CHAPTER EIGHT
GRAVITATION
8.1 INTRODUCTION
Early in our lives, we become aware of the tendency of all
material objects to be attracted towards the earth. Anything
thrown up falls down towards the earth, going uphill is lot
more tiring than going downhill, raindrops from the clouds
above fall towards the earth and there are many other such
phenomena. Historically it was the Italian Physicist Galileo
(1564-1642) who recognised the fact that all bodies,
irrespective of their masses, are accelerated towards the earth
with a constant acceleration. It is said that he made a public
demonstration of this fact. To find the truth, he certainly did
experiments with bodies rolling down inclined planes and
arrived at a value of the acceleration due to gravity which is
close to the more accurate value obtained later.
A seemingly unrelated phenomenon, observation of stars,
planets and their motion has been the subject of attention in
many countries since the earliest of times. Observations since
early times recognised stars which appeared in the sky with
positions unchanged year after year. The more interesting
objects are the planets which seem to have regular motions
against the background of stars. The earliest recorded model
for planetary motions proposed by Ptolemy about 2000 years
ago was a ‘geocentric’ model in which all celestial objects,
stars, the sun and the planets, all revolved around the earth.
The only motion that was thought to be possible for celestial
objects was motion in a circle. Complicated schemes of motion
were put forward by Ptolemy in order to describe the observed
motion of the planets. The planets were described as moving
in circles with the centre of the circles themselves moving in
larger circles. Similar theories were also advanced by Indian
astronomers some 400 years later. However a more elegant
model in which the Sun was the centre around which the
planets revolved – the ‘heliocentric’ model – was already
mentioned by Aryabhatta (5
th
century A.D.) in his treatise. A
thousand years later, a Polish monk named Nicolas
8.1 Introduction
8.2 Kepler’s laws
8.3 Universal law of
gravitation
8.4 The gravitational
constant
8.5 Acceleration due to
gravity of the earth
8.6 Acceleration due to
gravity below and above
the surface of earth
8.7 Gravitational potential
energy
8.8 Escape speed
8.9 Earth satellites
8.10 Energy of an orbiting
satellite
8.11 Geostationary and polar
satellites
8.12 Weightlessness
Summary
Points to ponder
Exercises
Additional exercises
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184 PHYSICS
B
A
C
P
S
S'
2b
2a
Copernicus (1473-1543) proposed a definitive
model in which the planets moved in circles
around a fixed central sun. His theory was
discredited by the church, but notable amongst
its supporters was Galileo who had to face
prosecution from the state for his beliefs.
It was around the same time as Galileo, a
nobleman called Tycho Brahe (1546-1601)
hailing from Denmark, spent his entire lifetime
recording observations of the planets with the
naked eye. His compiled data were analysed
later by his assistant Johannes Kepler (1571-
1640). He could extract from the data three
elegant laws that now go by the name of Kepler’s
laws. These laws were known to Newton and
enabled him to make a great scientific leap in
proposing his universal law of gravitation.
8.2 KEPLER’S LAWS
The three laws of Kepler can be stated as follows:
1. Law of orbits : All planets move in elliptical
orbits with the Sun situated at one of the foci
Fig. 8.1(a) An ellipse traced out by a planet around
the sun. The closest point is P and the
farthest point is A, P is called the
perihelion and A the aphelion. The
semimajor axis is half the distance AP.
Fig. 8.1(b) Drawing an ellipse. A string has its ends
fixed at F
1
and F
2
. The tip of a pencil holds
the string taut and is moved around.
of the ellipse (Fig. 8.1a). This law was a deviation
from the Copernican model which allowed only
circular orbits. The ellipse, of which the circle is
a special case, is a closed curve which can be
drawn very simply as follows.
Select two points F
1
and F
2
. Take a length
of a string and fix its ends at F
1
and F
2
by pins.
With the tip of a pencil stretch the string taut
and then draw a curve by moving the pencil
keeping the string taut throughout.(Fig. 8.1(b))
The closed curve you get is called an ellipse.
Clearly for any point T on the ellipse, the sum of
the distances from F
1
and F
2
is a constant. F
1
,
F
2
are called the focii. Join the points F
1
and
F
2
and extend
the line to intersect the ellipse at
points P and A as shown in Fig. 8.1(b). The
midpoint of the line PA is the centre of the ellipse
O and the length PO = AO is called the semi-
major axis of the ellipse. For a circle, the two
focii merge into one and the semi-major axis
becomes the radius of the circle.
2. Law of areas : The line that joins any planet
to the sun sweeps equal areas in equal intervals
of time (Fig. 8.2). This law comes from the
observations that planets appear to move slower
when they are farther from the sun than when
they are nearer.
Fig. 8.2 The planet P moves around the sun in an
elliptical orbit. The shaded area is the area
∆
A swept out in a small interval of time
∆
t.
3. Law of periods : The square of the time period
of revolution of a planet is proportional to the
cube of the semi-major axis of the ellipse traced
out by the planet.
Table 8.1 gives the approximate time periods
of revolution of eight* planets around the sun
along with values of their semi-major axes.
* Refer to information given in the Box on Page 182
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GRAVITATION 185
t
Table 8.1 Data from measurement of
planetary motions given below
confirm Kepler’s Law of Periods
(a ≡ Semi-major axis in units of 10
10
m.
T ≡ Time period of revolution of the planet
in years(y).
Q ≡ The quotient ( T
2
/a
3
) in units of
10
-34
y
2
m
-3
.)
Planet a T Q
Mercury 5.79 0.24 2.95
Venus 10.8 0.615 3.00
Earth 15.0 1 2.96
Mars 22.8 1.88 2.98
Jupiter 77.8 11.9
3.01
Saturn 143 29.5 2.98
Uranus 287 84 2.98
Neptune 450 165 2.99
Pluto* 590 248 2.99
The law of areas can be understood as a
consequence of conservation of angular
momentum whch is valid for any central force .
A central force is such that the force on the
planet is along the vector joining the Sun and
the planet. Let the Sun be at the origin and let
the position and momentum of the planet be
denoted by r and p respectively. Then the area
swept out by the planet of mass m in time
interval ∆t is (Fig. 8.2) ∆A given by
∆
A
= ½ (r × v∆t) (8.1)
Hence
∆
A
/∆t =½ (r × p)/m, (since v = p/m)
= L / (2 m) (8.2)
where v is the velocity, L is the angular
momentum equal to ( r × p). For a central
force, which is directed along r, L is a constant
as the planet goes around. Hence,
∆
A
/∆t is a
constant according to the last equation. This is
the law of areas. Gravitation is a central force
and hence the law of areas follows.
Example 8.1 Let the speed of the planet
at the perihelion P in Fig. 8.1(a) be v
P
and
the Sun-planet distance SP be r
P
. Relate
{r
P
, v
P
} to the corresponding quantities at
the aphelion {r
A,
v
A
}. Will the planet take
equal times to traverse BAC and CPB ?
Answer The magnitude of the angular
momentum at P is L
p
= m
p
r
p
v
p
, since inspection
tells us that r
p
and v
p
are mutually
perpendicular. Similarly, L
A
= m
p
r
A
v
A
. From
angular momentum conservation
m
p
r
p
v
p
= m
p
r
A
v
A
or
v
v
p
A
=
r
r
A
p
t
Since r
A
> r
p
,
v
p
> v
A
.
The area SBAC bounded by the ellipse and
the radius vectors SB and SC is larger than SBPC
in Fig. 8.1. From Kepler’s second law, equal areas
are swept in equal times. Hence the planet will
take a longer time to traverse BAC than CPB.
8.3 UNIVERSAL LAW OF GRAVITATION
Legend has it that observing an apple falling
from a tree, Newton was inspired to arrive at an
universal law of gravitation that led to an
explanation of terrestrial gravitation as well as
of Kepler’s laws. Newton’s reasoning was that
the moon revolving in an orbit of radius R
m
was
subject to a centripetal acceleration due to
earth’s gravity of magnitude
22
2
4
m
m
m
R
V
a
R
T
π
= =
(8.3)
where V is the speed of the moon related to the
time period T by the relation
2 /
m
V R T
π
=
. The
time period T is about 27.3 days and R
m
was
already known then to be about 3.84 × 10
8
m. If
we substitute these numbers in Eq. (8.3), we
get a value of a
m
much smaller than the value of
acceleration due to gravity g on the surface of
the earth, arising also due to earth’s gravitational
attraction.
Johannes Kepler
(1571–1630) was a
scientist of German
origin. He formulated
the three laws of
planetary motion based
on the painstaking
observations of Tycho
Brahe and coworkers. Kepler himself was an
assistant to Brahe and it took him sixteen long
years to arrive at the three planetary laws. He
is also known as the founder of geometrical
optics, being the first to describe what happens
to light after it enters a telescope.
* Refer to information given in the Box on Page 182
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186 PHYSICS
Central Forces
We know the time rate of change of the angular momentum of a single particle about the origin
is
d
d
t
= ×
l
r F
The angular momentum of the particle is conserved, if the torque
= ×
r F
τ
ττ
τ
due to the
force F on it vanishes. This happens either when F is zero or when F is along r. We are
interested in forces which satisfy the latter condition. Central forces satisfy this condition.
A ‘central’ force is always directed towards or away from a fixed point, i.e., along the position
vector of the point of application of the force with respect to the fixed point. (See Figure below.)
Further, the magnitude of a central force F depends on r, the distance of the point of application
of the force from the fixed point; F = F(r).
In the motion under a central force the angular momentum is always conserved. Two important
results follow from this:
(1) The motion of a particle under the central force is always confined to a plane.
(2) The position vector of the particle with respect to the centre of the force (i.e. the fixed point)
has a constant areal velocity. In other words the position vector sweeps out equal areas in
equal times as the particle moves under the influence of the central force.
Try to prove both these results. You may need to know that the areal velocity is given by :
dA/dt = ½ r v sin
α
.
An immediate application of the above discussion can be made to the motion of a planet
under the gravitational force of the sun. For convenience the sun may be taken to be so heavy
that it is at rest. The gravitational force of the sun on the planet is directed towards the sun.
This force also satisfies the requirement F = F(r), since F = G m
1
m
2
/r
2
where m
1
and m
2
are
respectively the masses of the planet and the sun and G is the universal constant of gravitation.
The two results (1) and (2) described above, therefore, apply to the motion of the planet. In fact,
the result (2) is the well-known second law of Kepler.
Tr is the trejectory of the particle under the central force. At a position P, the force is directed
along OP, O is the centre of the force taken as the origin. In time
∆
t, the particle moves from P to P
′
,
arc PP
′
= ∆s = v ∆t. The tangent PQ at P to the trajectory gives the direction of the velocity at P. The
area swept in
∆
t is the area of sector POP
′
(
)
sinr
α
≈
PP
′
/2 = (r v sin a) ∆t/2.)
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GRAVITATION 187
t
This clearly shows that the force due to
earth’s gravity decreases with distance. If one
assumes that the gravitational force due to the
earth decreases in proportion to the inverse
square of the distance from the centre of the
earth, we will have a
m
α
2
m
R
−
; g α
2
E
R
−
and we get
2
2
m
m
E
R
g
a
R
=
3600 (8.4)
in agreement with a value of g
9.8 m s
-2
and
the value of a
m
from Eq. (8.3). These observations
led Newton to propose the following Universal Law
of Gravitation :
Every body in the 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 the distance
between them.
The quotation is essentially from Newton’s
famous treatise called ‘Mathematical Principles
of Natural Philosophy’ (Principia for short).
Stated Mathematically, Newton’s gravitation
law reads : The force F on a point mass m
2
due
to another point mass m
1
has the magnitude
1 2
2
| |
m m
G
r
=F
(8.5)
Equation (8.5) can be expressed in vector form as
$
(
)
$
1 2 1 2
2 2
– –
m m m m
G G
r r
= =
F r r
$
1 2
3
–
m m
G=
r
r
where G is the universal gravitational constant,
$
r
is the unit vector from m
1
to m
2
and r = r
2
– r
1
as shown in Fig. 8.3.
The gravitational force is attractive, i.e., the
force F is along – r. The force on point mass m
1
due to m
2
is of course – F by Newton’s third law.
Thus, the gravitational force F
12
on the body 1
due to 2 and F
21
on the body 2 due to 1 are related
as F
12
= – F
21
.
Before we can apply Eq. (8.5) to objects under
consideration, we have to be careful since the
law refers to point masses whereas we deal with
extended objects which have finite size. If we have
a collection of point masses, the force on any
one of them is the vector sum of the gravitational
forces exerted by the other point masses as
shown in Fig 8.4.
Fig. 8.4 Gravitational force on point mass m
1
is the
vector sum of the gravitational forces exerted
by m
2
, m
3
and m
4
.
The total force on m
1
is
2 1
1
2
21
Gm m
r
=F
$
3 1
21
2
31
Gm m
r
+r
$ $
4 1
31 41
2
41
Gm m
r
+
r r
Example 8.2 Three equal masses of m kg
each are fixed at the vertices of an
equilateral triangle ABC.
(a) What is the force acting on a mass 2m
placed at the centroid G of the triangle?
(b) What is the force if the mass at the
vertex A is doubled ?
Take AG = BG = CG = 1 m (see Fig. 8.5)
Answer (a) The angle between GC and the
positive x-axis is 30° and so is the angle between
GB and the negative x-axis. The individual forces
in vector notation are
Fig. 8.3 Gravitational force on m
1
due to m
2
is along
r where the vector r is (r
2
– r
1
).
O
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188 PHYSICS
(b) Now if the mass at vertex A is doubled
then
t
For the gravitational force between an extended
object (like the earth) and a point mass, Eq. (8.5) is not
directly applicable. Each point mass in the extended
object will exert a force on the given point mass and
these force will not all be in the same direction. We
have to add up these forces vectorially for all the point
masses in the extended object to get the total force.
This is easily done using calculus. For two special
cases, a simple law results when you do that :
(1) The force of attraction between a hollow
spherical shell of uniform density and a
point mass situated outside is just as if
the entire mass of the shell is
concentrated at the centre of the shell.
Qualitatively this can be understood as
follows: Gravitational forces caused by the
various regions of the shell have components
along the line joining the point mass to the
centre as well as along a direction
prependicular to this line. The components
prependicular to this line cancel out when
summing over all regions of the shell leaving
only a resultant force along the line joining
the point to the centre. The magnitude of
this force works out to be as stated above.
Newton’s Principia
Kepler had formulated his third law by 1619. The announcement of the underlying universal law of
gravitation came about seventy years later with the publication in 1687 of Newton’s masterpiece
Philosophiae Naturalis Principia Mathematica, often simply called the Principia.
Around 1685, Edmund Halley (after whom the famous Halley’s comet is named), came to visit
Newton at Cambridge and asked him about the nature of the trajectory of a body moving under the
influence of an inverse square law. Without hesitation Newton replied that it had to be an ellipse,
and further that he had worked it out long ago around 1665 when he was forced to retire to his farm
house from Cambridge on account of a plague outbreak. Unfortunately, Newton had lost his papers.
Halley prevailed upon Newton to produce his work in book form and agreed to bear the cost of
publication. Newton accomplished this feat in eighteen months of superhuman effort. The Principia
is a singular scientific masterpiece and in the words of Lagrange it is “the greatest production of the
human mind.” The Indian born astrophysicist and Nobel laureate S. Chandrasekhar spent ten years
writing a treatise on the Principia. His book, Newton’s Principia for the Common Reader brings
into sharp focus the beauty, clarity and breath taking economy of Newton’s methods.
Fig. 8.5 Three equal masses are placed at the three
vertices of the ∆ ABC. A mass 2m is placed
at the centroid G.
(
)
GA
2
ˆ
1
Gm m
=
F j
(
)
( )
GB
2
ˆ ˆ
cos 30 sin 30
1
Gm m
ο ο
= − −F i j
(
)
( )
GC
2
ˆ ˆ
cos 30 sin 30
1
Gm m
ο ο
= + −F i j
From the principle of superposition and the law
of vector addition, the resultant gravitational
force F
R
on (2m) is
F
R
= F
GA
+ F
GB
+ F
GC
(
)
οο
−−+= 30 sin
ˆ
30 cos
ˆ
2
ˆ
2
22
R
j ij F GmGm
(
)
030 sin
ˆ
30 cos
ˆ
2
2
=−+
οο
j i Gm
Alternatively, one expects on the basis of
symmetry that the resultant force ought to be
zero.
2020-21
GRAVITATION 189
(2) The force of attraction due to a hollow
spherical shell of uniform density, on a
point mass situated inside it is zero.
Qualitatively, we can again understand this
result. Various regions of the spherical shell
attract the point mass inside it in various
directions. These forces cancel each other
completely.
8.4 THE GRAVITATIONAL CONSTANT
The value of the gravitational constant G
entering the Universal law of gravitation can be
determined experimentally and this was first
done by English scientist Henry Cavendish in
1798. The apparatus used by him is
schematically shown in figure.8.6
Fig. 8.6 Schematic drawing of Cavendish’s
experiment. S
1
and S
2
are large spheres
which are kept on either side (shown shades)
of the masses at A and B. When the big
spheres are taken to the other side of the
masses (shown by dotted circles), the bar
AB rotates a little since the torque reverses
direction. The angle of rotation can be
measured experimentally.
The bar AB has two small lead spheres
attached at its ends. The bar is suspended from
a rigid support by a fine wire. Two large lead
spheres are brought close to the small ones but
on opposite sides as shown. The big spheres
attract the nearby small ones by equal and
opposite force as shown. There is no net force
on the bar but only a torque which is clearly
equal to F times the length of the bar,where F is
the force of attraction between a big sphere and
its neighbouring small sphere. Due to this
torque, the suspended wire gets twisted till such
time as the restoring torque of the wire equals
the gravitational torque . If
θ
is the angle of twist
of the suspended wire, the restoring torque is
proportional to
θ
, equal to
τθ.
Where
τ
is the
restoring couple per unit angle of twist.
τ
can be
measured independently e.g. by applying a
known torque and measuring the angle of twist.
The gravitational force between the spherical
balls is the same as if their masses are
concentrated at their centres. Thus if d is the
separation between the centres of the big and
its neighbouring small ball, M and m their
masses, the gravitational force between the big
sphere and its neighouring small ball is.
2
Mm
F G
d
=
(8.6)
If L is the length of the bar AB , then the
torque arising out of F is F multiplied by L. At
equilibrium, this is equal to the restoring torque
and hence
2
Mm
G L
d
τ θ
=
(8.7)
Observation of
θ
thus enables one to
calculate G from this equation.
Since Cavendish’s experiment, the
measurement of G has been refined and the
currently accepted value is
G = 6.67×10
-11
N m
2
/kg
2
(8.8)
8.5 ACCELERATION DUE TO GRAVITY OF
THE EARTH
The earth can be imagined to be a sphere made
of a large number of concentric spherical shells
with the smallest one at the centre and the
largest one at its surface. A point outside the
earth is obviously outside all the shells. Thus,
all the shells exert a gravitational force at the
point outside just as if their masses are
concentrated at their common centre according
to the result stated in section 8.3. The total mass
of all the shells combined is just the mass of the
earth. Hence, at a point outside the earth, the
gravitational force is just as if its entire mass of
the earth is concentrated at its centre.
For a point inside the earth, the situation
is different. This is illustrated in Fig. 8.7.
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190 PHYSICS
Fig. 8.7 The mass m is in a mine located at a depth
d below the surface of the Earth of mass
M
E
and radius R
E
. We treat the Earth to be
spherically symmetric.
Again consider the earth to be made up of
concentric shells as before and a point mass m
situated at a distance
r from the centre. The
point P lies outside the sphere of radius r. For
the shells of radius greater than r, the point P
lies inside. Hence according to result stated in
the last section, they exert no gravitational force
on mass m kept at P. The shells with radius
≤
r
make up a sphere of radius r for which the point
P lies on the surface. This smaller sphere
therefore exerts a force on a mass m at P as if
its mass M
r
is concentrated at the centre. Thus
the force on the mass m at P has a magnitude
r
2
( )
Gm M
F
r
=
(8.9)
We assume that the entire earth is of uniform
density and hence its mass is
3
E
4
3
E
M R
π
ρ
=
where M
E
is the mass of the earth R
E
is its radius
and
ρ
is the density. On the other hand the
mass of the sphere M
r
of radius r is
3
4
3
r
π
ρ
and
hence
E
3
E
G m M
r
R
=
(8.10)
If the mass m is situated on the surface of
earth, then r = R
E
and the gravitational force on
it is, from Eq. (8.10)
2
E
E
M m
F G
R
=
(8.11)
The acceleration experienced by the mass m,
which is usually denoted by the symbol g is
related to F by Newton’s 2
nd
law by relation
F = mg. Thus
2
E
E
GM
F
g
m
R
= =
(8.12)
Acceleration
g is readily measurable.
R
E
is a
known quantity. The measurement of G by
Cavendish’s experiment (or otherwise), combined
with knowledge of
g and R
E
enables one to
estimate M
E
from Eq. (8.12). This is the reason
why there is a popular statement regarding
Cavendish : “Cavendish weighed the earth”.
8.6 ACCELERATION DUE TO GRAVITY
BELOW AND ABOVE THE SURFACE OF
EARTH
Consider a point mass m at a height h above the
surface of the earth as shown in Fig. 8.8(a). The
radius of the earth is denoted by R
E
. Since this
point is outside the earth,
Fig. 8.8 (a) g at a height h above the surface of the
earth.
its distance from the centre of the earth is (R
E
+
h ). If F (h) denoted the magnitude of the force
on the point mass m , we get from Eq. (8.5) :
2
( )
( )
E
E
GM m
F h
R h
=
+
(8.13)
The acceleration experienced by the point
mass is
( )/ ( )
F h m g h
≡
and we get
M
r
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GRAVITATION 191
2
( )
( ) .
( )
E
E
GM
F h
g h
m
R h
= =
+
(8.14)
This is clearly less than the value of g on the
surface of earth :
2
.
E
E
GM
g
R
=
For
,
E
h R
<<
we can
expand the RHS of Eq. (8.14) :
( )
2
2 2
( ) 1 /
(1 / )
E
E
E E
GM
g h g h R
R h R
−
= = +
+
For
1
E
h
R
<<
, using binomial expression,
g h g
h
R
E
( )
≅ −
1
2
. (8.15)
Equation (8.15) thus tells us that for small
heights h above the value of g decreases by a
factor
(1 2 / ).
E
h R−
Now, consider a point mass m at a depth d
below the surface of the earth (Fig. 8.8(b)), so
that its distance from the centre of the earth is
( )
E
R d
−
as shown in the figure. The earth can
be thought of as being composed of a smaller
sphere of radius (R
E
– d ) and a spherical shell
of thickness d. The force on m due to the outer
shell of thickness d is zero because the result
quoted in the previous section. As far as the
smaller sphere of radius ( R
E
– d ) is concerned,
the point mass is outside it and hence according
to the result quoted earlier, the force due to this
smaller sphere is just as if the entire mass of
the smaller sphere is concentrated at the centre.
If M
s
is the mass of the smaller sphere, then,
M
s
/M
E
= (
R
E
– d)
3
/ R
E
3
( 8.16)
Since mass of a sphere is proportional to be
cube of its radius.
Fig. 8.8 (b) g at a depth d. In this case only the
smaller sphere of radius (R
E
–d)
contributes to g.
Thus the force on the point mass is
F (d) = G M
s
m / (R
E
– d )
2
(8.17)
Substituting for M
s
from above , we get
F (d) = G M
E
m ( R
E
– d ) / R
E
3
(8.18)
and hence the acceleration due to gravity at
a depth d,
g(d) =
( )
F d
m
is
3
( )
( ) ( )
E
E
E
GM
F d
g d R d
m
R
= = −
(1 / )
E
E
E
R d
g g d R
R
−
= = −
(8.19)
Thus, as we go down below earth’s surface,
the acceleration due gravity decreases by a factor
(1 / ).
E
d R−
The remarkable thing about
acceleration due to earth’s gravity is that it is
maximum on its surface decreasing whether you
go up or down.
8.7 GRAVITATIONAL POTENTIAL ENERGY
We had discussed earlier the notion of potential
energy as being the energy stored in the body at
its given position. If the position of the particle
changes on account of forces acting on it, then
the change in its potential energy is just the
amount of work done on the body by the force.
As we had discussed earlier, forces for which
the work done is independent of the path are
the conservative forces.
The force of gravity is a conservative force
and we can calculate the potential energy of a
body arising out of this force, called the
gravitational potential energy. Consider points
close to the surface of earth, at distances from
the surface much smaller than the radius of the
earth. In such cases, the force of gravity is
practically a constant equal to mg, directed
towards the centre of the earth. If we consider
a point at a height h
1
from the surface of the
earth and another point vertically above it at a
height h
2
from the surface, the work done in
lifting the particle of mass m from the first to
the second position is denoted by W
12
W
12
= Force × displacement
= mg (h
2
– h
1
) (8.20)
M
s
M
E
2020-21
192 PHYSICS
t
If we associate a potential energy W(h) at a
point at a height h above the surface such that
W(h) = mgh + W
o
(8.21)
(where W
o
= constant) ;
then it is clear that
W
12
= W(h
2
) – W(h
1
) (8.22)
The work done in moving the particle is just
the difference of potential energy between its
final and initial positions.Observe that the
constant W
o
cancels out in Eq. (8.22). Setting
h = 0 in the last equation, we get W ( h = 0 ) =
W
o.
. h = 0 means points on the surface of the
earth. Thus, W
o
is the potential energy on the
surface of the earth.
If we consider points at arbitrary distance
from the surface of the earth, the result just
derived is not valid since the assumption that
the gravitational force mg is a constant is no
longer valid. However, from our discussion we
know that a point outside the earth, the force of
gravitation on a particle directed towards the
centre of the earth is
2
E
G M m
F
r
=
(8.23)
where M
E
= mass of earth, m = mass of the
particle and r its distance from the centre of the
earth. If we now calculate the work done in
lifting a particle from r = r
1
to r = r
2
(r
2
> r
1
) along
a vertical path, we get instead of Eq. (8.20)
W
G M m
r
r
r
r
12
2
1
2
=
∫
d
= − −
G M m
r r
E
1 1
2 1
(8.24)
In place of Eq. (8.21), we can thus associate
a potential energy W(r) at a distance r, such that
E
1
( ) ,
G M m
W r W
r
=− +
(8.25)
valid for r > R ,
so that once again W
12
= W(r
2
) – W(r
1
).
Setting r = infinity in the last equation, we get
W ( r = infinity ) = W
1
. Thus, W
1
is the
potential energy at infinity. One should note that
only the difference of potential energy between
two points has a definite meaning from Eqs.
(8.22) and (8.24). One conventionally sets W
1
equal to zero, so that the potential energy at a
point is just the amount of work done in
displacing the particle from infinity to that point.
We have calculated the potential energy at
a point of a particle due to gravitational forces
on it due to the earth and it is proportional to
the mass of the particle. The gravitational
potential due to the gravitational force of the
earth is defined as the potential energy of a
particle of unit mass at that point. From the
earlier discussion, we learn that the gravitational
potential energy associated with two particles
of masses m
1
and m
2
separated by distance by a
distance r is given by
1 2
–
Gm m
V
r
=
(if we choose V = 0 as r
→ ∞
)
It should be noted that an isolated system
of particles will have the total potential energy
that equals the sum of energies (given by the
above equation) for all possible pairs of its
constituent particles. This is an example of the
application of the superposition principle.
Example 8.3 Find the potential energy of
a system of four particles placed at the
vertices of a square of side l. Also obtain
the potential at the centre of the square.
Answer Consider four masses each of mass m
at the corners of a square of side l; See Fig. 8.9.
We have four mass pairs at distance l and two
diagonal pairs at distance
2
l
Hence,
2 2
G
( ) 4 2
2
m G m
W r
l
l
= − −
Fig. 8.9
2020-21
GRAVITATION 193
t
l
mG
l
mG
22
5.41
2
1
2
2
−=
+−=
The gravitational potential at the centre of
the square
(
)
2 2
=
r l/
is
G m
( ) 4 2 = −
U r
l
. t
8.8 ESCAPE SPEED
If a stone is thrown by hand, we see it falls back
to the earth. Of course using machines we can
shoot an object with much greater speeds and
with greater and greater initial speed, the object
scales higher and higher heights. A natural
query that arises in our mind is the following:
‘can we throw an object with such high initial
speeds that it does not fall back to the earth?’
The principle of conservation of energy helps
us to answer this question. Suppose the object
did reach infinity and that its speed there was
V
f
. The energy of an object is the sum of potential
and kinetic energy. As before W
1
denotes that
gravitational potential energy of the object at
infinity. The total energy of the projectile at
infinity then is
2
1
( )
2
f
mV
E W∞ = +
(8.26)
If the object was thrown initially with a speed
V
i
from a point at a distance (h+R
E
) from the
centre of the earth (R
E
= radius of the earth), its
energy initially was
2
1
1
( ) –
2 ( )
E
E i
E
GmM
E h R mV W
h R
+ = +
+
(8.27)
By the principle of energy conservation
Eqs. (8.26) and (8.27) must be equal. Hence
2
2
–
2 ( ) 2
f
i E
E
mV
mV
GmM
h R
=
+
(8.28)
The R.H.S. is a positive quantity with a
minimum value zero hence so must be the L.H.S.
Thus, an object can reach infinity as long as V
i
is such that
2
– 0
2 ( )
i
E
E
mV
GmM
h R
≥
+
(8.29)
The minimum value of V
i
corresponds to the
case when the L.H.S. of Eq. (8.29) equals zero.
Thus, the minimum speed required for an object
to reach infinity (i.e. escape from the earth)
corresponds to
( )
2
min
1
2
E
i
E
GmM
m V
h R
=
+
(8.30)
If the object is thrown from the surface of
the earth, h = 0, and we get
( )
min
2
E
i
E
GM
V
R
=
(8.31)
Using the relation
2
/
E E
g GM R
=
, we get
(
)
min
2
i E
V gR
=
(8.32)
Using the value of g and R
E
, numerically
(V
i
)
min
≈11.2 km/s. This is called the escape
speed, sometimes loosely called the escape
velocity.
Equation (8.32) applies equally well to an
object thrown from the surface of the moon with
g replaced by the acceleration due to Moon’s
gravity on its surface and r
E
replaced by the
radius of the moon. Both are smaller than their
values on earth and the escape speed for the
moon turns out to be 2.3 km/s, about five times
smaller. This is the reason that moon has no
atmosphere. Gas molecules if formed on the
surface of the moon having velocities larger than
this will escape the gravitational pull of the
moon.
Example 8.4 Two uniform solid spheres
of equal radii R, but mass M and 4 M have
a centre to centre separation 6 R, as shown
in Fig. 8.10. The two spheres are held fixed.
A projectile of mass m is projected from
the surface of the sphere of mass M directly
towards the centre of the second sphere.
Obtain an expression for the minimum
speed v of the projectile so that it reaches
the surface of the second sphere.
Fig. 8.10
Answer The projectile is acted upon by two
mutually opposing gravitational forces of the two
2020-21
194 PHYSICS
spheres. The neutral point N (see Fig. 8.10) is
defined as the position where the two forces
cancel each other exactly. If ON = r, we have
( )
22
rR6
m M G
r
m M G
−
=
4
(6R – r)
2
= 4r
2
6R – r = ±2r
r = 2R or – 6R.
The neutral point r = – 6R does not concern
us in this example. Thus ON = r = 2R. It is
sufficient to project the particle with a speed
which would enable it to reach N. Thereafter,
the greater gravitational pull of 4M would
suffice. The mechanical energy at the surface
of M is
R
mM G
R
m M G
vmE
2
i
5
4
2
1
−−=
.
At the neutral point N, the speed approaches
zero. The mechanical energy at N is purely
potential.
R
m M G
R
m M G
E
N
4
4
2
−−=
.
From the principle of conservation of
mechanical energy
1
2
4
2
v
GM
R
GM
R
GM
R
GM
R
2
− − = − −
5
or
−=
2
1
5
4 2
2
R
M G
v
2/1
5
3
=
R
MG
v
t
A point to note is that the speed of the projectile
is zero at N, but is nonzero when it strikes the
heavier sphere 4 M. The calculation of this speed
is left as an exercise to the students.
8.9 EARTH SATELLITES
Earth satellites are objects which revolve around
the earth. Their motion is very similar to the
motion of planets around the Sun and hence
Kepler’s laws of planetary motion are equally
applicable to them. In particular, their orbits
around the earth are circular or elliptic. Moon
is the only natural satellite of the earth with a
near circular orbit with a time period of
approximately 27.3 days which is also roughly
equal to the rotational period of the moon about
its own axis. Since, 1957, advances in
technology have enabled many countries
including India to launch artificial earth
satellites for practical use in fields like
telecommunication, geophysics and
meteorology.
We will consider a satellite in a circular orbit
of a distance (R
E
+ h) from the centre of the earth,
where R
E
= radius of the earth. If m is the mass
of the satellite and V its speed, the centripetal
force required for this orbit is
F(centripetal) =
2
( )
E
mV
R h
+
(8.33)
directed towards the centre. This centripetal force
is provided by the gravitational force, which is
F(gravitation) =
2
( )
E
E
G m M
R h
+
(8.34)
where M
E
is the mass of the earth.
Equating R.H.S of Eqs. (8.33) and (8.34) and
cancelling out m, we get
2
( )
E
E
G M
V
R h
=
+
(8.35)
Thus V decreases as h increases. From
equation (8.35),the speed V for h = 0 is
2
( 0) /
E E
V h GM R gR
= = =
(8.36)
where we have used the relation
g =
2
/
E
GM R
. In every orbit, the satellite
traverses a distance 2π(R
E
+ h) with speed V. Its
time period T therefore is
3 / 2
2 ( ) 2 ( )
E E
E
R h R h
T
V
G M
π π
+ +
= =
(8.37)
on substitution of value of V from Eq. (8.35).
Squaring both sides of Eq. (8.37), we get
T
2
= k ( R
E
+ h)
3
(where k = 4 π
2
/ GM
E
) (8.38)
which is Kepler’s law of periods, as applied to
motion of satellites around the earth. For a
satellite very close to the surface of earth h can
be neglected in comparison to R
E
in Eq. (8.38).
Hence, for such satellites, T is T
o
, where
0
2 /
E
T R g
π
=
(8.39)
If we substitute the numerical values
g
9.8 m s
-2
and R
E
= 6400 km., we get
6
0
6.4 10
2
9.8
T
π
×
=
s
Which is approximately 85 minutes.
2020-21
GRAVITATION 195
t
t
t
Example 8.5 The planet Mars has two
moons, phobos and delmos. (i) phobos has
a period 7 hours, 39 minutes and an orbital
radius of 9.4 ×10
3
km. Calculate the mass
of mars. (ii) Assume that earth and mars
move in circular orbits around the sun,
with the martian orbit being 1.52 times
the orbital radius of the earth. What is
the length of the martian year in days ?
Answer (i) We employ Eq. (8.38) with the sun’s
mass replaced by the martian mass M
m
T
GM
R
2
m
=
4
2
3
π
M
m
G
R
T
=
4
2 3
2
π
( ) ( )
( )
=
× × ×
× × ×
4 3.14
6.67 10 459 60
-11
2
2 3
18
9 4 10.
( ) ( )
( )
M
4 3.14
6.67 4.59 6 10
2
-5
m
=
× × ×
× × ×
2 3
18
9 4 10.
= 6.48 × 10
23
kg.
(ii) Once again Kepler’s third law comes to our
aid,
T
T
R
R
M
2
E
2
MS
3
ES
3
=
where R
MS
is the mars -sun distance and R
ES
is
the earth-sun distance.
∴ T
M
= (1.52)
3/2
× 365
= 684 days
We note that the orbits of all planets except
Mercury, Mars and Pluto* are very close to
being circular. For example, the ratio of the
semi-minor to semi-major axis for our Earth
is, b/a = 0.99986. t
Example 8.6 Weighing the Earth : You
are given the following data: g = 9.81 ms
–2
,
R
E
= 6.37×10
6
m, the distance to the moon
R = 3.84×10
8
m and the time period of the
moon’s revolution is 27.3 days. Obtain the
mass of the Earth M
E
in two different ways.
Answer From Eq. (8.12) we have
G
R g
M
2
E
E
=
(
)
=
× ×
×
9.81 6.37 10
6.67
10
6
2
-11
= 5.97× 10
24
kg.
The moon is a satellite of the Earth. From
the derivation of Kepler’s third law [see Eq.
(8.38)]
E
M G
R
T
32
2
4
π
=
2
32
4
T G
R
M
E
π
=
( )
( )
=
× × × ×
× × × × ×
4 3.14 3.14 3.84 10
6.67 10 27.3 24 60 60
3
24
-11
2
= ×6.02 10
24
kg
Both methods yield almost the same answer,
the difference between them being less than 1%.
t
Example 8.7 Express the constant k of
Eq. (8.38) in days and kilometres. Given
k = 10
–13
s
2
m
–3
. The moon is at a distance
of 3.84 × 10
5
km from the earth. Obtain its
time-period of revolution in days.
Answer Given
k = 10
–13
s
2
m
–3
=
( )
( )
10
1
d
1
km
−
× ×
13
2
2
3
3
24 60 60 1 1000
/
= 1.33 ×10
–14
d
2
km
–3
Using Eq. (8.38) and the given value of k,
the time period of the moon is
T
2
= (1.33 × 10
-14
)(3.84 × 10
5
)
3
T = 27.3 d t
Note that Eq. (8.38) also holds for elliptical
orbits if we replace (R
E
+h) by the semi-major
axis of the ellipse. The earth will then be at one
of the foci of this ellipse.
8.10 ENERGY OF AN ORBITING SATELLITE
Using Eq. (8.35), the kinetic energy of the satellite
in a circular orbit with speed v is
2
1
2
K E m v
=i
2( )
E
E
Gm M
R h
=
+
, (8.40)
* Refer to information given in the Box on Page 182
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196 PHYSICS
t
Considering gravitational potential energy at
infinity to be zero, the potential energy at distance
(R
e
+h) from the centre of the earth is
.
( )
E
E
G m M
P E
R h
= −
+
(8.41)
The K.E is positive whereas the P.E is
negative. However, in magnitude the K.E. is half
the P.E, so that the total E is
. .
2( )
E
E
G m M
E K E P E
R h
= + = −
+
(8.42)
The total energy of an circularly orbiting
satellite is thus negative, with the potential
energy being negative but twice is magnitude of
the positive kinetic energy.
When the orbit of a satellite becomes
elliptic, both the K.E. and P.E. vary from point
to point. The total energy which remains
constant is negative as in the circular orbit case.
This is what we expect, since as we have
discussed before if the total energy is positive or
zero, the object escapes to infinity. Satellites
are always at finite distance from the earth and
hence their energies cannot be positive or zero.
Example 8.8 A 400 kg satellite is in a circular
orbit of radius 2R
E
about the Earth. How much
energy is required to transfer it to a circular
orbit of radius 4R
E
? What are the changes in
the kinetic and potential energies ?
Answer Initially,
E
E
i
R
mMG
E
4
−=
While finally
E
E
f
R
mMG
E
8
−=
The change in the total energy is
∆E = E
f
– E
i
8
8
2
E
E
E
E
E
R m
R
MG
R
mMG
==
J10 13.3
8
10 37.6 400 81.9
8
9
6
×=
×××
==∆
E
Rmg
E
The kinetic energy is reduced and it mimics
∆E, namely, ∆K = K
f
– K
i
= – 3.13 × 10
9
J.
The change in potential energy is twice the
change in the total energy, namely
∆V = V
f
– V
i
= – 6.25 × 10
9
J t
8.11 GEOSTATIONARY AND POLAR
SATELLITES
An interesting phenomenon arises if in we
arrange the value of (R
E
+ h) such that T in
Eq. (8.37) becomes equal to 24 hours. If the
circular orbit is in the equatorial plane of the
earth, such a satellite, having the same period
as the period of rotation of the earth about its
own axis would appear stationery viewed from
a point on earth. The (R
E
+ h) for this purpose
works out to be large as compared to R
E
:
1/ 3
2
2
4
E
E
T G M
R h
π
+ =
(8.43)
and for T = 24 hours, h works out to be 35800 km.
which is much larger than R
E
. Satellites in a
circular orbits around the earth in the
equatorial plane with T = 24 hours are called
Geostationery Satellites. Clearly, since the earth
rotates with the same period, the satellite would
appear fixed from any point on earth. It takes
very powerful rockets to throw up a satellite to
such large heights above the earth but this has
been done in view of the several benefits of many
practical applications.
Fig. 8.11 A Polar satellite. A strip on earth’s surface
(shown shaded) is visible from the satellite
during one cycle. For the next revolution of
the satellite, the earth has rotated a little
on its axis so that an adjacent strip becomes
visible.
It is known that electromagnetic waves above
a certain frequency are not reflected from
ionosphere. Radio waves used for radio
broadcast which are in the frequency range 2
MHz to 10 MHz, are below the critical frequency.
They are therefore reflected by the ionosphere.
2020-21
GRAVITATION 197
Thus radio waves broadcast from an antenna
can be received at points far away where the
direct wave fail to reach on account of the
curvature of the earth. Waves used in television
broadcast or other forms of communication have
much higher frequencies and thus cannot be
received beyond the line of sight. A Geostationery
satellite, appearing fixed above the broadcasting
station can however receive these signals and
broadcast them back to a wide area on earth.
The INSAT group of satellites sent up by India
are one such group of Geostationary satellites
widely used for telecommunications in India.
Another class of satellites are called the Polar
satellites (Fig. 8.11). These are low altitude (h l
500 to 800 km) satellites, but they go around
the poles of the earth in a north-south direction
whereas the earth rotates around its axis in an
east-west direction. Since its time period is
around 100 minutes it crosses any altitude many
times a day. However, since its height h above
the earth is about 500-800 km, a camera fixed
on it can view only small strips of the earth in
one orbit. Adjacent strips are viewed in the next
orbit, so that in effect the whole earth can be
viewed strip by strip during the entire day. These
satellites can view polar and equatorial regions
at close distances with good resolution.
Information gathered from such satellites
is extremely useful for remote sensing,
meterology as well as for environmental studies
of the earth.
8.12 WEIGHTLESSNESS
Weight of an object is the force with which the
earth attracts it. We are conscious of our own
weight when we stand on a surface, since the
surface exerts a force opposite to our weight to
keep us at rest. The same principle holds good
when we measure the weight of an object by a
spring balance hung from a fixed point e.g. the
ceiling. The object would fall down unless it is
subject to a force opposite to gravity. This is
exactly what the spring exerts on the object. This
is because the spring is pulled down a little by
the gravitational pull of the object and in turn the
spring exerts a force on the object vertically upwards.
Now, imagine that the top end of the balance
is no longer held fixed to the top ceiling of the
room. Both ends of the spring as well as the
object move with identical acceleration g. The
spring is not stretched and does not exert any
upward force on the object which is moving down
with acceleration g due to gravity. The reading
India’s Leap into Space
India started its space programme in 1962 when Indian National Committee for Space Research was set
up by the Government of India which was superseded by the Indian Space Research Organisation (ISRO)
in 1969. ISRO identified the role and importance of space technology in nation’s development and
bringing space to the service of the common man. India launched its first low orbit satellite Aryabhata in
1975, for which the launch vehicle was provided by the erstwhile Soviet Union. ISRO started employing its
indigenous launching vehicle in 1979 by sending Rohini series of satellites into space from its main
launch site at Satish Dhawan Space Center, Sriharikota, Andhra Pradesh. The tremendous progress in
India’s space programme has made ISRO one of the six largest space agencies in the world. ISRO
develops and delivers application specific satellite products and tools for broadcasts, communication,
weather forecasts, disaster management tools, Geographic Information System, cartography, navigation,
telemedicine, dedicated distance education satellite etc. In order to achieve complete self-reliance in
these applications, cost effective and reliable Polar Satellite Launch Vehicle (PSLV) was developed in
early 1990s. PSLV has thus become a favoured carrier for satellites of various countries, promoting
unprecedented international collaboration. In 2001, the Geosynchronous Satellite Launch Vehicle (GSLV)
was developed for launching heavier and more demanding Geosynchronous communication satellites.
Various research centers and autonomous institutions for remote sensing, astronomy and astrophysics,
atmospheric sciences and space research are functioning under the aegis of the Department of Space,
Government of India. Success of lunar (Chandrayaan) and inter planetary (Mangalyaan) missions along
with other scientific projects has been landmark achievements of ISRO. Future endeavors of ISRO in-
clude human space flight projects, the development of heavy lift launchers, reusable launch vehicles,
semi-cryogenic engines, single and two stage to orbit (SSTO and TSTO) vehicles, development and use of
composite materials for space application etc. In 1984 Rakesh Sharma became the first Indian to go into
outer space aboard in a USSR spaceship. (www.isro.gov.in)
2020-21
198 PHYSICS
SUMMARY
1. Newton’s law of universal gravitation states that the gravitational force of attraction
between any two particles of masses m
1
and m
2
separated by a distance r has the magnitude
F G
m m
r
2
=
1 2
where G is the universal gravitational constant, which has the value 6.672 ×10
–11
N m
2
kg
–2
.
2. If we have to find the resultant gravitational force acting on the particle m due to a
number of masses M
1
, M
2
, ….M
n
etc. we use the principle of superposition. Let F
1
, F
2
, ….F
n
be the individual forces due to M
1
, M
2
, ….M
n,
each
given by the law of gravitation. From
the principle of superposition each force acts independently and uninfluenced by the
other bodies. The resultant force F
R
is then found by vector addition
F
R
= F
1
+ F
2
+ ……+ F
n
=
F
i
i
n
=
∑
1
where the symbol ‘Σ’ stands for summation.
3. Kepler’s laws of planetary motion state that
(a) All planets move in elliptical orbits with the Sun at one of the focal points
(b) The radius vector drawn from the Sun to a planet sweeps out equal areas in equal
time intervals. This follows from the fact that the force of gravitation on the planet is
central and hence angular momentum is conserved.
(c) The square of the orbital period of a planet is proportional to the cube of the semi-
major axis of the elliptical orbit of the planet
The period T and radius R of the circular orbit of a planet about the Sun are related
by
3
2
2
4
R
M G
T
s
π
=
where M
s
is the mass of the Sun. Most planets have nearly circular orbits about the
Sun. For elliptical orbits, the above equation is valid if R is replaced by the semi-major
axis, a.
4. The acceleration due to gravity.
(a) at a height h above the earth’s surface
(
)
2
( )
E
E
G M
g h
R h
=
+
≈ −
1
2
2
G M
R
h
R
E
E
E
for h << R
E
g h g( ) 1
2
=
(
)
−
(
)
=0
2
h
R
g
G M
R
E
E
E
where 0
recorded in the spring balance is zero since the
spring is not stretched at all. If the object were
a human being, he or she will not feel his weight
since there is no upward force on him. Thus,
when an object is in free fall, it is weightless and
this phenomenon is usually called the
phenomenon of weightlessness.
In a satellite around the earth, every part
and parcel of the satellite has an acceleration
towards the centre of the earth which is exactly
the value of earth’s acceleration due to gravity
at that position. Thus in the satellite everything
inside it is in a state of free fall. This is just as if
we were falling towards the earth from a height.
Thus, in a manned satellite, people inside
experience no gravity. Gravity for us defines the
vertical direction and thus for them there are no
horizontal or vertical directions, all directions are
the same. Pictures of astronauts floating in a
satellite show this fact.
2020-21
GRAVITATION 199
(b) at depth d below the earth’s surface is
g gd
G M
R
d
R
d
R
E
E
E E
( )
= −
=
( )
−
1 1
2
0
5. The gravitational force is a conservative force, and therefore a potential energy function
can be defined. The gravitational potential energy associated with two particles separated
by a distance r is given by
r
mmG
V
21
−=
where V is taken to be zero at r → ∞. The total potential energy for a system of particles
is the sum of energies for all pairs of particles, with each pair represented by a term of
the form given by above equation. This prescription follows from the principle of
superposition.
6. If an isolated system consists of a particle of mass m moving with a speed v in the
vicinity of a massive body of mass M, the total mechanical energy of the particle is given by
r
m M G
vmE
2
1
2
−=
That is, the total mechanical energy is the sum of the kinetic and potential energies.
The total energy is a constant of motion.
7. If m moves in a circular orbit of radius a about M, where M >> m, the total energy of the system is
a
mMG
E
2
−=
with the choice of the arbitrary constant in the potential energy given in the point 5.,
above. The total energy is negative for any bound system, that is, one in which the orbit
is closed, such as an elliptical orbit. The kinetic and potential energies are
a
m
M
G
K
2
=
a
m
M
G
V −=
8. The escape speed from the surface of the earth is
E
E
e
R
M G
v
2
=
=
2
E
gR
and has a value of 11.2 km s
–1
.
9. If a particle is outside a uniform spherical shell or solid sphere with a spherically symmetric
internal mass distribution, the sphere attracts the particle as though the mass of the
sphere or shell were concentrated at the centre of the sphere.
10. If a particle is inside a uniform spherical shell, the gravitational force on the particle is zero.
If a particle is inside a homogeneous solid sphere, the force on the particle acts toward the
centre of the sphere. This force is exerted by the spherical mass interior to the particle.
11. A geostationary (geosynchronous communication) satellite moves in a circular orbit in
the equatorial plane at a approximate distance of 4.22 × 10
4
km from the earth’s centre.
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200 PHYSICS
POINTS TO PONDER
1. In considering motion of an object under the gravitational influence of another object
the following quantities are conserved:
(a) Angular momentum
(b) Total mechanical energy
Linear momentum is not conserved
2. Angular momentum conservation leads to Kepler’s second law. However, it is not special
to the inverse square law of gravitation. It holds for any central force.
3. In Kepler’s third law (see Eq. (8.1) and T
2
= K
S
R
3
. The constant K
S
is the same for all
planets in circular orbits. This applies to satellites orbiting the Earth [(Eq. (8.38)].
4. An astronaut experiences weightlessness in a space satellite. This is not because the
gravitational force is small at that location in space. It is because both the astronaut
and the satellite are in “free fall” towards the Earth.
5. The gravitational potential energy associated with two particles separated by a distance
r is given by
V
G m m
r
= +–
1 2
constant
The constant can be given any value. The simplest choice is to take it to be zero. With
this choice
V
G m m
r
= –
1 2
This choice implies that V → 0 as r → ∞. Choosing location of zero of the gravitational
energy is the same as choosing the arbitrary constant in the potential energy. Note that
the gravitational force is not altered by the choice of this constant.
6. The total mechanical energy of an object is the sum of its kinetic energy (which is always
positive) and the potential energy. Relative to infinity (i.e. if we presume that the potential
energy of the object at infinity is zero), the gravitational potential energy of an object is
negative. The total energy of a satellite is negative.
7. The commonly encountered expression m g h for the potential energy is actually an
approximation to the difference in the gravitational potential energy discussed in the
point 6, above.
8. Although the gravitational force between two particles is central, the force between two
finite rigid bodies is not necessarily along the line joining their centre of mass. For a
spherically symmetric body however the force on a particle external to the body is as if
the mass is concentrated at the centre and this force is therefore central.
9. The gravitational force on a particle inside a spherical shell is zero. However, (unlike a
metallic shell which shields electrical forces) the shell does not shield other bodies outside
it from exerting gravitational forces on a particle inside. Gravitational shielding is not
possible.
EXERCISES
8.1 Answer the following :
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor.
Can you shield a body from the gravitational influence of nearby matter by putting
it inside a hollow sphere or by some other means ?
(b) An astronaut inside a small space ship orbiting around the earth cannot detect
gravity. If the space station orbiting around the earth has a large size, can he hope
to detect gravity ?
(c) If you compare the gravitational force on the earth due to the sun to that due
to the moon, you would find that the Sun’s pull is greater than the moon’s pull.
(you can check this yourself using the data available in the succeeding exercises).
However, the tidal effect of the moon’s pull is greater than the tidal effect of sun.
Why ?
2020-21
GRAVITATION 201
8.2 Choose the correct alternative :
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the
earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula –G Mm(1/r
2
– 1/r
1
)
is more/less accurate than the formula
mg(r
2
– r
1
) for the difference of potential energy between two points r
2
and r
1
distance
away from the centre of the earth.
8.3 Suppose there existed a planet that went around the Sun twice as fast as the earth.
What would be its orbital size as compared to that of the earth ?
8.4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius
of the orbit is 4.22 × 10
8
m. Show that the mass of Jupiter is about one-thousandth
that of the sun.
8.5 Let us assume that our galaxy consists of 2.5 × 10
11
stars each of one solar mass. How
long will a star at a distance of 50,000 ly from the galactic centre take to complete one
revolution? Take the diameter of the Milky Way to be 10
5
ly.
8.6 Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite
is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of earth’s gravitational
influence is more/less than the energy required to project a stationary object at
the same height (as the satellite) out of earth’s influence.
8.7 Does the escape speed of a body from the earth depend on (a) the mass of the body, (b)
the location from where it is projected, (c) the direction of projection, (d) the height of
the location from where the body is launched?
8.8 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a)
linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential
energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when
it comes very close to the Sun.
8.9 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen
feet, (b) swollen face, (c) headache, (d) orientational problem.
8.10 In the following two exercises, choose the correct answer from among the given ones:
The gravitational intensity at the centre of a hemispherical shell of uniform mass
density has the direction indicated by the arrow (see Fig 8.12) (i) a, (ii) b,(iii)c, (iv) 0.
Fig. 8.12
8.11 For the above problem, the direction of the gravitational intensity at an arbitrary
point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.
8.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s
centre is the gravitational force on the rocket zero ? Mass of the sun = 2×10
30
kg,
mass of the earth = 6×10
24
kg. Neglect the effect of other planets etc. (orbital radius =
1.5 × 10
11
m).
8.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of
the earth around the sun is 1.5 × 10
8
km.
8.14 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the
earth is 1.50 × 10
8
km away from the sun ?
8.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it
due to the earth at a height equal to half the radius of the earth ?
8.16 Assuming the earth to be a sphere of uniform mass density, how much would a body
2020-21
202 PHYSICS
weigh half way down to the centre of the earth if it weighed 250 N on the surface ?
8.17 A rocket is fired vertically with a speed of 5 km s
-1
from the earth’s surface. How far
from the earth does the rocket go before returning to the earth ? Mass of the earth
= 6.0 × 10
24
kg; mean radius of the earth = 6.4 × 10
6
m; G = 6.67 × 10
–11
N m
2
kg
–2
.
8.18 The escape speed of a projectile on the earth’s surface is 11.2 km s
–1
. A body is
projected out with thrice this speed. What is the speed of the body far away from the
earth? Ignore the presence of the sun and other planets.
8.19 A satellite orbits the earth at a height of 400 km above the surface. How much
energy must be expended to rocket the satellite out of the earth’s gravitational
influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×10
24
kg; radius of
the earth = 6.4 × 10
6
m; G = 6.67 × 10
–11
N m
2
kg
–2
.
8.20 Two stars each of one solar mass (= 2×10
30
kg) are approaching each other for a head
on collision. When they are a distance 10
9
km, their speeds are negligible. What is
the speed with which they collide ? The radius of each star is 10
4
km. Assume the
stars to remain undistorted until they collide. (Use the known value of G).
8.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart
on a horizontal table. What is the gravitational force and potential at the mid point
of the line joining the centres of the spheres ? Is an object placed at that point in
equilibrium? If so, is the equilibrium stable or unstable ?
Additional Exercises
8.22 As you have learnt in the text, a geostationary satellite orbits the earth at a height of
nearly 36,000 km from the surface of the earth. What is the potential due to earth’s
gravity at the site of this satellite ? (Take the potential energy at infinity to be zero).
Mass of the earth = 6.0×10
24
kg, radius = 6400 km.
8.23 A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a
speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as
neutron stars. Certain stellar objects called pulsars belong to this category). Will an
object placed on its equator remain stuck to its surface due to gravity ? (mass of the
sun = 2×10
30
kg).
8.24 A spaceship is stationed on Mars. How much energy must be expended on the
spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg;
mass of the sun = 2×10
30
kg; mass of mars = 6.4×10
23
kg; radius of mars = 3395 km;
radius of the orbit of mars = 2.28 ×10
8
km; G = 6.67×10
-11
N m
2
kg
–2
.
8.25 A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s
–1
. If 20%
of its initial energy is lost due to martian atmospheric resistance, how far will the
rocket go from the surface of mars before returning to it ? Mass of mars = 6.4×10
23
kg;
radius of mars = 3395 km; G = 6.67×10
-11
N m
2
kg
–2
.
2020-21
GRAVITATION 203
APPENDIX 8.1 : LIST OF INDIAN SATELLITES
S.No. Name Launch Date Launch Vehicle Application
1. Aryabhata Apr. 19, 1975 C-1 Intercosmos
a
Experimental
2. Bhaskara-I Jun. 07, 1979 C-1 Intercosmos
a
Earth Observation,
Experimental
3. Rohini Technology Payload (RTP) Aug. 10, 1979 SLV-3E1
b
Experimental
4. Rohini Satellite RS-1 Jul. 18, 1980 SLV-3E2
b
Experimental
5. Rohini Satellite RS-D1 May 31, 1981 SLV-3D1
b
Earth Observation
6. APPLE Jun. 19, 1981 Ariane -1(V-3)
c
Communication,
Experimental
7. Bhaskara-II Nov. 20, 1981 C-1 Intercosmos
a
Earth Observation,
Experimental
8. INSAT-1A Apr. 10, 1982 Delta
d
Communication
9. Rohini Satellite RS-D2 Apr. 17, 1983 SLV-3
b
Earth Observation
10. INSAT-1B Aug. 30, 1983 Shuttle [PAM-D]
d
Communication
11. SROSS-1 Mar. 24, 1987 ASLV-D1
b
Experimental
12. IRS-1A Mar. 17, 1988 Vostok
e
Earth Observation
13. SROSS-2 Jul. 13, 1988 ASLV-D2
b
Earth Observation,
Experimental
14. INSAT-1C Jul. 22, 1988 Ariane-3
c
Communication
15. INSAT-1D Jun. 12, 1990 Delta 4925
d
Communication
16. IRS-1B Aug. 29, 1991 Vostok
e
Earth Observation
17. SROSS-C May 20, 1992 ASLV-D3
b
Experimental
18. INSAT-2A Jul. 10, 1992 Ariane-44L H10
c
Communication
19. INSAT-2B Jul. 23, 1993 Ariane-44L H10
+c
Communication
20. IRS-1E Sep. 20, 1993 PSLV-D1
b
Earth Observation
21. SROSS-C2 May 04, 1994 ASLV-D4
b
Experimental
22. IRS-P2 Oct. 15, 1994 PSLV-D2
b
Earth Observation
23. INSAT-2C Dec. 07, 1995 Ariane-44L H10-3
c
Communication
24. IRS-1C Dec. 28, 1995 Molniya
e
Earth Observation
25. IRS-P3 Mar. 21, 1996 PSLV-D3/IRS-P3
b
Earth Observation
26. INSAT-2D Jun. 04, 1997 Ariane-44L H10-3
c
Communication
27. IRS-1D Sep. 29, 1997 PSLV-C1/IRS-1D
b
Earth Observation
28. INSAT-2E Apr. 03, 1999 Ariane-42P H10-3
c
Communication
29. Oceansat (IRS-P4) May 26, 1999 PSLV-C2/IRS-P4
b
Earth Observation
30. INSAT-3B Mar. 22, 2000 Ariane-5G
c
Communication
31. GSAT-1 Apr. 18, 2001 GSLV-D1/GSAT-1
b
Communication
32. The Technology Experiment Oct. 22, 2001 PSLV-C3/TES
b
Earth Observation
Satellite (TES)
33. INSAT-3C Jan. 24, 2002 Ariane5-V147
c
Climate &
Environment,
Communication
34. KALPANA-1 Sep. 12, 2002 PSLV-C4/ Climate &
KALPANA-1
b
Environment,
Communication
35. INSAT-3A Apr. 10, 2003 Ariane5-V160
c
Climate &
Environment,
Communication
36. GSAT-2 May 08, 2003 GSLV-D2/GSAT-2
b
Communication
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204 PHYSICS
37. INSAT-3E Sep. 28, 2003 Ariane5-V162
c
Communication
38. IRS-P6 / RESOURCESAT-1 Oct. 17, 2003 PSLV-C5/ Earth Observation
RESOURCESAT-1
b
39. EDUSAT Sep. 20, 2004 GSLV-F01/ Communication
EDUSAT(GSAT-3)
b
40. HAMSAT May 05, 2005 PSLV-C6/ Communication
CARTOSAT-1/HAMSAT
b
41. CARTOSAT-1 May 05, 2005 PSLV-C6/ Earth Observation
CARTOSAT-1/HAMSAT
b
42. INSAT-4A Dec. 22, 2005 Ariane5-V169c Communication
43. INSAT-4C Jul. 10, 2006 GSLV-F02/INSAT-4C
b
Communication
44. CARTOSAT-2 Jan. 10, 2007 PSLV-C7/CARTOSAT-2 Earth Observation
/SRE-1
b
45. SRE-1 Jan. 10, 2007 PSLV-C7/CARTOSAT-2 Experimental
/SRE-1
b
46. INSAT-4B Mar. 12, 2007 Ariane5
c
Communication
47. INSAT-4CR Sep. 02, 2007 GSLV-F04/INSAT-4 Communication
CR
b
48. IMS-1 Apr. 28, 2008 PSLV-C9/ Earth Observation
CARTOSAT-2A
b
49. CARTOSAT - 2A Apr. 28, 2008 PSLV-C9/ Earth Observation
CARTOSAT-2A
b
50. Chandrayaan-1 Oct. 22, 2008 PSLV-C11
b
Planetary
Observation
51. RISAT-2 Apr. 20, 2009 PSLV-C12/RISAT-2
b
Earth Observation
52. ANUSAT Apr. 20, 2009 PSLV-C12/RISAT-2
b
University/
Academic Institute
53. Oceansat-2 Sep. 23, 2009 PSLV-C14/ Climate &
OCEANSAT-2
b
Environment, Earth
Observation
54. GSAT-4 Apr. 15, 2010 GSLV-D3 / GSAT-4
b
Communication
55. CARTOSAT-2B Jul. 12, 2010 PSLV-C15/ Earth Observation
CARTOSAT-2B
b
56. STUDSAT Jul. 12, 2010 PSLV-C15/ University/
CARTOSAT-2B
b
Academic Institute
57. GSAT-5P Dec. 25, 2010 GSLV-F06/GSAT-5P
b
Communication
58. RESOURCESAT-2 Apr. 20, 2011 PSLV-C16/ Earth Observation
RESOURCESAT-2
b
59. YOUTHSAT Apr. 20, 2011 PSLV-C16/ Student Satellite
RESOURCESAT-2
b
60. GSAT-8 May 21, 2011 Ariane-5 VA-202
c
Communication
61. GSAT-12 Jul. 15, 2011 PSLV-C17/GSAT-12
b
Communication
62. Megha-Tropiques Oct. 12, 2011 PSLV-C18/Megha- Climate &
Tropiques
b
Environment, Earth
Observation
63. SRMSat Oct. 12, 2011 PSLV-C18/ University/
Megha-Tropiques
b
Academic Institute
64. Jugnu Oct. 12, 2011 PSLV-C18/ University/Academic
Megha-Tropiques
b
Institute
65. RISAT-1 Apr. 26, 2012 PSLV-C19/RISAT-1
b
Earth Observation
66. GSAT-10 Sep. 29, 2012 Ariane-5 VA-209
c
Communication,
Navigation
67. SARAL Feb. 25, 2013 PSLV-C20/SARAL
b
Climate &
Environment, Earth
Observation
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GRAVITATION 205
68. IRNSS-1A Jul. 01, 2013 PSLV-C22/IRNSS-1A
b
Navigation
69. INSAT-3D Jul. 26, 2013 Ariane-5 VA-214
c
Climate &
Environment,
Disaster
Management System
70. GSAT-7 Aug. 30, 2013 Ariane-5 VA-215
c
Communication
71. Mars Orbiter Mission Spacecraft Nov. 05, 2013 PSLV-C25
b
Planetary
(Mangalyaan-1) Observation
72. GSAT-14 Jan. 05, 2014 GSLV-D5/GSAT-14
b
Communication
73. IRNSS-1B Apr. 04, 2014 PSLV-C24/IRNSS-1B
b
Navigation
74. IRNSS-1C Oct. 16, 2014 PSLV-C26/IRNSS-1C
b
Navigation
75. GSAT-16 Dec. 07, 2014 Ariane-5 VA-221
c
Communication
76. Crew module Atmospheric Dec. 18, 2014 LVM-3/CARE Mission
b
Experimental
Reentry Experiment
77. IRNSS-1D Mar. 28, 2015 PSLV-C27/IRNSS-1D
b
Navigation
78. GSAT-6 (INSAT-4E) Aug. 27, 2015 GSLV-D6
b
Communication
79. Astrosat Sep. 28, 2015 PSLV-C30
b
Space Sciences
80. GSAT-15 Nov. 11, 2015 Ariane-5 VA-227
c
Communication,
Navigation
81. IRNSS-1E Jan. 20, 2016 PSLV-C31/IRNSS-1E
b
Navigation
82. IRNSS-1F Mar. 10, 2016 PSLV-C32/IRNSS-1F
b
Navigation
83. IRNSS-1G Apr. 28, 2016 PSLV-C33/IRNSS-1G
b
Navigation
84. Cartosat-2 Series Satellite Jun. 22, 2016 PSLV-C34/CARTOSAT-2 Earth Observation
Series Satellite
b
85. SathyabamaSat Jun. 22, 2016 PSLV-C34/CARTOSAT-2 University/
Series Satellite
b
Academic
Institute
86. Swayam Jun. 22, 2016 PSLV-C34/CARTOSAT-2 University/
Series Satellite
b
Academic
Institute
87. INSAT-3DR Sep. 08, 2016 GSLV-F05/ Climate &
INSAT-3DR
b
Environment,
Disaster
Management System
88. ScatSat-1 Sep. 26, 2016 PSLV-C35/ Climate &
SCATSAT-1
b
Environment
89. Pratham Sep. 26, 2016 PSLV-C35/ University/
SCATSAT-1
b
Academic Institute
90. PiSat Sep. 26, 2016 PSLV-C35/ University/
SCATSAT-1
b
Academic Institute
91. GSAT-18 Oct. 06, 2016 Ariane-5 VA-231
c
Communication
92. ResourceSat-2A Dec. 07, 2016 PSLV-C36/ Earth Observation
RESOURCESAT-2A
b
93. Cartosat -2 Series Satellite Feb. 15, 2017 PSLV-C37/Cartosat -2 Earth Observation
Series Satellite
b
94. INS-1A Feb. 15, 2017 PSLV-C37/Cartosat -2 Experimental
Series Satellite
b
95. INS-1B Feb. 15, 2017 PSLV-C37/Cartosat -2 Experimental
Series Satellite
b
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206 PHYSICS
India has so far also launched 239 foreign satellites of 28 countries from Satish Dhawan Space Center, Sriharikota,
Andhra Pradesh: May 26, 1999 (02); Oct. 22, 2001 (02); Jan. 10, 2007 (02); Apr. 23, 2007 (01); Jan. 21, 2008
(01); Apr. 28,2008 (08); Sep. 23,2009 (06); July 12, 2010 (03); Jan. 12,2011 (01);
Apr. 20, 2011 (01) Sep. 09, 2012 (02); Feb. 25, 2013 (06); June 30, 2014 (05); July 10, 2015 (05); Sep. 28,
2015 (06); Dec. 16, 2015 (06); June 22, 2016 (17); Sep. 26, 2016 (05); Feb. 15, 2017 (101) and thus setting a
world record; June 23, 2017 (29). Jan 12, 2018 (28); Sep. 16, 2018 (02).
Details can be seen at www.isro.gov.in.
aLaunched from Kapustin Yar Missile and Space Complex, Soviet Union (now Russia)
b Launched from Satish Dhawan Space Centre, Sriharikota, Andhra Pradesh
c Launched from Centre Spatial Guyanais, Kourou, French Guiana
d Launched from Air Force Eastern Test Range, Florida
e Launched from Baikonur Cosmodrome, Kazakhstan
96. GSAT-9 May 05, 2017 GSLV-F09/GSAT-9
b
Communication
97. GSAT-19 Jun. 05, 2017 GSLV Mk III-D1/ Communication
GSAT-19 Mission
b
98. Cartosat-2 Series Satellite Jun. 23, 2017 PSLV-C38/Cartosat-2 Earth Observation
Series Satellite
b
99. NIUSAT Jun. 23, 2017 PSLV-C38/Cartosat-2 University/
Series Satellite
b
Academic Institute
100. GSAT-17 Jun. 29, 2017 Ariane-5 VA-238
c
Communication
101. IRNSS-1H Aug. 31, 2017 PSLV-C39
/
IRNSS- Navigation
1H Mission
b
102. INS-1C Jan. 12, 2018 PSLV-C40/Cartosat-2 Experimental
Series Satellite Mission
b
103. Mircosat Jan. 12, 2018 PSLV-C40/Cartosat-2 Experimental
Series Satellite Mission
b
104. Cartosat-2 Series Satellite Jan. 12, 2018 PSLV-C40/Cartosat-2 Earth Observation
Series Satellite Mission
b
105. GSAT-6A Mar. 29, 2018 GSLV-F08/GSAT-6A Communication
Mission
b
106. IRNSS-1I Apr. 12, 2018 PSLV-C41/IRNSS-1I
b
Navigation
107. GSAT-29 Nov. 14, 2018 GSLV Mk III-D2/ Communication
GSAT-29 Mission
108. GSAT-11 Mission Dec. 05, 2018 Ariane-5 VA-246 Communication
109. GSAT-7A Dec. 19, 2018 GSLV-F11/GSAT-7A Communication
Mission
110. GSAT-31 Feb 06, 2019 Ariane-5 VA-247 Communication
111. HysIS Nov. 29, 2018 PSLV-C43/HysIS Earth Observation
Mission
112. RISAT-2B May 22, 2019 PSLV-C46 Mission Disaster
Management
System, Earth
Observation
113. Kalamsat-V2 Jan. 24, 2019 PSLV-C44 University/
Academic
Institute
GSLV MkIII-MI Successfully Launches Chandrayaan-2 spacecraft on July 22, 2019
2020-21
