unit - 6 gravitational ug physics... · gravitational force between two point masses is 1 2 2 ......

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Unit - 6Gravitational

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SUMMARY1. Gravitational force between two point masses is

1 22

Gm mF

r

2. Acceleration due to Gravity

(I) on the surface of earth = 22 9.81GM ms

R

(II) At a height h from surface of earth

12

2(1 )1

g hg gRh

R

if h << R

(III) At a depth d form the surfce of earth

1 /(1 )dg gR

g1 = g if d = R i. e. on the surface of earth(IV) Effect of rotation of earth at latitude

g1 = g – R2 cos2 - at the equator = 0g1 = g – R2 = minimum value- At the pole = 900

g1 = g – R2 = maximum value- At the equator effect of rotation of earth is maximum and value of g is minimum.- At the pole effect of rotation of earth is zero and value of g is maximum.

3. Field Strength Gravitational field strength at a point in gravirtational field is defined as,

FE

m

= gravitational force per unit mass

Due to point mass

2

GMEr

(towards the mass) 2r

1 E

Due to solid sphere

inside points 3i

GME rR

At r = 0, E = 0 at the center

At r = R, 2

GMER

i.e. on the surface

out side points Eo = 2

GM orr 2o r

1E



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At r , E 0 Due to a sphericell shell -

inside points E=0

outside points E0 = 2

GMr

just outside surface 2

GMER

on the surface E – r graph is discontinuous on the axis of a ring

32 2 2( )

r

GMrER r

At r = 0, E = 0 i.e. at the center

If r >> R, 2GMEr

, i.e. ring behaves as a points mass

As 0 r E4. Gravitational potential :-

(i) Gravitational potential at a point in a gravitational field is defined as the negative of workdone in moving a unit mass from infinity to that point per unit mass, thus

pp

w wV

m

(ii) Due to point mass

GmVrm

0 0 v as r and v as r(iii) Due to solid sphere

inside points 2 2 – (1.5 0.5 )GMVi R rR

as r = R GMV

R i.e. on the surface

V - r graph is parabola for inside points

out side points GMvr

(iv) Due to sphercal shell

inside points GMViR

outside points GMVi

R



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(v) on the axis of a ring

2 2

GMVrR r

at r = 0, GMV

R i.e. at center

5. Gravitational potential Energy(i) This is the work done by gravitational forces in arranging the system from infinite seprationin the present position(ii) Gravitational potential energy of two point massess is

1 2Gm mUr

(iii) To find the gravitational points energy of more than two points masses we have to makepairs of masses. Neighter of the pair should be repeated. For example in case of four pointmasses.

4 3 4 2 4 1 3 2 3 1 2 1

43 23 41 32 31 21

m m m m m m m m m m m mU G

r r r r r r

for n point masses total number of pairs will be

( 1)2

n n

(iv) If a point mass m is plaled on the surface of earth the potential energy here is Uo

oGMmU

R

and potential energy at height h is

hGMmU(R h)

the difference in potential energy would be

U = Uh – Uo = mgh

1 + h/rIf h << R, U = mgh

6. Relation between field strength

E and potential V(i) if V is a function of only one variable (Say r) then

dVEdr

slope of U r graph

(ii) If V is funtion at three coordinates variable; e x, y, and z then

v v vˆ ˆ ˆE i j kx y z



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7. Escape velocity(i) From the surface of earth

2

22 GM GMVe gR as gR R

=11.2 km / sec(ii) Escape velocity does not depend upon the angle which particle is projected form the surfaceand the mass of body

8. Motion of satellites

(i) orbital speed GMVor

(ii) time period 3/ 2 2 32T r T rGM

(iii) Kinetic energy 2

GMmK

r

(iv) Potential energy GMmUr

...

(v) Total Mechanical energy. GMmEr

9. Kepler’s laws- First law :Each planet moves in an elliptical orbit with the sun at one focus of ellipse- Second law :The radius vectors drawn form the sun to a planet, sweeps out equal area in equal time intervali.e. areal Velocity is constant.this law is derived from the law of conservation of angular momentum

2dA Ldt m

= constant here L is the angular momentum and m is mass of planet

- Third law2 3T r

where r is semi-major axis of elliptical path- The gravitational force acting between two bodies is always attractive. It is independent of medium

between bodies. It holds good over a wide range of distance. It is an action and reaction pair.It is conservative force. It is a central force and obey inverse square law as 21/F r

- The value of G is never zero any where but the value of g is zero at the center of earth.- the acceleration due to gravity is independent of mass, shape, size etc of falling body.- the rate of decrease of the acceteration due to gravity with height is twice as compared to that

with depth.- It the rate of rotation of earth increases the value of acceleration due to gravity decreases at

all points on the surface of earth except at poles.



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- If the radius of planet decreasees by n% keeping its mas unchanged, the accelreotion due togravity on its surface increases by 2n%.

- If the mass of a planet increases by n% keeping its radius unchanged the acceleration due togravity on its surface increases by n%.

- The value of g at a location gives the value of intensity of gravitational field at the location.- The orbital velocity of a satellite is independent of mass of the satellite but depends upon the

mass and radius of planet around which the rotation is taking place- The value of orbital velocity for a satellite near the surface of earth is 7.92 kms–1.- The direction of orbital velocity of satellite at an instant is along the tangent to the orbital path

at that instant.- The work done by a satellite in a complete orbit is zero.- For a satellite orbiting close to the surface of earth (h << R), the time period of revolution

is 2 / 84.6R g minutes and angular velocity. 2 / 0.001237 / secg R radT

- A geostationary satellite revolves around the earth from west to east. Its period of revolutonis one day i.e. 24 hours. The orbital velocity of geostationary satellite is 3.08 kms–1. Its heightabove the surface of earth is about 36000 km. The relative angular velocity of geostationarysatellite w.r.t earth is zero.

- When a satelilte is orbiting in its orbit, no energy is required to keep it in its orbit.- When the total energy of a satellite is negative, it will be moving in either a circular or an elliptical

orbit.- When the total energy of a satellite is zero, it will escape away from its orbit and its path becomes

parabolic.- When the height of satellites is increased its potential energy will increase and K.E. will decrease.- When the velocity of satellite is increased, its total energy will increase and it will start orbiting

in a circular path of larger radius.- For a satellite orbiting in a circular orbit, the value of potenial energy is always greater than

its K.E.- If the velocity of a satellite orbiting the earth is increased by 41.4% or its K.E. is doubled,

then it will escape away from the gravitational field of earth- Escape velocity =. 2 orbital velocity- If the gravitational force is inversely proportional to the nth power of distance r, then the orbital

velocity of a satellite

/ 20

nV r and time period is ( 2)

2n

T r

- When a body is projected horizontally with velocity v, from any height from the surface of earth,

then the following possibilities are there.(i) If v < v0, the body fails to revolve around the earth and finally falls to the surface of earth.(ii) If v = v0, the body will revolve around the earth in circular orbit.(iii) If v < ve the body will revolve around the earth in elliptical orbit.(iv) If v = ve, the body will escape from the gravitational field of earth.(v) If v > ve the body will escape, following a hyperbolic path.



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- If earth suddently contracts to1 thn of its present size without any change in mass, the duration

of the day will be nearly 24/n2 hours.- Force function F(r) is related with potential energy function U(r) by a relaion

dUFdr

- A given planet will have atmosphere if the root mean square velocity of molecules in its atmosphere

(i.e. Vrms =3RTM

) is smaller than escape velocity for that planet.

- In the weightlessness state, the bodies donot have weight but they do possess inertia on accountof their mass. the bodies floating inside the space craft may collide with each other and crash.

- If a body is released from a height aqual to n times the radius of earth, then its striking velocityon the surface of earth is

21

ngRn

- If polar ice caps melt then moment of inertia, Angular velocity will decrease and period of rotaitonof earth increase.

- The, line joining the places on earth having same value of g are called isogams.- Gravity meter and Etvos gravity balance are used to measure changes in accelaration due to

gravity.



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Newton’s Law of GravitationFor the answer of the following questions choose the correct alternative from among thegiven ones.1. Two identical solid copper spheres of radius R are placed in contact with each other. The

gravitational force between them is proportional to(A) R2 (B) R-2 (C) R-4 (D) R4

2. The gravitational force Fg between two objects does not depend on(A) sum of the masses (B) product of masses(C) Gravitational constant (D) Distance between the masses

3. The atmosphere is held to the earth by(A) clouds (B) Gravity (C) Winds (D) None of the above

4. Two sphere of mass m1 and m2 are situated in air and the gravitational force between themis F. The space around the masses is now filled with liquid of specific gravity 3. The gravitationalforce will now be(A) F (B) 3F (C) F/3 (D) F/9.

5. A satellite of the earth is revolveing in a circular orbit with a uniform speed v. If the gravitationalforce suddennly disappears, the satellite will(A) Contineue to move with velocity v along the original orbit.(B) Move with a Velocity v, tangentially to the original orbit.(C) Fall down with increasing velocity.(D) Ultimately come to rest somewhere on the original orbit.

6. Correct form of gravitational law is

(A) 1 22

Gm m = –Fr (B) 1 2

2

Gm m = –Fr

(C) 1 2

3

Gm m ˆ = –F rr

(D) 1 2

3

Gm m = –F rr

7. Mass M is divided into two parts xM and (1 - x) M. For a given separation, the value ofx for which the gravitational force between the two pieces becomes maximum is(A) 1 (B) 2 (D) 1/2 (D) 4/5

8. The earth (mass = 6 x 1024 kg) revolves around the sun with angular velocity 2 10–7 rad/sec in a circular orbit of radius 1.5 x 108 km. The force exerted by the sun on the earth is= ................... N(A) 18 1025 (b) zero (C) 27 1039 (D) 36 1021

9. Two particle of equal mass go round a circle of radius r. Under the action of their mutualgravitational force.The speed of each particle is =..................

(A)1 12r Gm

(B) 2Gm

r (C)

12

Gmr

(D)4Gm

r

10. The distance of the moon and earth is D the mass of earth is 81 times the mass of moon.At what distance from the center of the earth, the gravitational force will be zero(A) D/2 (B) 12D/3 (C) 4D/3 (D) 9D/10

11. One can easily “ Weight the earth ” by calculating the mass of earth using the formula(in usual notation)

(A) RegG (B) g/G Re2 (C)

2ReGg (D)

3ReGg



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12. Three equal masses of m kg each are plced the vertices of an equilateral triangle PQR and

a mass of 2m kg is placed at the centroid 0 of the triangle which is at a distance of 2 mfrom each of vertices of triangle. The force in newton. acting on the mass 2m is = ............(A) 2 (B) 1 (C) 2 (D) zero

13. Which of the follwing statement about the gravitational constant is true(A) It is a force(B) It has no unit(C) It has same value in all system of unit(D) It depends on the value of the masses.

14. Two point masses A and B having masses in the ratio 4 : 3 are seprated by a distance oflm. When another point mass c of mass M is placed in between A and B the forces A andC is 1/3rd of the force between band C, Then the distance C form A is = ............... m(A) 23 (B) 1/3 (C) 1/4 (D) 2/7

15. The gravitational force between two point masses m1 ans m2 at separation r is given by

1 22

m mF Gr

The constant k ...............

(A) Depends on system of units only.(B) Depends on medium between masses only.(C) Depends on both (a) and (b)(D) is independent of both (a) and (b)

Acceleration Due to Gravity16. As we go from the equator to the poles, the value of g ...............

(A) Remains constant (B) Decreases (C) Increases (D) Decreases upro latitude of 45o

17. If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface,the mean density of the earth is = ...............

(A) 4 3/G gR (B) 3 4/R gG (C) 3 4/g RG (D) 12/RG g18. The radius of the earth is 6400 km and g=10ms-2. In order that a body of 5 kg weights zero

at the equator, the angular speed of the earth is = ............... rad/sec(A) 1/80 (B) 1/400 (C) 1/800 (D) 1/600

19. The time period of a simple pendulum on a freely moving artificial satellite is ............... sec(A) 0 (B) 2 (C) 3 (D) Infinite

20. A spherical planet far out in space has mas Mo and diameter Do. A particle of m falling nearthe surface of this planet will experience an acceleration due to gravity which is equal to

(A) 20 /GM Do (B) 24 /mGMo Do (C) 24 /GMo Do (D) 2/GmMo Do

21. A body weights 700 g wt on the surface of earth How much it weight on the surface of planetwhose mass is 1/7 and radius is half that of the earth(A) 200g wt (B) 1400g wt (C) 50 g wt (D) 300g wt.

22. The value of g on he earth surface is 980 cm/sec2. Its value at a height of 64 km from theearth surface is ............... cm5–2

(a) 960.40 (B) 984.90 (C) 982.45 (D) 977.55



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23. If earth rotates faster than its present speed the weight of an object will.(A) increases at the equator but remain unchanged of the poles.(B) Decreases at the equator but remain unchanged at poles.(C) Remain unchanged at the equator but decreases at poles.(D) Remain unchanged at the equator but increases at the poles.

24. The moon’s radius is 1/4 that of earth and its mass is 1/80 times that of the earth. If g representsthe acceleration due to gravity on the surface of earth, that on the surface of the moon is..............(A) g/4 (B) g/5 (c) g/6 (D) g/8

25. The depth of at which the value of acceleration due to gravity becomes 1/n the time the valueof at the surface is (R = radius of earth)

(A) R/n (B) ( 1)n

nR

(C) R/n2 (D) 1

nRn

26. If the density of smalll planet is that of the same as that of the earth while the radius of theplanet is 0.2 times that of lthe earth, the gravitational acceleration on the surface of the planetis ...............(A) 0.2g (B) 0.4g (c) 2g (D) 4g

27. If mass of a body is M on the earth surface, than the mass of the same body on the moonsurfae is(A) M/6 (B) 56 (C) M (D) None of these

28. An object weights 72 N on the earth. Its weight at a height R/2 from earth is = .............. N(A) 32 (B) 56 (C) 72 (D) zero

29. If the radius of earth is R then height ‘h’ at which value of ‘g’ becomes one - fourth is(A) R/4 (B) 3R/4 (C) R (D) R/8

30. If the mass of earth is 80 times of that of a planet and diameter is double that of planet and‘g’ on the earth is 9.8 ms-2 , then the value of ‘g’ on that planet is = ............... ms-2

(A) 4.9 (B) 0.98 (c) 0.49 (D) 4931. Assuming earth to be a sphere of a uniform density, what is value of gravitational acceleration

in mine 100 km below the earth surface = ............... ms-2

(A) 9.66 (B) 7.64 (C) 5.00 (D) 3.132. Let g be the acceleration due to gravity at earth’s surface and k be the rotational K.E. of earth

suppose the earth’s radius decreases by 2% keeping alt other quantities same then(A) g decreases by 2% and K decreases by 4 %(B) g decreases by 4% and K increases by 2%(C) g increases by 4% and K increases by 4%(D) g decreases by 4% and K increases by 4%

33. A body weight 500 N on the surface of the earth. How much would it weight half way belowthe surface of earth(A) 125N (B) 1250N (C) 500N (D) 1000N

34. The radii of two planets are respectively R1 and R2 and their densities are respectively 1 and2 the ratio of the accelerations due to gravity at their surface is ...............

(A) 1 21 2 2 2

1 2

: .g gR R

(B) 1 2 1 2 1 2: :g g R R

(C) 1 2 1 2 2 1: :g g R R (D) 1 2 1 1 2 2: :g g R R



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35. At what height over the earth’s pole, the free fall acceleration decreases by one percent= .................. km (Re = 6400 km).(A) 32 (B) 80 (C) 1.253 (D) 64

36. Weight of a body is maximum at(A) moon (B) poles of earth (C) Equator of earth (D) Center of earth

37. At what distance from the center of earth, the value of aceeleration due to gravity g will behalf that of the surfaces (R = Radius of earth)(A) 2R (B) R (C) 1.414 R (D) o.414 R

38. The acceleration due to gravity near the surface of a planet of radius R and density d is proportionalto(A) d/R2 (B) dR2 (C) dR (D) d/R

39. The acceleration due to gravity is g at a point distance r from the center of earth R. if r < Rthen(A) g r (B) g r2 (c) g r-2 (D) g r-1

40. The density of a newly discoverd planet is twice that of earth. The acceleration due to gravityat the surface of the planet is equal to that at the surface of earth. If the radius of the earthis R, the radius of planet would be .....................(A) 2R (B) 4R (C) 1/4 R (D) .......

41. Density of the earth is doubled keeping its radius constant then acceleration, due to gravity willbe .................. ms-2(g = 9.8 ms2)(A) 19.6 (B) 9.8 (C) 4.9 (D) 2.45

42. Weight of body of mass m decreases by 1% when it is raised to height h above the earth’ssurface. If the body is taken to a depth h in a mine. change in its weight is(A) 2% decreases (B) 0.5% decreases(C) 1% increases (D) 0.5% increases

43. If density of earth increased 4 times and its radius becomes half of then out weight will be...(A) Four times it present calue (B) doubled(C) Remain same (D) halved

44. A man can jump to a height of 1.5 m on a planet A what is the height ne may be able tojump on another planet whose density and radius are respectively one- quater and one- thirdthat of planet A(A) 1.5 m (B) 15 m (C) 18 m (D) 28 m

45. If the value of ‘g’ acceleration due to gravity, at earth surface fis 10ms–2. its value in ms–2

at the center of earth, which is assumed to be a sphere of Radius ‘R’ meter and uniform density is(A) 5 (B) 10/R (C) 10/2R (D) zero

46. A research satellite of mass 200 kg. circles the earth in an orbit of avrage radius 3R/2 whereR is radius of earth. Assuming the gravitational pull 10 N, the pull on the satellite will be =..........N(A) 880 (B) 889 (C) 890 (D) 892

47. Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the

ratio of densities of earth e and moon m is 5 3e

m

then radius of moon Rm in terms

of Re will be ...............

(A) 5 Re18 (B)

1 Re6 (C)

3 Re16 (D)

1 Re2 3



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48. The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R fromthe surface of the earth is = ...............(g = acceleration due to gravity at the surface of earth)(A) g/9 (B) g/3 (C) g/4 (D) 9

49. The height at which the weight of a body becomes 1/16 th its weight on the surface of(radius R) is(A) 3R (B) 4R (C) 5R (D) 15R

50. A spherical planet has a mass Mp and diameter Dp A particle of mass m falling freely nearthe surface of this planet will experience an acceleration due to gravity, equal to

(A) 2

4 GMpDp (B) 2

GMpmDp (C) 2

GMpDp (D) 2

4GMpmDp

51. Assuming the earth to have a constant density, point out which of following curves show thevariation acceleration due to gravity from center of earth to points far away from the surfaceof earth ...............

(D) None of these

Gravitational potential, Energy and Escape Velocity52. In a gravitational field, at a point where the gravitational potential is zero

(A) The gravitational field is necessarily zero(B) The gravitational field is not necessarily zero(C) Nothing can be said definetely, about the gravitational field(D) None of these

53. The mass of the earth is 6.001024 kg and that of the moon is 7.401022 kg. The constantof gravitation G = 6.6710-11 Nm2 kg–2. The potential energy of the system is -7.791028

Joules. the mean distance between the earth and moon is = ............. meter.(A) 3.80108 (B) 3.37108 (C) 7.60108 (D) 1.90102

54. The masses and radii of earth and moon are M1, R1 and M2, R2 respectively. Their centresare d distance of apart. The minimum velocity with which a particle of mass m should be projectedfrom a point midway between their centres so that it esacapes to infinity is...............

(A) 1 22 ( )G M Md

(B) 1 222 ( )G M Md

(C) 1 22 ( )Gm M Md

(D) 1 2

1 2

( )2

( )Gm M M

d R R



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55. A rocket is launched with velocity 10 kms-1. If radius of earth is R then maximum height attainedby it will be = ..............(A) 2R (B) 3R (C) 4R (D) 5R

56. What is the intensity of gravitational field at the center of spherical shell

(A) 2

Gmr (B) g (C) zero (D) None of these

57. Escape velocity of a body of 1 kg. on a planet is 100 ms-1. Gravitational potential energy ofthe body at the planet is =............... J(A) -5000 (B) -1000 (C) -2400 (D) 5000

58. A body of mass m kg starts falling from a point 2R above the earth’s surface. Its K.E. whenit has fallen to a point ‘R’ above the Earth’s surface = .....................[R - Radius of Earth, M-mass of Earth G-Gravitational constant]

(A) 12

GMmR (B)

16

GMmR (C)

23

GMmR (D)

13

GMmR

59. The Gravitational P.E. of a body of mas m at the earth’s surface is -mgRe. Its gravitationalpotential energy at a height Re from the earth’s surface will be = ............ here(Re is the radius of the earth)

(A) –2 mgRe (B) 2 mgRe (C) 1 Re2

mg (D)1 Re2

mg

60. A body is projected vertically upwards from the surface of a planet of radius R with a velocityequal to half the escape velocity for that planet. The maximum height attained by the body is..........(A) R/3 (B) R/2 (C) R/4 (D) R/5

61. Energy required to move a body of mass m from an orbit of radius 2R to 3R is ................

(A) 2

GMm 12R (B) 23

G mR

(C) 8G m

R

(D) 6G m

R

62. Radius of orbit of satellite of earth is R. Its K.E. is proportional to

(A) 1R (B)

1R (C) R (D) 3/2

1R

63. A particle falls towards earth from infinity. It’s velocity reaching the earth would be.............

(A) infinity (B) 2gR (C) 2 gR (D) zero64. The escape velocity of a sphere of mass m from earth having mass M and Radius R is given by

(A) 2GM

R(B) 2 GM

R(C)

2GMmR

(D) GM

R65. The escape velocity for a rocket from earth is 11.2 kms–1 value on a planet where acceleration

due to gravity is double that on earth and diameter of the planet is twice that of earth will be= ........... kms–1

(A) 11.2 (B) 22.4 (C) 5.6 (D) 53.666. The escape velocity from the earth is about 11 kms–1. The escape velocity from a planet having

twice the radius and the same mean density as the earth is =............ kms–1.(A) 22 (B) 11 (C) 5.5 (C)15.5



160

67. If g is the acceleration due to gravity at the earth’s surface and r is the radius of the earth,the escape velocity for the body to escape out of earth’s gravitational field is.................

(A) gr (B) 2gr (C) g/r (D) r/g

68. The escape velocity of a projectile from the earth is approximately(A) 11.2 kms–1 (B) 112 kms–1 (C) 11.2 ms–1 (D) 1120 kms–1

69. The escape velocity of a particle of mass m varies as...................

(A) m2 (B) m (C) m0 (D) m-1

70. The escape velocity of an object from the earth depends upon the mass of earth (M), its meandensity (), its radius (R) and gravitational constant (G), thus the formula for escape veloctiy is

(A) 83

R G (B) 83

M GR (C) 2GMR (D) 2

2

GMR

71. Two small and heavy sphere, each of mass M, are placed distance r apart on a horizontal surfacethe gravitational potential at a mid point on the line joining the center of spheres is

(A) zero (B) GM

r (C)

2GMr

(D) 4

GMr

72. The escape velocity of a body from earth’s surface is Ve. The escape velocity of the samebody from a height equal to 7 R from earth’s surface will be

(A) 2Ve

(B) 2Ve

(C) 2 2Ve

(D) 4Ve

73. An artificial satellite is revolving round the earth in a circular orbit. its velocity is half the escapevelocity. Its height from the earth surface is = ............... km

(A) 6400 (B) 12800 (C) 3200 (D) 1600

74. The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is..........

(A) 3 Ve (B) 3 Ve (C) 2 Ve (D) 2Ve

75. There are two planets, the ratio of radius of two planets is k but the acceleration due to gravityof both planets are g what will be the ratio of their escape velocity.

(A) 1/ 2kg (B) 1/ 2kg (C) 2kg (D) 2kg

76. The escape velocity of a body on the surface of the earth is 11.2 km/sec. If the mass of theearth is increases to twice its present value and the radius of the earth becomes half, the escapevelocity becomes = ............... kms–1

(A) 5.6 (B) 11.2 (C) 22.4 (D) 494.8

77. Given mass of the moon is 181 of the mass of the earth and corresponding radius is

14 of

the earth, If escape velocity on the earth surface is 11.2 kms–1 the value of same on the surfaceof moon is = ....... kms–1.(A) 0.14 (B) 0.5 (C) 12.5 (D) 5



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78. 3 particle each of mass m are kept at vertices of an equilateral triangle of side L. The gravitationalfield at center due to these particies is ................

(A) zero (B) 2

3GML (C) 2

9GmL (D) 2

123

GmL

79. Escape velocity on the surface of earth is 11.2 kms–1 Escape velocity from a planet whose massesthe same as that of earth and radius 1/4 that of earth is = ....... kms–1

(A) 2.8 (B) 15.6 (C) 22.4 (D) 44.880. The velocity with which a projectile must be fired so that it escapes earth’s gravitational does

not depend on ..................(A) mass of earth(B) Mass of the projectile(C) Radius of the projectiles’s orbit(D) Gravitational constant

81. The escape velocity for a body projected vertically upwards from the surface of earth is 11kms-1. If the body is projected at an angle of 450 with the vertical, the escape velocity willbe ............kms-1.

(A) 112 (B) 11 2 (C) 22 (D) 11

82. The acceleration due to gravity on a planet is same as that on earth and its radius is four timesthat of earth. What will be the value of escape velocity on that planet if it is Ve on the earth(A) Ve (B) 2 Ve (C) 4 Ve (D) Ve/2

83. A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radiius10 cm. Find the work to be done aginst the gravitational force between them to take the particleis away from the sphere (G = 6.67 10-11 SI unit)(A) 6.6710-9 J (B) 6.6710-10 J (C) 13.3410-10 J (D) 3.3310-10 J

84. A particle of mass M is situated at the center of a spherical shell of same mass and radiusa the magnitude of gravitational potential at a point situated at a / 2 distance from the centerwill be

(A) 4GM

a (B) GM

a (C) 2GM

a (D) 3GM

a

85. The mass and radius of the sun are 1.99 x 1030 kg and R = 6.96 x 108 m. The escapevelocity of rocket from the sun is =.......... km/sec(A) 11.2 (B) 12.38 (C) 59.5 (D) 618

86. The mass of a space ship is 1000 kg. It is to be lauched from earth’s surface out into freespace the value of g and R (radius of earth) are 10ms-2 and 6400 km respectively the requiredenergy for this work will be = ................. J(A) 6.41011 (B) 6.4108 (C) 6.4109 (D) 6.41010



162

87. The diagram showing th variation of gravitational potential of earth with distance from the centerof earth is

(A) (B)

(C) (D)

88. A shpere of mass M and Radius R2 has a concentric cavity of Radius R1as shown in figure.The force F exerted by the shpere on a particle of mass m located at a distance r from thecenter of shhere varies as )r(O

(A) (B)

(C) (D)



163

89. Which one of the following graphs represents correctly the variation of the gravitational fieldwith the distance (r) from the center of spherical shell of mass M and radius a

(A) (B)

(C) (D)

90. Which of the following graphs represents the motion of a planet moving about the sun.

(A) (B)

(C) (D)

91. The curves for P.E. (U) K.E. (Ek) of two particle system ae shown in figure . At what pointssystemwill be bound.(A) Only at point D(B) Only at point A(C) At point D and A(D) At points A, B and C

Ek



164

92. The correct graph representing the variation of total energy (E) kinetic energy (K) and potentialenergy (U) of a satellite with its distance from the centre of earth is...........

(A) (B)

(C) (D)

93. A shell of mass M and radius R has a point mass m placed at a distance r from its center.The gravitational potential energy U(r) –v will be

(A) (B)

(C) (D)

Motion of satellite94. If Ve and Vo are represent the escape velocity and orbital velocity of satelllite correspoinding

to a circular orbit of radius r, then

(A) Ve = Vo (B) 2 Vo Ve (C) / 2Ve Vo (D)ve and vo are not related

Ener

gyEn

ergy

Ener

gyEn

ergy



165

95. If r represents the radius of the orbit of a saltellite of mass m moving around a planet of masM, the velocity of the satellite is given by

(A) 2 gMr

(B) 2 GMmr

(C) GMr

(D) 2 GMr

96. Two satellites of mass m1 and m2 (m1 > m2 ) are revolving round the earth in cirular orbitsof r1 and r2 ( r1 > r2) respectively. Which of the following statement is true regarding theirspeeds V1 and V2

(A) V1 = V2 (B) V1 < V2 (C) V1 > V2 (D) 1 2

1 2

V Vr r

97. A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance fromthe centere of earth in new orbit is two times of the earlier orbit. The time period in secondorbit is .........hours.

(a) 4.8 (B) 48 2 (C) 24 (D) 24 2

98. As astronaut orbiting the earth in a circular orbit 120 km above the surface of earth,gently dropsa spoon out of space-ship. The spoon will(A) Fall vertically down to the earth(B) move towards the moon(C) Will move along with sapace - ship(D) Will move in an irregualr way then fall down to earth

99. The period of a satellite in cirular orbit around a planet is independent of(A) the mass of the planet (B) the radius of the planet(C) mass of the satellite (D) all the three parameters (A), (B) and (C)

100. Two satellites A and B go round a planet p in circular orbits having radii 4 R and R respectivelyif the speed of the satellite A is 3V, the speed if satellite B will be(A) 12V (B) 6V (C) 4/3 V (D) 3/2 V

101 A small satellite is revolveing near earth’s surface. Its orbital velocity will be nearly= .......... kms–1.

(A) 8 (B) 4 (C) 6 (D) 11.2102 A satellite revolves around the earth in an elliptical orbit. Its speed

(A) is the same at all points in the orbit(B) is greatest when it is closest to the earth(C) is greatest when it is farthest to the earth(D) goes on increasing or decresing continuously depending upon the mass of the satellite

103. If the height of a satellite from the earth is negligible in comparison of the radius of the earthR, the orbital velocity of the stellite is ............

(A) gR (B) gR/2 (C) /g R (D) gR



166

104 A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbitradius is decreasd by 1% its speed will(A) increase by 1% (B) increase by 0.5%(C) decrrease by 1% (C) Decrrease by 0.5%

105. orbital velocity of an artifiicial satellite does not depend upon(A) mass of earth (B) mass of satellite(C) radius of earth (D) acceleration due to gravity

106. orbital velocity of eath’s satellite near the surface is 7 kms–1. when the radius of orbit is 4 timesthat of earth’s radius, then orbital velocity in that orbit is =........kms-1

(A) 3.5 (B) 17 (C) 14 (D) 35107. Two identical satellites are at R and 7R away from each surface, the wrong statement is

(R - Radius of earth)(A) ratio of total energy will be 4(B) ratio of kinetic energes will be 4(C) ratio of potential energies will be 4(D) ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2

108. Which one of follwoing sttements regarding artificial satellite of earth is incorrect(A) The orbital velocity depends on the mass of the satellite(B) A minimum velocity of 8kms-1 is required by a satellite to orbit quite close to the earth.(C) The period of revolution is large if the radius of its orbit is large(D) The height of geostationary satellite is about 36000 km from earth

109. The weight of an astronuat, in an artificial satellite revolving around the earth is(A) zero(B) Equal to that on the earth(C) more than that on earth(D) less than that on the earth

110. The distance of a geo-stationary satellite from the center of the earth (Radius R= 6400km)is nearest to(A) 5R (B) 7R (C) 10R (D) 18R

111. A geo-stationary satellite is orbiting the earth of a height of 6R above the surface of earth, Rbeing the radius of earth. The time period of another satellite at a height of 2.5 R from thesurface fo earth is = ...............hr(A) 6 (B) 6 2 (C) 10 (D) 6 / 2

112. If the gravitational force between two objects were proportional to 1R (and not as 2

1R )where

R is separation between them, then a particle in circular orbit under such a force would haveits orbital speed v proportional to

(A) 2

1R (B) R0 (C) R1 (D)

1R

113. A satellite moves around the earth in a circular orbit of radius r with speed v, If mass of thesatellite is M , its total energy is

(A) 212

MV (B) 212

MV (C) 232

MV (D) MV2



167

114. A satellite with K.E. Ek is revolving round the earth in a circular orbit. How much more K.E.should be given to it so that it may just escape into outerspace ?

(A) Ek (B) 2 Ek (C) 1 E2 k (D) 3 Ek

115. Potential energy of a satellite having mass m and rotating at a height of 6.4 x 106 m fromthe surface of earth(A) -0.5 mg Re (B) -mg Re (C) -2mgRe (D) 4 mgRe

116. When a satellite going round the earth in a circular obrit of radius r and speed v loses someof its energy, then r and v changes as(A) r and v bothe will increase(B) r and v both will decease(C) r will decrease and v will increase(D) r will increase and v will decrease

117. The time period of a satellite of earth is 5 hours If the sepration between the earth and thesatellite is increased to four times the previous value, the new time period will become ...... hours(A) 10 (B) 120 (C) 40 (D) 80

118. A person sitting in a chair in a satellite feels weightless because(A) the earth does not attract the objects in a satellite(B) the normal force by the chair on the person balances the earth’s attraction(C) the normal force is zero(D) the person in satellite is not accelerated

119. Two satellites A and B go round a planet in cirular orbits having radii 4R and R respectivelyIf the speed of satellite A is 3v, then speed of satellite B is(A) 3v/2 (B) 4v/2 (C) 6v (D) 12v

120. A satellite moves in a circle around the earth, the radius of this circlr is equal to one half ofthe radius of the moon’s orbit the satellite completes one revolution in .........lunar month(A) 1/2 (B) 2/3 (C) 2–3/2 (D) 123/2

121. The additional K.E. to be provided to a satellite of mass m revolving around a planet of massM, to transfer it from a circular orbit of radius R1 to another radius R2 (R2 > R1) is

(A) 2 21 2

1 1

G M m

R R(B)

1 2

1 1

G M m

R R

(C) 1 2

1 12

G M mR R

(D) 1 2

1 1 12

G M m

R R

122. Rockets are launched in eastward direction to take advantage of(A) the clear sky on eastem side(B) the thiner atmosphere on this side(C) earth’s rotation(D) earth’s tilt



168

123. A satellite of mass m is orbiting close to the surface of the earth (Radius R = 6400 km) hasa K.E. K. The corresponding K.E. of satellite to escape from the earth’s gravitational field is(A) k (B) 2 k (C) mg R (D) m k

124. A planet moving along an elliptical orbit is closest to the sun at a distance r1 and farthest awayat a distance of r2. If v1 and v2 are the liner velocities at these points respectively, then the

ratio 1

2

vv is ............

(A) 1

2

rr (B)

2

1

2

rr

(C)2

1

rr (D)

2

1

2

rr

125. The time period T of the moon of planet Mars(Mm) is related to its orbital radius R as(G = Gravitational constant)

(A) 2

2 4

RTGMm (B)

2 22 4π GRT

Mm

(C) 2

2 2

R GTMm (d) T2 = 4 Mm GR2

126. A geostationary satellite is orbiting the earth at a height of 5 R above that of surface of theearth. R being the radius of the earth. The time period of another satellite in hours at a heightof 2R from the surface of earth is ............ hr

(A) 5 (B) 10 (C) 6 2 (D) 6 / 2127. The figure shows ellipticall orbit of a planet m about the sun s. the shaded area SCD is twice

the saded area SAB. If t1 is the time for the planet to move from C and D and t2 is the timeto move from A to B then

(A) t1 > t2 (B) t1 > 4t2 (C) t1 = 2t2 (D) t1 = t2128. The period of a satellite in a circular orbit of radius R is T. the period of another satellite

in a circular orbit of radius 4R is(A) 4T (B) T/4 (C) 8T (D) T/8

129. If the earth is at one- fourth of its present distance from the sun the duration of year will be(A) half the pesent Year(B) one-eight the present year(C) one-fourth the present year(D) one-sixth the present year

130. The orbital speed of jupiter is(A) greater than the orbital speed of earth (B) less than the orbital speed of earth(C) equal to the orbital speed of earth (D) zero



169

131. Kepler’s second law regarding constancy of aerial velocity of a palnet is consequence of thelaw of conservation of(A) energy (B) angular Momentum(C) linear momentum (D) None of these

132. The largest and shortest distance of the earth from the sun are r1 and r2 its distance fromthe sun when it is at the perpendicular to the major axis of the orbit drawn from the sun

(A) 1 2

4r r

(B) 1 2

1 2

r rr r (C)

1 2

1 2

2r rr r (D) 1 2

3r r

133. According to keplar, the period of revolution of a planet (T) and its mean distance from thesun (r) are related by the equation

(A) 3 3 tanT r cons t (B) 2 3 tanT r cons t

(C) 3 tanTr cons t (D) 2 tanT r cons t

134. A satellite of mass m is circulating around the earth with constant angular velocity. If radius ofthe orbit is Ro and mass of earth M , the angular momentum about the center of earth is

(A) m GMRo (B) M GMRo

(C) GMmRo

(D) GMMRo

135. The earth E moves in an elliptical orbit with the sun s at one of the foci as shown in figure.Its speed of motion will be maximum at a point ..........

(A) C (B) A (C) B (D) D136. The period of revolution of planet A around the sun is 8 times that of B. The distance of A

from the sun is how many times greater than that of B from the sun.(A) 2 (B) 3 (C) 4 (D) 5

137. The earth revolves round the sun in one year. If distace between then becomes double the newperiod will be .......... years.

(A) 0.5 (B) 2 2 (C) 4 (D) 8138. The maximum and minimum distance of a comet from the sun are 8 x 1012m and 1.6 x 1012

m. If its velocity when nearest to the sun is 60 ms-1, What will be its velocity in ms-1 whenit is farthest ?(A) 6 (B) 12 (C) 60 (D) 112

139 The period of moon’s rotation around th earth is nearly 29 days. If moon’s mass were 2 foldits present value and all other things remained unchanged the period of moons’s rotation wouldbe nearly .....days

(A) 29 2 (B) 29 / 2 (C) 29 2 (D) 29



170

140. If the velocity of planet is given by a b cG M R then

(A) a =1/3 b = 1/3 c = -1/3(B) a =1/2 b = 1/2 c = -1/2(C) a =1/2 b = -1/2 c = 1/2(D) a =1/2 b = -1/2 c = -1/2

141. The radius of orbit of a planet is two times that of earth. The time period of planet is.........years.(A) 4.2 (B) 2.8 (C) 5.6 (D) 8.4

142. If r denotes the distance between the sun and the earth, then the angular momentum of theearth around the sun is proportional to(A) r3/2 (B) r (C) r1/1 (D) r2

143. What does not change in the field of central force(A) potential energy (B) Kinetic energy(C) linear momentum (D) Angular momentum

144. A thin uniform annular disc (See figure) of mass M has outerradius 4R and inner radius 3R. The work required to take aunit mass from point P on its axix to infinity is..............

(A) 2 (4 2 5)

7GM

R (B)

2 (4 2 5)7GM

R

(C) 4GM

R (D) 1)2(5R

2GM

145. Suppose the gravitational force varies inversely as the nth power of distance the time periodof planet in circular orbit of radius R around the sun will be proportional to

(A) 1

2n

R

(B)

12

n

R (C) Rn (D) 2

2

n

R

146. If the radius of the earth were to shrink by 1% its mass remaing the same, the accelration dueto gravity on the earth’s surface would(A) decrease by 2% (B) remain Unchanged(C) increase by 2% (D) increases by 1%

147. A body of mass m is taken from earth surface to the height h equal to radius of earth, theincrease in potential energy will be

(A) mg R (B) 12

mgR (C) 2 mgR (D)14

mgR

148. An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential lenergy)Eo, its potential energy is(A) -Eo (B) 1.5 Eo (C) 2 Eo (D) Eo



171

149. 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 agravitational attraction Their relative velocityof apporach at sepration distance r between them is

(A) 12

1 22 ( )

G m mr

(B) 12

1 22 ( )

G m mr

(C)

12

1 22

rG m m

(D) 12

1 22 r

G m m

150. A geostationary satellite orbits around the earth in a circular orbit of radus 3600 km the timeperiod of a satellite orbiting a few hundred kilometers above the earth’s surface (Rearth = 6400km) will approximately be = ...... h(A) 1/2 (B) 1 (C) 2 (D) 4

KEY NOTE1 . C 25. B 49. C 73. A 97. B 121 D 145. A2 . A 26. A 50. A 74. A 98. C 122. C 146. C3. B 27. C 51. C 75. A 99. C 123 B 147. B4. A 28. A 52. A 76. C 100. B 124. C 148. C5. B 29. C 53. A 77. C 101. A 125. A 149. B6. C 30. C 54. A 78. A 102. B 126. C 150. C7. A 31. A 55. C 79. C 103. D 127. C8. D 32. C 56. C 80. B 104. B 128. C9. C 33. B 57. A 81. D 105. B 129. B10. D 34. D 58. B 82. B 106. A 130. B11. B 35. A 59. D 83. B 107. D 131. B12. D 26. B 60. A 84. D 108. A 132. C13. A 27. C 61. D 85. D 109. A 133. B14. A 38. C 62. A 86. D 110. B 134. A15. A 39. A 63. B 87. C 111. D 135. B16. C 40. D 64. A 88. B 112. B 136. C17. C 41. A 65.B 89. D 113. A 137. B18. C 42. B 66. A 90. C 114. A 138. A19. D 43. B 67. B 91. D 115. A 139. D20. C 44. C 68. C 92. C 116. C 140. B21. B 45. D 69. C 93. C 117. D 141. C22. A 46. B 70. A 94. B 118. C 142. C23. B 47. A 71. D 95. D 119. C 143. D24. B 48. A 72. C 96. B 120. C 144.A

1 . C 25. B 49. C 73. A 97. B 121 D 145. A 2 . A 26. A 50. A 74. A 98. C 122. C 146. C3. B 27. C 51. C 75. A 99. C 123 B 147. B4. A 28. A 52. A 76. C 100. B 124. C 148. C5. B 29. C 53. A 77. C 101. A 125. A 149. B6. C 30. C 54. A 78. A 102. B 126. C 150. C7. A 31. A 55. C 79. C 103. D 127. C8. D 32. C 56. C 80. B 104. B 128. C9. C 33. B 57. A 81. D 105. B 129. B10. D 34. D 58. B 82. B 106. A 130. B11. B 35. A 59. D 83. B 107. D 131. B12. D 26. B 60. A 84. D 108. A 132. C 13. A 27. C 61. D 85. D 109. A 133. B14. A 38. C 62. A 86. D 110. B 134. A15. A 39. A 63. B 87. C 111. D 135. B16. C 40. D 64. A 88. B 112. B 136. C17. C 41. A 65.B 89. D 113. A 137. B18. C 42. B 66. A 90. C 114. A 138. A19. D 43. B 67. B 91. D 115. A 139. D20. C 44. C 68. C 92. C 116. C 140. B21. B 45. D 69. C 93. C 117. D 141. C22. A 46. B 70. A 94. B 118. C 142. C23. B 47. A 71. D 95. D 119. C 143. D24. B 48. A 72. C 96. B 120. C 144.A



172

HINT

(1) (c) 3. 2

2 2 42 2

4( )( )( ) 43(2 ) 4 3

G RG m mF R

R R

(4) (A) Gravitational force does not depend upon the medium(5) (B) Due to inertia to direction(6) (D) the force exerted by sun on the earth

F = m2R 24 7 2 11(6 10 ) (2 10 ) (1.5 10 )

2136 10 N

(7) 2( 1) ( 1)F xm x m m x x

for maximum force 0dFdx

2 22 0 1/ 2 dF m m x xdx

(9) (C) cenripetal force provided by the gravitational force of attraction between two particites2

2

( )( )(2 )

m G m m

r r

12

Gmr

(10) (D)

For will be zero at the point of zero intensity

1

1 2

81 91081

m mx D Dm m m m

11. B 2Re eGM mmg where Me and Re is the mass and radius of the earth respectivel

2Re egMG

12. (D) Here FOA = FOB = FOC = 2

( ) (2 )G m mr

OA OB OCF F F F

(D - X)



173

=

= 014. (A) let a point mass C is placed at a distance of x m from the point mass A as show in figure

43a

b

mherem

2

( )( ) .........( )G m mAFAC i

x

2

( )( ) .........( )(1 )

G m mBFBC iix

According to given problem 14

FAC FBC

with the help of equ (i) and (ii)

2 2

( )( ) 1 ( )( )3 (1 )

G m mA G m mBx x

2 2 2

2 2 2

4 43(1 ) 3 3(1 ) (1 )

A

B

m x x xm x x x

2 2 21

x x xx

23 2 3x x m

17. (c) 32

43

GMg and M RR

32

4 3.3 4

G gg RR RG

18. (c) for condition of weight lessness at equator

3

10 1 sec6400 10 800

gR

rad

19. (D) Time peripd of simple pendylym 2 'T l g

In artificial satellite g’ = 0 T

mA m mB(1–x)C

x



174

20. (C) 2 2 2

4( 2)

GM GMo GMogR Do Do

21. (B) we know that 2

GMgR

on the planet gp 2

7 4 74

GM gR

Hence the weight on the planet = 700 47 = 400 gm t

22.22

2' 6400 ' 960.400( ) 6400 64

g R g msg R h

23. 1 2 2 g g R Cos Rotation of the earth results in the decreased weight apparently.this decrease in weight is notfelt at the poles as the angle of latitude is 90’

24. (B) using 2

GMgR

we get gm = g/5

25. (B) 1 d ( 1)g g (1 )R

nd Rn

26. (A) 4g πGR and g' 4/3πGR'3

g' R' 0.2 g' 0.2gg R

27. (C) mass does not vary from place to place

28. (a) 22

2

4'9

R

R Rg g g gR h R

4W' 4/9 w (72) 32 N9

29. (c) RhsolvingbyhR

Rgg

4/9'

2

30. (c) 2

2 2p e

Mp Re 1 9.8g g 9.8 (2) 0.49msMe Rp 80 20

31. (A) 2d 100g' g 1 9.8 1 9.66 ms

R 6400



175

32. (C) 2RGMg and k =

2L2I

If mass of the earth and its angular momentum ramains constant then 2

1gR

and 2

1kR

i.e. if radius of earth decreases by 2% then g and k both increase by 4%33. (B) Weight on surface of earth, mg = 500N and weight below the surface of earth at

d = R2 , mg1 = d mgmg 1 mg 1 1/2 250N

R 2

34. (D) 1 1 1

2 2 2

g4 Rg π GR 3 g R

35. (A) 2rGMαg g

1αrorr1αg 2

If g decreases by one percent then r should be increase by 1/2 % i.e. 1 6400 32

2 100R km

37. (C) RhRhR

RhR

Rgg 22

1'2

h 2 1 R 0.414 R

Hence distance form center = R + 0.414R = 1.414R

38. (C) 4g π GR g α d R ( d given in the problem)3

39. (A) insided the earth π34g' Gr rg '

40. (D) 4g π GR3

1(1)

Re 2

p e

e p

gRpg

Re

22RRp

41. (A) g

42. (B) For height 100424100%

gΔg

%

For depth 0.51/2Rh

Rd100%

gΔg

43. (B) g R



176

44. (C) 2

B A

A B

V 1 H gH Hα2g g H g

Now AB

gg12

as g R

B AB A

A B

H g 12 H 12 H 12 1.5 18 mH g

46. (B) 2

2R Rg' g g 4/9.g(g 10 ms )RR h 3 2

47. (A) e e

m m

g4 Reg π GR g α R .3 g Rm

RmRe.

35

16

Re185Rm

48. (A) ' /92 2

R Rg g g gR R R R

49. (C) 2)/1('

Rhgg

2)/1(16 Rhgg

1612

Rh

41 Rh

3Rh

Rh 3

50. (A) Gravitational attraction fore on particle B

2)2/(DpGMpmFg

Acceleration of paritcle due to particle due to gravity 24DpGMp

mFga



177

51. (C) 2

1g α r it(r r) and g α (it r R)r

52. (A) dxdvI

53. (A) rGMmU

r

24221128 1061071067.61079.7

m103.8r 8

54. (A) 2/2/21

dGM

dGMV

1 22GMNow P.E mv (m m )

d (m mass of particie)

K.E = P.E

21 2

1 2 ( )2

GMm m md

1 2( )2 G M Md

55. (C) If the body is projected with velocity Ve)(υυ then height up to where it rises

2 2

2

4 ( )11.2110

R Rh R approxVeV

57. (A) 2GM GMmVe P.E. U 5000 JR R

RGM

100

5000R

GM

58. (B) P.E.

GMm GMmUr R h

Uinitial = – 3GMm

R and Ufinal = – 2GMm

R

loss of P.E. = gain in K.E = 2GMm

R – 3GMm

R = 6GMm

R



178

59. (D) 2 1mgh mgRe mgReΔU U U Re1 h/Re 21

Re

2mg ReU ( mg Re)

2 mgRe

21U2

60. (A) If body is projected with velocity Ve)(υυ then height up to which it will rise

R/314

R

1Ve/2Ve

Rhve/2ubut 2

61. (D) change in potential energy in displacing a body from r1and r2 is given by

1 2

1 1 1 12 3 6

GMmU GMm GMmr r R R R

62. (A) K.E = 2GMm

R R1αK.E.

63. (B) this should be equal to escape velocity i.e. = 2gR64. (A) A Escape velocity does not depend on the mass of the projectile

65 (B) p

e

gVp Rp. 2 2 2Ve g Re

122.4Kms11.222VeVe

66. (A) 83

2GMVe R GR

Ve α R if constant

since the planet is having double radius incomparision to earth therefore the escape velocitybecomes twice i.e. 22 kms-1

69. (C) Because it does not depend on the mass of projetile

71. (D)

Gravitational potential of A at 0 rGM

rGM 2

2/

of B at 0 rGM

rGM 2

2/

Total potential at 0 4GM

r



179

72. (C) r1αVe where r is the position of body from the surface

1 2 12

2 1

V r R 7R V2 2 VV r R 2 2

73. (A) R

2GM21

hRGMV

h)2(R4R

h R 6400 km

74. (A) 3216

RpRe

MeMp

VeVp

Ve3Ve

75. (A)1/2(kg)gk

R2R1.

g2g1

V2V12gRυ

76. (C) RMαVe

R2GMVe

If M becomes double and R becomes half then escape velocity; becomes two times

77. (C) on earth Ve = 12GM 11.2 kmsR

on moon 12GM 4 2 2GM 2 11.2Vm . 2.5 kmsR R 9 R 9

78. (A) Due to three particles net intensity at the center

A B CI I I I 0

because out of these three intensities ARE equal in magnitudeand between each other is 120’

79. (C) R1αVe if R becomes 4

1 then Ve will be 2 times

80. (D) Escape velocity does not depends upon the angle of projection

82. (B) p

e

gVp Rp2gR . 1 4 2Ve g Re

Ve = 2Ve83. (B) potential energy of system of two mass

J106.671010

1010100106.67R

GMmU 102

311

so, the amount of work done to take the particle up to infinte will be 6.6710-10 J



180

84. (D) Vp = Vsphere + Vpartical aGM

aGM

aGM 3

2/

85. (D) Ve = R

GM2=

11 3012 6.67 10 1.99 10 618 kms

6.98 108

86. (D)2GMm GMm mW 0- - gR mgR

R R R

3106400101000 91064

J106.4 10

87. (C) GM GM GMVin Vsurface VoutR R r

88. (B) 1F 0 When 0 r R

beccause intensity is zero inside Caviy

Fincrease When 21 RrR

2RrwhenR1Fα

89. (D) Intensity will zero inside the spherical shell

I = upto r = a and I 2

1r when r > a

90. (C) kepler’s law T2 α r3

91. (D) System will be bound at points where total energy is negative. In the given curve at pointA, B, and C the P.E. is more than K.E.

92. (C) 2rGMmEand

2rGMmK

rGMm-U

For satellite U, K, and E varies with r and also U and E remains negativw where K reaminalways positive

93. (C) Gravitational P.E. = m x gravitational potential U = mv so the graph of U will be sameas that of V for a spherical shell

94. (B) 0 0Ve 2gR and V gR Ve 2 V

96. (B) 2121 VVthenrritr

GMυ



181

97. (B) 3

2T α r if r becomes double then time period will become (2)3/2 times so new time period

will be 24 x 2 2 hr i.e. = 48 298. (C) The velocity of the spoon will be equal to the orbital velocity when dropped out of the

space ship

100. (B) A B

B A

GM V R R 1υR V R 4R 2

AB

B

V 3V V 6VV VA

104. (B) r1αυ

% increase in speed = 1 %2 decrease in radius

= 12 (1%) = 0.5 %

105. (B)r

GMυ

106. (A) r1αυ it orbital radius becomes 4 times then orbital velocity will becomes halt

107. (D) orbital rudius of satellites r1 = R+R = 2Rr2 = R+7R = 8R

22

11 r

GMmUandr

GMmU

1 21 2

GMm GMmK and K2r 2r

4EE

KK

UU

2

1

2

1

2

1

110. (B) 6R from the surface of earth and 7R from the center111. (D) Distance of satellite from the center are 7R and 3.5R respectively

2

3/2 3/22 2

1 1

T r 3.5RT 24 6 24 hrT r 7R

112. (B Gravitational force provides the required centripetal force for orbiting the satellite

RK

Rmυ2

becuse R1αF

oυ α R



182

113. (A) Total energy = -K.E. = 21

2m

114. (A) B.E. = -K.E.And it this amount of energy(Ek) given to satellite it will escape into outer space

115. (A) P.E. = GMmr

= –Re GMm

h = –

2 eGMm

R = –

2Re2 Re

g m = –

12 mg Re = 0.5 mg Re

116. (C) B. E. = GMm

r if B.E. decreases the r also decreases and V increases as 1r

117. (D) 3/2

3/222 1 1 1

1

RT T T (4) 8T 40 hrR

118. (C) 4 2 B A

A B

V r RV r R

120. time period of revolution of moon around the earth = 1 lunar month3/2 3/2

2/3e ee

m m

T r 1 T 2T r 2

lunar month

121 (D) 21 2R

GMmK.E.2R

GMm

21

112

.RR

GMmEK

122. Because Earth rotation from west to east direction

123. (B) 2GMmk

R

124. (C) v1 r1 = v2r2 (angular momentum is constant)

125. (A) Time period 2/322

GMR

RGMm

RT

GMmRT

322 4

126. (C) 2 3 3

1 12 3 3

2 2

T R (6R ) 8(3R )T R

22

24 24 728

T

2 6 2T



183

127. (C) ΔA L Δ A α Δ tΔt 2m

SCD 11 2

SAS 2

ΔA t 2A t 2tΔA t A

128. (C)

3 32 21 1

2 2

T R RT R 4 R

2 18T T

129. (B) 32 rαT since T81T

41

TT 1

321

130. (B) orbital radius of Jupiter > orbital radius of Earth J e

e j

V rV r

As rj < re there fore Vj < Vee

131. (B) 2LdA

dt m = constant

132. (C) The earth moves around the sun in elliptical path, so by using the properties of ellipse

e)(1r1 a and r2 = (1-e) r2, a = 1 2

2r r

2221 )1( aerr

where a= semi major axisb= semi minor axise= eccentircity

Now required distance = sem latysrectum = b2/9

21

21

21

212

2

rrr2r

)/2r(rrr

a)e(1a

133. (B) T2/ r3 = constant T2 r-3 = constant134. (A) Angular momentum = Mass orbital velocity x Radius

00

RR

GMm

0m GMR

135. (B) speed at the earth will be maximum when its distance from the sun is minimum becausem r = constant

136 (C) 3 3

2 2 3A A A 2A B B

B B B

T r 48 r 4 8 4rT r 4

137. (B) years22T22(2)rr

TT

22

323

1

2

1

2



184

138. (A) By conservation of angular momentum (mr) = constantVmin X rmax = Vmax X rmin

121

12

60 1.6 10 60Vmin 12 ms8 10 5

139. (D) Time period does not depends upon the mass of satellite

140. (B) 1 1 1

2 2 2GMυ G M RR

141. (B) yar2.821RRTT 2

32

3

1

212

142. (C) Angular momentum of earth around the sunL = MEVo r

2SE E S

GMM .r M .GM .rr

ME = mass of earthMS = mass of Sunr = Distance between sun and the earth

rαL143. (D) For central force toraqe is zero

dL 0 L Constantdt

144. (A) Wext FUU

Gdm0 (1)

x

2 2 2

M 2π r drGπ 7R 16R r

2 2 2

2 GM rdr7R 16 R r

2 2

GM π d2 GM (2)7R a 7R

4R

3R22

2 r16RR7GM2

5RR247RGM2

5247RGM2



185

145. (A) 2

2 2 n 1n 2 n

1 4π 1m Rα m R α T α RR T R

n+12T α R

146. (C) 2RGMg is mass ramains constant then

2

1R

147. (B) mgR21

Rh1

mghΔU

148. (C) P.E. = 2. total energy = 2E0

Because we know U = – GMM

r and Eo = – 2GMM

r149. (B) let velocities of these masses at r distance from each other be v1 and v2 respectively

By conservation of momentum 1 1 2 2 1 1 2 2m V m V 0 m V m V ____________ 1

By conservation of energychange in P.E. = change in K.E.

222

211

21 Vm21Vm

21

rmGm

ii_______r

mGm2m

VmmVm 21

2

222

1

211

on solving (i) and (ii)

21

2

122mmr

Gm

and 21

2

2 mmr2Gm

υ 1

1 21 2

2G m mV app V V

r

150. (C)

3 32 22 2

21 1

T r 6400T 24 2 hoursT r 3600



186

Assertion - Reason Type QuestionsDirection (Read the following uestions and choose)(A) If both Assertion and Reason are true and the Reason is correct explanation of assertion(B) If both Assertion and Reason are true, but reason is not correcte explanation of the Assertion(C) If Assertion is true, but the Reason is false(D) If Assertion is faluse, but the Reason is true

1. Assertion : The value of acc. due to gravity (g) does not depend upon mass of the body

Reason : This follows from 2

GMgR

, where M is mass of planet (earth) and R is radius of

planet (earth)(a) A (b) B (c) C (d) D

2. Assertion : Unit of gravitational field intensity is N/kg or ms-2

Reason : Gravitational field intensity

22

mskg

kg.m/seckgN

massForce

(a) A (b) B (c) C (d) D3. Assertion : The time period of a geostationary satellite is 24 hours

Reason : Such a satellite must have the same time period as the time taken by the earth tocomplete one revolution about its axis

(a) A (b) B (c) C (d) D4. Assertion : Even when orbit of a satellite is elliptical, its plane of rotaiion passes through the

center of earthReason : This is in accordance with the principle of conservation of angular momentum(a) A (b) B (c) C (d) D

5. Assertion : The time Period of pendulum, on a satlellite orbiting the earth is infinity

Reason : Time period of a pendulum is inversely proportional to g(a) A (b) B (c) C (d) D

6. Assertion : The escape velocity on the surface of a planet of the same mass but 14 times the

radius of earth is 5.6 kms-1

Reason : The escape velocity Ve = 2gR(a) A (b) B (c) C (d) D

7. Assertion : The comet does not obey kepler’s law of planetaty motionReason : The comet does not have ellitical orbit(a) A (b) B (c) C (d) D

8. Assertion : The square of the period of revolution of a planet is proportional to the cube ofits distance from the sun.

Reason : Sun’s gravitational field is inversely proportional to the square of its distance fromthe planet

(a) A (b) B (c) C (d) D9. Assertion : Space ship while entering the earth’s atmoshere is likely to catch fire



187

Reason : Temperature of upper atomosphere is very high(a) A (b) B (c) C (d) D

10 Assertion : The earth is slowing down and as a result the moon is coming nearer to itReason : The angular momentum of earth - moon system is not conserved(a) A (b) B (c) C (d) D



188

Assertion - Reason Type question :1. A Both the asseration and reason are true and the latter is correct expalnation of the former.2. A Both the asseration and reason are true and the latter is correct expalnation of the former.3. A As the satellite is to be stationary over a particular place, its time period of revolution = 24

hours= time peroid of revolution of earth about its axis.

4. A g2πT on a satellite, there is a weightless -ness, so g= 0 hence T

Thus both the asseration and reason are correct5. A As no torque is acting on the planet, its angular momentum must stay constant in a magnitude

as well as direction there for planet of rotation must pass throught the center of earth

6. DR1αVei.e.

RGM2Ve

1RVe' = Ve 2 Ve 2 11.2 22.4 kmsR4

thus assertion is wrong but Reason is correct7. A Both the asseration and reason are correct and the Reason is correct explanation of Asseration8. A A Both the asseration and reason are correct and the Reason is correct explanation of Asseration9. C Here Asseration is correct but Reason is wrong because the space ship while entering the earth’s

atmosphere may catch fire due to atmoshperic air friction10. D Here Both the asseration and reason are wrong because the angular momentum of earth

moon system is conserved in the absence of extermal touque.



189

Comperehensions Type Questions(1) If a smooth tunnel is dug across a diameter of earth and a particle is related from the surface

of earth, the particle oscillates simple harmonically along it.(1) Time period of the particle is not equal to

(A) gR2π (B) 2

3R

GM2π

(C) 84.0 min (D) Nome of these(2) Maximum speed of the these

(A) RGM2

(B) R

GM(C)

R 2GM3

(D) R 2

GM

2. When a paricle is projected from the surface of earth, it mechanincal energy and angular momentumabout center of earth at all time is constant(i) A particle of mass m is projected from the surface of earth with velocity vo at angle

with horizontal suppose h be the maximum height of particle from surface of earth and vits speed at that point them v is(A) v0coso (B) >v0coso(C) <v0coso (D) zero

(ii) Maximum height h of the paritcle is

(A) 2gθsinVo 22

(C) 2gθsinVo 22

(B) 2gθsinVo 22

(D) can be greater than or less than 2gθsinVo 22

3. A solid sphere of mass M and radius R is surrounding by a sphericalshell of same mass M and raius 2R as shown. A small particleof mass m is relased from rest from a height (h <<R) aobove theshell there is a hole in the shell.(i) in what time will it enter the hole at A

(A) GM

hR2 2

(C) GMhR 2

(B) GM

2hR 2

(D) None of these

(ii) what time will it take to move from A to B ?

(A) 2R

GMh (C)

2RGMh

(B) 2R

GMh (D) None of these



190

(iii) with what apporximate speed will it colide at B ?

(A) 2 GM

R(C)

2RGM

(B) 2RGM3

(D) R

GM

4. A planet is revolving round the sun in elliptical orbit. Velocity at perigee position (nearest) isv1| and at apogee position (farthest) is v2 Both these velocities are perpendicular to the joingingcenter of sun and planet r is the minimum distance and r2 the maximum distance.(i) when the planet is at perigee position, it wants to revolve in a circular orbit by itself. For

this value of G(A) Should increase (B) Should decrease(C) data is in sufficeint (D) will not depend on the value of G

(ii) At apogee position suppose speed of planer is slightly decreased from v2, then what willhappen to minimum distance r1 in the subsequent motion(A) r1 and r2 bothe will dicreases(B) r1 and r2 bothe will increases(C) r2 will remain as it is while r1 will increase(D) r2 will remain as it is while r1 will decrease

5. Garvitational potential at any point inside a spherical shall is uniform

and is given by GM

R where M is the mass of shell and R its

radius. At the center solid sphere, potential is 3

2

GMR

(i) There is a concentric hole of radius R in a solid sphere of radius 2R Mass of the remainingportion is M What is the gravitation potential at center?

(A) 7R3GM

(B) 5GM7R

(C) 14GM7

(D) R 14GM9



191

Solution

(1) (i) haveweRGMgputting

gR2πT 2

322πT R

Gm

(ii)maximum speed is at centre from conservation mechanicall energy(from cerface to center)increase in K.E. = decrease in P.E.

2i f i f

1 m U U m(V - V )2

n

i fυ 2(v v )

RGM

12GM3

RGM2

(2) (i) From conservation of angular momentum at A and B

mVocosθ m R h

0R V cosθ

R h

oV cosθ

(ii) From consercation of mechanical energy, Decrease in K.E. = increase in P.E.

R

h1mghVVm

21 22

0

θcosVVherebutR

h1gh2VV 0

220

xVVletsoθsinVVV 220

220

220

θsinVxhere 220

Rh1

2ghx

Rx2g

xh

i.e. h >x

2g i.e. h >2 2

O sin2g



192

(3) (i) Acceleration due to gravity near the surface of shell can be assumed to be uiform

22 2RGM

(2R)G(2M)g

From 21h2

gt

t = GMhR

gh 2

22

(ii) 2 22 22AGM GMhV gh h

R R

From A to B field due to shell is zero, but field due to sphere is not-zero

hence tAB

2

A

R RV GMh

(iii) KA=0 polential between A and B due to shell isFrom energy conservationKA + UA = KB + UBKB = (UA = UB) = m(VA – VB)

12 mVB

2 = m(VA – VVB)

B A BV 2(V V ) = GM

R(4) At perigee position v1 > v0 where v0 is the orbital velocity for circular motion

0GMV G

r

so value of G should incease , so that v0 will increase for this position and whichwill become equal to v1(ii) path will becomemore elliptical , keeping r2 constant and r1 to decrease

(5) Density of given material

3 3 343

M 3Mπ[(2R) R ] 28πR

Vwhole = Vhole + VremaingVremaing = Vwhole – Vhole

1 23 GM GM2 2R R



193

here 31

4 8M π (2R) M3 7

32

4 MM ( ) R3 7

14R9GMVremaining

Match the Column(1) on the surface of earth acceleration due to gravity is g and gravitational potential is V match

the followingTable - 1 Table -2

(A) At height h = R value of g (P) decreases by a factor 14

(B) At height h = R/2 value of g (Q) decreases by a factor 12

(C) At height h = R value of v (R) increases by a factor 11/8(D) At depth h = R/2 value of V (S) increases by a factor 2

(T) None(2) Density of planet is two times the density of earth Radius of this planet is half (As compared

to earth)Match the following

Table-1 Table-2(A) Acceleration due to gravity on (P) Half

this planet’s surface(B) Gravitational potential (Q) same

on the surface(C) Gravitational potential (R) Two times

at centre(D) Gravitational field strength (S) four times

at centre(3) let V and E denote the gravitational potential and gravitational field at a point. Then the match

the follwingTable-1 Table-2

(A) E = 0, V = 0 (P) At center of Spherical shell(B) E 0, V= o (Q) At centre of solid sphere(C) V 0, E -0 (R) At centre of circular ring(D) V 0, E 0 (S) At centre of two point masses of equal magnitude

(T) None



194

(4) Match the followingTable-1 Table -2

(A) time period of an earth satellite in circular orbit (P) Independent of mas of satellite(B) Orbital velocity of satellite (Q) independent of radius of orbit(C) Mechnical energy of stellite (R) independent of mass of earth

(S) none(5) Match the following (for a satellite in circular orbit)

Table-1 Table-2

(A) kinetic energy (P) 2GMm

r

(B) potential energy (Q)GM

r

(C) Total energy (R)GMm

r

(D) orbital velocity (S) 2GMm

r

solution :(1) A-p, B-Q, C-s D-t(2) A-Q, B-p, C-p D-o(3) A-t, B-t, C-p,q,r,s, D-t(4) A-q, B-t, C-r, D-s(5) A-s, s-B, c-p, D-q

• • •



unit - 6 gravitational ug physics... · gravitational force between two point masses is 1 2 2 ......

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