www.NetBadi.com 1 GRAVITATION Newton’s Law of Universal Gravitation Newton’s law of universal gravitation states that every body in the universe attracts every other body with a force which is directly proportional tot he 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. Consider two bodies of masses m 1 and m 2 that have a distance r between their centres as shown in fig. 15.1. According to Newton’s law of gravitation, the magnitude of attractive force (F) between the two bodies is 1 2 F mm ... (i) and 2 1 F r ... (ii) Combining eqs. (i) and (ii), we have, 1 2 2 mm F r or 1 2 2 mm FG r ... (iii) where G is a constant of proportionality and is called universal gravitational constant. In SI units, the value of G is 11 2 2 6.67 10 G Nm kg As shown in fig. 15.1 mass m 2 attracts mass m 1 with a gravitational force F that is directed toward m 2 . And mass m 1 attracts mass m 2 with a force F that is directed toward m 1 . The forces F and F from action-reaction pair and are opposite in direction but equal in magnitude. (i) The gravitational force between two bodies is always attractive. (ii) The value of G does not depend upon the nature and size of the masses. It also does not depend upon the nature of the medium between the two bodies. Indeed, G is truly universal; both observation and theory suggest that it has the same value throughout the universe. (iii) The gravitational force of attraction between two bodies is not altered by the presence of other bodies. Thus one mass will experience the same attractive force from a second body, independently of whether or not a third mass is placed between them. (iv) G is measured experimentally. The small value of G 11 2 2 ( 6.67 10 ) N m kg tells us tat the force of attraction between ordinary sized objects is very small. For example, if m 1 = 10 kg, m 2 = 100 kg and r = 1 m, then magnitude of gravitational attraction force between the two bodies is 11 8 1 2 2 2 (10)(100) (6.67 10 ) 6.67 10 (1) mm FG N r Note that the force is extremely small. We can measure the gravitational force if one of the two interacting objects is quite massive e.g., you and earth. The attractive force of earth on you can be easily measured. It is your weight. Dimensional formula of G = [M –1 L 3 T –2 ] NetBadi.com - Exclusive www.NetBadi.com AP's Largest Eamcet Resources Website Eamcet Rank Estimator, Mock Counselling, College Information System - www.NetBadi.com
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GRAVITATION
Newton’s Law of Universal Gravitation
Newton’s law of universal gravitation states that every body in the universe attracts every other body with a force which is
directly proportional tot he 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.
Consider two bodies of masses m1 and m
2 that have a distance r between their centres as shown in fig. 15.1. According to
Newton’s law of gravitation, the magnitude of attractive force (F) between the two bodies is
1 2F m m� ... (i)
and 2
1F
r� ... (ii)
Combining eqs. (i) and (ii), we have,
1 2
2
m mF
r�
or1 2
2
m mF G
r� ... (iii)
where G is a constant of proportionality and is called universal gravitational constant. In SI units, the value of G is
11 2 26.67 10G Nm kg� �� �
As shown in fig. 15.1 mass m2 attracts mass m
1 with a gravitational force F
� that is directed toward m
2. And mass m
1 attracts
mass m2 with a force F�
� that is directed toward m
1. The forces F and F�
� � from action-reaction pair and are opposite in
direction but equal in magnitude.
(i) The gravitational force between two bodies is always attractive.
(ii) The value of G does not depend upon the nature and size of the masses. It also does not depend upon the nature of the
medium between the two bodies. Indeed, G is truly universal; both observation and theory suggest that it has the same
value throughout the universe.
(iii) The gravitational force of attraction between two bodies is not altered by the presence of other bodies. Thus one mass
will experience the same attractive force from a second body, independently of whether or not a third mass is placed
between them.
(iv) G is measured experimentally. The small value of G 11 2 2( 6.67 10 )N m kg� �� � tells us tat the force of attraction between
ordinary sized objects is very small. For example, if m1 = 10 kg, m
2 = 100 kg and r = 1 m, then magnitude of gravitational
attraction force between the two bodies is
11 81 2
2 2
(10)(100)(6.67 10 ) 6.67 10
(1)
m mF G N
r
� �� � � � � �
Note that the force is extremely small. We can measure the gravitational force if one of the two interacting objects is
quite massive e.g., you and earth. The attractive force of earth on you can be easily measured. It is your weight.
Dimensional formula of G = [M–1 L3 T–2]
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(v)1 2
2
m mF G
r�
If m1 = m
2 = 1 and r = 1, then F = G
Hence universal gravitational constant is equal to the force of attraction between two bodies each of unit mass placed
unit distance (centre-to-centre) apart.
(vi) This law fails if the distance between the objects is less than 10–9 m i.e. of the order of intermolecular distances.
Note. Strictly speaking, Newton’s law of gravitation is valid for point masses. However, in the circumstances listed below,
it can be used for real objects of masses m1 and m
2 whose centres are a distance r apart.
(a) It is value for two bodies of any size provided they each have spherical symmetry e.g., the sun and the earth is a good
approximation.
(b) It is valid when one body has spherical symmetry and the other is small compared with the separation of their centres
e.g., the earth and the brick.
(c) It is a good approximation when neither body has spherical symmetry but where both are small compared with the
separation of their centres e.g., two bricks a few metres apart.
Vector form of newton’s law of gravitation
We can write Newton’s law of universal gravitation in vector form as :
1 2
12 212
21
ˆm m
F G rr
� ��
... (i)
where 12F�
is the force on mas m1 exerted by mas m
2 which is a distance r
21 away ;
21r̂ is a unit vector that points from m
2 to m
1
along the line joining their centres. Therefore, 21 21 21ˆ /r r r�
� where
21r�
is the displacement vector as shown in fig. 15.3. (i). The
minu sign in eq. (i) is necessary since the force on m1 due to m
2 points in the direction opposite to
21r̂ . The displacement vector
21r�
but it points in the opposite direction so that
12 21r r� �� �
By Newton’s third law, the force 12F�
acting on m2 due to m
1 must have the same magnitude as
12F�
but acts in the opposite
direction [see fig. 15.3 (ii)] so that
1 2
21 12 212
21
ˆm m
F F G rr
� � �� �
�1 2
21 122
12
ˆm m
F G rr
� ��
Note. The law of universal gravitation should not be confused with Newton’s second law of motion F ma�� �
. The former
describes a particular force - the gravitational force of attraction - and how its value varies with the distance and masses
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involved. On the other hand, Newton’s second law relates the force on a body - it can be any force - to the mass and acceleration
of that body.
Intertial and gravitational masses
Mass is the basic property of matter. We normally define the mass of a body as the quantity of matter contained in the body.
This notion of the mass of the body is not precise because the concept “quantity of matter” is itself not well defined. Mass
makes itself evident in two very different ways viz.
(i) According to Newton’s second law of motion. F = mi a. Here mass m
i of an object is the proportionality factor between
the net force F applied on the object and the object’s acceleration a. Thus the mass of an object is that property of the
object which causes it to resist a change in its linear velocity. In other words, the mass of a body is a measure of its
inertia. For this reason, the mass that appears in Newton’s second law of motion is called inertial mass. Suppose a
runaway grocery cart loaded with groceries is rolling out of control in a supermarket and you must stop the cart before
it crashes into a stack of cans. The force required to stop the cart depends on its inertial mass.
�Applied force
Inertial massAcceleration produced
�
or i
Fm
a�
(ii) We have see that the mass of an object also appears in Newton’s law of universal gravitation. The magnitude F of the
gravitational force on an object of mass mg due to another object of mass M is
2
gm M
F Gr
�
�2
,Fr
Gravitational mass mgGM
�
In this expression, the mass of an object is that property of the object that causes it to be attracted to another body by the
gravitational force. In other words, mass of a body is a measure of gravitational force on the body. For this reason, the
mass that appears in Newton’s law of gravitation is often called the gravitational mass. Suppose you are holding a bag
of groceries while waiting for your friend. The force you must exert while holding the bag depends on the gravitational
mass of the bag of groceries.
Conclusion. From the above discussion, we find that mass of a body makes itself evident in two very different ways. On the
one hand, it is a measure of an object’s resistance to a change of velocity (inertial mass) and on the other hand, it is a measure
of an object’s gravitational attraction to other objects in its environment (gravitational mass). Thus the difficulty you encountered
in stopping the runaway cart has nothing to do with its gravitational mass. Similarly, the effort you exerted in holding the bag
of groceries has nothing to do with its inertial mass.
Now it is not at all obvious that the inertial mass of a body should be equal to its gravitational mass. However, careful experiments
show that they are equal. For example, if we have to push twice as hard on body A as we do on body B to produce a given
acceleration, then the force of gravitational attraction between body A and some third body C is twice as great as the gravitational
attraction between body B and C, at the same distance. Therefore, we need not maintain the distinction between mi and m
g and
can put mi = m
g.
Note. The following points may be noted about inertial mass of gravitational mass:
(a) Both are measured in the same units.
(b) Both are scalar quantities.
(c) Inertial mass is not affected due to the presence of other bodies. However, gravitational mass is affected due to the
presence of other bodies.
Expression for Acceleration due to gravity
Consider earth to be a sphere of mass ME and radius R
E. Suppose a body of mass m is placed on the surface of the earth as
shown in fig. 15.4. It is *legitimate to consider that the whole mass ME of the earth is concentrated at its centre. It is reasonable
to assume that the distance between m and ME is equal to radius R
E of the earth. According to the law of gravitation, the force
of attraction acting on the body due to earth is given by ;
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2
E
E
M mF G
R� ... (i)
This attractive force which the earth exerts on the object is simply the weight (= mg) of the object i.e.,
F = mg ... (ii)
From eqs. (i) and (ii), 2
E
E
GM mmg
R�
or 2
E
E
GMg
E� ... (iii)
Eq. (iii) gives the expression for acceleration due to gravity on the surface of the earth. The following points may be noted:
(i) Eq. (iii) shows that the value of go does not depend on the mass m of the body. Thus if two bodies of different masses are
allowed to fall freely, they will have the same acceleration. If they are allowed to fall from the same height, they will
reach the earth simultaneously.
(ii) Both G and ME are constant and R
E (= distance between the centres of body and earth) does not change appreciably for
small variation in height near the surface of the earth. Therefore, the acceleration due to gravity of an object near the
surface of earth is approximately constant and does not depend on the mass of the object.
(iii) We can easily extend eq. (iii) to find the gravitational acceleration at the surface of any planet of mass MP and radius R
P.
Mass and density of earth
Using the law of universal gravitation and the measured value of the acceleration due to gravity, we can find the mass and
density of earth.
(i) Mass of earth. The acceleration due to gravity on the surface of the earth is given by;
2
E
E
GMg
R�
� Mass of earth,
2
E
E
gRM
G� ... (i)
Now g = 9.8 ms–2; 6 11 2 26.37 10 ; 6.67 10ER m G N m kg� �� � � �
� Mass of earth
6 224
11
(9.8) (6.37 10 )6 10 ( )
6.67 10EM kg approximately�
� �� � �
�
(ii) Density of earth. Let � be the average density of the earth. Earth is a sphere of radius RE.
�34
3ER� �� � �
�
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�34
3E E
M R� �� ... (ii)
From eqs. (i) and (ii),
2
34
3
EE
gRR
G� � �
� Average density of earth, 3
4E
g
R G�
��
Now 2 6 11 29.8 ; 6.37 10 ; 6.67 10 2E
g ms R m G N m kg� � �� � � � � .
� Average density of earth, 6 11
3 9.8
4 (6.37 10 ) 6.67 10�
� �
��
� � � �
3 35.5 10 kg m�� �
This is the average density of the entire earth and is 5.5 times the density of water.
Variations in the value of g
The value of acceleration due to gravity (g) varies as we go above or below the surface of the earth. It also varies from place to
place on the surface of the earth.
1. Variation of g with altitude. Consider earth to be a sphere of radius R and mass M. The acceleration due to gravity on the
surface of earth (point Q in fig. 15.5) is
2
GMg
R� ... (i)
Consider a point P at a height h above the surface of the earth. The acceleration due to gravity at point P is
2( )
GMg
R h� �
� ... (ii)
Dividing eq. (ii) by eq. (i), we have,
2 2
2 2
2( )
1
g R R
g R h hR
R
�� �
� � � ��
or2
1
gg
h
R
� �� � ��
... (iii)
Therefore, g g� � . Thus as we go above the surface of earth, acceleration due to gravity goes on decreasing. For example, at
a height equal to the radius of the earth (i.e., h = R = 6400 km), we have,
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2 41
g gg
R
R
� � �� � ��
Eq. (iii) can be written as :
2
1h
g gR
�� � � � ��
21
h hg terms containing higher power of
R R
� � � � ��
If h is small compared to R, higher power of h/R can be neglected.
�2
1h
g gR
� � � � ��
Eq. (iv) may be used when h is small as compared to R. However, if h is comparable to R, then eq. (iii) may be used. Note that
acceleration due to gravity is maximum at the surface of the earth.
Note. We can calculate the %age decrease in the value of g with height h as under:
From eq. (iv), we have,
21
hg g
R
� � � � ��
or2hg
g gR
�� �
or2g g h
g R
��� � ... fractional decrease in g
� % age decrease in the value of g with height h is
2100 100
g g h
g R
��� � �
2. Variation of g with depth. Consider the earth to be a sphere of radius R and mass M. The acceleration due to gravity at point
Q on the surface of the earth is
2
GMg
R�
If � is the density of the earth, then,
Mass of earth, 34
3M R� ��
�
2
2
4
3G R
gR
� �� �� �
or4
3g RG� �� ... (v)
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Consider a point P which is inside the earth and at a depth h below the surface of the earth. Its distance from the centre O is
(R – h) as radius. The acceleration due to gravity g � at P is * only due to the sphere of radius (R – h).
� 2( )
GMg
R h
�� �
�
where M � = mass of inner solid sphere (shaded portion)
34( )
3R h� �� �
�
3
2
4( )
3
( )
G R h
gR h
� �� ��� �� �� ��
... (vi)
Dividing eq. (vi) by eq. (v), we have,
g R h
g R
� ��
or 1h
g gR
� � � � ��
... (vii)
There g g� � . Thus as we go below the surface of the earth, the acceleration due to gravity goes on decreasing and becomes
zero at the centre of the earth (where h = R).
Fig. 15.7 shows the variation of g � as a function of r where r is the distance from the centre of the earth.
**Outside the earth,
2
2
Rg g
r� �
For points outside the earth 21/g r� � . The maximum value of ( )g g� � is obtained at the surface of earth where r = R.
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***Inside the earth, r
g gR
� �
Thus inside the earth, g r� � . In order words, inside the earth (assuming uniform earth density), g � varies linearly with the
distance fro the centre of the earth. It is easy to see that the value of ( )g g� � is maximum at the surface of the earth and it is zero
at the centre of the earth. That is why the weight of a body at the centre of the earth is zero through its mass is constant.
Note. We can calculate the % age decrease in the value of g with depth h inside the earth as under.
From eq. (vii), we have,
1h
g gR
� � � � ��
orhg
g gR
�� �
org g hg
g R
��� ... fractional decrease in g
� % age decrease in the value of g with depth h inside the earth is
100 100g g h
g R
��� � �
3. Variation of g with latitude. The value of acceleration due to gravity (g) changes due to the change in latitude. This is due
to two reasons: (i) shape of the earth and (ii) rotation of the earth about its own axis. The latitude at a place on the surface of
earth is defined as the angle which the line joining the place to the centre of the earth makes with the equatorial plane. It is
denoted by � . Thus referring to fig. 15.8, the latitude at place P POE �� � � . It is clear that 090� � at poles and 00� � at
the equator.
(i) Shape of the earth. The earth is not a perfect sphere. It flattens at the poles (where 090� � ) and bulges out at the equator
(where 090� � ). Its equatorial radius R
e is nearly 21 km larger than the polar radius R
p. We know that value of g depends upon
the radius (R) of the earth.
REquator
O RE
PN-pole
W
S-pole
Fig. 15.8
r
N
mg
P FA
(180 - )
EBROW
S
OE = R
C
Fig. 15.9
2
GMg
R�
Since G and M are constant, 21/g R�
Thus the value of g at a place on the surface of earth varies inversely as the square of the radius of earth at that place. The radius
of earth is least at the poles and greatest at the equator. Therefore, the value of g is maximum at the poles and least at the
equator. In fact, as we go from the equator toward the poles, the radius of earth goes on decreasing and hence the value of g
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goes on increasing. The value of g at the equator is 9.78 m s–2 whereas at poles, it is 9.83 ms–2.
(ii) Rotation of Earth. Earth is rotating about its own axis at an angular velocity � (= one revolution per 24 hours). The line
joining the north and south poles is the axis of rotation. As a result of rotation, every point on the earth moves along a circular
path with the same angular velocity � . A point at the equator moves in a circle of radius equal tot he radius of earth and the
centre of the circle is the same as the centre of the earth. For any other point on the earth the circle of rotation is smaller than
this.
Consider earth to be sphere of mass M and radius R. Suppose a particle of mass m is situated at point P on the surface of the
earth [see fig. 15.9]. Let � be the latitude of the point P i.e., POE �� � . Suppose g is the acceleration due to gravity in the
absence of rotational motion of the earth. In that case, the particle at P would have been attracted toward the centre O of the
earth. Therefore true weight mg of the particle is directed toward O and is represented by the vector PO����
.
Due to rotation of earth, the particle at P moves along a circular path whose centre is C and radius r = (= CP). Since
, cosOPC r R� �� � � . The centrifugal force on the particle due to rotational motion of the each acts along the radius of the
circular path and in outward direction. The magnitude of centrifugal force is given by;
2 2 cosCF mr mR� � �� �
The centrifugal force is represented by the vector PA����
. The apparent weight mg� of the particle is equal to the resultant of
actual weight (= mg) and the centrifugal force 2( cos )CF mr� �� . Complete the parallelogram PABO. The diagonal PB
���� of the
parallelogram represents the apparent weight mg� of the particle.
Now 2 2 2 0( ) 2 cos (180 )PB PO PA PO PA �� � � � � �
or 2 2 2 2 2( ) ( ) ( cos ) 2 cos ( cos )mg mg mR mg mR� � � � �� � � � � � �
or
1/ 22 4 2 2 2
2
cos 2 cos1R R
g ggg
� � � �� �� � � �� �
� �
Now 6 22
6.4 10 ; / sec 9.824 60 600
R m rad and g ms�
� �� � � �� �
�
22 636.4 10 2
3.45 109.8 24 60 60
R
g
� � �� � � � � � �� ��
As 2 /R g� is very small, 4 4 2/R g� will be still smaller. Therefore, neglecting the factor containing 2 4 2/R g� , we get,
1/ 22 22 cos
1R
g gg
� �� �� � �� �
� �
2 21 2 cos1
2
Rg higher terms
g
� �� �� � � �� �
� �
Neglecting the terms containing higher powers of 2 2cos /R g� � , we have,
22
1 cosR
g gg
��
� �� � �� �
� �
or 2 2cosg g R� �� � � ... (viii)
From eq. (viii), it is clear that acceleration due to gravity :
(a) decreases on account of rotation of the earth.
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(b) increases with the increases in the latitude of the place ( cos�� decrease as � increases). This means that value of g
increases as we go from equator to the poles.
At equator. At equator 00� � so that 0cos cos 0 1� � �
� 2
eg g R�� � ... minimum
Therefore, value of acceleration due to gravity is minimum at the equator. This is expected because the particle at the equator
executes a circle of maximum radius. Therefore, the centrifugal force is maximum.
At poles. At poles, 090� � so that 0
cos cos90 0� � �
� pg g�
Hence the value of g is maximum at the poles. This is expected because the particle at the pole moves in a circle of zero radius.
Therefore, no centrifugal force acts on the particle. As a result, the value of gp remains the same whether the earth is at rest or
rotating.
Notes (i)2 2( )
p eg g g g R R� �� � � � �
�2
p eg g R�� �
Note that6 2
6.4 10 / .24 60 60
R m and rad s�
�� � �� �
(ii) When a body of mass m is moved from equator to pole, the increase in weight
2( )p em g g mR�� � �
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Orbital Velocity and Time Period of Earth Satellite
Satellite can be launched from earth’s surface to circle to earth. These are called earth satellites. In practice, the satellite is
carried by a rocket to a certain height and then given a suitable velocity in a direction parallel to the tangent of the orbit. The
satellite is kept in its orbit by the gravitational attraction of the earth.
(i) Orbital Velocity. The velocity required to put a satellite into a given orbit is called orbital velocity. This velocity is imparted
in a direction parallel to the tangent of the orbit.
Suppose a satellite of mass m i to be put into circular orbit around the earth at a height h above its surface as shown in fig. 16.4.
Consider the earth to be a sphere of mass M and radius R. Then the radius of the orbit of the satellite is (R + h). Suppose v is
the magnitude of the orbital velocity required by the satellite at this height. This velocity will be imparted in a direction parallel
to the tangent of the orbit.
The centripetal force required to keep the satellite in the circular orbit is
2
( )C
mvF
R h�
� ... (i)
The centripetal force is provided by the gravitational force exerted by the earth on the satellite. According to Newton’s law of
universal gravitation, the magnitude of this force is
2( )
g
GMmF
R h�
� ... (ii)
�
2
2( )
mv GMm
R h R h�
� �
orGM
vR h
��
... (iii)
The satellite at this speed will continuously move in a circular orbit of radius (R + h). Note that orbital speed of an earth satellite
is independent of the mass of the satellite and depends only upon its height h above the earth’s surface. Greater is the height h
of the satellite above the earth’s surface, smaller i the orbital speed of the satellite, two satellites of different masses revolving
in the same orbit around the earth will have the same orbital speed.
The orbital speed can be written in another equivalent form. The acceleration due to gravity (g) on the earth’s surface is
2
2
GMg or GM gR
R� �
Putting GM = gR2 in eq. (iii), we have,
2
( )
gRv
R h�
�
or( )
gv R
R h�
� ... (iv)
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Both eqs. (iii) and (iv) give the orbital speed of the satellite circling the earth at a height h above the earth’s surface.
Special Case. When the satellite orbits very close to the surface of the earth, h = 0. Therefore, eqs. (iii) and (iv) becomes:
GMv
R� ... (v)
and v gR� ... (vi)
Both eqs. () and (vi) can be used to find the orbital speed of the satellite close to the surface of the earth.
Now, v gR�
6 26.4 10 ; 9.8R m g ms�� � �
� 6 3 1 19.8 6.4 10 8 10 8 ( .)v ms km s approx
� �� � � � � �
Thus the orbital speed of the satellite close to earth’s surface is about 8 km s–1. In practice, the satellite is carried by a rocket to
the rocket to the height of the orbit and then given an impulse, by firing jets, to deflect it in a direction parallel to the tangent of
the orbit. Since this motion may continue indefinitely, we may say that the orbit is stable.
Note. The orbital velocity of a satellite revolving around any planet is
( )
PP
P
GMv
R h�
�
Here MP = mass of the planet; R
P = radius of the planet.
Note that vP depends upon M
P, R
P and h ( = height of satellite above the planet’s surface).
(ii) Time period of satellite. The time taken by the satellite to complete one revolution around the earth is called time period
of the satellite. It is denoted by T.
Time period,Circumference of the orbit
TOrbital speed
�
or2 ( )R h
Tv
� �� ... (vii)
Putting the value of v given by eq. (iii), we have,
3
1/ 2
2 ( ) ( )2
[ / ]
R h R hT
GMGm R h
��
� �� �
�
�3( )
2R h
TGM
��
� ... (viii)
Putting the value of v given by eq. (iv) in eq. (vii)
32 ( ) 2 ( )R h R hT
R ggRR h
� �� �� �
�
�32 ( )R h
TR g
� �� ... (ix)
Time period of the satellite can also be given in yet another form.
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Mass of earth, M = volume � Density = 34
3R� ��
Putting the value of M in eq. (viii), we have,
3 3
33
( ) 1 3 ( )2
4
3
R h R hT
G G RR
��
�� �
� �� � �
�3
3
3 ( )R hT
G R
���
� ... (x)
Any of the three eqs. (viii), (ix) and (x) can be used to find the time period of the satellite. It is clear from these equations that
the time period of the satellite depends only upon its height above the earth’s surface. The greater is the height of a satellite
above the earth’s surface, the greater is its period of revolution. It is for this reason that the moon, which is at a height of
53.84 10 km� above earth, completes one revolution in about 27 days while an artificial satellite circling close to the earth’ss
surface completes 10 to 20 revolutions in a day.
Special case. When the satellite orbits very close to the surface of the earth, 0h � . Therefore, eqs. (viii), (ix) and (x) respectively
become:
3
2R
TGM
�� ... (xi)
2R
Tg
�� ... (xii)
3T
G
��
� ... (xiii)
Now 2R
Tg
��
Here 6 26.4 10 ; 9.8R m g ms�� � �
�66.4 0
2 5075 84 .9.8
T seconds minutes��
� � �
Thus the orbital speed of a satellite revolving very near to the earth’s surface is about 8 km s–1 and its period of revolution is
nearly 84 minutes.
(iii) Height of Satellite above earth’s surface. The time period of the satellite is given by eq. (ix) as:
32 ( )R h
TR g
� ��
Squaring both sides, we get,
2 32
2
4 ( )R hT
R g
� ��
or
2 23
2( )
4
gR TR h
�� �
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or
1/32 2
24
gR TR h
��
� � ��
or
1/32 2
24
gR Th R
��
� � ��
If we know the time period T of the satellite, radius R of earth and acceleration due to gravity (g) at earth’s surface, the height
h of the satellite above earth’s surface can be calculated.
Note. For potential energy and kinetic energy of a satellite, the reader should refer to chapter 17, article 17.8
Stationary Satellites
We have seen that time period of an artificial satellite circling the earth depends only upon its height above earth’s surface. The
time period of an artificial earth satellite orbiting very close to the surface of earth is 84 minutes i.e., the satellite will take 84
minutes to complete one revolution around the earth. Since the period of revolution of earth about its own axis is 24 hours, such
an artificial satellite will appear to the amoving w.r.t. the earth. 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 relative to the earth. In other words, the satellite will stay over the same place above
the earth while the earth rotates. Such a satellite is called geostationary or simply stationary satellite. The orbit of geostationary
satellite is sometimes called parking orbital. Most communication satellites are geostationary satellites so that information can
be transmitted from one part of the world to another.
Fig 16.5. shows the orbit of geometry satellite. Note that the geostationary satellite revolves around the earth from *west to east
in a close circular orbit and coplanar with the equatorial plane.
(i) Height of parking orbit above earth’s surface. We know that time period T of an artificial earth satellite is given by ;
22 ( )R h
TR g
� �� ... (i)
Here R = Radius of earth = 6400 km
T = Time period = 24 hours = 24 60 60 s� �
h = Height of the artificial satellite above earth’s surface, corresponding to T = 24 hours
g = 9.8 ms–2 = 0.0098 km s–2
�3
2 (6400 )24 60 60
6400 0.0098
h� �� � �
or 36000h km�
Thus a stationary satellite orbits around the earth at a height 36000 km above earth’s surface. Since the motion of the satellite
is synchronous with earth’s rotation about its axis, the satellite will always appear to be over the same place relative to an
observed on the earth. It is interesting to note [see eq. (i) above], that height of the parking orbit above earth’s surface does not
depend upon the mass of the satellite.
(ii) Speed of Satellite in parking orbit. Radius of parking orbit, RP = R + h = 6400 + 3600 = 42400 km. Speed of the satellite
in the parking orbit is
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Circumference of parking orbitv
Time period of satellite�
or12 2 42400
3.124 24 60 60
PRv km shours
� � ��� � �
� �
Thus a geostationary satellite revolves around the earth at a height of 36000 km above earth’s surface with an orbital speed of
3.1 km s–1.
Placing the satellite in parking orbit. When a satellite is to be placed in the parking orbit, it is first carried to a height of
36000 k above earth’s surface. It is then given the necessary tangential velocity v (= 3.1 km s–1) by firing rocket engines which
ar aligned parallel to earth’s surface. 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) Its motion should be synchronous with the axial rotation of the earth. In other words, the time period of the satellite should
be 24 hours.
(iii) It should rotate in the same direction as the earth is rotating.
(iv) The height of the parking orbit should be 36000 km above the earth’s surface.
Escape Velocity
If a ball is thrown upwards from the surface of the earth, its speed decreases from the moment it is projected due to the retarding
effect of earth’s gravitational field. It would simply rise to a certain height, reverse direction and then fall back to earth. The
height which the ball ultimately attains depends upon the speed with which it is projected - the greater the speed, the greater the
height. Ultimately, at a certain velocity of projection, the body will go out of the gravitational field and will never return to the
earth. The minimum projection velocity that achieves this is known as the escape velocity.
The escape velocity of a body is the minimum velocity with which it is to be projected so that it just overcomes the gravitational
pull of the earth (or any other planet). Since earth’s gravitational field extends to infinity (however weak it may be at large
distances), escape velocity is obviously the velocity that must be given to an object for it to escape all the way to infinity.
In order to project a body with escape velocity, we give kinetic energy to it. Let us calculate this energy. Consider earth to be
a sphere of mass M and radius R. Suppose a body of mass m is projected upward with escape velocity Ve. When the body is at
point P at a distance x from the centre of the earth, the gravitational force of attraction exerted by the earth on the body [See fig.
16.8] is
2
MmF G
x�
In moving a small distance x� against this gravitational force, the small work done at the expense of the kinetic energy of the
body is given by :
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2
MmW F x G x
x� � � � �
Therefore, total work done (W) in moving the body from earth’s surface (where x = R) to infinity (where x � � ) is given by :
2
2
R R
GMmW dx GMm x dx
x
� ��� �� �
1
R
GMmGMm
x R
�� �� � �� �� �
�GMm
WR
�
If the body is to be able to do this amount of work (and so escape), it needs to have atleast this amount of kinetic energy at the
moment it is projected. Therefore, escape velocity Ve is given by:
21
2e
GMmmV
R�
or2
e
GMV
R� ... (i)
The escape velocity can be written in another equivalent form. The acceleration due to gravity at the surface of earth is
2
2
GMg or GM gR
R� �
Putting 2GM gR� in eq. (i), we have,
222
e
gRV gR
R� �
� 2eV gR� ... (ii)
The escape velocity can also be expressed in terms of density and radius of the earth.
Mass of earth, 34
3M R� ��
Putting the value of M in eq. (i), we have,
3 22 4 8 2
3 3 3e
G GV R GR D
R
��� � ��� � � �
�2
3e
GV D
��� ... (iii)
Here D = 2R = diameter of earth.
Note that denser planet have greater escape velocities.
The escape speed from earth’s surface can be calculated from any of the eqs. (i), (ii) and (iii). Note that escape speed does not
depend on the mass of the body. It is the same for all masses. It may be noted that escape velocity has been derived by
neglecting resistance due to air. The following points may be noted:
(i) The escape velocity does not depend upon the direction of projection. This is because the kinetic energy a body
loses in reaching any particular height depends only on the height concerned and not the path taken to reach it.
(ii) The escape velocity is related in a simple way to the orbital velocity close to the surface of earth.
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2 2 1.41eV gR v v� � �
where v = orbital velocity close to surface of earth = gR
The *escape speed is only 41% grater than the speed of a satellite close to the surface of the earth.
(iii) The escape velocity from the earth’s surface is