CHAPTER SEVEN

GRAVITATION

7.1 INTRODUCTION
Early in our lives, we become aware of the tendency of all
material objects to be attracted towards the earth. Anything
7.1 Introduction
thrown up falls down towards the earth, going uphill is lot
7.2 Kepler’s laws
more tiring than going downhill, raindrops from the clouds
7.3 Universal law of
above fall towards the earth and there are many other such
gravitation
phenomena. Historically it was the Italian Physicist Galileo
7.4 The gravitational
constant (1564-1642) who recognised the fact that all bodies,
7.5 Acceleration due to
irrespective of their masses, are accelerated towards the earth
gravity of the earth with a constant acceleration. It is said that he made a public
7.6 Acceleration due to demonstration of this fact. To find the truth, he certainly
gravity below and above did experiments with bodies rolling down inclined planes and
the surface of earth arrived at a value of the acceleration due to gravity which is
7.7 Gravitational potential close to the more accurate value obtained later.
energy A seemingly unrelated phenomenon, observation of stars,
7.8 Escape speed planets and their motion has been the subject of attention
7.9 Earth satellites in many countries since the earliest of times. Observations
7.10 Energy of an orbiting since early times recognised stars which appeared in the
satellite sky with positions unchanged year after year. The more
Summary interesting objects are the planets which seem to have regular
Points to ponder motions against the background of stars. The earliest
Exercises 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
(5th century A.D.) in his treatise. A thousand years later, a
Polish monk named Nicolas Copernicus (1473-1543)

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128 PHYSICS

proposed a definitive model in which the planets of the ellipse (Fig. 7.1a). This law was a
moved in circles around a fixed central sun. His deviation from the Copernican model which
theory was discredited by the church, but allowed only circular orbits. The ellipse, of
notable amongst its supporters was Galileo who which the circle is a special case, is a closed
had to face prosecution from the state for his curve which can be drawn very simply as
beliefs. follows.
It was around the same time as Galileo, a Select two points F1 and F2. Take a length
nobleman called Tycho Brahe (1546-1601) of a string and fix its ends at F1 and F2 by
hailing from Denmark, spent his entire lifetime pins. With the tip of a pencil stretch the string
recording observations of the planets with the taut and then draw a curve by moving the
naked eye. His compiled data were analysed
pencil keeping the string taut throughout.(Fig.
later by his assistant Johannes Kepler (1571-
7.1(b)) The closed curve you get is called an
1640). He could extract from the data three
ellipse. Clearly for any point T on the ellipse,
elegant laws that now go by the name of Kepler’s
laws. These laws were known to Newton and the sum of the distances from F1 and F2 is a
enabled him to make a great scientific leap in constant. F1, F 2 are called the focii. Join the
proposing his universal law of gravitation. points F 1 and F 2 and extend the line to
intersect the ellipse at points P and A as shown
7.2 KEPLER’S LAWS in Fig. 7.1(b). The midpoint of the line PA is
The three laws of Kepler can be stated as the centre of the ellipse O and the length PO =
follows: AO is called the semi-major axis of the ellipse.
1. Law of orbits : All planets move in elliptical For a circle, the two focii merge into one and
orbits with the Sun situated at one of the foci the semi-major axis becomes the radius of the
circle.
B 2. Law of areas : The line that joins any planet
to the sun sweeps equal areas in equal
intervals of time (Fig. 7.2). This law comes from
the observations that planets appear to move
2b slower when they are farther from the sun
P S S' A
than when they are nearer.

C
2a
Fig. 7.1(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. 7.2 The planet P moves around the sun in an

elliptical orbit. The shaded area is the area
Fig. 7.1(b) Drawing an ellipse. A string has its ends ∆A swept out in a small interval of time ∆t.
fixed at F1 and F2. The tip of a pencil holds
the string taut and is moved around.

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GRAVITATION 129

3. Law of periods : The square of the time period the law of areas. Gravitation is a central force
of revolution of a planet is proportional to the and hence the law of areas follows.
cube of the semi-major axis of the ellipse traced
out by the planet. ⊳ Example 7.1 Let the speed of the planet
at the perihelion P in Fig. 7.1(a) be vP and
Table 7.1 gives the approximate time periods
the Sun-planet distance SP be rP. Relate
of revolution of eight* planets around the sun
{rP, vP} to the corresponding quantities at
along with values of their semi-major axes. the aphelion {rA, vA}. Will the planet take
Table 7.1 Data from measurement of equal times to traverse BAC and CPB ?
planetary motions given below
confirm Kepler’s Law of Periods Answer The magnitude of the angular
(a ≡ 10
Semi-major axis in units of 10 m. momentum at P is Lp = mp rp vp, since inspection
T ≡ Time period of revolution of the planet tells us that r p and v p are mutually
in years(y). perpendicular. Similarly, LA = mp rA vA. From
Q ≡ The quotient ( T2/a3 ) in units of
angular momentum conservation
10 -34 y2 m-3.)
mp rp vp = mp rA vA
Planet a T Q vp rA
or = ⊳
Mercury 5.79 0.24 2.95 vA rp
Venus 10.8 0.615 3.00
Earth 15.0 1 2.96 Since rA > rp, vp > vA .
Mars 22.8 1.88 2.98 The area SBAC bounded by the ellipse and
Jupiter 77.8 11.9 3.01 the radius vectors SB and SC is larger than SBPC
Saturn 143 29.5 2.98 in Fig. 7.1. From Kepler’s second law, equal areas
Uranus 287 84 2.98 are swept in equal times. Hence the planet will
Neptune 450 165 2.99 take a longer time to traverse BAC than CPB.
7.3 UNIVERSAL LAW OF GRAVITATION
The law of areas can be understood as a Legend has it that observing an apple falling from
consequence of conservation of angular a tree, Newton was inspired to arrive at an
momentum whch is valid for any central universal law of gravitation that led to an
force . A central force is such that the force explanation of terrestrial gravitation as well as
on the planet is along the vector joining the of Kepler’s laws. Newton’s reasoning was that
Sun and the planet. Let the Sun be at the the moon revolving in an orbit of radius Rm was
origin and let the position and momentum subject to a centripetal acceleration due to
of the planet be denoted by r and p earth’s gravity of magnitude
respectively. Then the area swept out by the
V2 4π 2 Rm
planet of mass m in time interval ∆ t is (Fig. am = = (7.3)
Rm T2
7.2) ∆ A given by
∆A = ½ (r × v∆t) (7.1)
where V is the speed of the moon related to the
Hence
time period T by the relation V = 2π Rm / T . The
∆A /∆t =½ (r × p)/m, (since v = p/m)
= L / (2 m) (7.2) time period T is about 27.3 days and Rm was
where v is the velocity, L is the angular already known then to be about 3.84 × 108m. If
momentum equal to ( r × p). For a central we substitute these numbers in Eq. (7.3), we get
force, which is directed along r, L is a constant a value of am much smaller than the value of
acceleration due to gravity g on the surface of
as the planet goes around. Hence, ∆ A /∆t is a
the earth, arising also due to earth’s gravitational
constant according to the last equation. This is attraction.

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130 PHYSICS

This clearly shows that the force due to The gravitational force is attractive, i.e., the
earth’s gravity decreases with distance. If one force F is along – r. The force on point mass m1
assumes that the gravitational force due to the due to m2 is of course – F by Newton’s third law.
earth decreases in proportion to the inverse Thus, the gravitational force F12 on the body 1
square of the distance from the centre of the
due to 2 and F21 on the body 2 due to 1 are related
earth, we will have am α Rm
−2
; g α R−
E
2
and we get as F12 = – F21.
Before we can apply Eq. (7.5) to objects under
g R2
= m2 3600 (7.4) consideration, we have to be careful since the
am RE law refers to point masses whereas we deal with
in agreement with a value of g 9.8 m s-2 and extended objects which have finite size. If we have
the value of am from Eq. (7.3). These observations a collection of point masses, the force on any
led Newton to propose the following Universal Law one of them is the vector sum of the gravitational
of Gravitation : forces exerted by the other point masses as
Every body in the universe attracts every other shown in Fig 7.4.
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 m2 due
to another point mass m1 has the magnitude
m1 m 2
|F | = G (7.5)
r2
Equation (7.5) can be expressed in vector form as

F= G
m1 m 2
r 2 ( )
– rɵ = – G
m1 m 2 ɵ
r2
r
Fig. 7.4 Gravitational force on point mass m1 is the
vector sum of the gravitational forces exerted
m1 m 2 ɵ by m2, m3 and m4.
= –G 3
r
r
The total force on m1 is
where G is the universal gravitational constant,
Gm 2 m1 ɵ Gm 3 m1 ɵ Gm 4 m1 ɵ
F1 = r 21 + r 31 + r 41
rɵ is the unit vector from m1 to m2 and r = r2 – r1 2
r21 2
r31 2
r41
as shown in Fig. 7.3.
⊳ Example 7.2 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
O vertex A is doubled ?
Take AG = BG = CG = 1 m (see Fig. 7.5)
Answer (a) The angle between GC and the
positive x-axis is 30° and so is the angle between
Fig. 7.3 Gravitational force on m1 due to m2 is along GB and the negative x-axis. The individual forces
r where the vector r is (r2– r1). in vector notation are

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GRAVITATION 131

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
Fig. 7.5 Three equal masses are placed at the three prependicular to this line cancel out when
vertices of the ∆ ABC. A mass 2m is placed
summing over all regions of the shell leaving
at the centroid G.
only a resultant force along the line joining
the point to the centre. The magnitude of
Gm (2m ) ˆ
FGA = j this force works out to be as stated above.
1 (2) The force of attraction due to a hollow
Gm (2m ) ˆ spherical shell of uniform density, on a
FGB =
1
(
−i cos 30ο − ˆj sin 30ο ) point mass situated inside it is zero.
Qualitatively, we can again understand this
Gm (2m ) ˆ
FGC =
1
(
+ i cos 30ο − ˆj sin 30ο ) result. Various regions of the spherical shell
attract the point mass inside it in various
From the principle of superposition and the law directions. These forces cancel each other
of vector addition, the resultant gravitational completely.
force FR on (2m) is
FR = FGA + FGB + FGC 7.4 THE GRAVITATIONAL CONSTANT
(
FR = 2Gm 2 ˆj + 2Gm 2 −ˆi cos 30ο −ˆj sin 30ο ) The value of the gravitational constant G entering
the Universal law of gravitation can be
( )
+ 2Gm 2 ˆi cos 30ο − ˆj sin 30ο = 0 determined experimentally and this was first done
Alternatively, one expects on the basis of by English scientist Henry Cavendish in 1798.
symmetry that the resultant force ought to be The apparatus used by him is schematically
zero. shown in Fig.7.6
(b) Now if the mass at vertex A is doubled
then

For the gravitational force between an extended Fig. 7.6 Schematic drawing of Cavendish’s
object (like the earth) and a point mass, Eq. (7.5) is not experiment. S1 and S2 are large spheres
directly applicable. Each point mass in the extended which are kept on either side (shown
object will exert a force on the given point mass and shades) of the masses at A and B. When
the big spheres are taken to the other side
these force will not all be in the same direction. We
of the masses (shown by dotted circles),
have to add up these forces vectorially for all the point the bar AB rotates a little since the torque
masses in the extended object to get the total force. reverses direction. The angle of rotation can
This is easily done using calculus. For two special be measured experimentally.

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132 PHYSICS

The bar AB has two small lead spheres all the shells exert a gravitational force at the
attached at its ends. The bar is suspended from point outside just as if their masses are
a rigid support by a fine wire. Two large lead concentrated at their common centre according
spheres are brought close to the small ones but to the result stated in section 7.3. The total mass
on opposite sides as shown. The big spheres of all the shells combined is just the mass of the
attract the nearby small ones by equal and earth. Hence, at a point outside the earth, the
opposite force as shown. There is no net force gravitational force is just as if its entire mass of
on the bar but only a torque which is clearly the earth is concentrated at its centre.
equal to F times the length of the bar,where F is For a point inside the earth, the situation
the force of attraction between a big sphere and is different. This is illustrated in Fig. 7.7.
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 Mr
concentrated at their centres. Thus if d is the
separation between the centres of the big and Fig. 7.7 The mass m is in a mine located at a depth
its neighbouring small ball, M and m their d below the surface of the Earth of mass
masses, the gravitational force between the big ME and radius RE. We treat the Earth to be
sphere and its neighouring small ball is. spherically symmetric.
Mm Again consider the earth to be made up of
F =G (7.6)
d2 concentric shells as before and a point mass m
If L is the length of the bar AB , then the situated at a distance r from the centre. The
torque arising out of F is F multiplied by L. At point P lies outside the sphere of radius r. For
equilibrium, this is equal to the restoring torque the shells of radius greater than r, the point P
and hence lies inside. Hence according to result stated in
the last section, they exert no gravitational force
Mm on mass m kept at P. The shells with radius ≤ r
G L =τ θ (7.7)
d2 make up a sphere of radius r for which the point
Observation of θ thus enables one to P lies on the surface. This smaller sphere
calculate G from this equation. therefore exerts a force on a mass m at P as if
Since Cavendish’s experiment, the its mass Mr is concentrated at the centre. Thus
measurement of G has been refined and the the force on the mass m at P has a magnitude
currently accepted value is Gm (M r )
G = 6.67×10-11 N m2/kg2 (7.8) F = (7.9)
r2
We assume that the entire earth is of uniform
7.5 ACCELERATION DUE TO GRAVITY OF
THE EARTH 4π 3
density and hence its mass is M E = RE ρ
3
The earth can be imagined to be a sphere made
where ME is the mass of the earth RE is its radius
of a large number of concentric spherical shells
and ρ is the density. On the other hand the
with the smallest one at the centre and the
largest one at its surface. A point outside the 4π
mass of the sphere Mr of radius r is ρ r 3 and
earth is obviously outside all the shells. Thus, 3

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GRAVITATION 133

hence 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
G m ME Eq. (7.5) :
= r (7.10)
RE 3
If the mass m is situated on the surface of GM E m
earth, then r = RE and the gravitational force on F (h ) = (7.13)
(R E + h )2
it is, from Eq. (7.10)
M Em The acceleration experienced by the point
F =G (7.11) mass is F (h )/ m ≡ g(h ) and we get
R E2
The acceleration experienced by the mass F (h ) GM E
g(h ) = = . (7.14)
m, which is usually denoted by the symbol g is m (R E + h )2
related to F by Newton’s 2nd law by relation
This is clearly less than the value of g on the
F = mg. Thus
GM E
F GM E surface of earth : g = R 2 . For h << RE , we can
g= = (7.12) E
m R E2 expand the RHS of Eq. (7.14) :
Acceleration g is readily measurable. RE is a GM E
= g (1 + h / R E )
−2
g(h ) = 2
known quantity. The measurement of G by R E (1 + h / R E )2

Cavendish’s experiment (or otherwise), combined h

with knowledge of g and RE enables one to For R <<1 , using binomial expression,
E
estimate ME from Eq. (7.12). This is the reason
 2h 
why there is a popular statement regarding g (h ) ≅ g 1 −
R E 
. (7.15)
Cavendish : “Cavendish weighed the earth”. 

7.6 ACCELERATION DUE TO GRAVITY BELOW Equation (7.15) thus tells us that for small
AND ABOVE THE SURFACE OF EARTH heights h above the value of g decreases by a
factor (1 − 2h / RE ).
Consider a point mass m at a height h above the
surface of the earth as shown in Fig. 7.8(a). The Now, consider a point mass m at a depth
radius of the earth is denoted by RE . Since this d below the surface of the earth (Fig. 7.8(b)),
point is outside the earth, so that its distance from the centre of the
earth is ( RE − 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 Ms is the mass of the smaller
sphere, then,
Ms/ME = ( RE – d)3 / RE3 ( 7.16)
Since mass of a sphere is proportional to be
Fig. 7.8 (a) g at a height h above the surface of the cube of its radius.
earth.

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134 PHYSICS

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
Ms ME point at a height h1 from the surface of the earth
and another point vertically above it at a height
h2 from the surface, the work done in lifting the
particle of mass m from the first to the second
position is denoted by W12
W12 = Force × displacement
Fig. 7.8 (b) g at a depth d. In this case only the smaller
sphere of radius (RE–d) contributes to g. = mg (h2 – h1) (7.20)
Thus the force on the point mass is If we associate a potential energy W(h) at a
point at a height h above the surface such that
2
F (d) = G Ms m / (RE – d ) (7.17)
W(h) = mgh + Wo (7.21)
Substituting for Ms from above , we get
(where Wo = constant) ;
F (d) = G ME m ( RE – d ) / RE3 (7.18) then it is clear that
and hence the acceleration due to gravity at W12 = W(h2) – W(h1) (7.22)
a depth d, The work done in moving the particle is just
the difference of potential energy between its
F (d )
g(d) = is final and initial positions.Observe that the
m constant Wo cancels out in Eq. (7.22). Setting h
F (d ) GM E = 0 in the last equation, we get W ( h = 0 ) = Wo.
g (d ) = = (RE − d ) . h = 0 means points on the surface of the earth.
m R E3 Thus, Wo is the potential energy on the surface
RE − d of the earth.
=g = g(1 − d / R E ) (7.19) If we consider points at arbitrary distance
RE
from the surface of the earth, the result just
Thus, as we go down below earth’s surface, derived is not valid since the assumption that
the acceleration due gravity decreases by a factor the gravitational force mg is a constant is no
(1 − d / RE ). The remarkable thing about longer valid. However, from our discussion we
know that a point outside the earth, the force of
acceleration due to earth’s gravity is that it is
gravitation on a particle directed towards the
maximum on its surface decreasing whether you
centre of the earth is
go up or down.
G ME m
F= (7.23)
7.7 GRAVITATIONAL POTENTIAL ENERGY r2
We had discussed earlier the notion of potential where ME = mass of earth, m = mass of the
energy as being the energy stored in the body at particle and r its distance from the centre of the
its given position. If the position of the particle earth. If we now calculate the work done in
lifting a particle from r = r1 to r = r2 (r2 > r1) along
changes on account of forces acting on it, then
a vertical path, we get instead of Eq. (7.20)
the change in its potential energy is just the
amount of work done on the body by the force. r2 GMm
W12 = ∫ dr
As we had discussed earlier, forces for which the r1 r2
work done is independent of the path are the
conservative forces.  1 1
= −G ME m  −  (7.24)
The force of gravity is a conservative force  r2 r1 
and we can calculate the potential energy of a In place of Eq. (7.21), we can thus associate
body arising out of this force, called the a potential energy W(r) at a distance r, such that
gravitational potential energy. Consider points

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GRAVITATION 135

G ME m
W (r ) = − + W1 , (7.25)
r
valid for r > R ,
so that once again W12 = W(r 2 ) – W(r 1).
Setting r = infinity in the last equation, we get
W ( r = infinity ) = W1 . Thus, W1 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. (7.22) and
(7.24). One conventionally sets W1 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 Fig. 7.9
on it due to the earth and it is proportional to
the mass of the particle. The gravitational The gravitational potential at the centre of
potential due to the gravitational force of the
earth is defined as the potential energy of a
the square (r = 2 l/2 is)
particle of unit mass at that point. From the Gm
U (r ) = − 4 2 . ⊳
earlier discussion, we learn that the gravitational l
potential energy associated with two particles
of masses m1 and m2 separated by distance by a 7.8 ESCAPE SPEED
distance r is given by
If a stone is thrown by hand, we see it falls back
Gm1m 2 to the earth. Of course using machines we can
V =– (if we choose V = 0 as r → ∞ )
r shoot an object with much greater speeds and
It should be noted that an isolated system of with greater and greater initial speed, the object
particles will have the total potential energy that scales higher and higher heights. A natural
equals the sum of energies (given by the above query that arises in our mind is the following:
equation) for all possible pairs of its constituent ‘can we throw an object with such high initial
particles. This is an example of the application speeds that it does not fall back to the earth?’
of the superposition principle. The principle of conservation of energy helps
us to answer this question. Suppose the object
⊳ Example 7.3 Find the potential energy of did reach infinity and that its speed there was
a system of four particles placed at the Vf . The energy of an object is the sum of potential
vertices of a square of side l. Also obtain and kinetic energy. As before W1 denotes that
the potential at the centre of the square. gravitational potential energy of the object at
infinity. The total energy of the projectile at
infinity then is
Answer Consider four masses each of mass m
at the corners of a square of side l ; See Fig. 7.9.
We have four mass pairs at distance l and two mV f2
E ( ∞) = W1 + (7.26)
diagonal pairs at distance 2 l
2
Hence, If the object was thrown initially with a speed
Vi from a point at a distance (h+RE) from the
G m2 G m2 centre of the earth (RE = radius of the earth), its
W (r ) = − 4 −2
l 2l energy initially was
1 GmM E
2G m2  1  G m2 E (h + R E ) = mVi2 – + W1 (7.27)
=−  2 +  = − 5.41 2 (h + R E )
l  2 l

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136 PHYSICS

By the principle of energy conservation ⊳

Eqs. (7.26) and (7.27) must be equal. Hence Example 7.4 Two uniform solid spheres
2 of equal radii R, but mass M and 4 M have
mVi2 GmM E mV
= a centre to centre separation 6 R, as shown
f
– (7.28)
2 (h + R E ) 2 in Fig. 7.10. The two spheres are held fixed.
The R.H.S. is a positive quantity with a A projectile of mass m is projected from the
minimum value zero hence so must be the L.H.S. surface of the sphere of mass M directly
Thus, an object can reach infinity as long as Vi towards the centre of the second sphere.
is such that Obtain an expression for the minimum
speed v of the projectile so that it reaches
mVi2 GmM E the surface of the second sphere.
– ≥0 (7.29)
2 (h + R E )

The minimum value of Vi corresponds to the

case when the L.H.S. of Eq. (7.29) equals zero.
Thus, the minimum speed required for an object
to reach infinity (i.e. escape from the earth)
corresponds to Fig. 7.10

Answer The projectile is acted upon by two

1
2
( )
m Vi2
min
=
GmM E
h + RE (7.30) mutually opposing gravitational forces of the two
spheres. The neutral point N (see Fig. 7.10) is
defined as the position where the two forces
If the object is thrown from the surface of cancel each other exactly. If ON = r, we have
the earth, h = 0, and we get
GMm 4G M m
=
2GM E r2 (6 R −r )2
(Vi )min = (7.31) (6R – r)2 = 4r2
RE
6R – r = ±2r
r = 2R or – 6R.
The neutral point r = – 6R does not concern
Using the relation g = GM E / RE2 , we get us in this example. Thus ON = r = 2R. It is
sufficient to project the particle with a speed
(Vi )min = 2gR E (7.32) which would enable it to reach N. Thereafter,
the greater gravitational pull of 4M would
suffice. The mechanical energy at the surface
Using the value of g and RE, numerically of M is
(Vi) min≈11.2 km/s. This is called the escape
1 GMm 4G M m
speed, sometimes loosely called the escape Ei = m v2 − − .
velocity. 2 R 5R
Equation (7.32) applies equally well to an At the neutral point N, the speed approaches
object thrown from the surface of the moon with zero. The mechanical energy at N is purely
g replaced by the acceleration due to Moon’s potential.
gravity on its surface and rE replaced by the G M m 4G M m
EN = − − .
radius of the moon. Both are smaller than their 2R 4R
values on earth and the escape speed for the From the principle of conservation of
moon turns out to be 2.3 km/s, about five times mechanical energy
smaller. This is the reason that moon has no
atmosphere. Gas molecules if formed on the
surface of the moon having velocities larger than 1 2 GM 4GM GM GM
v − − =− −
this will escape the gravitational pull of the 2 R 5R 2R R
moon. or

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GRAVITATION 137

2G M  4 1  traverses a distance 2π(RE + h) with speed V. Its

v2 =  −  time period T therefore is
R  5 2
2π ( R E + h ) 2π (R E + h )3 / 2
 3G M 
1/2 T = = (7.37)
v =  ⊳ V G ME
 5R  on substitution of value of V from Eq. (7.35).
A point to note is that the speed of the projectile Squaring both sides of Eq. (7.37), we get
is zero at N, but is nonzero when it strikes the
T 2 = k ( RE + h)3 (where k = 4 π2 / GME) (7.38)
heavier sphere 4 M. The calculation of this speed
is left as an exercise to the students. which is Kepler’s law of periods, as applied to
motion of satellites around the earth. For a
7.9 EARTH SATELLITES satellite very close to the surface of earth h can
be neglected in comparison to RE in Eq. (7.38).
Earth satellites are objects which revolve around Hence, for such satellites, T is To, where
the earth. Their motion is very similar to the
motion of planets around the Sun and hence T0 = 2π R E / g (7.39)
Kepler’s laws of planetary motion are equally If we substitute the numerical values
applicable to them. In particular, their orbits g ≃ 9.8 m s-2 and RE = 6400 km., we get
around the earth are circular or elliptic. Moon is
the only natural satellite of the earth with a near 6.4 × 106
T0 = 2π s
circular orbit with a time period of approximately 9.8
27.3 days which is also roughly equal to the Which is approximately 85 minutes.
rotational period of the moon about its own axis.
⊳ Example 7.5 The planet Mars has two
Since, 1957, advances in technology have enabled
many countries including India to launch artificial moons, phobos and delmos. (i) phobos has
earth satellites for practical use in fields like a period 7 hours, 39 minutes and an orbital
telecommunication, geophysics and meteorology. radius of 9.4 ×103 km. Calculate the mass
We will consider a satellite in a circular orbit of mars. (ii) Assume that earth and mars
of a distance (RE + h ) from the centre of the earth, move in circular orbits around the sun,
where RE = radius of the earth. If m is the mass with the martian orbit being 1.52 times
of the satellite and V its speed, the centripetal the orbital radius of the earth. What is
force required for this orbit is the length of the martian year in days ?

mV 2 Answer (i) We employ Eq. (7.38) with the sun’s

F(centripetal) = (7.33)
(RE + h ) mass replaced by the martian mass Mm
2
directed towards the centre. This centripetal force 2 4π 3
is provided by the gravitational force, which is T = R
GM m
2 3
G m ME 4π R
F(gravitation) = ( R + h )2 (7.34) Mm = 2
E G T
where ME is the mass of the earth.
4 × ( 3.14) × ( 9.4) × 10
2 3 18
Equating R.H.S of Eqs. (7.33) and (7.34) and
=
cancelling out m, we get × ( 459 × 60)
-11 2
6.67 × 10
G ME
V2 = (7.35) 4 × ( 3.14 ) × ( 9.4 ) × 10
2 3 18
(RE + h ) Mm =
6.67 × ( 4.59 × 6) × 10
2 -5
Thus V decreases as h increases. From = 6.48 × 1023 kg.
equation (7.35),the speed V for h = 0 is (ii) Once again Kepler’s third law comes to our
V 2 (h = 0) = GM / R E = gRE (7.36) aid,
2 3
where we have used the relation TM R MS
2
= 3
g = GM / RE 2 . In every orbit, the satellite TE R ES

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138 PHYSICS

where RMS is the mars -sun distance and RES is  1  1 

−13 2
the earth-sun distance. = 10  d  
 ( 24 × 60 × 60) 2   (1 / 1000) 3 km3 
∴ TM = (1.52)3/2 × 365
= 1.33 ×10–14 d2 km–3
= 684 days
Using Eq. (7.38) and the given value of k,
We note that the orbits of all planets except
the time period of the moon is
Mercury and Mars are very close to being
T 2 = (1.33 × 10-14)(3.84 × 105)3
circular. For example, the ratio of the semi-
T = 27.3 d ⊳
minor to semi-major axis for our Earth is,
Note that Eq. (7.38) also holds for elliptical
b/a = 0.99986. ⊳
orbits if we replace (RE+h) by the semi-major axis
⊳ Example 7.6 Weighing the Earth : You of the ellipse. The earth will then be at one of
the foci of this ellipse.
are given the following data: g = 9.81 ms–2,
RE = 6.37×106 m, the distance to the moon R
7.10 ENERGY OF AN ORBITING SATELLITE
= 3.84×108 m and the time period of the
moon’s revolution is 27.3 days. Obtain the Using Eq. (7.35), the kinetic energy of the satellite
mass of the Earth ME in two different ways. in a circular orbit with speed v is
1
Answer From Eq. (7.12) we have K iE = m v2
2
2
g RE
ME = Gm M E
G =
2(RE + h ) , (7.40)

=
(
9.81 × 6.37 × 10 )
6 2 Considering gravitational potential energy at
infinity to be zero, the potential energy at distance
-11
6.67 × 10 (R +h) from the centre of the earth is
= 5.97× 1024 kg. e

The moon is a satellite of the Earth. From G m ME

P .E = − (7.41)
the derivation of Kepler’s third law [see Eq. (R E + h )
(7.38)]
The K.E is positive whereas the P.E is
2 3
4π R negative. However, in magnitude the K.E. is half
T2 =
G ME the P.E, so that the total E is

4π2R 3
G m ME
ME = E = K .E + P .E = − (7.42)
GT 2 2(R E + h )
4 × 3.14 × 3.14 × ( 3.84 ) × 10 The total energy of an circularly orbiting
3 24
= satellite is thus negative, with the potential
× ( 27.3 × 24 × 60 × 60)
-11 2
6.67 × 10 energy being negative but twice is magnitude of
24
= 6.02 × 10 kg the positive kinetic energy.
Both methods yield almost the same answer, When the orbit of a satellite becomes
the difference between them being less than 1%. 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
Example 7.7 Express the constant k of Eq.
(7.38) in days and kilometres. Given discussed before if the total energy is positive or
k = 10–13 s2 m–3. The moon is at a distance zero, the object escapes to infinity. Satellites
of 3.84 × 105 km from the earth. Obtain its are always at finite distance from the earth and
time-period of revolution in days. hence their energies cannot be positive or zero.

Answer Given
k = 10–13 s2 m–3

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GRAVITATION 139

The change in the total energy is

⊳ Example 7.8 A 400 kg satellite is in a circular
∆E = Ef – Ei
orbit of radius 2RE about the Earth. How much
energy is required to transfer it to a circular
orbit of radius 4RE ? What are the changes in
G M E m  G M E m R E

the kinetic and potential energies ? = =
8 RE  R2  8
 E 
Answer Initially, g m RE 9.81 × 400 × 6.37 × 106
∆E = = = 3.13 × 109 J
G ME m 8 8
Ei = −
4 RE The kinetic energy is reduced and it mimics
While finally ∆E, namely, ∆K = Kf – Ki = – 3.13 × 109 J.
G ME m
The change in potential energy is twice the
Ef = − change in the total energy, namely
8 RE
∆V = Vf – Vi = – 6.25 × 109 J ⊳

SUMMARY

1. Newton’s law of universal gravitation states that the gravitational force of attraction between
any two particles of masses m1 and m2 separated by a distance r has the magnitude
m 1m 2
F =G 2
r
where G is the universal gravitational constant, which has the value 6.672 ×10–11 N m2 kg–2.
2. If we have to find the resultant gravitational force acting on the particle m due to a number of
masses M1, M2, ….Mn etc. we use the principle of superposition. Let F1, F2, ….Fn be the individual
forces due to M1, M2, ….Mn, 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 FR is
then found by vector addition
n
FR = F1 + F2 + ……+ Fn = ∑ Fi
i =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
 4 π2  3
T2 =  R
G M 
 s

where Ms 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
G ME
g(h ) =
(R E + h)
2

G ME  2h 
≈ 1 − R  for h << RE
R E2 E

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140 PHYSICS

 2h  G ME
g(h ) = g (0) 1 − where g (0) =
 R E  R E2
(b) at depth d below the earth’s surface is
G ME  d   d 
g (d ) = 1 − R  = g (0 ) 1 − R 
R E2 E E

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
G m1 m 2
V =−
r
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
1 GMm
E= m v2−
2 r
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
GMm
E =−
2a
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
GMm
K=
2a
GMm
V =−
a
8. The escape speed from the surface of the earth is
2 G ME
ve = = 2 gRE
RE
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.

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GRAVITATION 141

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. (7.1) and T2 = KS R3. The constant KS is the same for all
planets in circular orbits. This applies to satellites orbiting the Earth [(Eq. (7.38)].
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
G m1 m 2
V =– + constant
r
The constant can be given any value. The simplest choice is to take it to be zero. With
this choice
G m1 m 2
V =–
r
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
7.1 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 ?

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142 PHYSICS

7.2 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(r2 – r1) for the difference of potential energy between two points r2 and r1 distance
away from the centre of the earth.
7.3 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 ?
7.4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius
of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth
that of the sun.
7.5 Let us assume that our galaxy consists of 2.5 × 1011 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 105 ly.
7.6 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.
7.7 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?
7.8 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.
7.9 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen
feet, (b) swollen face, (c) headache, (d) orientational problem.
7.10 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 7.11) (i) a, (ii) b, (iii) c, (iv) 0.

Fig. 7.11

7.11 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.
7.12 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 = 2×1030 kg,
mass of the earth = 6×1024 kg. Neglect the effect of other planets etc. (orbital radius
= 1.5 × 1011 m).
7.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of
the earth around the sun is 1.5 × 108 km.
7.14 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the
earth is 1.50 × 108 km away from the sun ?
7.15 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 ?

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GRAVITATION 143

7.16 Assuming the earth to be a sphere of uniform mass density, how much would a
body weigh half way down to the centre of the earth if it weighed 250 N on the
surface ?
7.17 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 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2.
7.18 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.
7.19 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 = 6.0×1024 kg; radius of
the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2.
7.20 Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a
head on collision. When they are a distance 109 km, their speeds are negligible.
What is the speed with which they collide ? The radius of each star is 104 km.
Assume the stars to remain undistorted until they collide. (Use the known value
of G).
7.21 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 ?

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