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Chapter: 07

Gravitation
Prepared by: Kishor Khatiwoda
Department of Physics

© KK Physics
1

Newton's law of Gravitation

Newton's law of gravitation states that “everybody in this universe attracts every other body with a
force which is directly proportional to the product of their masses and inversely proportional to
the square of distance between their centres. The force acts along the line joining the centres of the
two bodies.”
According to Newton’s law of gravitation, the magnitude of attractive force F between the two
bodies is
F  m1m2 ………………………….. (i) m1 m2

1 F -F
and F  2 ……………………………… (ii)
r
Combining equation (i) and (ii) C1 C2
mm
F  12 2 r
r
mm
∴ F = G 12 2
r
Where G is a constant of proportionality, called universal gravitational constant. In SI unit, the value of G
is 6.67 × 10–11 Nm2/kg2

© KK Physics [2]

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Define gravitational constant (G)

m1m2
We have, F=G
r2
If m1 = m2 = 1 kg and r = 1 m, then F = G
Hence universal gravitational constant is defined as the force of attraction between two bodies
of unit masses placed unit distance apart.

1. The gravitational force between two bodies is always attractive.

2. The value of G does not depend upon the nature and size of masses.

3. The gravitational force of attraction between two bodies is not altered by the presence of other
bodies.

4. G is measured experimentally.

5. This law fails if the distance between the objects is less than 10-9 m i.e of the order of intermolecular
distance.

6. Mainly Newton’s law of gravitation is valid for point masses.

© KK Physics [3]

Gravity and Acceleration due to Gravity:

The force by which the earth attracts any object towards its centre is known as gravity.
The acceleration produced on an object due to earth’s gravity is called acceleration due to
gravity. It is denoted by g.
m
If air resistance is neglected, the standard value of g is 9.8 ms-2.
Consider an object of mass m is placed on the surface of earth. Let M and R R
be the mass and radius of earth. M
From Newton’s law of gravitation, the force of attraction between the earth C
GMm . . . . . . (i)
and the object is F=
R2
The weight of the object is F = mg . . . . . . (ii)
From eqn (i) and (ii) we get
GMm
mg =
R2
∴ g= GM . . . . . . (iii)
R2
This gives the expression for acceleration due to gravity.
© KK Physics [4]

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This relation shows that acceleration due to gravity is independent of the mass of the falling
object (m).
So, all the bodies regardless their masses, will fall with the same acceleration due to force of
gravity near the earth's surface. The value of g depends upon the mass of the earth and its
radius.

Gravitational and inertial mass

The mass defined by Newton’s law of motion is called inertial mass. According to Newton’s
second law of motion, if a force F acting on a body produces an acceleration a then F = ma,
where m is the inertial mass. It gives the measure of a body.

The mass defined by Newton’s law of gravitation is called gravitational mass.

GMm
F=
R2
m = gravitational mass
The mass m of the body thus determined is the gravitational mass.

© KK Physics [5]

Inertial mass Gravitational mass

(i) It is the mass of the body which measures (i) It is the mass of the body which determines
the inertia of the body. the gravitational pull acting on it.

(ii) It is measured when the body is in motion. (ii) It is measured at the rest potion of the body.

(iii) The inertial mass of a body is measured (iii) The gravitational mass of a body is
according to Newton's 2nd law of motion. measured by Newton's law of gravitation.

F FR2
i.e. mi = i.e. mG =
a GM

© KK Physics [6]

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Q)Assuming the mean density of the earth is 5500kgm–3, that G is 6.7 × 10–11 Nm2 kg –2

and that the earth's radius is 6400 km. Find a value for the acceleration due to gravity
at the earth's surface.
Solution:
Given,  = 5500kg/m3
G = 6.7 × 10–11 Nm2/kg2
R = 6400km = 6400000 m
g =?
GM
Now, we have, g = 2
R
G 4
g = 2 × πR3ρ
R 3
4
g = G. R
3
= 6.7 ×10-11× ×× 6400000×5500
g = 9.88m/s

© KK Physics [7]

Variation of Acceleration due to Gravity

i) Due to shape of the earth
ii) Due to height from earth’s surface
iii) Due to depth from earth’s surface
iv) Due to rotation of the earth (effect of latitude)

i) Due to shape of the earth

The earth is not perfectly spherical. It is flattered at the pole and bulged at the equator. So
equatorial radius Re is more thatn the polar radius RP
GM
We have, g = 2 R
1
i.e g ∝
R2
We have, Re > Rp
So, gp > ge
Thus, the acceleration due to gravity is maximum at the pole and minimum at the equator.

© KK Physics [8]

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ii) Due to Height from earth’s surface

Consider the earth of mass M and radius R. let g be the acceration due to
gravity at a point A on the surface of earth and is given by
GM
g= . . . . . . . . (i)
R2
Let g’ be the acceration due to gravity at point B at a height of h above the surface
of earth as shown in fig. M
∴ The acceleration due to gravity at point B is given by
2
GM R
g’ = . . . . . . . . (ii) ∴ g’ = g × R+h
R+h 2
Since (R+h)>R, then we have g’ < g.
Dividing eqn (ii) by (i) we get
As the height increases, the acceleration due to
g’ GM R2
or, = × gravity decreases
g R+h 2 GM
−2 −2
R+ h h
g’ R2 ∴ g’ = g × = g 1+ R
or, = R
g R+h 2

© KK Physics [9]

−2
h 2h h
Using binomial expansion, 1+ =1− + higer power of
R R R

h h
Sicne the value of is very small. So higher power of can be neglected.
R R
2h
∴ g’ = g 1 −
R
This shows that g’ < g.
Thus, the value of acceleration due to gravity decreases with increase in height.

© KK Physics [10]

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Q) At what distance above the surface of the earth is acceleration due to gravity 0.98 ms-2,
if the acceleration due to gravity at the surface of earth has magnitude 9.8 ms-2?

2h
We have g’ = g 1 − R Ans: 2.87 × 106 m

then find h = ….?

Q. Find the height at which the acceleration due to gravity becomes one fourth of value of
g.

© KK Physics [11]

iii) Due to Depth from earth’s surface M

Consider earth to be a perfect sphere of radius R and Mass M. Let  be its average x B
density.
R-x
Then, value of acceleration due to gravity on the surface of earth is given by
O R
GM
g= . . . . . . . . (i)
R2
We know,
Mass (M) = Volume (V) × density ()
4
i.e. M = R3 
3
Then equation (i) becomes
4
g=  G R  . . . . . . . . (ii)
3
Let g ‘ be the acceleration due to gravity at point B at a depth x below the surface of earth. A body at point
B will experience force only due to the material body of the earth whose radius is (R – x). Thus, value of g'
at point B is given by GM’
g’ = . . . . . . . . (iii)
R−x 2

© KK Physics [12]

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We know,
M' = V' × 

4
=  (R – x)3 × 
3
Thus, equation (iii) becomes,
g' =  G (R – x)  ………………….. (iv)
Dividing eqn (ii) by (i) we get
g’ R–x
=
g R
x
g’ = g 𝟏− . . . . . . . (v)
R
This relation shows that g' < g.
Thus, the value of acceleration due to gravity decreases with depth. In other word, as we go inside the earth,
acceleration due to gravity goes on decreasing.

At the Centre of the earth, x = R, hence equation (v) gives

g' = 0.

© KK Physics 13

The variation of g' inside and outside the earth has been plotted in the following graph

fig: Variation of g' inside and outside the earth

© KK Physics 14

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Q.) How does 'g' at a point vary with the distance from the centre of the
earth? Where is the highest value of g? Explain.
Ans: Acceleration due to gravity at any point below the surface of earth is
x
g' = g 1 − R

At depth g' < g

At centre of earth, x = R So that g' = 0
Hence, value of acceleration due to gravity increases as one moves above from the
surface of earth.

© KK Physics 15

iv) Due to Rotation of the Earth

Fc cos 𝛟
Consider the earth to be a sphere of radius R and mass M with centre O.
𝛚
The earth is rotating about its axis with angular velocity 𝛚. Let a body of 𝛟
r Fc
mass m is placed on the surface of earth at point P at latitude angle 𝛟 . C 𝛟
P S
We know the earth is rotating, So the body of mass m at P makes the circle
with centre C and radius r. 𝛟
CP r
From fig, Cos 𝛟 = = r O
OP R
∴ r = R Cos 𝛟 . . . . . (ii)

The weight of body is W = mg . . . . . (i)

acting towards the centre of earth along PO

Due to rotation of earth, the body experiences a centrifugal force FC which acts along the radius of circular
path and in outward direction along PS

The magnitude of centrifugal force is given by

FC = m𝛚2r = m 𝛚2 R Cos 𝛟 . . . . . . . (ii)

© KK Physics 16

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Now the body experiences two forces: one due to gravity (mg) and the other Fc due to rotation of the earth.

∴ Net force (apparent force) acting on the body is

mg' = mg - Fc Cos
or, mg' = mg - m𝛚2RCos 𝛟 (Cos)
or, g' = g - 2 Rcos2

∴ g’ = g - 𝛚2 R cos2

This relation shows that g‘ < g

© KK Physics 17

Special cases
(a) At the poles: When the object is at the poles i.e.  = 90°, then cos  = 0

Thus, g'p = g – R2 × O

g'p = g

It indicates that there is no effect on the value of acceleration due to gravity at the poles due to
the rotation of earth.

(b) At the equator: When the object is at the equator of the earth i.e.  = 0°, then cos = 1

Thus, gE′ = g – R2 × 1

∴ g'E = g – R2

It indicates that acceleration due to gravity is minimum at equator.

From above discussion, we can conclude that acceleration due to gravity gradually increases
when an object goes from equator to pole due to the rotation of the earth.

© KK Physics 18

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Q.)What will happen to the value of g if the earth stops rotating?

Ans: The variation of g due to the rotation of the earth is given by
g’ = g – R 2 cos2𝛟
Where  is the angular velocity of the earth.
If the earth stops rotating, its angular velocity becomes zero i.e. 𝛚 = 0
 g’ = g
This shows the value of acceleration due to gravity increases (i.e. becomes equal to the
value of g on earth's surface).

© KK Physics 19

Gravitational Field
The gravitational field due to a material body is the space around the body in which any other
mass experiences a force of attraction. Theoretically, the gravitational field due to a material
body extends up to infinity. However, the effect of gravitational field decreases as the
distance from the body is increased.

The space around a material body in which its gravitational pull can be experienced is called
its gravitational field.

© KK Physics 20

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Gravitational Field Intensity (E)

The gravitational intensity (or strength) at a point inside the gravitational field is defined as
the force experienced by a unit mass placed at that point. It measures the strength of the
gravitational field

It is a vector quantity. It is denoted by E and given by, F

R m
F
E = r
m M
If M and m be the masses of an object and unit mass, respectively such that which are
separated at a distance 'r'.
The gravitational force of attraction is given by,
GMm
F=
r2
Thus, intensity of gravitational field at that point will be

F GMm GM
E= = =
m r2m r2
© KK Physics 21

Intensity of gravitational field on the surface of earth

The gravitational force on a body of mass m placed on the surface of earth is
GMm
F=
R2
Thus, intensity of gravitational field at that point will be
F GMm GM
E= = = = g
m R2m R2

Thus, intensity of gravitational field at any point near the surface of earth is equal to the
acceleration due to gravity at that point. Its unit is N kg-1 or m s-2 in SI units.

© KK Physics 22

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M
Gravitational Potential Energy
The gravitational P.E. of a body at a point inside R B F C m
∞
the gravitational field is defined as the amount of O
A
work done in bringing a body from infinity to that dx
r
point.
Consider an earth to be a uniform sphere of radius R and x
Fig
mass M. We have to determine the gravitational P.E. of
the body at point A at a distance r from center of earth.
Let a body of mass 'm' is initially at infinity. This mass 'm' is to be brought to the position 'A' in
gravitational field.

Suppose at any instant, the body is at point C such that OC = x . Then, the gravitational force of
attraction on the body at C is given by GMm
F=
x2
When the body is moved through a small distance CB = dx, then the small amount of work done has
to be done. This work done is given by
© KK Physics 23

GMm
dW = F dx = x2 dx
Therefore, total amount of work done in bringing the body from infinity to point A is given by
𝑟
𝑮𝑴𝒎 𝑟 𝟏 𝒓 𝟏 𝟏
W∞A = න 𝟐 𝒅𝒙 = GM m ‫𝒙 ∞׬‬−𝟐 𝒅𝒙 = - GM m 𝒙 = - GM m −
𝒙 ∞ 𝒓 ∞
∞
GMm
∴ W∞A = - r
Since this work done is stored in the body in the form of potential energy U.
Therefore, gravitational potential energy at point A is given by
GMm
∴ U = - r
This is the expression for gravitational potential energy of a body of mass m at a distance r from
center of earth. –ve sign show that the potential energy is due to attractive force of earth.

On the surface of earth; r = R

GMm
∴ U = -
R
© KK Physics 24

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Gravitational Potential (V)

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.
W
Gravitational potential, Vg =
m
Gravitional potential is a scalar quantity. Its unit is Joule per kg.

© KK Physics 25

M
Escape Velocity (ve)
R P Q
The minimum velocity with which a body must be A
O ∞
projected upwards from the surface of earth to x dx

overcome its gravitational pull so that it can escape

into space is called escape velocity .

In this case, to escape the body from earth's surface, K.E. gain by it must be equal to the work done
against gravity going from the earth surface to infinity.
i.e., K.E. gained by the body = Work done to bring the body from surface of earth to infinity.

© KK Physics 26

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M
Expression of Escape Velocity (ve) R
A P Q
Consider an earth to be perfect sphere of radius R O ∞
x dx
and mass M. Let a body of mass m is to be
projected from a point A (surface of earth) as
shown in fig.

At any instant, the body be at a point P at a distance x form the center of earth.
Then, gravitational force of attraction is
GMm
F =
x2
If we displace this mass through small displacement dx from P to Q, small work done dW is given by
GMm
dW = F dx = x2 dx

© KK Physics 27

M
The total amount of workdone in taking the body R
A P Q
from surface of earth to infinity is calculated by O ∞
x dx
integrating this equation form limit x = R to x = ∞,
we get
∞ 𝟏 ∞ 𝟏 𝟏
𝑮𝑴𝒎 ∞ −𝟐
WA∞ = න 𝒅𝒙 = GM m ‫׬‬
𝑹
𝒙 𝒅𝒙 = - GM m 𝒙 𝑹
= - GM m −
𝒙𝟐 ∞ 𝑹
𝑹

GMm
∴ WA ∞ = 𝑹
. . . . . . . (i)

The kinetic energy by which the body is projected so that it just overcome the gravitational pull of
the earth is given by
𝟏 Where ve is the escape velocity
K.E. = m ve2 . . . . . . . (ii)
𝟐

© KK Physics 28

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Equating equation (i) and (ii) we get,

𝟏 GMm
∴ m ve2 =
𝟐 𝑹
2GM
Or, ve =
𝑹

We have GM = g R2

2
∴ ve = 2gR
𝑹

Or, ve = 2gR
For the earth:
This is the expression of the escape velocity.
g = 9.8 m/s2 and R = 6.4 × 106 m

Then the escape velocity of earth:

ve = 𝟐 × 𝟗. 𝟖 × 6.4 × 106 = 11.2 × 103 m/s

= 11.2 km/s

© KK Physics 29

Q) With what velocity must a rocket be fired from the earth's surface so that it just
escape from gravitational influence of earth. (g = 10ms–2, r = 6.4 × 106m)

ve = 11.3 km/s

© KK Physics 30

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Q) Explain, why moon has no atmosphere?

Ans: Moon has smaller value of acceleration due to gravity and the value of escape
velocity on the moon's surface is small as well. The values of r.m.s velocity of
different gases of atmosphere is much above the value of escape velocity on
moon. This results the different gases to escape from moon surface easily. Hence,
moon has no atmosphere.

© KK Physics 31

Motion of Satellite

Satellite: A satellite is a body which revolves continuously around a bigger body in a stable
orbit. For simplicity, it is assumed that satellites move in circular orbits.

The centripetal force required by a satellite to move in a circular orbit is provided by the

gravitational force of attraction between the satellites and the body around which it revolves.

© KK Physics 32

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Types of Satellite:
1) Natural Satellite: A heavenly body that revolves around a planet in a close and stable orbit
is called a natural satellite.

For example, moon is the natural satellite of the earth. Earth is natural satellite of sun.

They are naturally made heavenly bodies which revolves around planets.

2) Artificial Satellite: A man-made satellite that orbits around the earth or some other
heavenly body is called an artificial satellite.

Artificial satellites circulating the earth. They are called earth satellites.

For example, communication satellite is used to transmit information around the globe.

© KK Physics 33

Principle of Launching of satellite

Artificial satellites revolves around the body in a circular path with high
speed. At this high speed, the atmospheric friction would probably burn
the satellite. Therefore, the satellite is taken by a rocket to a suitable
height above the surface of the earth, and then given the suitable
horizontal velocity so as to put it in a circular path around the earth. The
situation for various initial horizontal speeds in shown in fig.
If the speed is small, the satellite will fall to earth's surface at points P 1, P2
and P3 by the pull of gravity. However, if the speed is just large enough,
the satellite will simply circle the path with its orbital velocity (Vorb).
Although the satellite is falling continuously towards earth's centre, the
curvature of the earth is the same as the curvature of satellite's path. The
speed of the satellite in its circular path is just proper so that the required
centripetal force is equal to gravitational force of attraction provided by
the earth.

© KK Physics 34

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Orbital velocity and time period of Earth's satellite

(a) Orbital velocity
The minimum velocity required for a satellite to revolve into a
particular orbit at certain height is called orbital velocity. This velocity
is projected in a direction parallel to the tangent of the orbit.

Consider the earth to be sphere of radius ‘R’ and mass ‘M’. let a
satellite of mass m revolving around the earth at a heigh ‘h’ from the
surface of earth with velocity ‘v’ as shown in fig. Then the radius of the
orbit of the satellite is r = (R + h).
This gravitational force of attraction is given by
𝐆𝐌𝐦
Fg = . . . . . . (i)
(𝐑+𝐡)𝟐
The centripetal force required to keep the satellite in circular orbit of radius (R + h) is

mvo2
FC = . . . . . . (ii)
𝐑+𝐡

© KK Physics 35

This centripetal force is provided by the gravitational force exerted by the earth on the satellite.
i.e. FC = Fg
mvo2 𝐆𝐌𝐦
or, =
𝐑+𝐡 (𝐑+𝐡)𝟐
GM
or, Vo2 =
R+h
𝑮𝑴
∴ VO = ....... 𝒊𝒊𝒊
𝑹+𝒉
𝑮𝑴
Since, 𝒈 = 𝑹𝟐
𝑮𝑴 = 𝒈𝑹𝟐
Thus, equation (iii) can be written as

𝑮𝑹𝟐
∴ vo = 𝑹+𝒉
Here, equation (iii) and equation (iv) are the required
𝑮
∴ vo = 𝑹 ....... 𝒊𝒗 expression for the orbital velocity of the satellite.
𝑹+𝒉

© KK Physics 36

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Special case:

When the satellite orbits very close to the surface of earth i.e. h << R, then h = 0.

So, equation (iv) becomes,

vo = 𝒈𝑹

For earth, g = 9.8 ms–2, R = 6.4 × 106 m

 v = 𝟗. 𝟖 × 𝟔. 𝟒 × 𝟏𝟎𝟔 = 8000 m/s = 8 km/s

Thus, the orbital speed of the satellite close to earth's surface is about 8 kms–1. The
expressions (i) and (ii) indicate that orbital velocity of a satellite is independent of the mass of
a satellite.

© KK Physics 37

Q. Define escape velocity and orbital velocity. How they are

related?
Ans: Since the expression of escape velocity and orbital velocity are

ve = 𝟐𝒈𝑹 and vo = 𝒈𝑹

∴ ve = 𝟐 vo

© KK Physics 38

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Time period of satellite (T) We know, GM = gR2

The time taken by the satellite to complete 𝑹+𝒉 𝟑
∴ T = 2
one revolution around the earth is called time 𝑮𝑹𝟐

period of the satellite. 2 𝑹+𝒉 𝟑

or, T= 𝒈 ………………. (ii)
𝑪𝒊𝒓𝒄𝒖𝒎𝒇𝒆𝒓𝒆𝒄𝒏𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒐𝒓𝒃𝒊𝒕 𝑹
∴ Time period (T) =
𝒐𝒓𝒃𝒊𝒕𝒂𝒍 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 This gives the time period of a satellite.
𝟐 𝝅 (𝑹+𝒉)
or, T = …………… (i) Special case: When the satellite orbits very
vo
𝑮𝑴 close to the surface of the earth, h = 0. Then
We know, vo =
𝑹+𝒉 equation (iv) can be written as,
Thus, equation (i) becomes,
2 𝑹𝟑
𝟐 𝝅 (𝑹+𝒉) T= ×
T= 𝑹 𝒈
𝑮𝑴
𝑹+𝒉
𝑹
∴ T = 2
𝑹+𝒉 𝟑 𝒈
T = 2
𝑮𝑴

© KK Physics 39

Height of satellite above earth's surface

The time period of satellite is given by
2(R+h)3
T=
𝑹 g
Squaring both sides, we get,
42 (R+h)3 For earth,
T2 =
R2 g
g = 9.8 ms–2, R = 6.4 × 106 m
T = 24 × 3600 seconds
gR2T2
Or, (R+h)3 = 𝟏
42 ∴ h =
9.8 × (6.4 × 106)2 × (24 × 3600)2 𝟑
– 6.4 × 106
𝟏 42
g R2 T 2 𝟑
= 36000 km
Or, R+h =
4 2 ∴ Radius of the parking orbit = R+h
𝟏
= (6400 + 36000) km
g R2 T 2 𝟑
 h= – R
4 2 = 42,400 km

© KK Physics 40

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Q) Calculate the time period of a satellite moving in a circular orbit 200 km above the
earth's surface if the radius of earth is 6400 km and acceleration due to gravity of this
height is 9.8ms–2?
Given, g at height h g’ = 9.8 m s-2
R = 6400km = 6400,000 m
h = 200 km = 200,000 m
r = R + h = 6400 + 200 = 6600 km = 6600,000 m
g =?
𝑮𝑴
v =
r
we have, GM = g’ r2
g′ r2
∴v= = g′ r = . .. .. . .?
r
2πr
Again T = = . . . . . .????
v
5156.3 sec
© KK Physics 41

Q) Find the acceleration due to gravity on the surface of the earth and hence calculate
the period of revolution of a satellite moving in a circular orbit round the earth at a
height of 3.6 × 106 m above the earth's surface.
Solution:
Given, G = 6.7 × 10–11 Nm2/kg2
R = 6400km = 6400000 m
M =6 × 1024kg
h = 3.6 × 106m
g =?
GM
g= 2
R
T = ?
2 (R+h)3
We have T =
𝑹 g
......
9912.09 sec

© KK Physics 42

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Geostationary Satellite
If the height of an artificial satellite above earth's surface is such that its period of
revolution is 24 hours, then the motion of the satellite will be synchronous with the
earth's axial rotation. As a result, the satellite will remain stationary over the same place
relative to the earth while the earth rotates. Such a satellite is said Geostationary satellite
or simply stationary satellite.
The orbit of geostationary satellite is sometimes called parking orbit. Most
communication satellites are geostationary satellites so that information can be
transmitted from one part of the world to another. It should be noted that the
geostationary satellite revolves around the earth from west to east in a close circular orbit
and coplanar with the equatorial plane.

© KK Physics 43

In order to put a satellite in the parking orbit, the following conditions must be
satisfied.
(i) It should revolve in an orbit concentric and coplanar with the equatorial plane.

(ii) The time period of the satellite should be 24 hrs. i.e. its motion should be synchronous
with the axial rotation of the earth.

(iii) It should rotate in the same direction as the earth is rotating.

(iv) The height of the parking orbit (h) should be 36,000 km above the earth's surface which is
determined by
2 (R+h)3
T=
𝑹 g

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Q.) What do you mean by geostationary satellite? Explain

Ans: Artificial satellite of earth which appears to be stationary when viewed from earth
is called geostationary satellite. Its time period is 24 hrs and orbits the earth at a height
of 36000 Km from it.

© KK Physics 45

Total Energy of Satellite

Consider the earth to be a spher of radius R and mass M. let a satellite of mass m is moving
with velocity v at a height if h from earth surface in a circular path of radius (R+h) with orbital
velocity vo.
The total energy of satellite is the sum of kinetic energy and potential energy.
i.e. Total energy of a satellite is (E) = K.E. + P.E
GM
we have, orbital velocity of satellite is vo =
R+h
then K.E. of satellite is given by
2
1 1 GM GMm
K.E. = mvo2 = m = . . . . . (i)
2 2 R+h 2(R + h)
Now, gravitational potential energy of satellite at a height ‘h’ above the surface of earth is
given by
GMm
P.E. = - . . . . . . (ii)
R+h
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∴ Total energy of satellite is

GMm GMm
E= -
2(R + h) (R + h)

𝑮𝑴𝒎
∴ E=-
𝟐(𝑹+𝒉)
This is the required expression of total energy of satellitye.
Negative sign indicates that work is to be done in order to free satellite from its the orbit
i.e. the satellite is bounded to the earth.

© KK Physics 47

Q) 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?
Given, m = 200 kg
radius of orbit r = 2.2 × 104 km = 2.2 × 107 m
R = 6.37 × 106 m
M = 5.98 × 1024 kg
additional potential energy = ?
We have, additional P.E. = P.E. on the orbit – P.E. on the surface of earth
GMm GMm
= - - −
r R
GMm GMm
= -
R r
1 1
= GMm −
R r
=
4.47 × 107 J
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Q) Assuming the earth to be uniform sphere of radius 6400 km, calculate the total
energy needed to raise the satellite of mass 2000 kg to a height of 800 km above the
surface of the earth and to set it into circular orbit at that altitude. (Ans:7.12 × 1010 J )
Given, m = 200 kg
R = 6400 km
h = 800 km
Radius of orbit r = R + h = 6400 + 800 = 7200 km = 7.2 × 106 m
Total energy needed E = ?
total energy needed (E) = increase in P.E. + K.E
GMm GMm GMm
= − − − +
r R 2r
GMm GMm GMm
=− + +
r R 2r
1 1
= GMm −
R 2r

© KK Physics 49

Weightlessness
The situation in which the effective or apparent weight of the body becomes zero is called
weightlessness.
If a body of mass ‘m’ is placed at a point where the effective acceleration due to gravity is
g’, then the effective weight of the body = mg’
The effective weight of the body will be zero if g’ = 0

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In the following circumstances, the body will be in the state of weightlessness:

(a) At null points
Null points are the points in outer space where the gravitational forces due to heavenly
masses cancel out. Hence, the effective value of acceleration due to gravity and the
weight of a body at null points will be zero.
(b) At the Centre of the earth
As we know the value of acceleration due to gravity at the centre of the earth is zero i.e.
g’ = 0
Hence, the effective weight of the body at the centre of the earth = mg’ = m × 0 = 0

© KK Physics 51

(c) In a freely falling lift

If a body is in the state of free fall, the acceleration of falling body is equal to the value
of acceleration due to gravity i.e. a = g.
∴ mg’ = mg – ma = mg – mg = 0
Where reaction of the floor gives the apparent weight. Due to this reason, a body lying in
a freely falling lift feels weightlessness.
(d) Astronaut in a space capsule orbiting earth
In this case, the gravitational force on the body due to earth is balanced by the
centrifugal force on the body. Due to this reason, the effective weight of the body
becomes zero.

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Short Answer Questions

Q.) Define acceleration due to gravity (g).
Ans: The rate of change of velocity for a freely falling body due to the force of
gravity is called acceleration due to gravity. It is a vector quantity and depends
on the radius and mass of a massive body i.e. earth, moon etc. Its SI unit is
m/s2. It helps to determine the weight of an object lying on the surface of a
heavenly body.

GM
i.e. g = R2 and W = mg

© KK Physics 53

Short Answer Questions

1. State Newton's law of gravitation.
2. Mention the differences between inertial mass and gravitational mass.
3. 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]
4. One can jump higher on the moon's surface than on the earth's surface, why?
5. If earth suddenly stops rotating about its axis, what would be the effect on g?
6. Define escape velocity and orbital velocity. How they are related?
7. The space rockets are launched from west to east. Why?
8. When a satellite is suddenly stopped in its orbit, what will happen to it?
9. What do you mean by geo-stationary satellite? Explain.
10. What do you mean by weightlessness?
11.If the sun somehow collapsed to form a black hole, what effect would this event have on the
orbit of the earth? [HSEB 2067]
12.An artificial satellite moving round the earth in a circular orbit possesses a changing
acceleration. Is it true?

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Numerical Problems
1) 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 of radius 2R from the centre of the earth. [Ans: 250 N]
2) Find the period of revolution of a satellite moving in a circular orbit round the earth
at a height of 3.6 × 106 m above the earth's surface. Assume the earth is a uniform sphere
of radius 6.4 × 106 m, the earth's mass is 6 × 1024 kg and G is 6.7 × 10–11 Nm2 kg–2.
[Ans: 9910 Sec.]
3) Assuming the mean density of the earth is 5500 kg m–3, that G is 6.7 × 10–11 Nm2 kg–2,
and that the earth's radius is 6400 km, find a value for the acceleration of free fall at the
earth's surface. [Ans: 99 m/s2]

© KK Physics 55

4) 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. (Radius of the earth = 6370 km,
mass of the earth = 5.98 × 1024 kg)[Ans: 24 hrs]
5) Assuming that the earth is a uniform sphere of radius 6.4 × 106 m and mass 6.0 ×
1024 kg, calculate the work done in taking a 10 kg mass from the earth's surface to a
point where the earth's gravitational field is negligible.[G = 6.7 × 10–11 Nm2 kg–2]
[Ans: 62.8 × 107 Joule]
6) Taking the earth to be 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 it into
circular orbit at this altitude. [Ans: 7.12 × 1010 J]
© KK Physics 56

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