PHYSICS

CJP SHEETS FOR NEET

GRAVITATION

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 POINTS TO REMEMBER AND FORMULAE
1. Kepler’s laws are applicable not only to the solar system but to the moons going around the planets as
well as to the artificial satellites
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2. Kepler’s laws are valid whenever inverse-square law is involved.
3. Kepler’s laws, which are empirical laws (i.e., laws based on observations and not on theory), sum up
nearly how planets of the solar system behave without indicating why they do so
4. Newton’s laws are about motion and force in general and as such involve an interaction between
objects. Kepler’s laws describe the motion of only a single system, i.e., the planetary system and do
not involve interactions.
5. Newton’s laws are dynamic and relate force, mass, distance and time. Kepler’s laws are kinematic and
give a relation between distance and time.

6. Since 𝐹⃗12 and 𝐹⃗21 are directed towards the centre of mass of the two particles, the gravitational force is
a central force.
7. Gravitational force is always attractive while electric and magnetic forces can be attractive or repulsive
8. Gravitational force is independent of the medium between the particles whereas electric and magnetic
forces depend on the nature of the intervening medium
9. Gravitational force is a conservative force which means that work done by it is independent of path
followed. This fact can also be stated by saying that work done in moving a particle round a closed
path under the action of gravitational force is zero.
10. Newton’s law of gravitation is valid for objects lying at huge distances (interplanetary distances) and
also for very small distances (interactomic distances) i.e., it holds over a wide range of distances.
11. Newton’s law of gravitation is of universal application and it holds irrespective of the state and the
nature of the attracting bodies
1
12. From eqn. (3), 𝐹 ∝ 𝑟 2 ………(5)
This means that the force exerted on a planet by the sun varies inversely as the square of the
distance from the sun, i.e., gravitational force is inverse square force. Though we have taken the
help of all the three laws of Kepler to deduce Newton’s law of gravitation, eqn.(5) is a direct
outcome of Kepler’s third law. Thus, Kepler’s third law enables us to determine the way in which
the gravitational force varies with the distance, i.e., it established the inverse square nature of
gravitational force.
13. Although we have bot proved there, Kepler’s first law is also a direct consequence of the fact that
the gravitational force varies as 1/r2. It can be shown that under an inverse square force, the orbit of
a planet is a conic section (i.e., circle, ellipse, parabola or hyperbola) with the sun at one focus.
14. From Art. 14.17(Part-I)
⃗ ⃗ ⃗⃗
⃗⃗ = 2𝑚 𝑑𝐴 or 𝑑𝐴 = 𝐿
𝐿 𝑑𝑡 𝑑𝑡 2𝑚
𝑑𝐴⃗
According to Kepler’s second law , 𝑑𝑡 = a constant. Hence, this implies that the angular
momentum of the planet is constant, i.e., Kepler’s second law follows from conservation of
angular momentum. As 𝐿 ⃗⃗ is constant, 𝜏⃗ = 𝑟⃗ × 𝐹⃗ , 𝑟⃗ × 𝐹⃗ = ⃗0⃗or rf sin 𝜃 = 0 or 𝜃 = 00 or 1800 Thus, 𝑟⃗
and 𝐹⃗ must act along the same line. Such a force 𝐹⃗ , which acts along 𝑟⃗, is called the central force.
Thus, Kepler’s second law established that the gravitational force is central. In fact, this law
applies to any situation that involves central force whether inverse square or not.
15. We have derived Newton’s law of gravitation from Kepler’s laws on the assumption that Newton
was guided by these laws while formulating the law of gravitation. By comparing the acceleration
of Moon (a) with the acceleration due to gravity on the Earth’s surface (g), he only checked the
correctness of the inverse square nature of gravitational force on which his law was based.
There is another view point according to which it is believed that having discovered the (1/r2)
nature of gravitational force by comparing (a) and (g), newton formulated his universal law of

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 gravitation, Later on, he was able to derive Kepler’s laws using his laws of motion and universal
law of gravitation.
We have already talked about the derivation of first law (comment 2) and second law
[Art.14.17(part-I)] The third law can also be derived as discussed in Q.6(page 42). From there it
follows that
4𝜋 2 3
𝑟2 = ( ) 𝑟 = 𝐾𝑟 3
𝐺𝑀
Where K is a constant whose value depends on M. In case of planets moving around the sun, M=
MS (mass of the sun),
4𝜋 2
𝐾 = 𝐾𝑆 = ( ) = 2.97 × 10−19 𝑠 2 /𝑚3
𝐺𝑀𝑆
For Moon and other satellites around the Earth,
4𝜋 2
𝐾 = 𝐾𝐸 = (𝐺𝑀 ) = 10−13 𝑠 2 /𝑚3
𝐸
16. The historical connection between Kepler’s laws and Newton’s law of gravitation can best be
understood from the following two statements of Newton.
i) “If have seen farther from others, it is because I stand on the shoulders of giants”
ii) “From Kepler’s third law, I deduced the inverse-square property of gravitational force and
thereby compared the force requisite to keep the Moon in her orbit with the force of gravity at the
surface of the Earth, and found them answer pretty nearly” (Newton’ s nostalgic look back-a year
before his death)
But all this is now a part of glorious history of physics.

17. The component 𝑚𝑟𝜔2 𝑠𝑖𝑛 𝜆 changes the direction of 𝑚𝑔

⃗⃗⃗⃗⃗⃗⃗
18. If the Earth were to stop rotating, the weight would increase due to the absence of the centrifugal
force.
19. Gravitational intensity is a vector quantity with [LT2]as its dimensional formula. Its SI units is N/kg.
𝐺𝑀
20. For a point of mass M, gravitational field at a distance r from it is given by 𝐼 = 𝑟2

21. For a sphere of radius R,

𝐺𝑀
a) 𝐼 = (𝑟 > 𝑅, 𝑓𝑜𝑟𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙𝑝𝑜 𝑖𝑛𝑡)
𝑟2
𝐺𝑀
b) 𝐼 = (𝑟 = 𝑅, 𝑎𝑡𝑡ℎ𝑒𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝑅2
𝐺𝑀
c) 𝐼 = 𝑅3 𝑟(𝑟 < 𝑅, where r is the distance of the internal point from the centre of the sphere)
d) 𝐼 = 0(𝑟 = 0, 𝑎𝑡𝑡ℎ𝑒𝑐𝑒𝑛𝑡𝑟𝑒)
22. For a spherical shell of radius R,
𝐺𝑀
a) 𝐼 = (𝑟 > 𝑅)
𝑟2
𝐺𝑀
b) 𝐼 = 𝑅2 (𝑟 = 𝑅𝑎𝑡𝑡ℎ𝑒𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
c) 𝐼 = 0(𝑟 < 𝑅)
d) 𝐼 = 0(𝑟 = 0, 𝑎𝑡𝑡ℎ𝑒𝑐𝑒𝑛𝑡𝑟𝑒)
23. The expression for intensity of gravitational field in case of the Earth is the same as that for a sphere.

24. If the body is projected from a point above the Earth’s surface, at distance r from its centre, 𝜐𝑒 =
√2𝐺𝑀/𝑟 which is obviously less than √2𝐺𝑀/𝑅 (as r>R)
25. Escape speed does not depend upon the mass (m) of the body.
26. Escape sped depends upon the mass (M0 and radius (R) of the planet from which the body is
projected. Of all the planets, the escape speed is minimum for the planet Mercury, being only 4.2km/s.
Its value is 61 km/s on Jupiter and 618km/s for the sun. for Moon, the escape speed is 2.38km/s.
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 𝐺𝑀 2𝑔𝑅 2
27. As 𝑔 = , 𝑔𝑅 2 = 𝐺𝑀 from 𝑒𝑞𝑛. (4), 𝜐𝑒 = √ = √2𝑔𝑅. . . .. clearly, 𝜐𝑒2 = 2𝑔𝑅 = 2𝑅 × 𝑔i.e. the
𝑅2 𝑅
square of the escape speed is the product of diameter (i.e., 2R) of the Earth and the acceleration due to
gravity (g) on Earth’s surface
28. The escape speed does not depend upon the direction in which the body is projected.
29. Though we have ignored Earth’s rotation, it does play a role. Firing eastward has an advantage as in
that direction, the Earth’s tangential surface speed (which is 0.46m/s at cape Canaveral) can be
subtracted from the calculated value of escape speed.
30. If 𝜐 < 𝜐𝑒 ,then the body will attain a certain height and thereafter either it will move in an orbit around
the Earth or may stage a come-back to the Earth.
1 1 1
31. If 𝜐 > 𝜐𝑒 , the body will move in space with a speed 𝜐 ' = √𝜐 2 − 𝜐𝑒2 (As 2 𝑚𝜐 2 = 2 𝑚𝜐𝑒2 + 2 𝑚𝜐 '2 ,
where 𝜐 ' is the speed left after the body has escaped the gravitational pull of earth, 𝜐 '2 = 𝜐 2 −
𝜐𝑒2 𝑜𝑟𝜐 ' = √𝜐 2 − 𝜐𝑒2 )
32. Escape Speed and Atmosphere
If the rms speed of the molecules of a gas is comparable with the escape speed, the faster molecules
will escape from the upper regions of the atmosphere and with time, the gas will disappear from the
atmosphere. The rms speed of all the constituents of our atmosphere like O2, N2,CO2 and water
vapours lies between 0.4km/s to 0.8km/s. The escape speed in case of Moon being 2.8 km/s, is
comparable to these values and as such there is no atmosphere on Moon. Hydrogen and helium whose
rms speeds are about 2 km/s, apart from being absent from the Moon’s surface, are rare even on Earth.
But these two gases are present in abundance in the atmosphere of the Sun on account of the fact that
the escape speed for sun is much higher.
33. Gravitational Binding Energy
1 𝐺𝑀𝑚
We know that 2 𝑚𝜐𝑒2 = 𝑅
The energy GMm/R is called the binding energy of the Earth-mass system and is the minimum energy
needed to take a mass infinitely away from the Earth.
i) If the kinetic energy of the mass is less than the binding energy at the Earth’s surface, it will
not leave the Earth but will rise to some maximum separation and then fall back to earth.
ii) If the kinetic energy is greater than the binding energy, the mass will continue moving forever
without returning
iii) The escape speed is just that speed corresponding to a kinetic energy equal to the binding
energy
iv) The Earth-mass system is said to be bound or unbound according to whether the kinetic energy
at the Earth’s surface is less or greater than the binding energy.
34. Radius of a Black Hole
2𝐺𝑀
Since no light can escape a black hole, 𝜐𝑒 = 𝑐 (speed of light ) or √ =𝑐
𝑅
R is called the schewarzchild radius and is denoted by Rs. For a black hole with a mass equal to the
mass of the Sun, 𝑀 = 2 × 1030 𝑘𝑔,
2(6.67 × 10−11 𝑁𝑚2 /𝑘𝑔2 )(2 × 1030 𝑘𝑔)
𝑅𝑆 = = 2.96 × 103 𝑚 ≈ 3𝑘𝑚!
(3 × 108 𝑚/𝑠 2 )
𝐺𝑀𝑚
Owing to friction of the Earth’s atmosphere, the total energy (− 2𝑟 )of the satellite decreases, i.e., it
becomes more negative .This is possible only if r decreases. If r is the radius of the smaller orbit of the
satellite, then 𝑟 ' < 𝑟.Further, if 𝐸 ' and 𝐾 ' are the total energy and kinetic energy respectively of the satellite in
𝐺𝑀𝑚 𝐺𝑀𝑚 𝐺𝑀𝑚 𝐺𝑀𝑚
the smaller orbit, then 𝐸 ' < 𝐸 or − 2𝑟 ' < − 2𝑟 or 2𝑟 ' > − 2𝑟 , 𝑖. 𝑒. , 𝐾 ' > 𝐾

Which means that the kinetic energy of the satellite has increased when it is in its smaller orbit. As such it
starts moving with increased speed. Thus, the viscous forces due to atmosphere increase the speed of the
satellite.

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This apparent anamoly is explained by the fact that the potential energy of the satellite decreases twice as
much as the kinetic energy as U = - 2K. Thus, on the whole there is a loss of energy as expected. In fact, due
to atmospheric friction, the satellite spirals down towards the Earth with increasing speed till it finally burns in
the denser layers of the atmosphere.
35. The intensity of gravitational field (I) and acceleration due to gravity (g) are two separate physical
quantities though having equal magnitudes and same direction. Whereas g represents acceleration, I
represents force per unit mass.
36. The principle of equivalence is one of the best established of all the physical laws as the equivalence
of inertial and gravitational masses have been established to about 1 part in 1012
37. Whereas 'G ' is a universal constant, 'g ' varies from planet to planet. 'g for ceres (a most massive of
asteroids) is merely 0.53894 cm / s2 and for a neutron star its value is unimaginably large, i.e.. 2 x 1012
m /s2.
38. For a planet, 'g varies due to its shape, altitude, depth and rotation. 'g’ For the Earth's surface is 9.806
in/ s2 and at an altitude of 3.8x 108 m, g is merely 0.00271 m/s2 (at 450 latitude). Further, 'g' at equator
is 9.78(139 m/s2 and at the poles 'g' is 9.83217 in / s2, i.e., there is a difference of 5.178 cm/s2.

39. For very small heights, (Δ𝑔)ℎ𝑒𝑖𝑔ℎ𝑡 = 2(Δ𝑔)𝑑𝑒𝑝𝑡ℎ

40. At the centre of the Earth, g = O.

41. (a)For points inside the Earth (r< R),g ∝ 𝑟

(b) For points on and above the Earth's surface (𝑟 ≥ 𝑅),𝑔 ∝ 1/𝑟 2
where r is the distance of a point from Earth's centre.
Thus, 'g' decreases whether we go up the Earth's surface (r increases) or go down into Earth (r
decreases).

42. Due to change in shape of Earth, 𝑔𝑝 > 𝑔𝑒 𝑎𝑛𝑑(Δ𝑔)𝑠ℎ𝑎𝑝𝑒 = 𝑔𝑝 − 𝑔𝑒 = 1.8𝑐𝑚/𝑠 2

43. Rotation of earth results in a decrease in the value of g.

(Δ𝑔)𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 = (𝑔𝑝 ) − (𝑔𝑒 )𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 = 𝑅𝜔2 = 3.3𝑐𝑚/𝑠 2 clearly,
𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛
(Δ𝑔)𝑡𝑜𝑡𝑎𝑙 = (Δ𝑔)𝑠ℎ𝑎𝑝𝑒 + (Δ𝑔)𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 = 1.8𝑐𝑚/𝑠 2 + 3.3𝑐𝑚/𝑠 2 = 5.1𝑐𝑚/𝑠 2 ( as already mentioned
at number)
44. It is helpful to remember that 𝑅𝜔2 /𝑔 = 3.37 × 10−3
45. If the Earth stops, g at the equator would increase by 𝑅𝜔2 = 3.3𝑐𝑚/𝑠 2
[as 𝑅𝜔2 = (𝑅𝜔2 /𝑔) × 𝑔 = (3.37 × 10−3 ) × 9.8𝑚/𝑠 2 = 3.3𝑐𝑚/𝑠 2 ]
46. It the Earth were to rotate faster, 'g' would decrease at all points except at the poles.
47. If Earth were to start rotating 17 times faster than its present angular speed, objects lying on Earth at
the equator would fly off. But the value of 'g' at the poles will remain unchanged. The length of the
day in that case would be 1.4 h.
48. For solar system, g is minimum at the surface of Mercury and maximum at the surface of Jupiter.
49. Gravitational potential energy (U) and gravitational potential (V ) can be positive as well as negative
depending upon the choice of zero level of potential energy. Normally, we take U = 0 at r = ∞ and in
this case gravitational potential energy and gravitational potential are negative.
50. It can be shown that GM/R = 6.26x 1026 J/kg. This is a very commonly occurring constant and it
represents the magnitude of gravitational potential on Earth's surface.
51. While calculating the gravitational potential energy of a body in the gravitational field of the Earth, the
body should be moved by exerting a force F which is just sufficient to balance the gravitational
attraction of the Earth (GMm/ r2 ). This would move the body without acceleration and as a result
without increasing its velocity (i.e., KE). All the work done would, therefore, become the PE of the
body.
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52. Escape velocity can be as large as 2x 105 m/s for neutron star and can be as small as 0.64 km/s for
ceres.
53. A meteorite (or any other body).coming from infinity strikes a planet with a velocity equal to the
escape velocity for that planet. (Refer to Topic Based Problem for Practice 18, Page 44).
54. Whereas escape speed from Earth's surface is given by √2𝐺𝑀/𝑅 , i.e., √2𝑔𝑅 , its value from any
other point away from the Earth's surface and at a distance r from its centre is given by √2𝐺𝑀/𝑟.
55. The orbital speed of a satellite is independent of its mass (𝜐 = √𝐺𝑀/𝑟). It depends upon : (i) the
mass of the planet around which it orbits and (ii) the radius of the orbit.
56. Escape speed for a body is independent of : (i) the direction of projection and (ii) the mass of the body.
57. Escape speed depends upon the mass and radius of the planet.
58. The orbital speed for a satellite close to Earth's surface is 7.92 km/s and its time-period is 84 min 20s,
(5060 s). This time-period is the same as that of a body left in a tunnel dug through a hole between any
two points on Earth's surface (not necessarily along a diameter).
59. Whereas orbital speed for a satellite in an orbit close to Earth√𝐺𝑀/𝑅 = √𝑔𝑅, its value is√𝐺𝑀/𝑟 for a
distance r from Earth's centre.
1
60. 𝐾 = |𝑈|is true for any circular orbit in an inverse-square force field.
2
61. The minimum additional energy an orbiting satellite must be given to escape from the Earth possessed
1 2𝐾
by it. In other words, its kinetic energy should be doubled. Since, 𝐾 = 2 𝑚𝑣 2 , 𝑣 = √ 𝑚 , 𝑖. 𝑒. , 𝑣 ∝ √𝐾, 𝑖. 𝑒.,
its speed should be increased to√2 times, i.e. increased from 𝜐𝑡𝑜√2𝜐. This implies a percentage
( 2𝜐−𝜐)
increase of √ 𝜐 × 100 ≈ 42%
62. 𝜐𝑒 = √2𝜐𝑐 is applicable not only to a satellite orbiting around the Earth in an orbit close to Earth. It is
applicable to any orbit, i.e., escape speed from any orbit should be equal to √2 times the speed in that
orbit.
63. The angular velocity of a satellite close to Earth's surface is 2𝜋/𝑇 = (2 × 3.14𝑟𝑎𝑑)/5060𝑠 =
1.24 × 10−3 rad/s.
64. The time-period of a geostationary satellite is 24 h, orbital speed is 3.08 m/s and the height above
Earth's surface is 36,000 km while it revolves around the Earth from west to east (i.e., anticlockwise).
65. We need no energy to keep a satellite in an orbit around Earth.
66. When a satellite enters Earth's atmosphere, its (i) total energy (E) decreases (ii) potential energy (U)
decreases (iii) kinetic energy (K) increases which results in an increase in its velocity.
67. If E < 0, the satellite is bound to Earth. If E = O the satellite is no longer bound to the Earth and it
escapes away from its orbit along a parabolic path. If E> 0, the satellites escapes away from its orbit in
a hyperbolic path.
68. In an accelerating lift (or a spaceship), a pseudo-force acts on the body in the lift in a direction
opposite to the acceleration (a) of the lift. The magnitude of this force is ma. Of course, this force
should be taken into consideration while working from within the accelerated lift (or the spaceship),
i.e, while situated in a non-inertial frame of reference.
69. 𝑇 2 ∝ 𝑟 3 , (Kepler's third law)
i.e., square of the time period (T) is directly proportional to the cube of semi-major axis (r), i.e., the
mean distance of the planet from the Sun.
𝑚 𝑚
70. 𝐹 = 𝐺 𝑟1 2 2 (Newton’s law of gravitation)
where F is the gravitational force of attraction between two masses placed a distance r apart
𝐺 = 6.67 × 10−11 𝑁𝑚2 /𝑘𝑔2
𝑘𝜃𝑟 2
71. 𝐺 = 𝑀𝑚𝑙 (Cavendish’s experiment)
where k is the torsion constant, r is the distance between a big and a small mass, 𝜃 is the deflection of
the rod of length ℓ at the ends of which two small masses (m, m) are attached and M is the mass of
each of the two big spheres.
𝐺𝑀 4𝜋
72. 𝑔 = 𝑅2 = 3 𝐺𝑅𝜌
where g is the acceleration due to gravity at the surface of a planet of mass M and radius R. The mean
density of the planet is 𝜌.
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 𝑔𝜌 𝑅2
73. = 𝑅𝑒2
𝑔𝑒 𝑝
where gp ,ge are the accelerations due to gravity at poles and the equator respectively. Rp and Re are the
polar and equatorial radii. Note that gp —ge = 1.8 cm/s2.
𝑔ℎ 𝑅2
74. a) = (𝑅+ℎ)2 (applicable to all heights)
𝑔
𝑔ℎ 2ℎ
b) = (1 − 𝑅 ) (true only for small heights)
𝑔
where gh is the acceleration due to gravity at a height h.
𝑑
75. 𝑔𝑑 = (1 − 𝑅)
where gd is the acceleration due to gravity at a depth d.
76. a) 𝑔𝑟 = 𝑔 − 𝑅𝜔2 𝑐𝑜𝑠 2 𝜆
where gr is the acceleration due to gravity due to Earth point on Earth's surface when rotation is taken
account and g is its corresponding value when rotation not taken into account.
b) (𝑔𝑟 )𝑝𝑜𝑙𝑒𝑠 = 𝑔𝑎𝑛𝑑(𝑔𝑟 )𝑒𝑞𝑢𝑎𝑡𝑜𝑟 = 𝑔 − 𝑅𝜔2
c) (𝑔𝑟 )𝑝𝑜𝑙𝑒𝑠 − (𝑔𝑟 )𝑒𝑞𝑢𝑎𝑡𝑜𝑟 = 𝑅𝜔2 = 3.3𝑐𝑚/𝑠 2
𝐺𝑀
77. 𝐼 = 𝑅2
Where I is the intensity of the gravitational field near Earth’s surface
Also 𝐼⃗ = 𝑔⃗𝑎𝑛𝑑𝐼⃗ = ⃗⃗⃗⃗
𝐹𝑔 /𝑚where ⃗⃗⃗⃗
𝐹𝑔 is the gravitational force of attraction acting on a body of mass m
placed in a gravitational field of intensity 𝐼⃗.
𝑚𝑔 𝑚
78. = 𝑖'
𝑚'
𝑔 𝑚𝑖
Where 𝑚𝑔 and 𝑚𝑔' are the gravitational masses of two bodies and 𝑚𝑖 and 𝑚𝑖' are their corresponding
inertial masses.
1 1 𝐺𝑀𝑚 𝐺𝑀𝑚
79. 𝑈𝑓 − 𝑈𝑖 = −𝐺𝑀𝑚 (𝑟 − 𝑟 ) = − + is the general expression for gravitational PE.
𝑓 𝑖 𝑟𝑓 𝑟𝑖
𝐺𝑀𝑚
a) 𝑈 = − (at any point, where U = 0 at 𝑟 = ∞
𝑟
𝐺𝑀𝑚 𝐺𝑀𝑚
b) 𝑈 = − + (at any point, when U = 0 at Earth’s surface)
𝑟 𝑅
𝐺𝑀𝑚
c) 𝑈 = − 𝑅 (at earth’s surface, when 𝑈 = 0𝑎𝑡𝑟 = ∞
d) U = 0 ( at Earth’s surface, when U = 0 at Earth’s surface)
𝐺𝑀𝑚 𝐺𝑀𝑚
e) 𝑈 = − (𝑅+ℎ) = − 𝑅 + 𝑚𝑔ℎ
(at a height h above Earth’s surface when 𝑈 = 0𝑎𝑡𝑟 = ∞)
f) 𝑈 = 𝑚𝑔ℎ (at a height h above Earth’s surface, (h<<R) when U = 0 at Earth’s surface)
𝐺𝑀 𝐺𝑀
80. 𝑉𝑓 − 𝑉𝑖 = − 𝑟 + 𝑟
𝑓 𝑖
Is the general expression for gravitational potential
𝐺𝑀
a) 𝑉 = − 𝑟 (at any point, when 𝑈 = 0𝑎𝑡𝑟 = ∞)
𝐺𝑀
b) 𝑉 = − ( on Earth’s surface, when 𝑉 = 0𝑎𝑡𝑟 = ∞)
𝑅
𝐺𝑀
c) 𝑉 = − (𝑅+ℎ) (at height h above Earth’s surface, when 𝑉 = 0𝑎𝑡𝑟 = ∞)
Are the most commonly used equations for gravitational potential (V).
𝑈
81. 𝑈 = 𝑚𝑉𝑜𝑟𝑉 = 𝑚
2𝐺𝑀
82. 𝜐𝑒 = √ = √2𝑔𝑅
𝑅
When 𝜐𝑒 is the escape velocity at the surface of a planet of mass M and radius R. ‘g’ is the acceleration
due to gravity at its surface.
𝐺𝑀𝑚
83. 𝑊=− 𝑅
Where W is the amount of work done to take the body of mass m from Earth’s surface to infinity and
is called the binding energy of the Earth-mass system

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 𝐺𝑀 𝑔 𝐺𝑀 𝑔
84. 𝜐=√ = 𝑅√ 𝑟 = √(𝑅+ℎ) = 𝑅 √(𝑅+ℎ)
𝑟
Where 𝜐 is the orbital velocity of a satellite in an orbit of radius r around a planet of mass M. ‘g’ is the
acceleration due to gravity on the planet’s surface.
Here, H = r-R = height of the satellite.
𝐺𝑀
85. 𝜐𝑐 = √ 𝑅 = √𝑔𝑅
Where 𝜐𝑐 is the orbital velocity for an orbit close to the earth’s surface
2𝜋𝑟 3/2 2𝜋𝑟 𝑟
86. 𝑇= = √𝑔
√𝐺𝑀 𝑅
Where T is the time – period of a satellite in an orbit of radius r around a satellite of mass M.
2𝜋𝑅 3/2 𝑅
87. 𝑇𝑐 = = 2𝜋√𝑔
√𝐺𝑀
4𝜋 2 𝑅 3 4𝜋 2 𝑅 3
Where 𝑇𝐶 is the time period for an orbit close to earth’s surface. Also, as 𝑇𝑐2 = = 𝐺(4𝜋/3)𝑅3𝜌 =
𝐺𝑀
3𝜋
𝐺𝜌
3𝜋
𝑇𝑐 = √𝐺𝜌
1/3
𝑇 2 𝑅2𝑔
88. ℎ = ( 4𝜋2 ) − 𝑅
Where h gives the altitude of a satellite of time – period T
𝐺𝑀𝑚 𝐺𝑀𝑚 𝐺𝑀𝑚
89. a) 𝑈 = − 𝑟 b) 𝐾 = 2𝑟 c) 𝐸 = 2𝑟
Where U, K and E denote the potential energy, the kinetic energy and the total energy of a satellite of
mass m in an orbit of radius r around the Earth . Clearly, E = - K, U = 2E and U = -2K
𝐺𝑀𝑚
90. is the binding energy of Earth-satellite system
2𝑟
91. a) 𝑅 = 𝑚(𝑔 − 𝑎) where R measures the weight of a body in a lift accelerating downwards.
b) 𝑅 = 𝑚(𝑔 + 𝑎) measures the weight of a body in a lift accelerating upwards.

Topic-wise analysis of NEET 2014-2021

Topic name/ 2014 2015 2016 2017 2018 2019 2019 2020 2020 2021
year (Orissa) (covid-19)
T1, T2 & T3 : 1 3 2 2 1 1 1
Kepler's Laws of
Planetary Motion,
Newton's
Universal Law of
Gravitation and
Acceleration due
to Gravity
T4 : Gravitational 1 1 1 1
Field, Potential
and Potential
Energy
T5 : Motion of 1 1 1 1 2
Satellites, Escape
Speed and Orbital
Velocity
NEET Previous 8 Years Papers
1. If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times
larger in magnitude, which of the following is not correct? [2018]
(a) Raindrops will fall faster (b) Walking on the ground would become more difficult

9|Page
 (c) ‘g’ on the Earth will not change (d) Time period of a simple pendulum on the Earth would decrease
2. The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB
and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as
shown in the figure. Then [2018]

(a) KA < KB < KC (b) KA > KB > KC (c) KB > KA > KC (d) KB < KA < KC
3. The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the
surface of earth. Then [2017]
3 1
(a) d = 1 km (b)𝑑 = 2 𝑘𝑚 (c) d = 2 km (d) 𝑑 = 2 𝑘𝑚
4. Two astronauts are floating in gravitation free space after having lost contact with their spaceship. The
two will [2017]
(a) move towards each other. (b) move away from each other.
(c) become stationary (d) keep floating at the same distance between them.
5. At what height from the surface of earth the gravitational potential and the value of g are –5.4 × 107Jkg–
1
and 6.0 ms–2 respectively? Take the radius of earth as 6400 km: [2016]
(a) 2600 km (b) 1600 km (c) 1400 km (d) 2000 km
6. The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean
density are twice as that of earth is : [2016]
(a) 1 : 2 (b) 1 : 2√2 (c) 1 : 4 (d) 1 : 2
7. Kepler's third law states that square of period of revolution (T) of a planet around the sun, is
proportional to third power of average distance r between sun and planet i.e. T2 = Kr3 here K is constant.
If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force
𝐺𝑀𝑚
of attraction between them is𝐹 = 𝑟 2 , here G is gravitational constant. The relation between G and K is
described as [2015]
2 1 2
(a) GMK = 4𝜋 (b) K = G (c) 𝐾 = 𝐺 (d) GK = 4𝜋
8. Two spherical bodies of mass M and 5 M and radii R and 2R released in free space with initial
separation between their centres equal to 12 R. If they attract each other due to gravitational force only,
then the distance covered by the smaller body before collision is [2015]
(a) 4.5 R (b) 7.5 R (c) 1.5 R (d) 2.5 R
9. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small
compared to the mass of the earth. Then, [2015]
(a) the total mechanical energy of S varies periodically with time.
(b) the linear momentum of S remains constant in magnitude.
(c) the acceleration of S is always directed towards the centre of the earth.
(d) the angular momentum of S about the centre of the earth changes in direction, but its magnitude
remains constant.
10. A remote - sensing satellite of earth revolves in a circular orbit at a height of 0.25 × 106 m above the
surface of earth. If earth's radius is 6.38 × 106 m and g = 9.8 ms–2, then the orbital speed of the satellite
is:
[2015]
(a) 8.56 km s–1 (b) 9.13 km s–1 (c) 6.67 km s–1 (d) 7.76 km s–1
11. A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To
what approximate radius would earth (mass = 5.98 × 1024 kg) have to be compressed to be a black hole?
[2014]
(a) 10–9 m (b) 10–6 m (c) 10–2 m (d) 100 m

12. Dependence of intensity of gravitational field (E) of earth with distance (r) from centre of earth is
correctly represented by: [2014]

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 (a) (b) (c) (d)
13. A projectile is fired from the surface of the earth with a velocity of 5 ms–1 and angle q with the
horizontal. Another projectile fired from another planet with a velocity of 3 ms–1 at the same angle
follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value
of the acceleration due to gravity on the planet is (in ms–2) given g = 9.8 m/s2 [2014]
(a) 3.5 (b) 5.9 (c) 16.3 (d) 110.8
14. A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre
of the earth? [NEET – 2019]
(1) 150 N (2) 200 N (3) 250 N (4) 100 N
15. At a point A on the earth's surface the angle of dip, 𝛿 = +25°. At a point B on the earth's surface the
angle
of dip, 𝛿 = –25°. We can interpret that : [NEET – 2019]
(1) A and B are both located in the northern hemisphere.
(2) A is located in the southern hemisphere and B is located in the northern hemisphere.
(3) A is located in the northern hemisphere and B is located in the southern hemisphere.
(4) A and B are both located in the southern hemisphere
16. The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius
of the earth, is : [NEET – 2019]
1 3
(1) mgR (2) 2 mgR (3) 2 mgR (4) 2 mgR
17. The time period of a geostationary satellite is 24 h, at a height 6RE (RE is radius of earth) from surface of
earth. The time period of another satellite whose height is 2.5 RE from surface will be,
[NEET – 2019 (ODISSA)]
24 12
1) 6√2ℎ 2) 12√2ℎ 3) 2.5 ℎ 4) 2.5 ℎ
18. Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential
energy (final – initial) of an object of mass m, when taken to a height h from the surface of earth (of
radius R), is given by, [NEET – 2019 (ODISSA)]
𝐺𝑀𝑚 𝐺𝑀𝑚ℎ 𝐺𝑀𝑚
1) − 𝑅+ℎ 2) 𝑅(𝑅+ℎ) 3) mgh 4) 𝑅+ℎ
19. What is the depth at which the value of acceleration due to gravity becomes 1/n times the value that at
the surface of earth? (radius of earth = R) NEET-2020(COVID-19)
(1) R/n2 (2) R(n . 1)/n (3) Rn/(n . 1) (4) R/n
20. A body weight 72N on the surface of the earth. What is the gravitational force on it at a height equal to
half the radius of the earth? [NEET-2020]
1) 24N 2) 48 N 3) 32 N 4) 30 N
21. The escape velocity from the Earth’s surface is 𝜐.The escape velocity from the surface of another planet
having a radius , four times that of Earth and same mass density is : [NEET-2021]
1. 2𝜐 2. 3𝜐 3. 4𝜐 4. 𝜐
22. A particle of mass ‘m’ is projected with a velocity u = kVe (k < 1) from the surface of the earth.
(Ve = escape velocity) The maximum height above the surface reached by the particle is [NEET-2021]
𝑘 2 𝑅2𝑘 𝑅𝑘 2 𝑘 2
1) 𝑅 (1+𝑘) 2) 1+𝑘 3) 1−𝑘 2 4) 𝑅 (1−𝑘)

LEVEL-1
Topic 1: Kepler’s Laws of Planetary Motion
1. Kepler’s second law regarding constancy of areal velocity of a planet is a consequence of the law of
conservation of
(a) energy
(b) angular momentum
(c) linear momentum
(d) None of these

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

(a) (b) (c) (d)

3. Two satellites revolve round the earth with orbital radii 4R and 16R, if the time period of first satellite is
T then that of the other is
(a) 4 T (b) 42/3 T (c) 8 T (d) None of these
4. A comet moves in an elliptical orbit with an eccentricity of e = 0.20 around a star. The distance between
the perihelion and the aphelion is 1.0 × 108 km. If the speed of the comet at perihelion is 81 km/s, then
the speed of the comet at the aphelion is:
(a) 182 km/s (b) 36 km/s (c) 121.5 km/s (d) 54 km/s
5. The period of moon's rotation around the earth is nearly 29 days. If moon's mass were 2 fold its present
value and all other things remain unchanged, the period of moon's rotation would be nearly
(a) 29 2 days (b) 29/ 2 days (c) 29 × 2 days (d) 29 days
6. 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. The time period of planet is proportional to
2/5
(a) 𝑟1 (b) (𝑟1 + 𝑟2 )3/2 (c) (𝑟1 − 𝑟2 )3/2 (d) 𝑟 3/2
7. A satellite moves round the earth in a circular orbit of radius R making one revolution per day. A second
satellite moving in a circular orbit, moves round the earth once in 8 days. The radius of the orbit of the
second satellite is
(a) 8 R (b) 4 R (c) 2 R (d) R
8. Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle cSa is1/4
the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semi minor axis. If t1 is
the time taken for planet to go over path abc and t2 for path taken over cda then:

(a) 𝑡1 = 4𝑡2 (b) t1 = 2t2 (c) t1 = 3t2 (d) t1 = t2

9. A planet moves around the sun. At a point P it is closest from the sun at a distance d1 and has a speed v1.
At another point Q, when it is farthest from the sun at a distance d2 its speed will be
(a) 𝑑12 𝑣1 /𝑑22 (b) 𝑑2 𝑣1 /𝑑1 (c) 𝑑1 𝑣1 /𝑑2 (d) 𝑑22 𝑣1 /𝑑12
10. The planet mercury is revolving in an elliptical orbit around the sun as shown in fig. The kinetic energy
of mercury will be greatest at

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

11. The distance of Neptune and Saturn from the sun is nearly 1013 and 1012 meter respectively. Assuming
that they move in circular orbits, their periodic times will be in the ratio
(a) 10 (b) 100 (c) 10√10 (d) 1000
12. A satellite is revolving round the earth in an orbit of radius r with time period T. If the satellite is
revolving round the earth in an orbit of radius r + Δr (Δr << r) with time period T +ΔT then,
Δ𝑇 3 Δ𝑟 Δ𝑇 2 Δ𝑟 Δ𝑇 Δ𝑟 Δ𝑇 Δ𝑟
(a) 𝑇 = 2 𝑟 (b) 𝑇 = 3 𝑟 (c) 𝑇 = 𝑟 (d) 𝑇 = − 𝑟
13. The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is
increased to 4 times the previous value, the new time period will become
(a) 10 hours (b) 80 hours (c) 40 hours (d) 20 hours
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14. The maximum and minimum distances of a comet from the sun are 8 × 1012 m and 1.6 × 1012 m. If its
velocity when nearest to the sun is 60 m/s, what will be its velocity in m/s when it is farthest
(a) 12 (b) 60 (c) 112 (d) 6

Topic 2: Newton’s Universal Law of Gravitation

15. Two masses m1 and m2 (m1 < m2) are released from rest from a finite distance. They start moving under
their mutual gravitational attraction, then
(a) acceleration of m1 is more than that of m2 (b) acceleration of m2 is more than that of m1
(c) centre of mass of system will remain at rest in all the reference frame
(d) total energy of system does not remains constant
16. Two bodies of masses 4 kg and 9 kg are separated by a distance of 60 cm. A 1 kg mass is placed in
between these two masses. If the net force on 1 kg is zero, then its distance from 4 kg mass is
(a) 26 cm (b) 30 cm (c) 28 cm (d) 24 cm
17. A body weighs 72 N on the surface of the earth. What is the gravitational force on it due to earth at a
height equal to half the radius of the earth from the surface?
(a) 32 N (b) 28 N (c) 16 N (d) 72 N
18. If masses of two point objects is doubled and distance between them is tripled, then gravitational force
of attraction between them will nearly
(a) increase by 225% (b) decrease by 44% (c) decrease by 56% (d) increase by 125%
19. The distance of the centres of moon and earth is D. The mass of earth is 81 times the mass of the moon.
At what distance from the centre of the earth, the gravitational force will be zero?
𝐷 2𝐷 4𝐷 9𝐷
(a) 2 (b) 3 (c) 3 (d) 10
20. Six stars of equal mass are moving about the centre of mass of the system such that they are always on
the vertices of a regular hexagon of side length a. Their common time period will be
𝑎3 4√3𝑎3 3𝑎3
(a)4𝜋√𝐺𝑚 (b) 2𝜋√𝐺𝑚(5√3+4) (c) 4𝜋√ 𝐺𝑚 (d) None of these
21. Two stars of mass m1 and m2 are parts of a binary system. The radii of their orbits are r1 and r2
respectively, measured from the C.M. of the system. The magnitude of gravitational force m1 exerts on
m2 is
𝑚 𝑚 𝐺 𝑚1 𝐺 𝑚2 𝐺 (𝑚 +𝑚 )
(a) (𝑟 1+𝑟2)2 (b) (𝑟 +𝑟 )2
(c) (𝑟 +𝑟 )2
(d) (𝑟 1+𝑟 )22
1 2 1 2 1 2 1 2
22. The percentage change in the acceleration of the earth towards the sun from a total eclipse of the sun to
the point where the moon is on a side of earth directly opposite to the sun is
𝑀 𝑟2 𝑀 𝑟 2 𝑟 2 𝑀 𝑟 2 𝑀
(a)𝑀 𝑠 × 100 (b) 𝑀 𝑠 (𝑟2 ) × 100 (c)2 (𝑟1) 𝑠
× 100 (d) (𝑟1) 𝑠
× 100
𝑚 𝑟1 𝑚 1 2 𝑀𝑚 2 𝑀𝑚
23. There are two bodies of masses 103 kg and 105 kg separated by a distance of 1 km. At what distance
from the smaller body, the intensity of gravitational field will be zero
(a) 1/9 km (b) 1/10 km (c) 1/11 km (d) 10/11 km
24. Two spheres of masses m and M are situated in air and the gravitational force between them is F. The
space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will
now be
(a) F/9 (b) 3F (c) F (d) F/3
Topic 3: Acceleration due to Gravity
25. The weight of an object in the coal mine, sea level and at the top of the mountain, are respectively W1,
W2 and W3 then
(a) W1< W2 > W3 (b) W1= W2 = W3 (c) W1< W2 < W3 (d) W1> W2 > W3
26. The ratio between the values of acceleration due to gravity at a height 1 km above and at a depth of 1
km
below the Earth’s surface is (radius of Earth is R)
𝑅−2 𝑅 𝑅−2
(a)𝑅−1 (b) 𝑅−1 (c) 𝑅 (d) 1
27. The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60°
latitude becomes zero is (Radius of earth = 6400 km, at the poles g = 10 ms–2)
(a) 2.5 × 10–3 rad/s (b) 5.0 × 10–1 rad/s (c) 10 × 101 rad/s (d) 7.8 × 10–2 rad/s
13 | P a g e
28. A (nonrotating) star collapses onto itself from an initial radius Ri with its mass remaining unchanged.
Which curve in figure best gives the gravitational acceleration ag on the surface of the star as a function
of the radius of the star during the collapse

(a) a (b) b (c) c (d) d

29. What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a
person becomes 3/5 of the present weight at the equator. Equatorial radius of the earth is 6400 km
(a) 8.7 × 10–7 rad/s (b) 7.8 × 10–4 rad/s (c) 6.7 × 10–4 rad/s (d) 7.4 × 10–3 rad/s
30. If the density of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that
of the earth, the gravitational acceleration on the surface of the planet is
(a) 0.2 g (b) 0.4 g (c) 2 g (d) 4 g
31. As we go from the equator to the poles, the value of g
(a) remains the same
(b) decreases
(c) increases
(d) decreases upto latitude of 45°
32. The radius of a planet is n times the radius of earth (R). A satellite revolves around it in a circle of radius
4nR with angular velocity w. The acceleration due to gravity on planet’s surface is
(a) R𝜔2 (b) 16 R𝜔2 (c) 32 nR𝜔2 (d) 64 nR𝜔2
33. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented
by (R = Earth's radius):

(a) (b) (c) (d)

34. The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A
man jumps to a height of 2m on the surface of A. What is the height of jump by the same person on the
planet B?
2 2
(a) 3 𝑚 (b) 9 𝑚 (c) 18 m (d) 6 m
35. A roller coaster is designed such that riders experience “weightlessness” as they go round the top of a
hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between:
(a) 14 m/s and 15 m/s (b) 15 m/s and 16 m/s (c) 16 m/s and 17 m/s (d) 13 m/s and 14 m/s
36. If earth is supposed to be a sphere of radius R, if g30 is value of acceleration due to gravity at lattitude
of 30° and g at the equator, the value of g – g30 is
1 3 1
(a) 4 𝜔2 𝑅 (b) 4 𝜔2 𝑅 (c) 𝜔2 𝑅 (d) 2 𝜔2 𝑅
37. In order to make the effective acceleration due to gravity equal to zero at the equator, the angular
velocity of rotation of the earth about its axis should be (g = 10 ms–2 and radius of earth is 64000 km)
1 1 1
(a) Zero (b)800 rad sec–1 (c)80rad sec–1 (d)8rad sec–1
38. If the mass of earth is eighty times the mass of a planet and diameter of the planet is one fourth that of
earth, then acceleration due to gravity on the planet would be
(a) 7.8 m/s2 (b) 9.8 m/s2 (c) 6.8 m/s2 (d) 2.0 m/s2
39. Explorer 38, a radio-astronomy satellite of mass 200 kg, circles the Earth in an orbit of average radius
3R/2 where R is the radius of the Earth. Assuming the gravitational pull on a mass of 1 kg at the earth's
surface to be 10 N, calculate the pull on the satellite
(a) 889 N (b) 89 N (c) 8889 N (d) 8.9 N
40. How many hours would make a day if the earth were rotating at such a high speed that the weight of a
body on the equator were zero?
(a) 6.2 h (b) 1.4 h (c) 28 h (d) 5.6 h

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41. Let g be the acceleration due to gravity at earth’s surface and K be the rotational kinetic energy of the
earth. Suppose the earth’s radius decreases by 2% keeping all 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 decreases by 4%
(d) g decreases by 4% and K increases by 4%
42. Let w be the angular velocity of the earth’s rotation about its axis. Assume that the acceleration due to
gravity on the earth’s surface has the same value at the equator and the poles. An object weighed at the
equator gives the same reading as a reading taken at a depth d below earth’s surface at a pole (d << R).
The value of d is
𝜔2𝑅2 𝜔2𝑅2 2𝜔 2 𝑅 2 √𝑅𝑔
(a) (b) (c) (d)
𝑔 2𝑔 𝑔 𝑔
Topic 4: Gravitational Field, Potential and Potential Energy
43. The magnitude of gravitational potential energy of earth-moon system is U which is zero at infinite
separation. If K is the K.E. of the moon with respect to earth, then
(a) |U| = K (b) |U| < K (c) |U| > K (d) either B or C
44. The gravitational potential due to a hollow sphere (mass M, radius R) varies with distance r from centre
as

(a) (b) (c) (d)

45. A planet is moving in an elliptical orbit around the sun. If T, V, E and L stand respectively for its kinetic
energy, gravitational potential energy, total energy and magnitude of angular momentum about the
centre of force, then which of the following is correct ?
(a) T is conserved (b) V is always positive (c) E is always negative
(d) L is conserved but direction of vector L changes continuously
46. If ‘g’ is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an
object of mass ‘m’ raised from the surface of the earth to a height equal to the radius ‘R' of the earth is
1 1
(a)4 𝑚𝑔𝑅 (b) 2 𝑚𝑔𝑅 (c) 2 mgR (d) mgR
47. In a certain region of space, gravitational field is given by I = –(K/r). Taking the reference point to be at
r = r0 with V =V0, find the potential.
𝑟 𝑟 𝑟 𝑟
(a) 𝐾 𝑙𝑜𝑔 𝑟 + 𝑉0 (b) 𝐾 𝑙𝑜𝑔 𝑟0 + 𝑉0 (c) 𝐾 𝑙𝑜𝑔 𝑟⬚ − 𝑉0 (d) 𝑙𝑜𝑔 𝑟0 − 𝑉0 𝑟
0 0
48. Taking the gravitational potential at a point infinite distance away as zero, the gravitational potential at a
point A is –5 unit. If the gravitational potential at point infinite distance away is taken as + 10 units, the
potential at point A is
(a) – 5 unit (b) + 5 unit (c) + 10 unit (d) + 15 unit
49. Two rings having masses M and 2M, respectively, having the same radius are placed coaxially as shown
in the figure. If the mass distribution on both the rings is non – uniform, then the gravitational potential
at point P is

𝐺𝑀 1 2 𝐺𝑀 2
(a) − [ + ] (b) − [1 + ] (c) zero (d)cannot be determined from the given information
𝑅 √2 √5 𝑅 √2
Topic 5: Motion of Satellites, Escape Speed and Orbital Velocity
50. A satellite revolves around the earth of radius R in a circular orbit of radius 3R. The percentage increase
in energy required to lift it to an orbit of radius 5R is
(a) 10 % (b) 20 % (c) 30 % (d) 40 %

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51. The mean radius of earth is R, its angular speed on its own axis is𝜔and the acceleration due to gravity at
earth's surface is g. What will be the radius of the orbit of a geostationary satellite?
(a) (𝑅 2 𝑔/𝜔2 )1/3 (b)(𝑅𝑔/𝜔2 )1/3 (c) (𝑅 2 𝜔2 /𝑔)1/3 (d)(𝑅 2 𝑔/𝜔)1/3
52. The moon has a mass of 1/81 that of the earth and a radius of 1/4 that of the earth. The escape speed
from the surface of the earth is 11.2 km/s. The escape speed from the surface of the moon is:
(a) 1.25 km/s (b) 2.5 km/s (c) 3.7 km/s (d) 5.6 km/s
53. A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times
smaller. Given that the escape velocity from the earth's surface is 11 km s–1, the escape velocity from the
surface of the planet would be
(a) 1.1 km s–1 (b) 11 km s–1 (c) 110 km s–1 (d) 0.11 km s–1
54. What is the minimum energy required to launch a satellite of mass m from the surface of a planet of
mass M and radius R in a circular orbit at an altitude of 3R?
7𝐺𝑚𝑀 2𝐺𝑚𝑀 𝐺𝑚𝑀 𝐺𝑚𝑀
(a) 8𝑅 (b) 3𝑅 (c) 2𝑅 (d) 𝑅
55. The orbital velocity of an artificial satellite in a circular orbit just above the centre’s surface is v0. For a
satellite orbiting at an altitude of half of the earth’s radius, the orbital velocity is
2 2 3 3
(a) (√(3)) 𝑣0 (b) 3 𝑣0 (c) 2 𝑣0 (d) √(2) 𝑣0
56. A satellite of mass m revolves around the earth of radius R at a height ‘x’ from its surface. If g is the
acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is
1/2
𝑔𝑅 2 𝑔𝑅 𝑔𝑅 2
(a) 𝑅+𝑥 (b) 𝑅−𝑥 (c) 𝑔𝑥 (d) (𝑅+𝑥)
57. Two satellites of masses m and 2m are revolving around a planet of mass M with different speeds in
orbits of radii r and 2r respectively. The ratio of minimum and maximum forces on the planet due to
satellites is

1 1 1
(a)2 (b) 4 (c) 3 (d) None of these
58. A satellite is revolving round the earth in a circular orbit of radius 'a' with velocity v0. A particle of mass
5
m is projected from the satellite in forward direction with relative velocity𝑉 = [√4 − 1] 𝑉0.During
subsequent motion of the particle total energy is
(a) –3G Me m/8a (b) zero (c) –5G Me m/6a (d) ∞
59. Two particles of equal mass ‘m’ go around a circle of radius R under the action of their mutual
gravitational attraction. The speed of each particle with respect to their centre of mass is
𝐺𝑚 𝐺𝑚 𝐺𝑚 𝐺𝑚
(a) √ 4𝑅 (b) √ 3𝑅 (c) √ 2𝑅 (d) √ 𝑅
60. The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of
the Earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the
surface of the Earth. The value of f is
1 1 1
(a)3 (b) 2 (c) √2 (d)
√2

LEVEL-2
1. Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence
(a) there will be no change in weight anywhere on the earth
(b) weight of the object, everywhere on the earth, will decrease
(c) weight of the object, everywhere on the earth, will increase
(d) except at poles, weight of the object on the earth will
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2. A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it
splits into two equal masses. The first mass moves in a circular orbit of radius R/2, and the other mass,
in a circular orbit of radius. The difference between the final and initial total energies is:
𝐺𝑀𝑚 𝐺𝑀𝑚 𝐺𝑀𝑚 𝐺𝑀𝑚
(a) − 2𝑅 (b) + 6𝑅 (c) − 6𝑅 (d) 2𝑅
3. A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius R(R < <
L). A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing
through its centre. If the time period of star is T and its distance from the galaxy’s axis is r, then:
(a) T∞r (b) T  √𝑟 (c) T∞r2 (d) T2∞r3
4. The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the
surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius
of the planet would be
(a) 1/2 R (b) 2 R (c) 4 R (d) 1/4 R
5. A central particle M is surrounded by a square array of other particles, separated by either distance d or
distance d/2 along the perimeter of the square. The magnitude of the gravitational force on the central
particle due to the other particles is

9𝐺𝑀𝑚 5𝐺𝑀𝑚 3𝐺𝑀𝑚 𝐺𝑀𝑚

(a) 𝑑2 (b) 𝑑2 (c) 𝑑2 (d) 𝑑2
6. An astronaut of mass m is working on a satellite orbiting the earth at a distance h from the earth's
surface. The radius of the earth is R, while its mass is M. The gravitational pull FG on the astronaut is :
𝐺𝑀𝑚 𝐺𝑀𝑚
(a) Zero since astronaut feels weightless (b) (𝑅+ℎ)2 < 𝐹𝐺 < 𝑅2
𝐺𝑀𝑚 𝐺𝑀𝑚
(c) 𝐹𝐺 = (𝑅+ℎ)2 (d) 0 < 𝐹𝐺 < 𝑅2
7. A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V. Due to
the rotation of planet about its axis the acceleration due to gravity g at equator is 1/2 of g at poles. The
escape velocity of a particle on the pole of planet in terms of V is
(a) Ve = 2V (b) Ve = V (c) Ve = V/2 (d) Ve = √3V
8. The gravitational field strength (E) due to a solid sphere (mass M, radius R) varies with distance r from
centre as [for r < R]
1 1
(a) E∞r (b) E∞ 𝑟 2 (c) E∞ r2 (d) E∞𝑟 2
9. The change in the value of acceleration of earth towards sun, when the moon comes from the position of
solar eclipse to the position on the other side of earth in line with sun is: (mass of the moon = 7.36 ×
1022 kg, radius of the moon’s orbit = 3.8 × 108 m).
(a) 6.73 × 10–5 m/s2 (b) 6.73 × 10–3 m/s2 (c) 6.73 × 10–2 m/s2 (d) 6.73 × 10–4 m/s2
10. If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed
with which the earth would have to rotate about its axis so that the person at the equator will weight
3/4W. Radius of the Earth is 6400 km and g =10 m/s2.
(a) 1.1×10–3 rad/s (b) 0.83×10–3 rad/s (c) 0.63 × 10–3 rad/s (d) 0.28×10–3 rad/s
11. Two point masses m1 and m2 are initially at rest and at infinite distance apart. They start moving
towards one another under their mutual gravitational field. Their relative speed when they are at a
distance d apart is
2𝐺(𝑚1 +𝑚2 ) 2𝐺(𝑚1 𝑚2 ) 2𝐺𝑑
(a)√[2𝐺(𝑚1 + 𝑚2 )] (b) √[ ] (c) √[ ] (d) √[(𝑚 ]
𝑑 𝑑 1 +𝑚2 )
12. Two rings each of radius ‘a’ are coaxial and the distance between their centres is a. The masses of the
rings are M1 and M2. The work done in transporting a particle of a small mass m from centre C1 to C2 is

17 | P a g e
 𝐺𝑚(𝑀2 −𝑀1 ) 𝐺𝑚(𝑀2 −𝑀1 )
(a) (b) (√2 + 1)
𝑎 𝑎√2
𝐺𝑚(𝑀2 −𝑀1 ) 𝐺𝑚(𝑀2 −𝑀1 )
(c) (√2 − 1) (d) 𝑎
𝑎√2 √2
13. A planet is revolving around the sun in an elliptical orbit. Its closest distance from the sun is rmin. The
farthest distance from the sun is rmax. If the orbital angular velocity of the planet when it is nearest to the
sun is𝜔, then the orbital angular velocity at the point when it is at the farthest distance from the sun is
22 22
(a) ( rmin / rmax ) (b) ( rmax / rmin ) 𝑚𝑖𝑛
(c) (𝑟𝑚𝑎𝑥 𝑚𝑎𝑥
𝜔) (d) (𝑟𝑚𝑖𝑛 𝜔)
14. The radii of two planets are respectively R1 and R2 and their densities are respectively𝜌1 and𝜌2 . The ratio
of the accelerations due to gravity at their surfaces is
𝜌 𝜌
(a) 𝑔𝑙 : 𝑔2 = 𝑅12 : 𝑅22 (b) 𝑔𝑙 : 𝑔2 = 𝑅1 𝑅2 : 𝜌1 𝜌2
1 2
(c) 𝑔𝑙 : 𝑔2 = 𝑅1 𝜌2 : 𝑅2 𝜌1 (d) 𝑔𝑙 : 𝑔2 = 𝑅1 𝜌1 : 𝑅2 𝜌2
15. Two blocks A and B of masses MA and MB respectively, are located 1.0 m apart on a horizontal surface.
The coefficient of static friction𝜇𝑠 between the block and the surface is 0.50. Block A is secured to the
surface and cannot move, What is the minimum mass of Block A that provides enough gravitational
attraction to move Block B? The universal gravitation constant is 6.67 × 10–11 Nm2 /kg2.
(a) 7.5 × 109 kg (b) 7.3 × 1010 kg (c) 14.7 × 1011 kg
(d) The problem cannot be solved without knowing the mass of Block B.
16. A cavity of radius R/2 is made inside a solid sphere of radius R. The centre of the cavity is located at a
distance R/2 from the centre of the sphere. The gravitational force on a particle of mass ‘m’ at a distance
R/2 from the centre of the sphere on the line joining both the centres of sphere and cavity is– (opposite
to the centre of gravity) [Here g = GM/R², where M is the mass of the sphere]
𝑚𝑔 3𝑚𝑔 𝑚𝑔
(a) 2 (b) 8 (c) 16 (d) None of these
17. The gravitational field in a region is given by𝑔⃗ = 5𝑁/𝑘𝑔𝑖̂ + 12𝑁/𝑘𝑔𝑗̂. The change in the gravitational
potential energy of a particle of mass 1 kg when it is taken from the origin to a point (7 m, – 3 m) is:
(a) 71 J (b) 13√58J (c) – 71 J (d) 1 J
18. A point particle is held on the axis of a ring of mass m and radius r at a distance r from its centre C.
When released, it reaches C under the gravitational attraction of the ring. Its speed at C will be
2𝐺𝑚 𝐺𝑀 2𝐺𝑀 1 2𝐺𝑀
(a) √ (√2 − 1) (b) √ (c) √ (1 − ) (d) √
𝑟 𝑟 𝑟 √2 𝑟
19. In order to simulate different values of g, aspiring astronauts are put on a plane which dives in a
parabola given by the equation :𝑥 2 = 500𝑦 where x is horizontal, y is vertically upwards; both being
measured in
meter. The x–component of the velocity of the plane is constant throughout, and has the value of 360
km/h. The effective g experienced by an astronaut on the plane equals

(a) 4g (b) 3g (c)g/5 (d) 5g

20. A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R; h < <
R). The minimum increase in its orbital velocity required, so that the satellite could escape from the
earth's gravitational field, is close to: (Neglect the effect of atmosphere.)
(a) √𝑔𝑅/2 (b) √𝑔𝑅(√2 − 1) (c) √2𝑔𝑅 (d) √𝑔𝑅
21. Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and lengthℓis

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 𝐺𝑚 𝐺𝑚 2𝜋𝐺𝑚 2𝜋𝐺𝑚
(a) ℓ2 along +x-axis (b) 𝜋ℓ2 along +y-axis (c) ℓ2 along +x-axis (d) ℓ2 along +y-axis
22. An artificial satellite of mass m of a planet of mass M, revolves in a circular orbit whose radius is n
times the radius R of the planet. In the process of motion, the satellite experiences a slight resistance due
to cosmic dust. Assuming resistance force on satellite depends on velocity as F = av2 where ‘a’ is
constant. Calculate how long the satellite will stay in orbit before it falls onto the planet’s surface.
𝑚√𝑅(√𝑛−1) 𝑚√𝑅(√𝑛+1) 2𝑚√𝑅(√𝑛−1) 𝑚√𝑅(√𝑛−1)
(a) 𝑎√𝐺𝑀 (b) 𝑎√𝐺𝑀 (c) (d) 𝑎√2𝐺𝑀
𝑎√𝐺𝑀
23. A skylab of mass m kg is first launched from the surface of the earth in a circular orbit of radius 2R
(from the centre of the earth) and then it is shifted form this circular orbit of radius 3R. The minimum
energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit
are
3 𝑚𝑔𝑅 3 𝑚𝑔𝑅
(a)4 𝑚𝑔𝑅, 6 (b) 4 𝑚𝑔𝑅, 12 (c) mgR, mgR (d) 2mgR, mgR
24. With what minimum speed should m be projected from point C in presence of two fixed masses M each
at A and B as shown in the figure such that mass m should escape the gravitational attraction of A and B

2𝐺𝑀 2√2𝐺𝑀 𝐺𝑀 𝐺𝑀
(a)√ (b) √ (c) 2√ 𝑅 (d) 2√2√
𝑅 𝑅 𝑅
25. From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in
the figure. Taking gravitational potential V = 0 at r =∞, the potential at the centre of the cavity thus
formed is :(G = gravitational constant)

−2𝐺𝑀 −2𝐺𝑀 −𝐺𝑀 −𝐺𝑀

(a) 3𝑅 (b) 𝑅 (c) 2𝑅 (d) 𝑅
26. A body starts from rest from a point distance R0 from the centre of the earth. The velocity acquired by
the body when it reaches the surface of the earth will be (R represents radius of the earth).
1 1 1 1 1 1 1 1
(a) 2𝐺𝑀 (𝑅 − 𝑅 ) (b) √2𝐺𝑀 (𝑅 − 𝑅 ) (c) 𝐺𝑀 (𝑅 − 𝑅 ) (d) 2𝐺𝑀√(𝑅 − 𝑅 )
0 0 0 0
27. Which of the following most closely depicts the correct variation of the gravitational potential V(r) due
to a large planet of radius R and uniform mass density ? (figures are not drawn to scale)

(a) (b) (c) (d)

28. A small satellite of mass m is revolving around earth in a circular orbit of radius r0 with speed v0. At
certain point of its orbit, the direction of motion of satellite is suddenly changed by angle𝜃 = cos–1(3/5)
by turning its velocity vector, such that speed remains constant. The satellite consequently goes to
elliptical orbit around earth. The ratio of speed at perigee to speed at apogee is

19 | P a g e
 (a) 3 (b) 9 (c) 1/3 (d) 1/9
29. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under
the action of their mutual gravitational attraction. The speed of each particle is:
𝐺𝑀 𝐺𝑀 𝐺𝑀 1 𝐺𝑀
(a) √ (b) √2√2 (c) √ (1 + 2√2) (d) 2 √ (1 + 2√2)
𝑅 𝑅 𝑅 𝑅
30. The gravitational potential of two homogeneous spherical shells A and B of same surface density at
their respective centres are in the ratio 3 : 4. If the two shells coalesce into single one such that surface
charge density remains same, then the ratio of potential at an internal point of the view shell to shell A is
equal to
(a) 3 : 2 (b) 4 : 3 (c) 5 : 3 (d) 5 : 6
ANSWER KEYS
NEET 8 Years at a Glance
1) 3 2) 2 3) 3 4) 1 5) 1 6) 2 7) 1 8) 2 9) 3 10) 4
11) 3 12) 1 13) 1 14) 4 15) 3 16) 3 17) 1 18) 2 19) 2 20) 3
21) 3 22) 3
LEVEL-1
1) 2 2) 3 3) 3 4) 4 5) 4 6) 2 7) 2 8) 3 9) 3 10) 1
11) 3 12) 1 13) 3 14) 1 15) 2 16) 4 17) 1 18) 3 19) 4 20) 2
21) 1 22) 3 23) 3 24) 3 25) 1 26) 1 27) 1 28) 2 29) 2 30) 1
31) 3 32) 4 33) 2 34) 3 35) 1 36) 2 37) 2 38) 4 39) 1 40) 2
41) 3 42) 1 43) 3 44) 2 45) 3 46) 2 47) 1 48) 2 49) 1 50) 2
51) 1 52) 2 53) 3 54) 1 55) 1 56) 4 57) 3 58) 1 59) 1 60) 4
LEVEL-2
1) 4 2) 3 3) 1 4) 1 5) 3 6) 3 7) 1 8) 1 9) 1 10) 3
11) 2 12) 3 13) 4 14) 4 15) 2 16) 2 17) 4 18) 3 19) 4 20) 2
21) 4 22) 1 23) 2 24) 2 25) 4 26) 2 27) 3 28) 2 29) 4 30) 3

Hints and Solutions

NEET Previous Years at a Glance:
1. (c) If universal gravitational constant becomes ten times, then𝐺 | = 10𝐺
𝐺𝑀
Acceleration due to gravity,𝑔 = 𝑅2
So, acceleration due to gravity increases.
2. (b) Speed of the planet will be maximum when its distance from the sun is minimum as mvr = constant.

Point A is perihelion and C is aphelion.

Clearly, VA > VB > VC
So, KA > KB > KC
2ℎ
3. (c) Acceleration due to gravity at height h, 𝑔𝑛 = 𝑔0 (1 − )ℎ = 1𝑘𝑚
𝑅
𝑑
Acceleration due to gravity at depth d, 𝑑𝑑 = 𝑔0 (1 − 𝑅
)
2ℎ 𝑑
𝑔ℎ = 𝑔𝑑 ; 𝑔0 (1 − 𝑅 ) = 𝑔0 (1 − 𝑅) ⇒ 𝑑 = 2ℎ; = 2 × 1𝑘𝑚 ⇒ 𝑑 = 2𝑘𝑚
4. (a) Both the astronauts are in the condition of weightlessness. Gravitational force between them pulls
towards each other. Hence Astronauts move towards each other under mutual gravitational force.
5. (a) As we know, gravitational potential (v) and acceleration due to gravity (g) with height
−𝐺𝑀
𝑉 = 𝑅+ℎ = −5.4 × 107 ----------(1)
𝐺𝑀
and 𝑔 = (𝑅+ℎ)2 = 6-----------(2)

20 | P a g e
 −𝐺𝑀
𝑅+ℎ −5.4×107 5.4×107
Dividing (1) by (2) (𝑅+ℎ)2
= ⇒ (𝑅+ℎ)
=6
6
⇒R + h = 9000 km so, h = 2600 km
6. (b) As we know, escape velocity,
2𝐺𝑀 2𝐺 4
𝑉𝑒 = √ = √ 𝑅 . (3 𝜋𝑅 3 𝜌) ∝ 𝑅√𝜌
𝑅
𝑉 𝑅 𝜌 𝑉 𝑅 𝜌 𝑉
∴ 𝑉𝑒 = 𝑅 𝑒 √𝜌𝑒 ⇒ 𝑉𝑒 = 2𝑅𝑒 √2𝜌𝑒 ; ∴Ratio 𝑉𝑒 = 1: 2√2
𝑝 𝑝 𝑝 𝑝 𝑒 𝑒 𝑝

𝐺𝑀
7. (a) As we know, orbital speed, 𝑉𝑜𝑟𝑏 = √ 𝑟
2𝜋𝑟 2𝜋𝑟
Time period𝑇 = 𝑣 = √𝑟
𝑜𝑟𝑏 √𝐺𝑀
Squaring both sides,
2
2𝜋𝑟√𝑟 4𝜋 2 𝑇2 4𝜋 2
𝑇2 = ( ) = . 𝑟3 ⇒ = =𝐾
√𝐺𝑀 𝐺𝑀 𝑟3 𝐺𝑀
2
⇒ 𝐺𝑀𝐾 = 4𝜋
8. (b)

Let the distance moved by spherical body of mass M is x1 and by spherical body of mass 5m is x2
As their C.M. will remain stationary
So, (M) (x1) = (5M) (x2) or, x1 = 5x2 and for touching x1 + x2 = 9R
So, x1 = 7.5 R
9. (c) The gravitational force on the satellite will be aiming towards the centre of the earth so acceleration
of the satellite will also be aiming towards the centre of the earth.
10. (d) Given: Height of the satellite from the earth's surface h= 0.25 × 106m
Radius of the earth R = 6.38 × 106m
Acceleration due to gravity g = 9.8 m/s2
Orbital velocity, V0 = ?
𝐺𝑀 𝐺𝑀 𝑅2 9.8×6.38×6.38 𝐺𝑀
𝑉0 = √(𝑅+ℎ) = √ 𝑅2 . (𝑅+ℎ) =√ = 7.76𝑘𝑚/𝑠 [∵ = 𝑔]
6.63×106 𝑅2
11. (c) From question,
Escape velocity
2𝐺𝑀 2𝐺𝑀 2×6.6×10−11 ×5.98×1024
=√ = 𝑐 =speed of light ⇒𝑅= = 𝑚 = 10−2 𝑚
𝑅 𝑐2 (3×108 )2
1
12. (b) First when (r < R) E∝r and then when r > R𝐸 ∝ 𝑟 2
Hence graph (b) correctly dipicts.

𝑢2 𝑠𝑖𝑛 2𝜃
13. (a) Horizontal range = 𝑠𝑜𝑔 ∝ 𝑢2
𝑔
2
𝑔𝑝𝑙𝑎𝑛𝑒𝑡 (𝑢𝑝𝑙𝑎𝑛𝑒𝑡 ) 3 2
or = (𝑢𝑒𝑎𝑟𝑡ℎ )2
; Therefore𝑔𝑝𝑙𝑎𝑛𝑒𝑡 = (5) (9.8𝑚/𝑠 2 ) = 3.5𝑚/𝑠 2
𝑔𝑒𝑎𝑟𝑡ℎ
14. g' = g (1-d/R)

mg' = mg(1/2) ; W' = 200(1/2) = 100 N

15. In northern hemisphere dip is +ve and in southern hemisphere dip is .ve.
16.

21 | P a g e
17. We know that square of time period is proportional to cube of the radius.
𝑇12 (𝑅 +6𝑅 )3
𝑇2 ∝ 𝑟3 ; 𝑇 2 ∝ (𝑅𝐸 + ℎ)3 ; = (𝑅 𝐸+2.5𝑅𝐸
𝑇22 𝐸 𝐸)
3

𝑇12 73 𝑇12 𝑇1
= 7 3
; = 8; 𝑇2 = 2√2
𝑇22 ( ) 𝑇22
2
24
𝑇2 = 2√2 ⇒ 𝑇2 = 6√2ℎ

18. Gravitational potential energy of the two particle system can be written as follows :
𝐺𝑚 𝑚
𝑈 = 𝑟1 2. Hence potential energies in two cases can be written as follows :
𝐺𝑀𝑚
(𝑃. 𝐸. )𝐴 =
𝑅
𝐺𝑀𝑚
(𝑃. 𝐸. )𝐵 =
𝑅+ℎ
∴ Δ𝑈 = (𝑃. 𝐸)𝐵 − (𝑃. 𝐸)𝐴
𝐺𝑀𝑚 𝐺𝑀𝑚 𝐺𝑀𝑚ℎ
= + =
𝑅+ℎ 𝑅 𝑅(𝑅+ℎ)
𝑑
19. At depth: 𝑔𝑒𝑓𝑓 = 𝑔 (1 − 𝑅)
𝑔 𝑑
⇒ 𝑛 = 𝑔 (1 − 𝑅) ⇒ 𝑑 = (𝑛 − 1)𝑅/𝑛
𝑅 2
20. 𝑔𝑛 = 𝑔 [𝑅+ℎ]
2
𝑅
𝑚𝑔ℎ = 𝑚𝑔 [ 𝑅 ]
𝑅+
2
2 2
𝑊ℎ = 72 (3) = 32𝑁
21.
2𝐺𝑀 4
𝑉𝑒 = √ ⇒ 𝑀 = 3 𝜋𝑅 3 𝐷
𝑅

4 3
√2𝐺 × 3 𝜋𝑅 𝐷
𝑉𝑒 =
𝑅
𝑉1 𝑅1 𝑉 𝑅
𝑉𝑒 𝛼𝑅; = ; = ; 𝑉 = 4𝑉
𝑉2 𝑅2 𝑉2 4𝑅 2

22
given v = kVe
where, k < 1
Thus, v < Ve
From conservation of mechanical energy,
1 2 𝐺𝑚𝑀 𝐺𝑚𝑀
𝑚𝑉 − = − (𝑅+ℎ)
2 𝑅

22 | P a g e
 LEVEL-1
⃗⃗
1. (b) Since areal velocity𝐴⃗& angular momentum𝐿 ⃗⃗of a planet are related by equation𝐴⃗ = 𝐿 , where M is
2𝑀
⃗⃗is(𝜏⃗𝑒𝑥𝑡 = 0), hence𝐴⃗is also constant
the mass of planet. Since in planetary motion𝐿
2. (c)
𝑇 𝑅 3/2 𝑇 4𝑅 3/2
3. (c) 𝑇1 = (𝑅1) ⇒ 𝑇 = (16𝑅) ⇒ 𝑇2 = 8𝑇
2 2 2
𝑣𝑝 1+𝑒 1+0.20 3
4. (d) = 1−𝑒 = 1−0.20 = 2
𝑣𝑎
5. (d) Time period does not depend upon the mass of satellite
𝑟 +𝑟
6. (b) T2𝛼r3, where r = mean radius= 1 2 2
7. (b) Given that𝑇1 = 1day and 𝑇2 = 8days
𝑇 𝑟 3/2
∴ 𝑇2 = (𝑟2)
1 1
𝑟2 𝑇2 2/3 8 2/3
⇒ 𝑟 = (𝑇 ) = (1) = 4 ⇒ 𝑟2 = 4𝑟1
1 1
8. (c) Since area of triangle csa is 41 of total area of ellipse, therefore:

1
Area of cdas = 3Area of abcs
Now that from Kepler's second law areal velocities of the planets are constant which essentially means
planets cover equal area in equal time interval.
Hence,
Time taken in covering path abc and path cda will be in proportion to their respective enclosed areas.
⇒t1=3t2
9. (c) In planetary motion𝜏⃗𝑒𝑥𝑡 = 0 ⇒ 𝐿⃗⃗ =constant
⃗⃗ = 𝑟⃗ × 𝑝⃗(= 𝑚𝑣⃗) = 𝑚𝑟𝑣(∵ 𝜃 = 900 )
𝐿
So m1d1v1 = m2d2v2 (here r = d)

𝑣 𝑑
⇒ 𝑣2 = 12 1
10. (a) Angular momentum is conserved. At A, the moment of inertia is least and hence angular speed is
maximum. Thus the K.E. at A is maximum.
11. (c) T 2∝R3 (According to Kepler’s law)
𝑇12 ∝ (1013 )3 and 𝑇22 ∝ (1012 )3
𝑇2 𝑇
∴ 𝑇12 = (10)3 or 𝑇1 = 10√10
2 2
12. (a) Since, T2 = kr3
Differentiating the above equation
Δ𝑇 Δ𝑟 Δ𝑇 3 Δ𝑟
⇒2 𝑇 =3 𝑟 ⇒ 𝑇 =2 𝑟
13. (c) According to Kepler’s law of planetary motion,

23 | P a g e
 𝑅 3/2 4𝑅 3/2
∴ 𝑇2 = 𝑇1 (𝑅2) = 5×[𝑅] = 40hours
1
14. (a) By law of conservation of angular momentum,
mvr = constant
𝑣𝑚𝑖𝑛𝑚𝑎𝑥𝑚𝑎𝑥𝑚𝑖𝑛
60×1.6×1012 60
∴𝑣
8×1012 5 𝑚𝑎𝑥
15. (b) Same force acts on both masses
1
Hence 𝑎 ∝ 𝑚 (𝐹 = 𝑚𝑎)
In absence of external force (remember mutual gravitational force is an internal force for the system)
total energy remains constant.
16. (d)

4×1 9×1 2 𝑥
∴𝐺 = 𝐺 (60−𝑥)2 𝑜𝑟 3 = (60−𝑥) ⇒ 𝑥 = 24𝑐𝑚
𝑥2
17. (a) Weight of body on the surface of the earth =mg=72N
2
𝑔𝑅𝐸
Acceleration due to gravity at height h is𝑔ℎ = (𝑅 2
𝐸 +ℎ)
𝑅𝐸
Substitute ℎ = in above expression:
2
2
𝑔𝑅𝐸 4
𝑔ℎ = 𝑅 2 = 9𝑔
(𝑅𝐸 + 𝐸 )
2
Gravitational force on body at height h is 𝐹 = 𝑚𝑔ℎ
4 4 4
= 𝑚 × 9 𝑔 = 9 × 𝑚𝑔 = 9 × 72𝑁 = 32𝑁
𝐺(2𝑚1 )(2𝑚2 ) 𝐺𝑚1 𝑚2 4
[ − ] −1
(3𝑟)2 𝑟2
18. (c) % change= 𝐺𝑚1 𝑚2 × 100 ; = 9 1 × 100 = −56%
𝑟2
–ve sign indicates that force of attraction decreases
𝐺𝑚 𝐺𝑚𝑚
19. (d) 𝑥 2𝑒 = (𝐷−𝑥)2

𝐺(81𝑚) 𝑚
or = (𝐷−𝑥)2
𝑥2
9𝐷
∴𝑥= 10
20. (b)𝐹 = 𝐹1 + 𝐹2 + 𝐹3 + 𝐹4 + 𝐹5

𝐺𝑚2 5 1
𝐹= (4 + ) = 𝑚𝜔2 𝑎
𝑎2 √3

24 | P a g e
 𝐺𝑚 5 1
𝜔 = √ 𝑎3 (4 + )
√3
𝐺𝑚1 𝑚2 𝐺𝑚 𝑚2
1
21. (a)𝐹 = = (𝑟 +𝑟
𝑟2 1 2)
2

22. (c) During total eclipse:

Total attraction due to sun and moon,
𝐺𝑀 𝑀 𝐺𝑀 𝑀
𝐹1 = 𝑟𝑠2 𝑒 + 𝑟𝑚2 𝑒
1 2
When moon goes on the opposite side of earth effective force of attraction
𝐺𝑀 𝑀 𝐺𝑀 𝑀
𝐹2 = 𝑟𝑠2 𝑒 − 𝑟𝑚2 𝑒
1 2
2𝐺𝑀𝑚 𝑀𝑒
Change in force, Δ𝐹 = 𝐹1 − 𝐹2 = 𝑟22
Δ𝐹 2𝐺𝑀𝑚
Change in acceleration of earth Δ𝑎 = 𝑀 =
𝑒 𝑟22
𝐹𝑎𝑣 𝐺𝑀𝑠
Average force on earth,𝐹𝑎𝑣 = =
𝑀𝑒 𝑟12
Percentage change in acceleration
Δ𝑎 2𝐺𝑀𝑚 𝑟2 𝑟 2𝑀
= 𝑎 × 100 = 1
× 𝐺𝑀 × 100 = 2 (𝑟1 ) 𝑚
× 100
𝑎𝑣 𝑟22 𝑠 2 𝑀𝑠
𝐺×103 𝐺×105 1 102 1 10 1
23. (c) (𝑟)2
= (1−𝑟)2; = (1−𝑟)2; = 1−𝑟 ⇒ 10𝑟 = 1 − 𝑟; ∴ 𝑟 = 11 𝑘𝑚
𝑟2 𝑟
24. (c) Gravitational force is independent of medium, Hence, this will remain same.
25. (a) At the surface of earth, the value of g = 9.8m/sec2. If we go towards the centre of earth or we go
above the surface of earth, then in both the cases the value of g decreases.
Hence W1=mgmine, W2=mgsea level, W3=mgmoun
So W1< W2 > W3 (g at the sea level = g at the surface of earth)
26. (a) We know that,
variation in g with height "h"
𝑅 2
𝑔| = 𝑔 (𝑅+ℎ)
𝑔| →gravity at height from surface of earth.
𝑟 →Radius of earth
ℎ →height above surface
1
Therefore, 𝑔| ∝ 𝑟 2
Acceleration due to gravity above the earth
2ℎ
(𝑔1 ) = 𝑔0 (1 − )-----------(i)
𝑟
Since h<<R acceleration due to gravity below earth surface
𝑑
𝑔2 = 𝑔0 (1 − 𝑅)--------------(ii)
𝑔 𝑅−2
Now, put d = h = 1 km Thus, 𝑔1 = 𝑅−1
2
27. (a) 𝑔| = 𝑔 − 𝜔2 𝑐𝑜𝑠 2 𝜆 ⇒ 0 = 𝑔 − 𝜔2 𝑅 𝒄𝒐𝒔2 6 00
𝜔2𝑅 𝑔 1 𝑟𝑎𝑑 𝑟𝑎𝑑
0=𝑔− ⇒ 𝜔 = 2√𝑅 = 400 𝒔𝒆𝒄 = 2.5 × 10−3 𝒔𝒆𝒄
4
1
28. (b)𝑔 ∝ 𝑅2
R decreasing g increase hence, curve b represents correct variation
29. (b) True weight at equator, W=mg
3
Observed weight at equator, 𝑊 | = 𝑚𝑔| = 5 𝑚𝑔

At equator, latitude λ=0 ; Using the formula, 𝑚𝑔| = 𝑚𝑔 − 𝑚𝑅𝜔2 𝒄𝒐𝒔2 𝜆

3 3 2
= 5 𝑚𝑔 = 𝑚𝑔 − 𝑚𝑅 2 𝜔2 𝒄𝒐𝒔2 𝜃 = 𝑚𝑔 − 𝑚𝑅𝜔2 ⇒ 𝑚𝑅𝜔2 = 𝑚𝑔 − 5 𝑚𝑔 = 5 𝑚𝑔
2 𝑔 1/2 2×9.8 1/2
∴ 𝜔 = (5 𝑅) = (5×6.4×106) = 7.8 × 10−4 𝑟𝑎𝑑/𝑠
30. (a) We know that,
25 | P a g e
 4
𝐺𝑀 𝐺( 𝜋𝑅 3 )𝜌 4
3
𝑔= = = 3 𝜋𝐺𝑅𝜌
𝑅2 𝑅2
𝑔| 𝑅| 0.2𝑅
= = = 0.2 ∴ 𝑔| = 0.2𝑔
𝑔 𝑅 𝑅
𝐺𝑀𝑒
31. (c) Since𝑔 = for earth.
𝑅𝑒2
At poles the earth is slightly flattened. It means that the radius of earth at poles is slightly less in
comparison to radius at equator. So from the above expression, the value of ‘g’ at poles is greater in
comparison to value of ‘g’ at equator.
𝐺𝑀𝑚
32. (d) 𝑚𝑟𝜔2 = 𝑟 2
𝐺𝑀 = 𝑟 3 𝜔2 (𝐺𝑀 = 𝑔𝑅 2 )
𝑟 3 .𝜔2
𝑔= 𝑅2
| (4𝑛𝑅)3 .𝜔 2
𝑔 = 𝑔| = 64𝑛𝑅𝜔2
𝑛2 .𝑅 2
𝑑
33. (b) With depth 𝑔1 = 𝑔 (1 − )
𝑅
As depth d goes on increasing g1 goes on decreasing, it remains maximum at the surface of Earth .The
above equation is in the form of straight line.
With height
2ℎ 2𝑔ℎ
𝑔2 = 𝑔 (1 − 𝑅 ) = 𝑔 − 𝑅
1
𝑔2 ∝ 𝑅 (Hyperbola)
Acceleration due to gravity goes on decreasing as the h above Earth surface increases.
34. (c) Applying conservation of total mechanical energy principle
1
𝑚𝑣 2 = 𝑚𝑔𝐴 ℎ𝐴 = 𝑚𝑔𝐵 ℎ𝐵
2
⇒ 𝑔𝐴 ℎ𝐴 = 𝑚𝑔𝐵 ℎ𝐵
𝑔
⇒ ℎ𝐵 = (𝑔𝐴 ) ℎ𝐴 = 9 × 2 = 18𝑚
𝐵
35. (a) For the riders to experience weightlessness at the top of the hill, the weight of the rider must be
balanced by the centripetal force.

𝑚𝑣 2
i.e., 𝑚𝑔 = ⇒ 𝑣 = √𝑔𝑅
𝑅
= √10 × 20 = 14.1𝑚𝑠 −1
Hence, the speed of the car should be between 14 ms–1 and 15 ms–1.
36. (b) Acceleration due to gravity at lattitude’𝜆’ is given by
𝑔𝜆 = 𝑔𝑒 − 𝑅𝑒 𝜔2 𝒄𝒐𝒔2 𝜆
At equator, 𝜆 = 900 ⇒ 𝑐𝑜𝑠 𝜆 = 𝑐𝑜𝑠 9 00 = 0
𝑔𝜆 = 𝑔𝑒 = 𝑔 (as given in question)
3
𝐴𝑡300 , 𝑔30 = 𝑔 − 𝑅𝜔2 𝑐𝑜𝑠 2 3 0 = 𝑔 − 4 𝑅𝜔2
3
or 𝑔 − 𝑔30 = 4 𝑅𝜔2
37. (b) 𝑔| = 𝑔 − 𝜔2 𝑅 𝒄𝒐𝒔2 𝜆
To make effective acceleration due to gravity zero at equator𝜆 = 0 and g ' = 0
𝑔 1 𝑟𝑎𝑑
∴ 0 = 𝑔 − 𝜔2 𝑅 ⇒ 𝜔 = √𝑅 = 800 𝑠
38. (d) Since gravitational acceleration on earth is defined as
𝐺𝑀
𝑔𝑒 = 𝑅2𝑒 ----------------(i)
𝑒
𝑀𝑒 𝑅𝑒
mass of planet is𝑀𝑝 = & radius𝑅𝑝 =
80 4
𝐺𝑀𝑝
So, 𝑔𝑝 = 2 -----------(ii)
𝑅𝑝
From (i) & (ii), we get
26 | P a g e
 𝑀𝑝 𝑅2 𝑔𝑒
𝑔𝑝 = 𝑔𝑒 𝑅2 × 𝑀𝑒 = = 2𝑚/ 𝑠𝑒𝑐 2 ⬚ (as g=10m/sec2)
𝑝 𝑒 5
2ℎ 4𝑔 3𝑅
39. (a)𝑔ℎ = 𝑔 (1 − )= (since ℎ = 𝑅 + )
𝑅 ℎ 2
4
Force on the satellite= 𝑚𝑔ℎ = 9 𝑚𝑔
4
= 9 × 200 × 10 ≈ 889𝑁
40. (b) mg = mR𝜔2
𝑔 𝑅
𝜔 = √𝑅 ⇒ 𝑇 = 2𝜋√𝑔 = 2𝜋√64000
2𝜋×800
= 2𝜋 × 800𝑠 = ℎ = 1.36 = 1.4ℎ
3600
𝐺𝑀 −2
41. (c) 𝑔 = = 𝐺𝑀𝑅
𝑅2

𝑑 𝜔2𝑅2
42. (a) 𝑔 (1 − 𝑅) = 𝑔 − 𝜔2 𝑅; 𝑑 = 𝑔
43. (c) The orbital velocity of moon is
𝐺𝑀𝑒
𝑣0𝑚 = √ --------------(i)
𝑟
𝐺𝑀𝑒 𝑀𝑚
--------------(ii)
2𝑟
𝐺𝑀𝑒 𝑀𝑚
𝑈=− -----------(iii)
𝑟
2 1
So kinetic energy of moon is𝐾 = 2 𝑀𝑚 𝑣0𝑚
where r is distance between the centres of earth & moon.
It is clear from (ii) & (iii) that U>K (in magnitudes term)
𝐺𝑀 𝐺𝑀
44. (b)𝑣𝑔 = − 𝑅 for 𝑟 ≤ 𝑅and𝑣𝑔 = − 𝑟 , for𝑟 > 𝑅, and so option (b) is correct.
45. (c) In a circular or elliptical orbital motion, torque is always acting parallel to displacement or velocity.
So, angular momentum is conserved. In attractive field, potential energy is negative. Kinetic energy
changes as velocity increase when distance is less. So, option (c) is correct.
−𝐺𝑀𝑚
46. (b) Gravitational potential energy on the earth surface𝑈𝑟 = 𝑅
−𝐺𝑀𝑚
Gravitational potential energy at a height h above the earth's surface,𝑈ℎ = 𝑅+ℎ
−𝐺𝑀𝑚 −𝐺𝑀𝑚
𝑈ℎ = =
𝑅+𝑅 2𝑅
Gain in gravitational potential energy= 𝑈ℎ − 𝑈𝑟
−𝐺𝑀𝑚 −𝐺𝑀𝑚 𝐺𝑀𝑚 𝐺𝑀𝑚
= 2𝑅 − ( 𝑅 ) = 𝑅 − 2𝑅
𝐺𝑀𝑚 1
= 2𝑅 = 2 𝑚𝑔𝑅
47. (a) We know that intensity is negative gradient of potential,
i.e., I = – (dV/dr) and as here I = – (K/r), so
𝑑𝑉 𝐾
= 𝑟 , i.e., ∫ 𝑑𝑉 = 𝐾
𝑑𝑟
𝑟
or 𝑉 − 𝑉0 = 𝐾 𝑙𝑜𝑔 𝑟
0
𝑟
so 𝑉 = 𝐾 𝑙𝑜𝑔 𝑟 + 𝑉0
0
48. (b) The gravitational potential V at a point distance ′r′ from a body a mass m is equal to the amount of
work done in moving a unit mass from infinity to that point

27 | P a g e
 𝑟 𝐺𝑀 𝑑𝑉
𝑉𝑟 − 𝑉∞ = − ∫∞ 𝐸⃗⃗ . 𝑑𝑟⃗ = −𝐺𝑀(1/𝑟 − 1/∞) = − [𝐴𝑠𝐸⃗⃗ = − 𝑑𝑟 ]
𝑟

(i) In the first case

−𝐺𝑀
When 𝑉∞ = 0, 𝑉𝑟 = = −5𝑢𝑛𝑖𝑡
𝑟

(ii) In the second case 𝑉∞ = +10𝑢𝑛𝑖𝑡

𝑉𝑟 − 10 = −5 or 𝑉𝑟 = +5unit

49. (a) As all the points on the periphery of either ring are at the same distance from point P, the potential at
point P due to the whole ring can be calculated as𝑉 = −(𝐺𝑀)/(√𝑅2 + 𝑥 2 )where x is the axial distance
from the centre of the ring. This expression is independent of the fact whether the distribution of mass of
uniform or non- uniform.
𝐺𝑀 𝐺×2𝑀 𝐺𝑀 1 2
So, at𝑃, 𝑉 = − − =− 𝑅 [ + ]
√2𝑅 √5𝑅 √2 √5
50. (b) Conceptual
2𝜋𝑟 2𝜋𝑟 2𝜋𝑟 3/2 2𝜋
51. (a) 𝑇 = = (𝑔𝑅2/𝑟)1/2 = =
𝑣0 √𝑔𝑅 2 𝜔
√𝑔𝑅 2 𝑔𝑅 2
Hence, 𝑟 3/2 = 𝜔 or 𝑟 3 = 𝜔2
or 𝑟 = (𝑔𝑅 2 /𝜔2 )1/3
𝑀
2𝐺𝑀𝑒 2𝐺 𝑒 2
52. (b) 𝑣𝑒 = √ ; 𝑣𝑚 = √ 𝑅𝑒
81
= 9 𝑣𝑒 = 2/9 × 11.2𝑘𝑚𝑠 −1 = 2.5𝑘𝑚𝑠 −1
𝑅𝑒
4

2𝐺𝑀𝑝
√ 𝑅
(𝑣𝑒 )𝑝 𝑝 𝑀𝑝 𝑅 10𝑀𝑒 𝑅𝑒
53. (c) (𝑣 = = √ 𝑀 × 𝑅𝑒 = √ ×𝑅 = 10
𝑒 )𝑒 2𝐺𝑀
√ 𝑅 𝑒 𝑒 𝑝 𝑀𝑒 𝑒 /10
𝑒
∴ (𝑣𝑒 )𝑝 = 10 × (𝑣𝑒 )𝑒 = 10 × 11 = 110𝑘𝑚/𝑠
54. (a) As we know,
−𝐺𝑀𝑚
Gravitational potential energy = 𝑟
and orbital velocity,𝑣0 = √𝐺𝑀/𝑅 + ℎ
1 𝐺𝑀𝑚 1 𝐺𝑀 𝐺𝑀𝑚
𝐸𝑓 = 2 𝑚𝑣02 − 3𝑅 = 2 𝑚 4𝑅 − 4𝑅
𝐺𝑀𝑚 1 −𝐺𝑀𝑚
= ( − 1) =
4𝑅 2 8𝑅
−𝐺𝑀𝑚
𝐸𝑖 = + 𝐾; 𝐸𝑖 = 𝐸𝑓
𝑅
7𝐺𝑀𝑚
Therefore minimum required energy, 𝐾 = 8𝑅
𝐺𝑀
55. (a) 𝑣 = √( )where r is radius of the orbit of the satellite
𝑟
𝑅𝑒 3
Here 𝑟 = 𝑅𝑒 + ℎ = 𝑅𝑒 + = 2 𝑅𝑒
2
2𝐺𝑀 2
So, 𝑣 = √ 3𝑅 = 𝑣0 √3
𝑒
where v0 is the orbital velocity of the satellite, which is moving in circular orbit of radius, r = Re
𝑚𝑣 2 𝐺𝑚𝑀 𝐺𝑀
56. (d)(𝑅+𝑥) = (𝑅+𝑥)2also𝑔 = 𝑅2
𝑚𝑣 2 𝐺𝑀𝑅2
∴ (𝑅+𝑥)
= 𝑚 ( 𝑅2 ) (𝑅+𝑥)2
𝑚𝑣 2 𝑅2
∴ (𝑅+𝑥)
= 𝑚𝑔 (𝑅+𝑥)2
1/2
𝑔𝑅 2 𝑔𝑅 2
∴ 𝑣 2 = 𝑅+𝑥 ⇒ 𝑣 = (𝑅+𝑥)

28 | P a g e
 𝐺𝑀𝑚 𝐺𝑀(2𝑚) 𝐺𝑀𝑚
57. (c) 𝐹 𝑟2 (2𝑟)2 2𝑟 2 𝒎𝒊𝒏
𝐺𝑀𝑚 𝐺𝑀(2𝑚) 3 𝐺Μ𝑚
and 𝐹 𝑟 2 (2𝑟)2 2 𝑟 2
𝒎𝒂𝒙
𝐹𝒎𝒊𝒏
∴ 1
𝐹𝒎𝒂𝒙 =
3
𝐺𝑀𝑒
58. (a) Angular momentum of particle=m(v0+v)a where𝑣0 = √ 𝑎
1 𝐺𝑀𝑒 𝑚
Total energy of particle= 2 𝑚(𝑣0 + 𝑣 2 ) − 𝑎
5 𝐺𝑀𝑒 𝑚 𝐺𝑀𝑒 𝑚 −3 𝐺𝑀𝑒 𝑚
=8 − =
𝑎 𝑎 8 𝑎
1 𝐺𝑀𝑒 𝑚
At any distance 'r', total energy== 2 𝑚𝑢2 − 𝑟
But angular momentum conservation gives,
5𝐺𝑀𝑒 5 𝐺𝑀𝑒 𝑎
𝑚𝑢𝑟 = 𝑚√ 𝑎 ⇒ 𝑢 = √4
4𝑎 𝑟2
1 5 𝐺𝑀𝑒 𝑎 𝐺𝑀 𝑚
Therefore total energy=2 𝑚 4 𝑟 2 − 𝑟𝑒
According to conservation of energy this is equal to the initial enegy.
1 5 𝐺𝑀 𝑎 𝐺𝑀 𝑚 3𝐺𝑀 𝑚
Hence, 2 𝑚 4 𝑟 2𝑒 − 𝑟𝑒 = − 8𝑎𝑒
5
Solving this gives𝑟 = 𝑎, 3 𝑎
59. (a) Here, centripetal force will be given by the gravitational force between the two particles.
𝐺𝑚2
(2𝑅)2
= 𝑚𝜔2 𝑅

𝐺𝑚 𝐺𝑚
⇒ = 𝜔2 ⇒ 𝜔 = √
4𝑅 3 4𝑅 3
If the velocity of the two particles with respect to the centre of gravity is v then𝑣 = 𝜔𝑅
𝐺𝑚 𝐺𝑚
𝑣=√ ×𝑅 =√
4𝑅 3 4𝑅

2𝐺𝑀 | 2𝐺𝑀 2𝐺𝑀 𝑣𝑒 1

60. (d)𝑣𝑒 = √ and 𝑣𝑒 = √𝑅+ℎ = √𝑅+𝑅 = ∴𝑓=
𝑅 √2 √2

LEVEL-2
1. (d) With rotation of earth or latitude, acceleration due to
gravity vary as𝑔| = 𝑔 − 𝜔2 𝑅 𝑐𝑜𝑠 2 𝜙
Where𝜙is latitude, there will be no change in gravity at poles as𝜙 = 90°
At all other points as𝜔increases g' will decreases hence, weight, W = mg decreases.
𝐺𝑀𝑚
2. (c) Initial gravitational potential energy,𝐸𝑖 = − 2𝑅
Final gravitational potential energy,
𝐺𝑀𝑚/2 𝐺𝑀𝑚/2 𝐺𝑀𝑚 𝐺𝑀𝑚
𝐸𝑓 = − 𝑅 − 3𝑅 = − 2𝑅 − 6𝑅
2( ) 2( )
2 2
4𝐺𝑀𝑚 2𝐺𝑀𝑚
= 6𝑅 = 3𝑅
∴Difference between initial and final energy,
𝐺𝑀𝑚 2 1 𝐺𝑀𝑚
𝐸𝑓 − 𝐸𝑖 = 𝑅 (− 3 + 2) = − 6𝑅
2𝐺𝑀 𝑚𝑣 2 2𝐺𝑀
3. (a)𝐹 = 𝑚or = 𝑚
𝐿𝑟 𝑟 𝐿𝑟
 2 
2
2GMm 2𝜋
mr   = [∴ 𝑣 = 𝑟𝜔𝑎𝑛𝑑𝜔 = 𝑇 ]
 T  Lr
⇒𝑇∝𝑟
𝐺𝑚 4 4
4. (a)𝑔 = 2 . Also, 𝑀 = 𝑑 × 𝜋𝑅 3 ∴ 𝑔 = 𝐺 𝑑𝜋𝑅
𝑅 3 3
29 | P a g e
 4
At the surface of planet,𝑔𝑝 = 3 𝐺(2𝑑)𝜋𝑅 |
4
At the surface of the earth𝑔𝑒 = 𝐺𝑑𝜋𝑅
3
𝑔𝑒 = 𝑔𝑝 ⇒ 𝑑𝑅 = 2𝑑𝑅 | ⇒ 𝑅 | = 𝑅/2
𝐺𝑀(3𝑚) 3𝐺𝑀𝑚
5. (c)𝐹 = 𝑑2 = 𝑑2
6. (c) According to universal law of Gravitation,
𝐺𝑀𝑚
Gravitational force𝐹 = (𝑅+ℎ)2

7. (a) 𝑉 = 𝜔𝑅
𝑔 = 𝑔0 − 𝜔2 𝑅 [g=at equator, g0 = at poles]
𝑔0 2 2 𝑔0 2 𝑔0 𝑅
= 𝑔0 − 𝜔 𝑅; 𝜔 𝑅 = ; 𝑉 =
2 2 2
𝑉𝑒 = √2𝑔0 𝑅 = √4𝑉 2 = 2𝑉
8. (a)
9. (a)
10. (c) We know,𝑔| = 𝑔 − 𝜔2 𝑅 𝑐𝑜𝑠 2 𝜃
3𝑔
= 𝑔 − 𝜔2 𝑅
4
3 𝑔
Given, 𝑔| = 4 𝑔 𝜔2 𝑅 = 4
𝑔 10 1
𝜔 = √4𝑅 = √4×6400×103 = 2×8×100
= 0.6 × 10−3 𝑟𝑎𝑑/𝑠
11. (b)
12. (c)𝑊 = 𝑚(𝑉2 − 𝑉1 )
𝐺𝑀 𝐺𝑀
When, 𝑉1 = − [ 𝑎 1 + 2 ]
√2𝑎
𝐺𝑀2 𝐺𝑀1
𝑉2 = − [ + ]
𝑎 √2𝑎
𝐺𝑚(𝑀2 −𝑀1 )
∴𝑊= (√2 − 1)
𝑎√2
13. (d)𝑣1 𝑟1 = 𝑣2 𝑟2or r121 = r222 (∵ 𝐿 = 𝑚𝑟𝑣 = 𝑐𝑜𝑛𝑠 𝑡𝑎𝑛 𝑡)
|
22 22
𝑚𝑎𝑥
or 𝑟𝑚𝑖𝑛 ∴ 𝜔 | = (𝑟𝑚𝑖𝑛
𝑚𝑎𝑥
𝜔)
𝐺𝑀1 𝐺𝑀2
14. (d) 𝑎1 = , 𝑎2 =
𝑅12 𝑅22
4 4
𝐺𝜌1 𝜋𝑟13 𝐺𝜌2 𝜋𝑟22
3 3
𝑎1 = , 𝑎2 =
𝑅12 𝑅22
𝑎1 𝜌1 𝑅1
=𝜌
𝑎2 2 𝑅2
15. (b) We need to determine the maximum force that static friction exerts on Block B. This is the force that must
be overcome by the gravitational force between Blocks A and B
Since motion is only along the x− direction, ΣFY=0
For block BΣFY=0=FN−FW gives FN=FW=MBG
By definition the fractional force is:
𝑓𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝜇𝑆 𝐹𝑁 = 𝜇𝑆 𝑀𝐵𝐺
From 𝐹𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 𝑓𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 we get,
𝐺𝑀𝐴 𝑀𝐵 /𝑟 2 = 𝜇𝑆 𝑀𝐵𝑔 and
𝑀𝐴 = 𝜇𝑆 𝑀𝐵 𝑔𝑟 2 /𝐺, 𝑀𝐵 = 𝜇𝑆 𝑔𝑟 2 /𝐺
𝑀𝐴 = 0.5(9.8𝑚/𝑠 2 )(1.0𝑚)2 /6.67 × 10−11 𝑁𝑚2 /𝑘𝑔2 = 7.3 × 1010 𝑘𝑔
30 | P a g e
16. (b) Gravitational field at mass m due to full solid sphere
𝜌𝑟⃗ 𝜌𝑅 1
𝐸⃗⃗1 = = -----------[𝜀0 = ]
3𝜀0 6𝜀0 4𝜋𝐺
Gravitational field at mass m due to cavity(−𝜌)

𝑑𝑣
17. (d) Gravitational field,𝐼 = (5𝑖̂ + 12𝑗̂)𝑁/𝑘𝑔; 𝐼 = − 𝑑𝑟
𝑥 𝑦
𝑣 = −[∫0 𝐼𝑥 𝑑𝑥 + ∫0 𝐼𝑦 𝑑𝑦] = −[𝐼𝑥 . 𝑥 + 𝐼𝑦 . 𝑦]
= −[5(7 − 0) + 12(−3 − 0)] = −[35 + (−36)] = 1𝐽/𝑘𝑔
i.e., change in gravitational potential 1 J/kg.
Hence change in gravitational potential energy 1 J.
18. (c) Let 'M' be the mass of the particle
Now, Einitial = Efinal

19. (d) Differentiating the equation of the curve w.r.t. t we get,

𝑑𝑥 2 𝑑2 𝑦
2 ( 𝑑𝑡 ) = 500∞𝑡 𝑑𝑡 2
𝑑2 𝑦 2×104
or 𝑑𝑡 2 = 500 = 4𝑔
the effective 'g' = 5g
20. (b) For h << R, the orbital velocity is√𝑔𝑅
Escape velocity =√2𝑔𝑅
∴The minimum increase in its orbital velocity
= √2𝑔𝑅 − √𝑔𝑅 = √𝑔𝑅(√2 − 1)
21. (d) Assume a small element of mass dM at angular position θ from x axis.
We get dM = M/π dθ
𝐺𝑑𝑀 𝐺𝑀
Gravitational field due to small element of mass 𝑑𝐸 = 𝑙2 = 𝜋𝑙2 𝑑𝜃
Components of dE along x-axis cancel out each other while that along y-axis add up.
𝜋/2
Total gravitational field 𝐸𝑛𝑒𝑡 = ∫0 2𝑑𝐸 𝑠𝑖𝑛 𝜃 𝑑𝜃 (along y axis)

31 | P a g e
 𝑚𝑑𝑣
22. (a) 𝐹 = = 𝑎𝑣 2
𝑑𝑡

𝐺𝑀
√ 𝑡
𝑅 𝑑𝑣
𝑚∫ = 𝑎 ∫0 𝑑𝑡
𝐺𝑀 𝑣 2
√
𝑛𝑅
𝑅 𝑛𝑅
[−√ +√ ]
𝐺𝑀 𝐺𝑀
=𝑡
𝑎
𝑚√𝑅(√𝑛−1)
𝑡 = 𝑎√𝐺𝑀
23. (b) Energy required to place the lab in first orbit is the change in potential energy+k0E0
−𝐺𝑀𝑚 𝐺𝑀𝑚
Δ𝑃. 𝐸. = 2𝑅 + 2𝑅 = Δ𝑢
Velocity given to lab is provided by gravitation force. So, centripetal force=gravitational force
𝑚𝑣 2 𝐺𝑀𝑚 𝐺𝑀
= (2𝑅)2 ⇒ 𝑣 = √ 2𝑅
2𝑅
1 1 𝐺𝑀 1 𝐺𝑀𝑚
𝐾. 𝐸. = 2 𝑚𝑣 2 = 2 2𝑅 = 4 𝑅
𝐺𝑀𝑚 1 𝐺𝑀𝑚 3 𝐺𝑀𝑚 𝐺𝑀𝑚
Total energy=△u+K.E0 = 2𝑅 + 4 =4 𝑔=
𝑅 𝑅 𝑅2
similarly to shift to orbit of 3R

𝐺𝑀𝑚 2𝐺𝑀√2
2 √
1 𝑅√2 𝑅 𝒎𝒊𝒏
24. (b) 2 𝑚𝑣𝒎𝒊𝒏

32 | P a g e
 −𝐺𝑀 𝑅 2 11𝐺𝑀
25. (d) By superposition principle𝑣1 = 2𝑅3
[3𝑅 2 − (2 ) ] = − 8𝑅3
3 𝐺(𝑀/8) −3𝐺𝑀
Also, 𝑣2 = − =
2 (𝑅/2) 8𝑅
11𝐺𝑀 3𝐺𝑀 𝐺𝑀
The required potential is, v=v1−v2 = − 8𝑅
− (− 8𝑅
);𝑉 =− 𝑅
𝑅 𝐺𝑀𝑚 1 1
26. (b)𝑃. 𝐸. = ∫𝑅 𝑑𝑟 = −𝐺𝑀𝑚 [ − ]
0 𝑟2 𝑅 𝑅0
The K.E. acquired by the body at the surface
1 1 1 1 1
∴ 2 𝑚𝑣 2 = −𝐺𝑀𝑚 [𝑅 − 𝑅 ] ; 𝑣 = √2𝐺𝑀 (𝑅 − 𝑅 )
0 0
𝐺𝑚
27. (c)𝑉 = − 2𝑅3 (3𝑅 2 −𝑟 2 );
Graph (c) most closely depicts the correct variation of v(r).
−𝐺𝑀𝑒 𝑚
28. (b) 𝑇𝐸 = Using conservation of angular momentum about O
2𝑟0
𝑚𝑣𝑝 𝑟𝑝 = 𝑚𝑣𝐴 𝑟𝐴 = 𝑚𝑣0 𝑟0 𝑐𝑜𝑠 𝜃
3𝑣0 𝑟0
𝑣𝐴 𝑟𝐴 = 𝑣𝑝 𝑟𝑝 = 5

Using conservation of energy

𝑀𝑣 2
29. (d)2𝐹 𝒄𝒐𝒔 4 50 + 𝐹 | = (From figure)
𝑅

𝐺𝑀2 𝐺𝑀2 2×𝐺𝑀2 𝐺𝑀2 𝑀𝑣 2 𝐺𝑀2 1 1

Where𝐹 = 2 and 𝐹 | = ⇒ 2 + = ⇒ [ + ] = 𝑀𝑣 2
(√2𝑅 ) 4𝑅 2 √2(𝑅√2) 4𝑅 2 𝑅 𝑅 4 √2
𝐺𝑀 √2+4 1 𝐺𝑀
∴𝑣=√ ( )= √ (1 + 2√2)
𝑅 4√2 2 𝑅

30. (c)𝑀𝐴 = 𝜎4𝜋𝑅𝐴2 , 𝑀𝐵 = 𝜎4𝜋𝑅𝐵2

where𝜎is surface density

for new shell of mass M and radius R

33 | P a g e
 2 2 2 +𝑅 2
𝑉 𝑀 𝑅𝐴 𝜎4𝜋(𝑅𝐴 +𝑅𝐵 ) 𝑅𝐴 𝑅𝐴 𝐵 5
then 𝑉 = = 2 +𝑅 2 )1/2
= 𝜎4𝜋𝑅 2 =√ =3
𝐴 𝑅 𝑅𝐵 (𝑅𝐴 𝐴 𝑅𝐴
𝐵

34 | P a g e