GRAVITATION

𝑚1 𝑚2 GM
1. (c) 𝐹=𝐺 𝑟2
= 6.675 × 10. (b) g= . If radius shrinks to half of its present value
1×1 R2
−11 −11
× 10 = 6.675 × 10 𝑁 then g will becomes four times.
12

2. (d) 11. (c) Acceleration due to gravity at poles is independent of the

angular speed of earth.
𝐺𝑀 6.67×10−11 ×7.34×1022
3. (d) 𝑔 = = = 1.62𝑁 ⥂/⥂ 12. (a) Mass of the ball always remain constant. It does not
𝑅2 (1.74×106 )2
depend upon the acceleration due to gravity
𝑘𝑔

13. (a)
4. (b) Actually gravitational force provides the centripetal
force.
14. (a) 𝑔′ = 𝑔 − 𝜔2 𝑅 𝑐𝑜𝑠 2 𝜆0 = 𝑔 − 𝜔2 𝑅 𝑐𝑜𝑠 2 6 0𝑜
5. (b) The value of g at the height h 𝜔2 𝑅 𝑔 1 𝑟𝑎𝑑 𝑟𝑎𝑑
0=𝑔− ⇒ 𝜔 = 2√ = = 2.5 × 10−3
4 𝑅 400 𝑠𝑒𝑐 𝑠𝑒𝑐
from the surface of earth
 2h 
g  = g 1 − 
 R 
The value of g at depth x below the surface of earth 15. (a)
 d  100  2
 x g' = g 1 −  = 9 .8 1 −  = 9 . 66 m / s
g  = g 1 −   R  6400 
 R
 2h   x
These two are given equal, hence  1 −  = 1 −  4 𝑔 𝑅 𝜌
 R   R 16. (a) 𝑔 = 3 𝜋𝜌𝐺𝑅 𝑔 ∝ 𝑟𝜌  𝑔 𝑒 = 𝑟 × 𝜌 𝑒
𝑚 𝑚
On solving, we get x = 2h
2
 M p   Re 
6. (a) 17. (c) g p = g e    = 9 .8  1  (2) 2
 
 Me  Rp   80 

7. (b) In pendulum clock the time period depends on the value = 9 . 8 / 20 = 0 . 49 m / s 2

of g, while in spring watch, the time period is independent of
the value of g.
u 2 sin 2
18. (d) Range of projectile R =
g
8. (b) Because value of g decreases with increasing height.
1
if u and  are constant then R 
2 2
g
𝑔′ 𝑅 6400
9. (a) =( ) =( ) ⇒ Rm g R 1 R
𝑔 𝑅+ℎ 6400+64 = e  m =  Rm = e  R m = 5 R e
𝑔′ = 960.40 𝑐𝑚 ⥂/⥂ 𝑠 2 Re gm Re 0 .2 0 .2
 Trust the process!!!
2
u 1 HB g 2GM GM
19. (c) H = H   = A 30. (a) ve = = 100  = 5000
2g g HA gB R R
gA GMm
Now g B = as g  R Potential energy U = − = −5000 J
12 R
HB g
 = A =12 H B = 12  H A = 12  1 . 5 = 18 m
HA gB
31. (a) If body is projected with velocity v (v  v e ) then
R
2 height up to which it will rise, h =
 R  g ve2
20. (a) g = g   = −1
R +h  h
2
v2
1 + 
 R ve R R R
v= (given)  h = = =
2  ve 
2
4 −1 3
 
2  v /2  − 1
 R  4 4  e 
21. (c) g' = g   = g  W' = W
R +h 9 9

vp gp Rp
32. (c) =  = 22 = 2
22. (d) ve ge Re

 v p = 2  v e = 2  11 . 2 = 22 .4 km /s
4 𝑔𝑒 𝜌𝑒 𝑅𝑒
23. (a) 𝑔 = 𝜋𝐺𝜌𝑅 ⇒ 𝑔 ∝ 𝜌𝑅 ⇒ = ×
3 𝑔𝑚 𝜌𝑚 𝑅𝑚
6 5 𝑅𝑒 5
⇒ = × ⇒ 𝑅𝑚 = 𝑅𝑒 2GM
1 3 𝑅𝑚 18
33. (b) ve =  v e  M if R = constant
R
24. (b) For height If the mass of the planet becomes four times then escape
g 2h velocity will become 2 times.
 100 % = = 1 %;
g R
g d h 1 34. (c) Potential energy of a body at the surface of earth
For depth  100 % = = = % = 0 .5 %
g R R 2 𝐺𝑀𝑚 𝑔𝑅 2 𝑚
𝑃𝐸 = − =− = −𝑚𝑔𝑅
𝑅 𝑅
= −500 × 9.8 × 6.4 × 106 = −3.1 × 1010 𝐽
25. (d)
2 2
So if we give this amount of energy in the form of kinetic
R 
 =     =  g m = g e
gm M 1 2 4 4 energy then body escape from the earth.
= m   e 
ge M e  Rm  9 1 9 9
4 4 2 GM vp Mp Re
 Wm =  We =  90 = 40 kg 35. (d) Escape velocity v =  = 
9 9 R ve Me Rp

 v p = 5 v e = 5  11 . 2 = 56 km /s
26. (a) g ' = g −  2 R, when  increases g' decreases.

2GM
27. (b) Acceleration due to gravity at latitude  is given by 36. (c) On earth v e = = 11 . 2 km / s
R
g  = g − R  2 cos 2 
2GM  4 2 2GM
3 On moon v m = =
o
At 30 , g 30  = g − R  cos 30 = g − R  2
2 2 o 81  R 9 R
4
2
=  11 .2 = 2 .5 km / s
3 9
 g − g 30 =  2 R.
4

37. (a)
28. (b) Potential energy of the 1 kg
GM
mass which is placed at the earth surface = − 2GM
R 38. (d) ve =
(R + h)
its potential energy at infinite = 0
GM
 Work done = change in potential energy = 39. (d)
R
2GM 2  6 .67  10 −11  6  10 24
ve = = = 3  10 8
29. (c) R R
By solving R = 9 mm

Winning Star Institute / winningstar.in / 7200037940

 Trust the process!!!
40. (b) Escape velocity is independent of mass of object. (R + h) 3 R
52. (c) (i) Tst = 2 = 2
GM g
8 v R p 1 [As h <<R and GM = gR 2 ]
41. (c) v=R G  p = p =2 =1 
3 ve Re e 4
R
8 8 
(ii) Tma = 2
ve = R G   = 2 R G  g
3 3 4
1 R
(iii) Tsp = 2 = 2
1 1  2g
−GM g + 
42. (a) Potential at the centre due to single mass = l R
L/ 2
[As l = R]
Potential at the centre due to
m m R
all four masses (iv) Tis = 2 [ As l = ]
L2 g
GM GM
= −4 −4 2 L
L/ 2 L
1
GM 53. (c) v , If r = R then v = V0
= − 32  . m m
r
L
V0
If r = R + h = R + 3 R = 4 R then v = = 0 . 5 V0
2
GM gR 2
43. (d) v0 = =
r R+h
54. (d) Distances of the satellite from the centre are 7R and 3.5R
respectively.
44. (b) T  r 3 / 2 . If r becomes double then time period will 3/2 3/2
T2  R 2   3 .5 R 
becomes (2)3/2 times. =   T2 = 24   = 6 2 hr
T1  R1 
  7R 
So new time period will be 24  2 2 hr i.e. T = 48 2

55. (a)
45. (c) The velocity of the spoon will
be equal to the orbital velocity when dropped out of the space-
ship. 56. (c)

57. (b) Gravitational force provides the required centripetal

46. (c)
force for orbiting the satellite
mv 2 K  1
1/3 = because  F  
r3
GMT 2  GMT 2  R R  R
47. (d) T = 2  r3 = r =  
4 2  4
2
GM   v  R

GMm
GM 58. (c) B. E. = − . If B.E. decreases then r also decreases
48. (b) v= r
r
1
and v increases as v 
r
49. (a)
59. (b) Angular momentum is conserved in central field.
50. (d) Orbital radius of satellites r1 = R + R = 2 R
r2 = R + 7 R = 8 R 𝑇2
60. (d) 𝑇 2 ∝ 𝑅3  =constant
𝑅3
−GMm −GMm
U1 = and U 2 =
r1 r2 3/2
𝑅
61. (d) 𝑇2 = 𝑇1 ( 2 ) = 𝑇1 (4)3/2 = 8𝑇1 = 40ℎ𝑟
GMm GMm 𝑅1
K1 = and K 2 =
2r1 2r2
GMm GMm 62. (b) Given that, 𝑇1 = 1day and 𝑇2 =8 days
E1 = and E 2 =
2r1 2r2 𝑇2 𝑟 3/2 𝑟2 𝑇 2/3 8 2/3
 = ( 2)  = ( 2) =( ) =4
U K E 𝑇1 𝑟1 𝑟1 𝑇1 1
 1 = 1 = 1 =4
U2 K2 E2  𝑟2 = 4𝑟1 = 4𝑅

51. (b) vB rA 4R
63. (c) = = =2
vA rB R
Winning Star Institute / winningstar.in / 7200037940
 Trust the process!!!
 v B = 2  v A = 2  3v = 6 v v e should be more than or equal to speed of light

2 Gm
i.e. c
64. (a) r

65. (b) v e = 2 v 0 = 1 . 414 v 0 79. (c): The acceleration due to gravity at a

 v  height h is given as
Fractional increase in orbital velocity  
 v 
 2h 
=
ve − v0
= 0 . 414
g h = g 1 − 
v0  Re 
 Percentage increase = 41.4%
where Re is radius of earth.

66. (b) The acceleration due to gravity at a depth d is

 d 
given a sg d = g 1 − 
67. (c)  Re 
Given, g h = g d
68. (b) Kinetic and potential energies varies with position of
earth w.r.t. sun. Angular momentum remains constant every  2h   d 
where. g 1 −  = g 1 − 
 Re   Re 
69. (c) d = Lh = 2  1 = 2km h = 1 km )

70. (c) Angular momentum remains constant

80. (b): Total energy of satellite at height h from
v1 d 1
mv 1 d 1 = mv 2 d 2  v 2 = the earth surface,
d2

E = PE + KE
71. (b) Orbital radius of Jupiter > Orbital radius of Earth
𝑣𝐽 𝑟𝑒 GMm 1
= . As 𝑟𝐽 > 𝑟𝑒 therefore 𝑣𝐽 < 𝑣𝑒 =− + m 2 (i)
𝑣𝑒 𝑟𝐽
( R + h) 2
72. (c) mv 2 GMm
Also, =
( R + h ) ( R + h) 2
73. (b) Mass of satellite does not affects on orbital radius.
GM
or, v 2 = (ii)
74. (a) By conservation of angular momentum mvr = constant R+h
v min  rmax = v max  rmin From eqns. (i) and (ii),
GMm 1GMm 1GMm
 v min =
60  1 .6  10 12 60 E=− + =−
8  10 12
=
5
= 12 m / s
( R + h) 2 ( R + h) 2 ( R + h)
𝑅 3/2
75. (b) 𝑇2 = 𝑇1 ( 2 ) = 1 × (2)3/2 = 2.8𝑦𝑒𝑎𝑟
𝑅1
1 GM mR 2
=− 
𝑇 2 𝑅 3 2𝑅𝐸 3
2 R2 ( R + h )
76. (a) 𝑇 2 ∝ 𝑅3 ( 𝑃) = ( 𝑃 ) = ( )
𝑇𝐸 𝑅𝐸 𝑅𝐸


𝑇𝑃
= (2)3/2 = 2√2 mg0 R 2  GM 
𝑇𝐸 =−  g0 = 2 
 𝑇𝑃 = 2√2 × 365 = 1032.37= 1032 days 2 ( R + h)  R 

77. (d)
81. (b): Here, RP = 2 RE , pE = pP
2Gm
78. (c) Escape velocity for that body v e = Escape velocity of the earth,
r

Winning Star Institute / winningstar.in / 7200037940

 Trust the process!!!
2GM 2 2G  4  8 T2 1 24 24
So, = 3/2 or T2 = 3/2 = = 6 2 hours
VE = =   RE  E  = RE  G  E ( i )
3

RE RE  3  3 24 2 2 2 2
Escape velocity of the planet
84. (a): The acceleration due to gravity at a depth
2GM P 2G  4 3  8
  RP  P  = RP  G  P ( ii )
VP = = d below surface of earth is
RP RP  3  3
GM  d  d
g = 1 −  = g 1 − 
Divide (i) by (ii), we get R2  R  R
VE RE E R E 1 g  = 0 at d = R.
= = E =
VP RP  P 2 RE E 2
i. e., acceleration due to gravity is zero at the
or VP = 2VE centre of earth.
Thus, the variation in value g with r is
82. (c): Acceleration due to gravity at a height h For, r  R,
from the surface of earth is g gR 2 1
g = = 2  g  2
g 1 + 
h
r r
g = (i)  
h 2  R 
(1 + )
R
Here, R + h = r
where g is the acceleration due to gravity at the
 d  gr
surface of earth and R is the radius of earth. For r  R, g  = g 1 −  =
 R R
Multiplying by m (mass of the body) on both
Here, R − d = r  g   r
sides in (i), we get
Therefore, the variation of g with distance from
mg
mg  = centre of the earth will be as shown in the figure.
h
(1 + ) 2
R
Weight of body at height h , W  = mg 
Weight of body at surface of earth, W = mg
1
According to question, W ' = W
16
85. (a)
1 1
=
16 (1 + h ) 2
R GMm GMm
86. (a): − 2
+ m 2 R = 0  2
= m 2 R
h 2 h R R
(1 + ) = 16 or 1 + = 4 1 1 GMm
R R K .E. = I  2 = mR 2 2 =
2 2 2R
h
or = 3 or h = 3R
R GMm
P.E. = − .
R
83. (c): According to Kepler’s third law T  r 3/ 2
P.E. K .E 1
3/2
K .E . = or, =
T2  r2   R + 2R 
3/2
1 2 P.E. 2
=  =  =
T1  r1   R + 5R  23/2
87. (d): From equation of acceleration due to
Since T1 = 24 hours gravity.

Winning Star Institute / winningstar.in / 7200037940

 Trust the process!!!
GM e G ( 4 / 3)  R dM
3

ge = = e
e 91. (d): Rate of change of mass =  v.
&2 Re2 dt
Retarding force = Rate of change of
g e  Re e
momentum
Acceleration due to gravity of planet dM
= Velocity  Rate of change in mass = −v 
g p  Rp  p dt
1 = −v   v = − v 2 . (Minus sign of v due to
Re e = R p  p  Re e = R p 2 e  R p = R
2 deceleration)
( Re = R )  v2
Therefore, acceleration = − .
M
88. (c): Gravitational potential energy on
earth’s 92. (d) According to question

GMm Gravitation potential at midpoint (O) due to both sphere

surface = − , where M and R are the mass −GM −GM
R VA = =
R r/2
and radius of the earth respectively, m is the mass
−GM −2GM
of the body and G is the universal gravitational VB = =
r/2 r
constant. −2GM −2GM
V = VA + VB = +
Gravitational potential energy at a height r r
h = 3R − 4 GM
V=
GMm GMm GMm r
=− =− =−
R+h R + 3R 4R
93. (c) The weight of body at the centre of Earth will be less than
Change in potential energy that at Earth’s surface

GMm  GMm  94. (a)The potential energy due to gravitational field of Earth will
=− −− 
4R  R  be maximum at infinite distance

GMm GMm 3GMm 3 95. (d)

=− + = = mgR 96. (a)
4R R 4R 4
𝐺𝑀𝑚
Initial Total energy, 𝐸𝑖 = −
4𝑅
89. (b) Final total energy, 𝐸𝑓 =
𝐺𝑀𝑚
− 8𝑅
𝑚𝑔𝑅
So, energy required, Δ𝐸 = 𝐸𝑓 − 𝐸𝑖 = 8
90. (c): Radius of earth ( Re ) = 6400 km; Radius 97. (c)
of mars ( Rm ) = 3200 km; Mass of earth Motion is under a conservative force.
( M e ) = 10 Mm and weight of the object on 98. (d)

earth (We ) = 200 N. 𝐺𝑚𝑀 𝑚𝑣 2

=
𝑟 𝑟
2
Wm mg m M m  Re  1 2 𝑣 = √𝐺𝑀
= =   =  (2) =
2
𝑣 ∝ 𝑟0
We mge M e  Rm  10 5
⇒ 𝑣 ∝ 𝑟0
2
or Wm = We  = 200  0.4 = 80 N.
5 99. (c)

𝑣𝑒 11.2
𝑣𝑜 = = ≈ 8 km/s
√2 √2
Winning Star Institute / winningstar.in / 7200037940
 Trust the process!!!
100. (a)

Winning Star Institute / winningstar.in / 7200037940