance from the centre of earth above and below its centre?

[HSEB 2064]

15. Suppose that a strong man can throw a stone so that it will never return to the surface of earth, How much

work does he have to do in throwing up the stone of mass 'm' and find an expression for its minimum

velocity? [HSEB 2063]

16. Explain what is meant by the universal gravitational constant. Discuss the variation of acceleration due to

gravity with attitude and derive an expression for its value at height h above the surface of the earth.

[HSEB 2061]

17. Explain what is meant by the universal gravitational constant. Discuss the variation of acceleration due to gravity

with depth and derive an expression for its value at depth d below the surface of the earth. [HSEB 2060]

18. Define escape velocity of a body on a planet. Derive an expression for it. [HSEB 2059]

19. Assuming the earth to be perfectly spherical, give sketch graphs to show how:

(a) The acceleration due to gravity, and

(b) The gravitational potential due to earth's mass vary with distance from the surface of the earth to

points external to it. [HSEB 2050]

20. Obtain an expression for acceleration due to gravity due to rotation of the earth

21. Explain the meaning of escape velocity based on the concept of gravitational potential. Hence derive

expression for escape velocity of a body thrown from the surface of the earth. [NEB 2074]

200 | Essential Physics

Practice Numerical Questions

1. An earth's satellite moves in a circular orbit with an orbital speed of 6200 m/s. (a) Find the time of one

revolution. (b) Find the radial acceleration of the satellite in its orbit. [Ans: 175 min, 3.6 ms–2]

2. Find the total energy required to launch a satellite of mass 2000 kg in an orbit of radius 7.2 × 106 m from the

centre of earth. [Ans: 7.2 × 1010 J]

3. At what height from the surface of earth will the value of g be reduced to 64% from the value at the surface?

Radius of earth = 6400 km. [Ans: 1600 km]

4. Calculate the earth's gravity force on a 75 kg astronaut who is repairing the Hubble Space Telescope 600 km

above the earth's surface, and then compare this value with his weight at the earth's surface. Take mass and

radius of earth as 5.97 × 1024 kg and 6380 km respectively. [Ans: 613 N, 613:750]

5. The typical adult human has a mass of 70 kg. What force does a full moon exert on such a human when it is

directly overhead with its centre 378,000 km away? Mass of moon is 7.35 × 1022 kg. [Ans: 2.4 × 10–3 N]

6. Ten days after it was launched towards mars in December 1998, the mars climate orbiter spacecraft (mass 629
kg) was 2.87 × 106km from the earth and travelling at 1.20 × 104 km/h relative to the earth. At this time, what

were (a) the spacecraft's kinetic energy relative to the earth and (b) the potential energy of the earth–

spacecraft system?
7. The acceleration due to gravity at the north pole of Neptune is approximately 10.7 m/s2. Neptune has mass

1.0 × 1026 kg and radius 2.5 × 104 km and rotates once around its axis in about 16 hr. What is the gravitational

force on a 5.0 kg object at the north pole of Neptune? [Ans: 53.4 N]

8. At what distance above the surface of earth is acceleration due to gravity 0.980 ms–2, if the acceleration due to

gravity at the surface has magnitude 9.80 ms–2? [Ans: 2.87 × 106]

9. Rhea, one of Saturn's moon, has a radius of 765 km and an acceleration due to gravity of 0.278 ms–2 at its

surface. Calculate its mass and average density. [Ans: 9.7 × 1021 kg, 5172 kgm–3]

10. The gravitational force on a mass of 1 kg at the earth's surface is 10 N. Assuming the earth is a sphere of

radius R, Calculate the gravitational force on a satellite of mass 100 kg in a circular orbit round the earth of

radius 2R? [Ans: 250 N]

11. A proposed communication satellite would revolve round the earth in a circular orbit in the equatorial plane,

at a height of 35880 km above the earth's surface. Find the period of revolution of the satellite in hours, and

comment on the result. (Radius of earth = 6370 km, mass of earth = 5.98 × 1024 kg and constant of gravitation

= 6.66 × 10–11 Nm2 kg–2) [Ans: 24 hrs]

12. Explore 38, a radio–astronomy research satellite of mass 200 kg circles the earth in an orbit of average radius

3R/2 where R is the radius of the earth. Assuming the gravitational pull on a mass of 1 kg at the earth's

surface to be 10 N, Calculate the pull on the satellite. [Ans: 889 N]

13. A man can jump 1.5 m high on the earth. Calculate the approx. height he might be able to jump on a planet

whose density is 1 quarter of the earth and whose radius of 1/3 of the radius of earth. [Ans: 18 m]

14. The radius of earth's orbit is 1.5 × 108km and that of mars is 2.5 × 1011 m. In how many years does the mars

complete its one revolution? [Ans: 2.15 years]

MBBS, B.E., B.Sc. Nursing Entrance Preparation Questions

1. If the value of 'g' is same at depth 'd' inside earth and height 'h' above earth, then

a. d = h b. d = 2h h d. d = 3h
c. d = 2

2. A body projected with a velocity equal to twice the escape velocity from the earth. The velocity of the body
in free space will be

a. ve b. 1.9 ve c. 2 ve d. 3 ve

g
3. A what height from earth, g becomes 2 ?

R b. 0.414 R c. 0.7 R d. R
a. 2