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 Solution. Both forces will be equal in magnitude as
gravitational force between two bodies is a mutual force.

Gravitation Problem 9. Earth is continuously pulling the moon

towards its centre, still it does not fall to the earth.
Reasoning Questions Why?
Solution. Gravitational force of attraction due to earth
provides the centripetal force, which keeps the moon in
Very Short Questions orbit around the earth. Moreover, this gravitational force
acts perpendicular to the velocity of the moon.

Problem 1. Why is the law of gravitation called the Problem 10. We cannot move finger without
universal law ? disturbing all stars. Why ?
Solution. This is because of the fact that the law of Solution. When we move our finger, the distance between
gravitation holds good for any pair of bodies in the the objects and our finger changes. Hence, the force of
universe whether microscopically small or astronomically attraction changes, disturbing the entire universe,
large in size. including the stars.

Problem 2. Why is 𝑮 called the universal gravitational Problem 11. According to Newton's law of gravi.
constant? tation, the apple and the earth experience equal and
Solution. This is because the value of G is same for any

h
opposite forces due to gravitation. But it is the apple
pair of the bodies in the universe. It does not depend on that falls towards the earth and not vice-versa. Why?
the nature of the medium between the two bodies or on Solution. According to Newton's third law of motion, the

es
the nature of the bodies themselves. force with which the earth is attracted towards the apple is
equal to the force with which earth attracts the apple.
Problem 3. What is the ratio of the force of attraction However, the mass of the earth is extremely large as
between two bodies kept in air and the same distance compared to that of apple. So acceleration of the earth is
apart in water ? very small and is not noticeable.
ak
Solution. 1:1, because the gravitational force does not
depend on the nature of the medium. Problem 12. According to Newton's law of gravitation,
Problem 4. If spheres of same material and same radius 𝑟 every particle of matter attracts every other particle.
are touching each other, then show that the gravitational But bodies on the surface of earth never move
force between them is directly proportional to 𝑟 . towards each other on account of this force of
.R

Solution. If 𝜌 is the density of the material, then attraction. Why ?

4 Solution. Because of their small masses, the gravitaional
𝑚 𝑚 3𝜋𝑟 𝜌 × 𝜋𝑟 𝜌 4
𝐹=𝐺 =𝐺 3 = 𝜋 𝜌 𝐺𝑟 attraction between two bodies is too small to produce any
(2𝑟) (2𝑟) 9 motion in them. But due to the large mass of the earth, the
Clearly, 𝐹∝𝑟 . gravitational attraction between the bodies and earth is
very large and so all bodies are attracted towards the
Er

Problem 5. If the density of planet is doubled without centre of the earth.

any change in its radius, how does 𝒈 change on the
planet? Problem 13. Does the gravitational force of attraction
Solution. It gets doubled, because 𝑔 ∝ 𝜌. of the earth become zero at some height above the
earth ? Give reason.
Problem 6. Is it possible to shield a body from Solution. No, the gravitational force of attraction of earth
y

gravitational effects ? on a body at height ℎ is

Solution. No, it is not possible to shield a body from 𝑀𝑚
B

gravitational effects because gravitational interaction does 𝐹=𝐺

(𝑅 + ℎ)
not depend upon the nature of the intervening medium. Clearly, 𝐹 will be zero only when ℎ is infinity.
Problem 7. If the force of gravity acts on all bodies in Problem 14. Which is more fundamental-mass or
proportion to their masses, why does a heavy body weight of a body?
not fall faster than a light body ? Solution. Mass is more fundamental than weight because
[Delhi 12] the mass of a body remains constant while its weight
Solution. If 𝐹 be the gravitational force on a body of mass changes from place to place due to change in the value of
𝑚, then 𝑔.
𝑀𝑚 𝐺𝑀
𝐹=𝐺 = 𝑚𝑔 or 𝑔 =
𝑅 𝑅 Problem 15. If the diameter of the earth becomes twice
Clearly, 𝐹 ∝ 𝑚 but 𝑔 does not depend on 𝑚. Hence all its present value but its mass remains unchanged,
bodies fall with same rapidness if there is no air resistance. then how would be the weight of an object on the
surface of the earth affected?
Problem 8 . The mass of the moon is nearly 𝟏𝟎% of the Solution. Weight of body,
mass of the earth. What will be the gravitational force 𝐺𝑀𝑚
of the earth on the moon, in comparison to the 𝑊 = 𝑚𝑔 =
𝑅
gravitational force of the moon on the earth ? When the diameter or radius of the earth becomes double
its present value, the weight of the body will be
 𝐺𝑀𝑚 1 Problem 22. Why a body weighs more at the poles
𝑊 =
= 𝑊
(2𝑅) 4 than at the equator?
i.e., weight will become one-fourth of the present value. Solution. At the poles 𝑔 > 𝑔 , hence, 𝑚𝑔 > 𝑚𝑔 .

Problem 16. If the diameter of the earth becomes half Problem 23. Where does a body weigh more ? At the
its present value but its average density remains sea level or on the mountains?
unchanged then how would be the weight of an object Solution. At the sea level because weight decreases with
on the surface of the earth affected ? altitude.
Solution. Acceleration due to gravity on the surface of the
earth Problem 24. Where will a body weigh more - 𝟏 𝐤𝐦
𝐺𝑀 𝐺 4 4 above the surface of earth or 𝟏 𝐤𝐦 below the surface
𝑔= = × 𝜋𝑅 𝜌 = 𝜋𝐺𝑅𝑝
𝑅 𝑅 3 3 of earth?
When the diameter or radius becomes half its present Solution. As 𝑔 = 𝑔 1 − and 𝑔 = 𝑔 1 −
value,
4 𝑅 𝑔 For ℎ = 𝑑 = 1 km, clearly 𝑔 > 𝑔 .
𝑔 = 𝜋𝐺 𝜌= . Hence the body will weigh more at a depth of 1 km below
3 2 2
Hence the weight of the object will be halved. the surface of earth.

Problem 17. The mass and diameter of a planet are Problem 25. Does the concentration of the earth's
twice those of the earth. What will be the time-period mass near its centre change the variation of 𝐠 with
height compared with a homogeneous sphere ? How?

h
of that pendulum on this planet, which is a second's
pendulum on the earth? Solution. Any change in the distribution of the earth's mass
will not affect the variation of acceleration due to gravity
Solution. Initially, 𝑔 = and 𝑇 = 2𝜋

es
=2s with height. This is because for a point outside the earth,
(For a second's pendulum) the whole mass of the earth is 'effective' and the earth
When 𝑀 and 𝑅 are doubled, behaves as a homogeneous sphere.
( )
𝑔 = = and 𝑇 = 2𝜋 = √2𝑇 = 2√2 s. Problem 26. The weight of a body is less inside the
( ) /
ak earth than on the surface. Why ?
Solution. As we go inside the earth, the value of the
Problem 18. If the radii of two planets be 𝑹𝟏 and 𝑹𝟐 and
attracting mass 𝑀 decreases and hence the value of 8
their mean densities be 𝝆𝟏 and 𝝆𝟐 , then prove that the
decreases. Therefore, the weight of the body 𝑚𝑔 is less
ratio of accelerations due to gravity on the planets will
inside the earth than on the surface.
be 𝑹𝟏 𝝆𝟏 : 𝑹𝟐 𝝆𝟐 .
.R

Solution. On the surface of any planet, 𝑔 = 𝜋𝐺𝑅𝜌 Problem 27. Why do you feel giddy while moving on a
𝑔 4 merry-go-round ?
∴ = = 𝑅 𝜌 :𝑅 𝜌 .
𝑔 3 Solution. When moving in a merry-go-round, our weight
𝜋𝐺𝑅 𝜌
4 appears to decrease when we move down and appears to
increase when we move up.
Er

Problem 19. The distance between two bodies 𝑨 and

𝑩 is 𝒓. Taking the gravitational force according to the Problem 28. At which place on earth's surface, the
law of inverse square of 𝒓, the acceleration of the body value of 𝒈 is largest and why ?
is 𝒂. If the gravitational force follows an inverse fourth Solution. The value of 𝑔 is largest at the poles due to
power law, then what would be the acceleration of the following two reasons :
body 𝑨 ? (a) Distance of poles from the centre of earth is smaller
y

Solution. For inverse square law, 𝐹 = than the distance of any other point on earth's surface from
its centre.
∴ Acceleration of body 𝐴 = = =𝑎
B

(b) At poles, no centrifugal force acts on the body.

For inverse fourth power law, 𝐹 =
Problem 29. When dropped from the same height a
∴ Acceleration of body 𝐴 = = = . body reaches the ground quicker at poles than at the
equator. Why ?
Problem 20. Why is centre of mass of a body often Solution. The acceleration due to gravity is more at the
called its centre of gravity ? poles than at the equator. When the initial velocities and
Solution. The torque due to gravity on a body acts as if its distances travelled are the same, the time taken by the
entire mass were concentrated at its centre of mass. That body is smaller if the acceleration due to gravity is large.
is why centre of mass of a body is often called its centre Hence, when dropped from the same height, a body
of gravity. reaches the ground quicker at the poles than at the
equator.
Problem 21. Why the value of 𝒈 is more at the poles
than at the equator? Problem 30. When a clock controlled by a pendulum
Solution. As 𝑔 = and the value of 𝑅 at equator is is taken from the plains to a mountain, it becomes
slow but a wrist-watch controlled by a spring remains
greater than that at poles, hence 𝑔 at poles is greater than unaffected. Explain the reason for the difference in the
𝑔 at equator. behaviour of the two watches.
Solution. Due to decrease in the value of 𝑔 at the Problem 39. Explain why tidal waves (high tide and
mountain, the time period of the pendulum of the clock low tide) are formed on seas.
increases. On the other hand, the spring watch remains Solution. The gravitational attraction of moon on sea water
unaffected by the variation in 𝑔. causes high tides. Tides at one place cause low tides
(ebbs) at another. Attraction by sun also causes tides but
Problem 31. A clock fitted with a pendulum and only of half the magnitude. Hence on new moon and full
another with a spring indicate correct time on earth. moon days, when both effects add, tides are very high.
Which shows correct time on the moon?
Solution. A clock fitted with a spring will show correct time Problem 40 . Why are we not thrown off the surface of
on the moon, because its time period is not affected by the the earth by the centrifugal force?
variation in 𝑔. Solution. The force of gravity exerted on our body by the
earth towards its centre is greater than centrifugal force
Problem 32 . Why does a tennis ball bounce higher on acting an our body away from the centre.
a hill than on plains ?
Solution. The value of 𝑔 is less on hills because they are Problem 41. A satellite does not need any fuel to circle
comparatively at a greater distance from the centre of the around the earth. Why ?
earth. Therefore, the gravitational pull on the tennis ball is Solution. The gravitational force between satellite and
less on hill tops and so it bounces higher on hills than on earth provides the centripetal force required by the
plains. satellite to move in a circular orbit.

Problem 42 . Why a multi-stage rocket is required to

h
Problem 33. A man can jump six times as high on the
moon as that on the earth. Justify. launch a satellite ?
Or Solution. The multistage rocket helps to economise the

es
Explain, why one can jump higher on the surface of consumption of fuel.
the moon than that on the earth.
Solution. The value of ' g ' on the surface of moon is about Problem 43. Why do different planets have different
th of its value on the surface of earth. escape velocities?
For the same gain of P.E. in both cases, we have Solution. As 𝑣 = , therefore escape velocities have
mg h = mg h
ak
different values on different planets which are of different
or
g h g h masses and different sizes.
h = = = 6h .
g g /6 Problem 44. An elephant and an ant are to be
.R

projected out of earth into space. Do we need different

Problem 34. Moon-travellers tie heavy weight at their velocities to do so?
back before landing on the moon. Why ?
Solution. The escape velocity, 𝑣 = 2𝑔𝑅, does not
Solution. Due to the small value of g on the moon, the
traveller would weigh less. To compensate for this loss in depend upon the mass of the projected body. Thus we
weight, the travellers load their backs with heavy weight. need the same velocity to project an elephant and an ant
Er

into space.
Problem 35. Why is earth flat at the poles ?
Solution. Due to its rotation about the polar axis. Problem 45. Does a rocket really need the escape
velocity of 𝟏𝟏. 𝟐𝐤𝐦𝐬 𝟏 initially to escape from the
Problem 36. What is the effect of rotation of the earth earth?
on the acceleration due to gravity ? Solution. No, rocket can have any velocity at the start. The
y

Solution. The acceleration due to gravity decreases due to rocket can continue to increase the velocity due to thrust
rotation of the earth. This effect is zero at the poles and provided by the escaping gases that will carry it to a
desired position.
B

maximum at the equator.

Problem 37. Name two factors which determine Problem 46. The moon has no atmosphere. Why ?
whether a planet would have an atmosphere or not. Solution. Due to the small value of ' 𝑔 ', the escape velocity
Solution. The two such factors are on the moon surface is small (2.38kms ). The air
(i) The value of acceleration due to gravity on the planet. molecules have thermal velocities greater than the escape
(ii) Surface temperature of the planet. velocity. Therefore, the air molecules escape away and
cannot form atmosphere on the moon.
Problem 38. If the earth stops rotating about its axis,
what will be the effect on the value of 𝒈 ? Will this Problem 47. The earth has atmosphere. Why ?
effect be same at all places ? Solution. Because the r.m.s. velocity of air molecules is
Solution. The value of 𝑔 will increase at all places except less than escape velocity on the earth, the air cannot
at the poles. The increase will be different at different escape from the surface of the earth. Hence the earth has
places, maximum at the equator. atmosphere.

Problem 38(a). If the earth stops rotating about its Problem 48. Lighter gases like 𝐇𝟐 , 𝐇𝐞, etc. are rare in
axis, then by what value will the acceleration due to the atmosphere of the earth. Why ?
gravity change at the equator? Solution. Usually the average velocity of the molecules of
Solution. The value of 𝑔 increases by 𝜔 𝑅. lighter gases is greater than the escape velocity at the
earth's surface. So lighter gases have escaped from the Solution. No. The gravitational acceleration of the
earth's atmosphere. astronaut relative to the satellite is zero, whatever be the
distance of the satellite from the earth.
Problem 49. The gravitational force exerted by the sun
on the moon is greater than (about twice as great as) Problem 57. Can we determine the gravitational mass
the gravitational force exerted by the earth on the of a body inside an artificial satellite?
moon. Why then doesn't the moon escape from the Solution. No, because both the body and the artificial
earth (during a solar eclipse, for example)? satellite are in a state of free fall.
Solution. The moon can escape only if the moon has no
orbital motion. In fact, while revolving around the earth, the Problem 58. A satellite is orbiting the earth with speed
moon has orbital motion around the sun also. The 𝒗𝟎 . To make the satellite escape, what should be the
gravitational attraction of the sun on the moon provides the minimum percentage increase in its velocity?
centripetal force required for the orbital motion around the Solution. Required percentage increase in velocity
sun. 𝑣 −𝑣 𝑣
= × 100 = − 1 × 100
𝑣 𝑣
Problem 50. The escape velocity for a satellite is = (√2 − 1) × 100 = 41.4%
𝟏𝟏. 𝟐𝐤𝐦𝐬 𝟏 . If the satellite is launched at an angle of
𝟔𝟎∘ with the vertical, what will be the escape velocity?
Solution. 11.2kms , because escape velocity 𝑣 =
2𝐺𝑀/𝑅 does not depend on the angle of projection. Short Answer

h
Problem 51. An astronaut, while revolving in a circular
Questions
orbit happens to throw a spoon outside. Will the

es
spoon reach the surface of the earth?
Solution. The spoon will continue to move in the same Problem 1. Answer the following :
circular orbit, and chase the astronaut. It will never reach (a) Among the known types of forces in nature, the
the surface of the earth. gravitational force is the weakest. Why then does it
play a dominant role for motion of bodies on the
ak
Problem 52. An artificial satellite revolves around the terrestrial, astronomical and cosmological scale?
earth without using any fuel. On the other hand, an (b) Do the forces of friction and other contact forces
aeroplane requires fuel to fly. Why ? arise due to gravitational attraction? If not, what is the
Solution. Air is essential for an aeroplane to fly. To origin of these forces?
overcome the resistive forces of air, aeroplane requires
.R

Solution. (a) The strong nuclear forces operate only over

fuel. The satellite orbits at a much higher height where air a range of distance of the order of 10 to 10 m.
resistance is negligible and so it needs no fuel. The forces involved in weak interactions operate only
during radioactive decay. Electrical forces are stronger
Problem 53. Why are space rockets usually launched than gravitational forces for a given distance, but they can
from west to east in the equatorial plane? be attractive as well as repulsive unlike gravitational force,
Er

Solution. Due to rotation of the earth about its polar axis, which is always attractive. As a result, the forces between
every particle on the earth has a linear velocity directed massive neutral bodies are predominantly gravitational.
from west to east. This velocity (𝑣 = 𝑅𝜔) is maximum at (b) No, the forces of friction and other contact forces have
the equator. When a rocket is launched from west to east, electromagnetic origin.
this maximum velocity gets added to the launching
velocity, so the launching becomes easier. Problem 2. Choose the correct alternatives :
y

(i) If the gravitational potential energy of two mass

Problem 54. A satellite of small mass burns during its points infinite distance away is taken to be zero, the
descent and not during ascent. Why ?
B

gravitational potential energy of a galaxy is

Solution. The speed of the satellite during descent is much (positive/negative/zero).
larger than that during its ascent. As the air resistance is (ii) The universe on the large scale is shaped by
directly proportional to velocity, so heat produced during (gravitational/electromagnetic forces), on the atomic
descent is very large and the satellite burns up. scale by (gravitational/electromagnetic) forces, on the
nuclear scale by (gravitational/
Problem 55. Is it possible to place an artificial satellite electromagnetic/strong nuclear) forces.
in an orbit such that it is always visible over New Solution. (i) The gravitational potential energy of a galaxy
Delhi? is negative because it is a bound system.
Solution. No. A satellite remains always visible only if it (ii) The universe on the large scale is shaped by
revolves in the equatorial plane with a period of revolution gravitational forces, on the atomic scale by
equal to that of the earth. New Delhi does not lie in the electromagnetic forces and on the nuclear scale by strong
region of equatorial plane. nuclear forces.

Problem 56 . The astronauts in a satellite orbiting the Problem 3. What is the difference between inertial
earth feel weightlessness. Does the weightlessness mass and gravitational mass of a body ?
depend upon the distance of the satellite from the earth? Solution. The inertial mass of a body is a measure of its
Give reason. inertia and is given by the ratio of the external force applied
on it to the acceleration produced in it. The gravitational
mass of a body, on the other hand, is a measure of the
gravitational pull acting on it due to the earth. The decreases maximum at the equator and is not affected at
gravitational mass is measured by a common balance. poles.
(iii) If the rotational speed of the earth is increased to
Problem 4. Which of the following observations point seventeen times its present value, the value of 𝑔 at the
to the equivalence of inertial and gravitational mass : equator will become zero. At the poles, the value of 𝑔
(a) Two spheres of different masses dropped from the remains unchanged.
top of a long evacuated tube reach the bottom of the
tube at the same time. Problem 8 . When a satellite moves to a lower orbit in
(b) The time-period of a simple pendulum is the atmosphere of the earth, it becomes hot. This
independent of its mass.
indicates that there is some dissipation in its
(c) The gravitational force on a particle inside a hollow
mechanical energy. But the satellite spirals down
isolated sphere is zero.
towards the earth with an increasing speed, why ?
(d) For a man in closed cabin that is falling freely
Solution. For a satellite orbiting the earth, Total energy =
under gravity, gravity 'disappears'.
(e) An astronaut inside a spaceship orbiting around − Kinetic energy i.e., 𝐸 = −𝐾, When the satellite enters
the Earth feels weightless. the atmosphere of the earth, it dissipates its mechanical
(f) Planets orbiting around the sun obey Kepler's Third energy (which is negative) against atmospheric friction.
Law (approximately). Energy 𝐸 becomes more negative. As a result, the kinetic
(g) The gravitational force on a body due to the Earth energy of the satellite increases and hence its speed
is equal and opposite to the gravitational force on the increases. But its orbital speed 𝑣 = 𝐺𝑀/(𝑅 + ℎ) can
Earth due to the body.[NCERT]

h
increase only if its height ℎ becomes smaller. Hence the
Solution. The observations (𝑎), (𝑏), (𝑑), (𝑒) and (𝑓) point satellite moves to a lower orbit with an increased speed.
to the equivalence of gravitational masses, because in Thus due to the atmospheric friction, the satellite spirals

es
these situations the bodies are in motion. down towards the earth with increasing speed and
The observations (𝑐) and (𝑔) do not establish the ultimately burns out in the lower dense atmosphere.
equivalence of inertial and gravitational masses because
the bodies are not in the state of motion. Problem 9. What are the conditions under which a
rocket fired from the earth becomes a satellite of the
ak
Problem 5. A body is taken from the centre of the earth earth and orbits in a circle ?
to the moon. What will be the changes in the weight of
Solution. (i) First the rocket should be given a sufficient
the body ?
vertical velocity so that it reaches a height at which it is
Solution. (i) At the centre of the earth, the weight of the
supposed to revolve around the earth.
body is zero (𝑔 = 0).
.R

(ii) At this height, the rocket must be given a horizontal

(ii) As the body is moved from centre to the earth surface,
its weight increases (due to increase in 𝑔 ) orbital velocity given by 𝑣 = .
(iii) As the body is moved from earth's surface towards the
moon, the weight decreases (due to the decrease in 𝑔 ), (iii) The air resistance should be negligible at the height of
becomes zero at the point where the gravitational forces its orbit.
of the earth and moon are equal and opposite.
Er

(iv) Again the weight increases, becoming 𝑚𝑔/6 on the Problem 10. What would happen if the force of gravity
moon's surface. were to disappear suddenly?
Solution. If the force of gravity suddenly disappears, then
Problem 6. Mention the conditions under which the (i) All bodies will lose weight.
weight of a person can become zero. (ii) We would be thrown away from the earth due to the
Solution. The weight of a person can become zero under centrifugal force.
y

the following conditions: (iii) Eating, drinking and in fact all operations would
(i) When the person is at the centre of the earth (as 𝑔 = 0 become impossible.
B

at the centre of the earth). (iv) Motion of satellites around the planets and the motion
(ii) When the person is at the null points in space (At these of planets around the sun would cease.
points, the gravitational forces due to different masses
cancel out). Problem 11. The radii of two planets are 𝑹 and 𝟐𝑹
(iii) When the person is standing in a freely falling lift. respectively and their densities 𝝆 and 𝝆/𝟐
(iv) When the person is inside a spacecraft which is respectively. What is the ratio of acceleration due to
orbiting around the earth. gravity at their surfaces? [Central Schools 05]
Solution. Here,
Problem 7. How will the value of 𝒈 be affected if (i) the 𝐺𝑀 𝐺 4 4
rotation of the earth stops (ii) the rotational speed of 𝑔= = ⋅ 𝜋𝑅 𝜌 = 𝜋𝐺𝑅𝜌
𝑅 𝑅 3 3
the earth is doubled (iii) the rotational speed of the or 𝑔 ∝ 𝑅𝜌
earth is increased to seventeen times its present 𝑔 𝑅𝜌
value? ∴ = 𝜌 = 1: 1.
𝑔 2𝑅 ⋅
Solution. (i) If the rotation of the earth stops, no centrifugal 2
force will act on the bodies lying on it. The value of 𝑔 Problem 12. The time period of the satellite of the earth
increases maximum at the equator. As no centrifugal force is 5 hours. If the separation between the earth and the
acts on body at poles, so the value 𝑔 is not affected. satellite is increased to 4 times the previous value,
(ii) If the rotational speed of the earth is doubled, the then what will be the new time period of the satellite?
centrifugal force on the bodies increases. The value of 𝑔 [AIEEE 03]
Solution. According to Kepler's law of periods,
𝑇 𝑅 𝑇 4𝑅
= or =
𝑇 𝑅 5 𝑅
or 𝑇 = √64 × 25 = 40 h.

Problem 13. Prove that acceleration due to gravity on

𝟒
the surface of the earth is given by 𝒈 = 𝝅𝝆𝑮𝑹, where
𝟑
is gravitational constant, 𝝆 is mean density and 𝑹 is
the radius of the earth.
Solution. Mass of the earth, 𝑀 = 𝜋𝑅 𝜌
Acceleration due to gravity on the earth's surface,
𝐺𝑀 𝐺 4 4
𝑔= = × 𝜋𝑅 𝜌 = 𝜋𝜌𝐺𝑅.
𝑅 𝑅 3 3

Problem 14. Why is the weight of a body at the poles

more than the weight at the equator ? Explain.
Solution. As 𝑔 = 𝐺𝑀/𝑅 and the value of 𝑅 at the poles is
less than that at the equator, so g at poles is greater that

h
𝑔 at the equator. Now, as 𝑔 > 𝑔 , hence 𝑚𝑔 > 𝑚𝑔 i.e.,
the weight of a body at the poles is more than the weight
at the equator.

es
Solution. As 𝑔 = 𝐺𝑀/𝑅 and the value of 𝑅 at the poles is
less than that at the equator, so 𝑔 at poles is greater that
𝑔 at the equator. Now, as 𝑔 > 𝑔 , hence 𝑚𝑔 > 𝑚𝑔 i.e.,
the weight of a body at the poles is more than the weight
at the equator.
ak
Problem 15. Why does the earth impart the same
acceleration to all bodies?
Solution. The force of gravitation exerted by the earth on
.R

a body of mass 𝑚 is
𝑀𝑚
𝐹=𝐺 = 𝑚𝑔
𝑅
Acceleration imparted to the body,
𝐺𝑀
𝑔=
𝑅
Er

Clearly, 𝑔 does not depend on 𝑚. Hence the earth imparts

same acceleration to all bodies.

Problem 16. If suddenly the gravitational force of

attraction between the earth and a satellite revolving
around it becomes zero, what will happen to the
y

satellite?
Solution. If the gravitational force of the earth suddenly
B

becomes zero, the satellite will stop revolving around the

earth and it will move in a direction tangential to its original
orbit with a speed with which it was revolving around the
earth.
Example 1. At what height from the surface of the earth, will the value of 𝒈 be reduced by 𝟑𝟔% from the
value at the surface ? Radius of the earth = 𝟔𝟒𝟎𝟎 𝐤𝐦.
Solution. Suppose at height ℎ, the value of 𝑔 reduces by 36% i.e., it becomes 64% of that at the surface. Thus
64
𝑔 = 64% of 𝑔 = 𝑔
100
But
𝑅
𝑔 =𝑔
(𝑅 + ℎ)
64 𝑅 8 𝑅
𝑔=𝑔 or =
100 (𝑅 + ℎ) 10 𝑅 + ℎ
or
𝑅 6400
ℎ= = = 1600 km.
4 4
Example 2. At what height above the earth's surface, the value of 𝒈 is half of its value on earth's surface?
Given its radius is 𝟔𝟒𝟎𝟎 𝐤𝐦
Solution. Here 𝑔 = 𝑔/2
But

h
𝑅
𝑔 =𝑔
𝑅+ℎ

es
𝑔 𝑅 𝑅 1
∴ =𝑔 or =
2 𝑅+ℎ 𝑅+ℎ 2
or = √2
or
ak
ℎ = (√2 − 1)𝑅 = 0.414𝑅 = 0.414 × 6400
= 2649.6 km.
Example 3. Find the percentage decrease in the weight of a body when taken to a height of 𝟑𝟐 𝐤𝐦 above
.R

the surface of the earth. Radius of the earth is 𝟔𝟒𝟎𝟎 𝐤𝐦

Solution. Here ℎ = 32 km, 𝑅 = 6400 km
As ℎ ≪ 𝑅, so
2ℎ 2𝑔ℎ
𝑔 =𝑔 1− =𝑔−
Er

𝑅 𝑅
or
2𝑔ℎ
𝑔−𝑔 =
𝑅
Percent decrease in weight
𝑚𝑔 − 𝑚𝑔 𝑔−𝑔
y

= × 100 = × 100
𝑚𝑔 𝑔
B

2𝑔ℎ 2ℎ
= × 100 = × 100
𝑔×𝑅 𝑅
2 × 32
= × 100 = 1%.
6400
Example 4. Find the percentage decrease in weight of a body, when taken 𝟏𝟔 𝐤𝐦 below the surface of the
earth. Take radius of the earth as 6400 km
Solution. Here 𝑅 = 6400 km, 𝑑 = 16 km
𝑑 16 399
𝑔 =𝑔 1− =𝑔 1− = 𝑔
𝑅 6400 400
399 1
∴ 𝑔−𝑔 =𝑔− 𝑔= 𝑔
400 400
The percentage decrease in weight of the body
𝑚𝑔 − 𝑚𝑔 𝑔−𝑔
= × 100 = × 100
𝑚𝑔 𝑔
(1/400)𝑔
= × 100 = 0.25%.
𝑔
Example 5. How much below the surface of the earth does the acceleration due to gravity become 𝟏% of its
value at the earth's surface? Radius of the earth = 𝟔𝟒𝟎𝟎 𝐤𝐦
Solution. Here 𝑔 = 1% of 𝑔 =
But 𝑔 = 𝑔 1 −
𝑔 𝑑
∴ =𝑔 1−
100 𝑅
or
or
𝑑 1 99
= 1− =
𝑅 100 100
99 99
𝑑= ×𝑅 = × 6400 = 6336 km.
100 100

EXAMPLE 6. A rocket is fired from the earth towards the sun. At what point on its path is the gravitational
force on it zero? The masses of the sun and that of the earth are respectively 𝑀 = 𝟐 × 𝟏𝟎𝟑𝟎 𝐤𝐠 and 𝑴𝐞 =
𝟔 × 𝟏𝟎𝟐𝟒 𝐤𝐠. The orbital radius of the earth is 𝒓 = 𝟏. 𝟓 × 𝟏𝟎𝟏𝟏 𝐦. Neglect the effect of other planets

h
present in the outer space.
Solution Let the gravitational force on the rocket be zero at a distance of 𝑥 from the surface of the earth. This

es
means that the force on the rocket at this point due to the earth and that due to the sun are equal and
opposite. If 𝑚 is the mass of the rocket, then at this distance 𝑥, we have
𝐺𝑀 𝑚 𝐺𝑀 𝑚
⇒ =
𝑥 (𝑟 − 𝑥)
ak
𝑟−𝑥 𝑀 2 × 10 10
⇒ = = =
𝑥 𝑀 6 × 10 √3
𝑟 10
.R

⇒ −1=
𝑥 √3
𝑟 10 √3 + 10
⇒ = 1+ =
𝑥 √3 √3
Er

√3𝑟 √3 × 1.5 × 10 m
⇒ 𝑥= = = 2.59 × 10 km.
√3 + 10 √3 + 10
EXAMPLE 7. A saturn-year is 29.4 times the earth-year. How far is saturn from the sun if the earth is
𝟏. 𝟓 × 𝟏𝟎𝟖 𝐤𝐦 away from the sun?
Solution Using Kepler's law, we have
y

𝑇earth 𝑅earth
=
𝑇saturn 𝑅saturn
B

/
𝑇saturn 29.4 × 𝑇earth /
𝑅saturn = 𝑅earth × = 1.5 × 10 km ×
𝑇earth 𝑇earth
𝑅saturn = 1.43 × 10 km.
EXAMPLE 8. The escape velocity of a projectile on the surface of the earth is 𝟏𝟏. 𝟐 𝐤𝐦 𝐬 𝟏 . If a body is
projected out with thrice this speed, find the speed of the body far away from the earth. Ignore the
presence of other planets and the sun.
Solution If 𝑣 is the escape velocity of the body on the surface of the earth, then the body is projected with a
speed of 𝑣 = 3𝑣 . If 𝑚 is the mass of the body and 𝑣 is the velocity of the body in outer space far away from
the earth, then the principle of conservation of energy gives
𝑚𝑣 − = 𝑚𝑣 −
𝑣 − =𝑣
(3𝑣 ) − 𝑣 = 𝑣
𝑣 = √8𝑣 = √8 × 11.2 km s
𝑣 = 31.68 km s