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GRAVITATION

CONCEPTUAL PROBLEMS

Q-01 Gravitational force is a weak force but still it is considered the most important force. Why ?

Ans Gravitational force plays an important role for initiating the birth of stars, for controlling the entire

structure of the universe and evolution of the universe. It helped to explain many natural phenomena.

Q-02 Is the value of g same every where on the surface of earth ? How has it been decided ?

Ans The value of g is different at different places on the surface on earth. The shape of the earth is not exactly

spherical ; it is flattened at the poles and is bulging out at the equator. Due to which the radius of the earth

is smaller at poles and is larger at equator. Since g ∝ 1/R2, therefore the acceleration due to gravity is

smaller at equator than that at poles.

Q-03 Two particles of equal mass move in a circle of radius r under the action of their mutual gravitational

attraction. Find the speed of each particle if its mass is m.

Ans The two particles will move on a circular path if they always remain diametrically opposite so that the

gravitational force on one particle due to other is directed along the radius. Taking into consideration the

circulation of one particle we have.

Q-04 The radii of two planets are respectively R1 and R2 and their densities are respectively and . What is

the ratio of the acceleration due to gravity at their surfaces.

Ans As g =

so g ∝ Rρ ∴

Q-05 Draw graphs showing the variation of acceleration due to gravity with (a) height above the earth’s surface,

(b) depth below the earth’s surface.

Ans (a) The variation of g with height h is related by relation g ∝ 1/r2 where r = R+ . Thus the variation of g

and r is a parabolic curve. i.e. part AB of the graph, Fig. 6.26. (b) The variation of g with depth is related by

equation g’ = g (1 – d/R) i.e.g’ ∝ (R - d). Thus the variation of g and d is a st. line i.e. part AC of the graph.

Fig. 6.26.

Q-06 Earth is continuously pulling moon towards its centre. Why does not moon fall on the earth.

Ans It is so because the gravitational attractions of earth provides the necessary centripetal force to the moon

for its orbital motion around the earth. Due to which the moon is revolving around the earth.

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Q-07 What are the conditions under which a rocket fired the earth, launches an artificial satellite of earth ?

Ans Following are the basic conditions ;

(i)The rocket must take the satellite to a suitable height above surface of earth.

(ii)From the desired height, the satellite must be projected with a suitable speed, called the orbital speed.

(iii)In the orbital path of satellite, the air resistance should be negligible so that its speed does not decrease

and it does not burn due to the heat produced.

Q-08 Two artificial satellites, one close the surface and the other away are revolving around the earth. Which has

larger speed ?

Ans The relation for orbital speed is where is the height of the satellite above the earth’s

surface. Clearly, the smaller is the value of , greater is the value of and vice versa. Hence satellite

revolving close to earth has large speed.

Q-09 Should the speed of two artificial satellites of the earth having different masses but the same orbital radius,

be the same ?

Ans Yes, it is so because the orbital speed of a satellite is independent of the mass of a satellite. Therefore, the

speeds of the artificial satellites of different masses but of the same orbital radius will be the same.

Q-10 Which has longer period of revolution, a satellite revolving close or away from surface of earth ?

Ans We know that (time period)2 ∝ (orbital radius)3. Therefore time period is larger in case of a satellite

revolving away from the surface of earth.

Q-11 What would happen if gravity suddenly disappear ?

Ans If gravity suddenly disappears,

(i) all bodies will lose their weights.

(ii) We shall be thrown away from the surface of earth due to centrifugal force.

(iii) The motion of planets around the sun will cease because centripetal force shall not be provided.

(iv) Motion of the satellite around earth will also be not possible as no centripetal force will be provided.

Q-12 Can a pendulum vibrate in an artificial satellite ?

Ans No, this is because inside the satellite, there is no gravity i.e. g = 0. As t = 2π , hence, = ∞ for g = 0.

Thus the pendulum will not vibrate.

Q-13 What is the total energy of a satellite revolving around earth ?

Ans The satellite revolving around the earth has types of energies.

(i) Gravitational potential energy U due to position of the satellite in the orbit. Where,

U = −GMm/r

(ii) Kinetic energy K due to orbital speed of satellite. Where,

Total energy = U + K =

Q-14 The mass and the diameter of a planet are twice those of earth. What will be the time period of that

pendulum on this planet, which is a second pendulum on the earth.

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Ans Since g = GM/R2, if mass M and radius R are doubled, then g will reduce to half. Since time period

t = 2π i.e t ∝ . Therefore, time period of a pendulum will be times to that on the

earth. Thus the time period of second of second pendulum of the earth will be 2 second on the planet.

Q-15 Will 1 kg sugar be more at poles or at the equator.

Ans The value of g is larger at the poles than at the equator. If the sugar is weighed in a physical balance then

there will be no difference. If it is weighed by a spring balance, calibrated at the equator, then 1 kg of sugar

will have a lesser amount at poles.

Q-16 Why do different planets have different escape speeds ?

Ans As, escape speed = , therefore its values are different for different planets which are of different

masses and different sizes.

Q-17 What are the two factors which determine why some bodies in solar system have atmosphere and others

don’t have ?

Ans The ability of a body (planet) to hold the atmosphere depends on : (i) acceleration due to gravity and (ii)

surface temperature.

Q-18 Why does moon have no atmosphere ?

Ans Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on surface of moon is small.

Therefore, the value of escape speed on the surface of moon is small (only 2∙5 kms-1). The molecules of the

atmospheric gases on the surface of the moon have thermal speeds greater than the escape speed. That is

why all the molecules of gases have escaped and there is no atmosphere on moon.

Q-19 Although gravitational pull of sun on earth is more than that of moon, yet moon’s contribution towards

tide formation on earth is greater than that due to sun. Why ?

Ans The distance between moon and earth is very small as compared to the distance, between earth and sun.

Since the tidal effect is inversely proportional to the cube of the distance, therefore tidal effect of moon is

larger than that due to sun.

Q-20 What is binding energy of a satellite ?

Ans The minimum energy required to free a satellite from the gravitational attraction is called binding energy.

Binding energy is the negative value of total energy of satellite. Let a satellite of mass be revolving

closely around earth of mass M and radius R.

Total energy of satellite = P.E. + K.E. = −

∴ Binding energy of satellite = −[total energy of satellite] =

Q-21 What are the main features of gravitational force ?

Ans Following are the main features of gravitational force :

(1) It is always an attractive force.

(2) It is independent of the medium between the particles.

(3) It holds good over a wide range of distances (i.e. from interplanetary distances to interatomic distances)

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(4) It is an action-reaction pair i.e. the force of attraction exerted by body A on body B is equal to the force

of attraction exerted by body B on body A. However, the acceleration of the two bodies will not be equal.

(5) The gravitational force between two particles is independent of presence or absence of other particles.

(6) The total gravitational force on one particle due to individual particles i.e.

…… It means the principle of superposition is valid.

(7) It expresses the force between point masses.

(8) It is a conservative force i.e. the work done in moving a particles once around a closed path under the

action of gravitational force is zero.

Q-22 If a body is projected with speed greater than escape speed e from the surface of earth, find its speed in

intersteller space.

Ans Let ’ be the speed of projected body in intersteller space, where gravitational field of earth is zero, hence

the potential energy of body is zero. Following law of conservation of energy we have,

Or

[∵ ]

Or

Q-23 Two identical copper spheres of radius R are incontact with each other. If the gravitational attraction

between them is F, find the relation between F and R.

Ans Let M be the mass of each copper sphere ; and ρ be the density of copper. Then

M =

Gravitational attraction between the two spheres is

∝ if ρ is constant.

It is not correct because the gravitational force obeys inverse square law is related with masses and not

with density. So F ∝ 1/R2 is correct answer.

Q-24 The magnitude of gravitational field at distances r1 and r2 from the centre of a uniform sphere of radius R

and mass M are I1 and I2 respectively. Find the ratio of (I1/I2) if r1 > R and r2 <R.

Ans When r1 > R, the point lies outside the sphere. Then sphere can be considered to be a point mass body

whose whole mass can be supposed to be concentrated at its centre. Then gravitational intensity at a point

distance r1 from the centre of sphere will be,

I1 = GM/ ….(i)

When r2 <R, the point P lies inside the sphere. The unit mass body placed at P, will experience gravitational

pull due to sphere of radius r2, whose mass is

Therefore the gravitational intensity at P will be I2 =

….(ii)

So

Q-25 A mass M is broken into two parts of masses m1 and m2. How are m1 and m2 related so that force of

gravitational attraction between the two parts is maximum.

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Ans Let m1 = m, then m2 = M – m. Gravitational force of attraction between them when placed distance r apart

will be

.

Differentiating it w.r.t.m, we get

=

If F is maximum, then

∴

or or

Q-26 The radius of a planet is double than that of the earth but their average densities are the same. If the escape

speeds at the planet and at the earth are and respectively, then prove that .

Ans

∴

or

Q-27 A rocket is fired with a speed v = 2 near the earth’s surface and directed upwards. Find its speed in

interstellar space.

Ans Let v be the speed of rocket in interstellar space. Using law of conservation of energy, we have

Or or .

Q-28 A person sitting in an artificial satellite of earth feels weightlessness but a person standing on moon has

weight through moon is also a satellite of earth.

Ans When a person is sitting in an artificial satellite of earth, the gravitational pull on the person due to earth

i.e. the weight of the person due to earth is counterbalanced by the centrifugal force acting on the person.

Due to which the net force acting on the person in satellite is zero and the person feels weightlessness. But

when the person standing on moon ; the gravitational pull on person due to moon (which is left

unbalanced) is responsible for his weight on moon.

Q-29 Why does a body lose weight at the centre of the earth ?

Ans The weight of the body is the force with which the body is attracted by the earth towards its centre.

Quantitatively, the weight of body of mass m is equal to mg. Where g is the acceleration due to gravity. At

the centre of earth, g = 0, so weight of body is zero at the centre of earth.

Q-30 Explain, why a tennis ball bounces higher on hills than in plains.

Ans Let u be the velocity of the tennis ball with which it bounces at a place, where acceleration due to gravity is

g’. The ball will go up till its velocity becomes zero. If h is the height upto which the ball rises up at a place,

then initial K.E. of ball is equal to final P.E. of ball at highest point. i.e.

As acceleration due to gravity (g’) on hill is less than that on the surface of earth (due to altitude effect), so

the tennis ball will bounce higher on hills than on plains.

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Q-31 A particle is projected vertically upwards from the surface of the earth (radius R) with a kinetic energy

equal to half of the minimum value needed for it to escape. Find the height to which it rises above the

surface of earth.

Ans We know that the value of escape velocity of a particle from the surface of earth is given by ;

So, the minimum kinetic energy needed for the particle to escape

As per question, the kinetic energy given to the particle from the surface of earth =

If is the height attained by the particle from the surface of earth, then its potential energy =

According to law of conservation of total energy;

(K.E + P.E.) on the surface of earth = (P.E.) on the height h.

So,

or

Or 2R = R+ or = R.

Q-32 Air friction increases the velocity of the satellite. Explain.

Ans If a satellite of mass is revolving in a circular orbit of radius , with speed around the earth of mass M,

the centripetal force is provided by the gravitational pull i.e.

or ( ) = GM m = constant

or L = constant

Where = L is the angular momentum of the satellite. As the air friction causes the retarding torque, it

will decreases the angular momentum L of satellite. Due to which the velocity of the satellite increases in

order to keep L = constant

Q-33 What is the escape velocity of the object. If the magnitude of the potential energy per unit mass of the

object at he surface of earth is E ?

Ans Magnitude of the potential per unit mass of the object at the surface of the earth is the magnitude of the

gravitational potential at the surface of the earth is =

E (given)

Escape velocity,

=

Q-34 Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of

revolution T. If the gravitational force of attraction between planet and the star is proportional to R−5/2,

then find the relation between T and R.

Ans Here, F = k R−5/2 = m R

or T2 =

Hence, ∝

Q-35 A planet of mass m moves around the sun of mass M in an elliptical orbit. The maximum and minimum

distance of the planet from the sun are r1 and r2 respectively. Find the relation between time period of the

planet in terms of r1 and r2.

Ans Here, semi major axis of the elliptical orbit of planet around the sun, r =( r1+ r2)/2. Therefore

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∝

or ∝

or ∝

Q-36 Suppose the gravitational force varies inversely as the nth power of the distance. Show that the time period

of a planet in circular orbit of radius R around the sun will be proportional to

Ans

or

=

or

or T ∝

Q-37 Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then allowed to

move towards each other under mutual gravitational attraction. Find their relative velocity of approach at

a separation distance r between them.

Ans Let be the relative velocity of approach of two bodies at distance r apart. The reduced mass of the

system of two particles is, μ =

According to law of conservation of mechanical energy.

Decrease in potential energy = increase in K.E.

or

=

or

.

VERY SHORT ANSWER QUESTIONS

Q-01 Why G is called a universal constant ?

Ans It is so because the value of G is same for all the pairs of the bodies (big or small) situated any where in the

universe.

Q-02 Does the force of attraction between two bodies depend upon the presence of other bodies and properties

of intervening medium ?

Ans The gravitational force of attraction between bodies is independent of presence of other bodies and

properties of intervening medium.

Q-03 Why is Newton’s law of gravitation called universal law ?

Ans Newton’s law of gravitation holds good irrespective of the nature of two bodies i.e. big or small, at all

times, at all locations and for all distances in the universe.

Q-04 On earth value of G = 6∙67 × 10-11 Nm2 kg-2. What is its value on moon, where g is nearly one-sixth than that

of earth ?

Ans The value of G is same on moon as on the surface of earth because G is a universal gravitational constant.

Q-05 The value of g on the moon is 1/6 th of that of earth. If a body is taken from the earth to the moon, then

what will be the change in its (i) weight, (ii) inertial mass and (iii) gravitational mass ?

Ans Weight of body on moon will become 1/6 th of that on the earth, but there will be no change in inertial and

gravitational masses.

Q-06 Moon travellers tie heavy weight at their back before landing on the moon. Why ?

Ans The value of g on moon is small, therefore, the weight of moon travellers will also be small.

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Q-07 If a man goes from the surface of earth to a height equal to the radius of the earth, then what will be his

weight relative to that on the earth ? What if he goes equally below the surface of earth ?

Ans – One fourth zero.

Q-08 The astronauts in a satellite orbiting the earth feel weightlessness. Does the weightlessness depend upon

the distance of the satellite from the earth ? If so how ? Explain your answer.

Ans No, the gravitational acceleration of the astronaut w.r.t. satellite is zero, whatever be the distance of the

satellite from the earth.

Q-09 Is it possible to put an artificial satellite on an orbit in such a way that it always remains visible directly

over Chandigarh ?

Ans No, because to put an artificial satellite in an orbit such that it always remains directly over a particular

place, its time period should be the same as that of the earth in the equatorial plane. As Chandigarh does

not lie on the equatorial plane, a geostationary satellite cannot be seen over Chandigarh.

Q-10 When a body falls towards earth, earth moves towards the body. Why is earth’s motion not noticed ?

Ans The motion of earth is not noticed because the acceleration produced in earth is negligible small, due to

large mass of the earth.

Q-11 The centripetal force on a satellite revolving around the earth is E. What is the gravitational force due to

earth on it ? Net force ?

Ans F ; F

Q-12 From where does a satellite revolving around a planet get the required centripetal force ?

Ans From gravitational attraction of planet exerted on satellite.

Q-13 The earth is acted upon by the gravitational force of attraction due to the sun. Then why does the earth not

fall towards sun ?

Ans Earth is orbiting around the sun in a stable. The gravitational attraction of sun exerted on earth provides

required centripetal force to earth. Moreover, the gravitational force of attraction of the sun is

perpendicular to the velocity of the earth.

Q-14 An artificial satellite revolves in its orbit around the earth without using any fuel. But an aeroplane

requires fuel to fly at a certain height. Why ?

Ans Air is necessary for the aeroplane to fly. Therefore, the aeroplane requires fuel against the frictional force

due to air. The satellite orbits around earth at such a higher height where air is very much rarer and its

friction is nearly negligible.

Q-15 Can we determine the gravitational mass of a body inside an artificial satellite ?

Ans No, artificial satellite is like a freely falling body and the weight of the body inside the satellite is zero.

Q-16 An artificial satellite is revolving around the earth at a height 200 km from the earth’s surface. If a packet is

released from the satellite, what will happen to it ? Will it reach the earth ?

Ans No, it will also revolve around the earth and become the satellites of earth.

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Q-17 A spring balance is suspended inside an artificial satellite revolving around the earth. If a body of mass 2

kg is suspended from it, what would be its reading ?

Ans Zero.

Q-18 Does the speed of a satellite remain constant in a particular orbit ?

Ans Yes, the speed of a satellite on a particular orbit remains constant. Since orbital velocity shows

that is constant if is constant, because GM is constant for earth.

Q-19 A satellite revolving around earth loses height. How will its time period be changed ?

Ans Time period of satellite is given by ;

Therefore, T will decrease, when decreases.

Q-20 What is more fundamental : mass or weight ?

Ans Mass is more fundamental than weight because mass of a body remains constant but weight of the body

changes due to change in the value of g.

Q-21 Under what condition, the gravitational potential energy of a body will be zero ?

Ans The gravitational potential energy of a body is zero when there is an infinite separation between two

interacting masses.

Q-22 Why is gravitational potential energy negative ?

Ans Because it arises due to attractive forces.

Q-23 What is the maximum value of gravitational potential energy and where ?

Ans The value of gravitational potential energy increases as we move away from the earth and becomes

maximum (infact zero) at infinity.

Q-24 The linear speed of a planet is not constant in its orbit. Comment.

Ans The areal velocity of the planet around the sun is constant but not the linear velocity. The linear velocity of

the planet is large while passing close to the sun and is small while passing away from sun.

Q-25 The escape speed from earth for a piece of 10 gram is 11.2 km s-1. What would it be for a piece of 100 gram ?

Ans The escape speed from the earth is 11.2 kms-1. Escape speed does not depend upon mass of the body to be

escaped.

Q-26 If the kinetic energy of a satellite revolving in an orbit close to the earth happens to be doubled, will the

satellite escape ?

Ans It will escape because its speed will become equal to the escape speed.

Q-27 What is the relation between gravitational intensity and gravitational potential at a point ?

Ans Gravitational intensity at a point is equal to negative value of gradient of potential at that point.

Q-28 Mass of a body cannot be changed without changing weight but weight can be changed without changing

mass. Explain.

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Ans Weight of a body = mass × acc. Due to gravity (i.e. W=mg). If mass m of a body is changed, its weight will

change. But it g is changed, Weight will change but mass will not change.

Q-29 Where will the true weight of the body be zero ?

Ans The true weight of a body will be zero where the gravitational effects are nil e.g. at the centre of earth or

deep in space for away from the given star or planet.

Q-30 If the force of gravity acts on all bodies in proportion to their masses, why does not a heavy body fall

corresponding faster than a light body.

Ans Acceleration = force / mass. If the force is due to gravity, then force = mass × acceleration due to gravity.

So acceleration of every body under gravity is equal to acceleration due to gravity, which is independent of

mass of the body.

Q-31 If the radius of the earth shrinks by one percent, its mass remaining the same by what percentage will the

acceleration due to gravity on its surface change ?

Ans Will increase by 2%.

Q-32 What is the acceleration due to gravity of earth at the surface of moon if the distance between earth and

moon is 3.8 × 105 km. and radius of earth is 6.4 × 103 km ?

Ans 0.00278 ms−2

Q-33 The escape speed on earth is 11.2 km s−1. What is its value for a planet having double the radius and 8

times the mass of the earth ?

Ans 22.4 km s−1

Q-34 What is the orbital period of revolution of an artificial satellite revolving in a geostationary orbit ?

Ans one day.

Q-35 According to kepler’s second law the radius vector to a planet form the sun sweeps out equal area in equal

interval of time. The law is a consequence of which conservation law.

Ans Law of conservation of angular momentum.

Q-36 Where is the gravitational field zero and where is the gravitational potential zero, in case of earth ?

Ans Gravitation field at a point is zero both at the centre of earth and infinity. Gravitational potential at a point

is zero at infinity.

Q-37 An astronaut, while revolving in a circular orbit happens to throw a ball outside. Will the ball reach the

surface of earth ?

Ans The ball will never reach the surface of earth as it will continue to move in the same circular orbit and will

chase the astronaut.

Q-38 A planet is revolving around the sun in an elliptical orbit. Which out of the following remains constant. (a)

linear speed (b) angular momentum (c) kinetic energy (d) potential energy (e) total energy throughout its

orbit.

Ans (b) Angular momentum and (e) total energy of the planet, remain constant.

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Q-39 Does a rocket really need the escape speed of 11.2 km/s initially to escape from the earth ?

Ans No, a rocket can have any initial speed at the smart but its speed should continue to increase. It will escape

from the earth only if its speed becomes 11.2 km/s

Q-40 The length of second’s pendulum on earth where the acceleration due to gravity is 10 ms−2 is

approximately 1 m. What will be the length of second’s pendulum on a planet where g = 5 ms−2.

Ans T = 2π

Q-41 A planet revolves in an elliptical orbit around the sun. The semimajor and semiminor axis are and b.

How time-period is related with them ?

Ans T2 ∝ 3.

Q-42 The gravitational potential energy of a body at a distance r from the centre of earth is U. What is the weight

of the body at that point ?

Ans

∴ Weight of body =

Q-43 The distance of two planets from the sun are 1011 and 1010 metres respectively. What is the ratio of time

periods of these two planets ?

Ans

Q-44 For a satellite escape speed is 11 km s-1. If the satellite is launched at an angle of 600 with the vertical, what

will be the escape speed ?

Ans 11 km/s ; It is so because the escape speed It means escape speed depends upon mass of

the earth, radius of the earth which are constant quantities and is independent of angle of launching.

Q-45 A satellite is orbiting around the earth with a speed . To make the satellite escape, what is the minimum

percentage increase in its speed ?

Ans % increase in speed =

Q-46 What is the sense of rotation of stationary satellite around the earth ?

Ans West to East.

Q-47 What is the time period of revolution of polar satellite of earth ?

Ans About 100 minutes.

Q-48 If a point mass m is at a distance from the centre of a spherical shell of mass M and radius R (> ), what is

the force between them ?

Ans In case of spherical shell, the gravitational intensity at a point inside the shell, I=0. So force on particle of

mass m inside the shell, F=mI=0.

Q-49 If earth be at one half its present distance from the sun, how many days will there be in year ?

12

Ans

days.

Q-50 If a satellite is revolving around a planet of density ρ, show that the entity ρ, T2 is a universal constant.

Ans When a satellite is orbiting close to earth, then

or

or

a constant.

Q-51 An astronaut on a satellite releases a glass out of the satellite into the space. Will the glass fall to the earth ?

Ans No. The glass released by astronaut from a satellite will never fall to earth but will become a satellite of

earth due to inertia of motion.

Q-52 Choose the correct alternative :

(a) If the gravitational potential energy of two mass points infinite distance away is taken to be zero, the

gravitational potential energy of a galaxy is (positive/negative/zero).

(b) The universe on the large scale is shaped by (Gravitational/electromagnetic) forces, on the atomic scale

by (gravitational/electromagnetic) force, on the nuclear scale by (gravitational/electromagnetic/strong

nuclear) forces.

Ans (a) negative (b) Gravitational, electromagnetic, strong nuclear.

Q-53 Which of the following observations point to the equivalence of inertial and gravitational mass

(a) Two spheres of different masses dropped from the top of a long evacuated tube reach the bottom of the

tube at the same time.

(b) The time-period of a simple pendulum is independent of its mass.

(c) The gravitational force on a particle inside a hollow isolated sphere is zero.

(d) For a mass in a closed cabin that is falling freely under gravity ‘disappears’.

(e) An astronaut inside a spaceship orbiting around the earth feels weightless.

(f) Plants orbiting around the sun obey Kepler’s third law (approximately).

(g) The gravitational force on a body due to the earth is equal and opposite to the gravitational force on the

earth due to the body.

Ans All except (c) and (g).

Q-54 If M is the mass and R is the radius of the earth, find the ratio of the acceleration due to gravity and

gravitational constant.

Ans g =

Q-55 Two satellites of masses 3 m and m orbit the earth in circular orbits of radii r and 3 r respectively. What is

the ratio of their orbital speeds ?

Ans

i.e. ∝

; so

Q-56 The escape speed of a body projected vertically upwards from the surface of the earth is . What is the

escape speed, if the body is projected in a direction making an angle θ with the vertical ?

Ans ; As the escape speed is independent of the direction of projection.

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Q-57 Two satellites A and B are orbiting around the earth in circular orbits of the same radius. The mass of A is

16 times that of B. What is the ratio of the period of revolution of B to that of A ?

Ans

and ; where M is the mass of the earth. It means T is independent of mass of the

satellite but depends upon the radius of the orbital path. So TB/TA = 1 : 1.

Q-58 Assuming that the earth is a sphere of radius R. At What altitude will the value of acceleration due to

gravity be half its value at the surface of the earth ?

Ans

or = 2R2 or or

Q-59 When a stone of mass m is falling in the earth of mass M, find the acceleration of earth if any ?

Ans Force exerted by falling stone on earth, F =

∴ Acceleration of earth =

SHORT ANSWER QUESTIONS

Q-01 Discuss Newton’s law of gravitation in vector form and state the important characteristics of gravitational

force.

Ans Refer to Art. 6.10.

Q-02 Give some evidences in support of Newton’s law of gravitation.

Ans Refer to Art.6.9.

Q-03 Distinguish between g and G.

Ans ‘g’ stands for acceleration due to gravity. It is defined as the acceleration set up in a body while falling

freely under the effect of gravity alone. The value of g changes with height depth and rotation of the earth

and is zero at the centre of the earth. The value of g on the surface of earth is taken 9.8 ms−2 or 980 cm/s2. g

is a vector quantity. ‘G’ stands for universal gravitational constant. It is defined as the force of attraction

acting between two bodies each of unit mass placed unit distance apart. The value of G is constant at all

location. The value of G is never zero any where.

The value of G = 6.67 × 10−11 N m2 kg−2 or 6.67 × 10−8 dyne − cm2 g−2. G is a scalar quantity.

Q-04 Distinguish between mass and weight.

Ans The mass of a body is defined as the quantity of matter constained in the body. It does not change with

height, depth and rotation of earth. The mass of a body is a scalar quantity and is measured in kilogram or

gram. The mass of a body is never zero anywhere. The weight of a body is defined as the force of attraction

on the body exerted by earth towards its centre. If m is the mass of the body placed on the surface of earth,

where the acceleration due to gravity is g, then weight of the body = mg. The weight of the body changes

with height, depth and rotation of earth. The weight of a body is a vector quantity and is measured in

newton or dyne or kg wt. or gwt. The weight of a body is zero at the centre of earth or in a lift falling down

with an acceleration, equal to the acceleration due to gravity.

Q-05 Show that weight of all bodies is zero at the centre of earth.

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Ans The value of acceleration due to gravity at a depth d below the surface of earth of radius R is given by g’ =

g (1 – d/R). At the centre of earth, d = R ; so, g’ = 0. The weight of a body of mass m at the centre of earth =

mg’ = m × 0 = 0.

Q-06 Prove that the value of the acceleration due to gravity at a point above the surface of the earth is inversely

proportional to the square of the distance of that point from the centre of the earth.

Ans Earth being a spherical body, its whole mass M can be supposed to be concentrated at its centre O. If g is

the acceleration due to gravity at a point above the surface of earth, where a body of mass m is lying and r

is the distance of the body from the centre of earth then acceleration of the body due to earth is

or g = GM/r2 or g ∝ 1/r2.

Q-07 Gravitational force between two bodies is 1 newton. If the distance between them is made twice, what will

be the force ?

Ans Since F ∝ 1/r2 ; so F2 / F1 =

or F2 = F1 / 4 = ¼ N.

Q-08 The mass of the moon is nearly 1.0% of the mass earth. What will be the gravitational force of the earth on

the moon, incomparison to the gravitational force of the moon on the earth ?

Ans Since the gravitational force is action and reaction pair, so the gravitational force of the earth on moon is

equal to the gravitational force of the moon on the earth.

Q-09 If a person goes to a height equal to radius of earth from its surface. What would be his weight relative to

that on the earth.

Ans At the surface of earth, weight W = g = GM m /R2

At height (=R), weight W’= g’ =

∴

or W’ =

It means the weight would reduce to one-fourth of the weight on the surface of earth.

Q-10 The gravitational force exerted by the sun on the moon is greater than that exerted by the earth on the

moon. Why then does not the moon escape from the earth, during solar eclipse ?

Ans The moon can escape only if there is no orbital motion. In fact, the moon while revolving around the earth,

has also orbital velocity around the sun. The gravitational force of attraction of the sun to the moon

provides the required centripetal force necessary for the orbital motion around the sun.

Q-11 If the change in the value of g at the height h above the surface of the earth is the same as at a depth x

below it, both x and h being much smaller than radius of the earth, find the relation between x and h.

Ans As per question ; g’ = g

Q-12 What will be the effect on the time period of a simple pendulum on taking to a mountain ?

Ans The time period of simple pendulum, T = 2 i.e. T ∝ 1/ . As the value of g is less at mountain

than at plane, hence time period of simple pendulum will be more at mountain than at plane.

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Q-13 The gravitational force acting on a rocket at a height h from the surface of earth is 1/3 of the force acting on

a body at see level. What is the ratio of h and R (radius of earth) ?

Ans

or = R or

Q-14 Does the gravitational force of attraction of the earth becomes zero at some height above the surface of

earth ? Explain.

Ans No, the gravitational force of attraction of earth on a body at height h is ;

F is zero only, when h is infinity.

Q-15 There is no atmosphere on the moon. Explain this on the basis of escape speed.

Ans The value of escape speed on the surface of moon is small (only 2.5 km s −1). It is so because the value of

acceleration due to gravity on the surface of moon is small. The molecules of the atmospheric gases on the

surface of the moon have thermal speeds greater than the escape speed. Due to which all the molecules of

gases have escaped and there is no atmosphere.

Q-16 The speed of revolution of an artificial satellite revolving very near the earth is 8 km s −1. What will be the

escape speed for a body on the earth ?

Ans Orbital speed,

Escape speed, So,

Q-17 Does speed of satellite remain constant in an orbit ? Explain.

Ans If the orbital path of a satellite is circular, then its speed is constant and if the orbital path of a satellite is

elliptical one then its speed in its orbit is not constant. In that case its areal velocity is constant.

Q-18 The mean distance of sun from a planet is four times the distance of the sun from the earth. In how many

years will that planet complete one revolution around the sun ?

Ans

or T2 = T1

8 years.

Q-19 Can centre of gravity and centre of mass coincide ? Explain with illustrations.

Ans Refer to Art.6.27.

Q-20 A satellite of mass m in a circular orbit of radius r round the earth. Calcualte its angular momentum with

respect to the centre of the orbit in terms of the mass M of the earth and G.

Ans Angular momentum of satellite, L =

Q-21 A satellite of mass m is orbiting just above the surface of the earth. Derive expressions for the orbital speed

and time period of the satellite in terms of acceleration due to gravity ‘g’ and radius R of the earth.

Ans Refer to Art.6.30.

Q-22 Suppose a hole is drilled completely through the earth along a diameter. Mass and radius of earth are M

and R. What is the force acting on a body of mass m at a distance r from the centre of earth ?

Ans If d is the depth of the body from the surface of earth then

r = R – d and mg’ = mg(1−d/R) = mg(R−d)/R = m g r/R.

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Q-23 A satellite is revolving around the earth, close to the surface of earth with a kinetic energy E. How much

kinetic energy should be given to it so that it escapes from the surface of earth ?

Ans Let be the orbital and escape speeds of the satellite, then .

Energy in the given orbit,

Energy for the escape speed,

∴ Energy required to be supplied =

Q-24 Taking the moon’s orbit around earth to be r and mass of earth 81 times the mass of the moon. Find the

position of the point from the earth, where the net gravitational field is zero.

Ans Let be the distance of a point from the earth where resultant gravitational field intensity is zero. So

or

or

Or 9 = 10 or = 9 /10 = 0.9

Q-25 Two satellites A and B go around a planet P in circular orbits having radius 4R and R respectively. If the

speed of the satellite A is 3 , find the speed of the satellite B.

Ans As,

and ’ =

∴

or

Q-26 Is it possible to put an artificial satellite into an orbit in such a way that it will always remain directly over

New Delhi, (a place which is not on the equatorial plane ?

Ans Not possible, because a satellite will appear stationary only when it revolves in an orbit concentric and

coplanar with the equatorial plane, having time period of revolution 24 hours and it should have the sense

of revolution from west to east that of earth. As New Delhi is not in the equatorial plane, hence it will not

be possible to put a geostationary satellite over New Delhi.

Q-27 Do the forces of friction and other contact forces arise due to gravitational attraction ? If not, what is the

origin of these forces ?

Ans The forces of friction and other contact forces do not arise due to gravitational attraction. These forces have

electromagnetic origin.

Q-28 An artificial satellite is moving in circular orbit around the earth with a speed equal to half of the escape

speed from the earth of radius R. What is the height of the satellite above the surface of the earth ?

Ans Here, orbital speed.

So R + h = 2R or h = R.

Q-29 A geostationary satellite is orbiting the earth at a height of 6 R above the surface of the earth ; R being the

radius of the earth. What will be the time period of another satellite at a height 2.5 R from the surface of the

earth ?

Ans As T2 = Kr3 or T ∝ r3/2

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∴

or

= 24

hour

Q-30 Is the potential energy of a system of bodies positive or negative ? Give reason in support of your answer.

Ans Since the gravitational force between the different bodies of a system are attractive in nature, so the

potential energy of the system is negative.

Q-31 A tennis ball and a cricket ball are to be projected out of gravitational field of the earth. Do we need

different velocities to achieve so ?

Ans We require the same velocity for the two balls, while projecting them out of the gravitational field. It is so

because, the value of escape velocity does not depend upon the mass of the body to be projected

.

Q-32 Why an astronaut in an orbiting space craft is not in zero gravity although weightless ?

Ans When a space craft is in orbit, then at any point on the orbit, there is some value of the acceleration due to

gravity. But the weight of astronaut is used up in providing the necessary centripetal force for the orbital

motion.

Q-33 If the earth be one half its present distance from the sun, how many days will the present one year on the

surface of earth will change ?

Ans Here, T1 = 365 days ; r1 = r ; T2 = ? ; r2 = r/2

As

or T2 = T1

days

Therefore, the decrease in number of days in one year will be = 365 – 129 = 236 days

Q-34 On a planet whose size is the same and mass 4 times as that of the earth, find the energy, find the energy

needed to lift a 2 kg mass vertically upwards through 2 m distance in joule. The value of g on the surface of

earth is 10 ms−2.

Ans On the earth, g =

On the planet, g’ =

So g’ = 4 g = 4 × 10 = 40 ms−2

∴ Energy needed = mg’ h = 2 × 40 × 2 = 160 J

Q-35 What do you understand by gravity and acceleration due to gravity. Establish a relation between g and G.

Q-36 Explain how the knowledge of g helps us to find (i) mass of earth and (ii) mean density of earth ?

Q-37 Explain Inertial mass and Gravitational mass. Compare their properties and establish their equivalence.

Q-38 What is a satellite ? Explain the principle of launching a satellite.

Q-39 Obtain the expressions for orbital velocity, time period and altitude of a satellite.

Q-40 What do you understand by geostationary and polar satellites ? Discuss their important uses.

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Q-41 What do you understand by ‘Escape velocity’ ? Derive an expression for it in terms of parameters of given

planet

Q-42 What do you understand by Gravitational field, Intensity of gravitational field. Prove that gravitational

intensity at a point is equal to the acceleration due to gravity at that point.

Q-43 Explain gravitational potential at a point and gravitational potential energy of a body in a gravitational

field. Establish a relation between them.

Q-44 Explain weightlessness. Discuss the problems of weightlessness.

LONG ANSWER QUESTIONS

Q-01 Explain Kepler’s laws of planetary motion and deduce Newton’s law of gravitation from them.

Q-02 State Kepler’s laws of planetary motion and explain the deduction of Kepler’s second law and third law of

planetary motion.

Q-03 Explain Newton’s law of gravitation. Define gravitational constant, and give its dimensional formula. Give

the evidences in support of the Newton’s law of gravitation.

Q-04 Discuss the variation of g with height and depth.

Q-05 Discuss the Principle of launching of a satellite and explain what do you understand by geostationary

satellites. Mention the essential conditions for geostationary satellite.