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SHAURYA 2.0
Gravitation DPP: 8.1

1. If a body is moved from the poles to the equator, its 7. Which of the following best describes the
weight gravitational force acting between two objects?
(a) Increases (a) It is a contact force.
(b) Decreases (b) It is a long-range force.
(c) Remains constant (c) It is an electromagnetic force.
(d) None of the above (d) It is a nuclear force.

2. Which of the following is the correct relation for the 8. If an object is in free fall near the surface of the
gravitational field inside a hollow sphere where Earth, which of the following quantities remains
r<R? (r is the distance of an object from the center constant?
and R is the radius of the hollow sphere) (a) Velocity
GM (b) Acceleration
(a) E = 2
r (c) Gravitational force
GM (d) Gravitational potential energy
(b) E = 2
R 9. Which of the following factors does NOT affect the
(c) E = 0 period of a satellite orbiting a planet?
(d) None of the above
(a) Mass of the satellite
(b) Mass of the planet
3. What is the universal law of gravitation?
(c) Radius of the orbit
(a) Objects attract each other with a force (d) Gravitational constant
proportional to their masses
(b) Objects attract each other with a force inversely 10. Which of the following is the correct relation for the
proportional to the square of the distance gravitational field of a solid sphere inside?
between them GM GM
(c) Objects repel each other with a force (a) E = 2 (b) E = 2
proportional to their masses r R
GMr
(d) Objects repel each other with a force inversely (c) E = 3 (d) None of the above
proportional to the square of the distance R
between them explain the correct option
11. What will be the gravitational field of a solid sphere
4. The gravitational force between two objects will at a point r = 3 m from the center of the sphere
decrease if: having a radius of 5m and a mass of 10 kg?
(a) The masses of the objects increase (Take G = 6.67 × 10–11 Nm2kg–2)
(b) The distance between the objects decreases (a) 1.2 × 1011 N/kg
(c) The distance between the objects increases (b) 1.6 × 10–6 N/kg
(d) The masses of the objects decrease (c) 1.6 × 10–11 N/kg
(d) 16 × 10–6 N/kg
5. Newton’s law of gravitation applies to
(a) Small bodies only 12. A satellite is orbiting Earth at an altitude of 500
(b) Plants only kilometers above the surface. Calculate the
(c) All bodies irrespective of their size acceleration due to gravity at this height. The mass
(d) For solar system of the Earth is 5.97 × 1024 kg, and the radius of the
Earth is 6400 kilometers. Assume the satellite is in a
6. The escape velocity from a planet depends on circular orbit.
(a) The mass of the planet (a) 8.43 m/s2
(b) The radius of the planet (b) 6.55 m/s2
(c) Both the mass and the radius of the planet (c) 8 m/s2
(d) None of the above (d) 7.05 m/s2
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13. An object is dropped from a height of 50 meters 17. Which of the following statements is true, according
above the ground. Calculate the acceleration due to to Kepler's laws of planetary motion?
gravity experienced by the object as it falls (a) The planets move in elliptical orbits with the
downward. (Assume no air resistance) Sun at one of the foci.
g (b) The speed of a planet is constant throughout its
(a) (b) 2g orbit.
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g (c) The square of the orbital period of a planet is
(c) g (d) directly proportional to the cube of its average
2 distance from the Sun.
(d) All planets have the same orbital eccentricity.
14. If a body goes down to a depth of 200 km from the
surface of the earth. Then calculate the acceleration 18. The orbital period of a satellite around a planet is
due to the gravity of the object at the depth. Take measured to be 5 hours. If the average distance
the radius of the earth as 6400 km.
between the satellite and the planet is 1.2  106 km,
(a) 4.49 m/s2
calculate the orbital period of another satellite that
(b) 9.49 m/s2
orbits the same planet at an average distance of
(c) 9 m/s2
3.6  106 km.
(d) 6.49 m/s2
(a) 26 hrs. (b) 22 hrs.
(c) 20 hrs. (d) 15 hrs.
15. Two objects, A and B, have masses of 5 kg and 10
kg, respectively. The distance between them is 5
19. Gravitational potential at a point in space is a
meters. Calculate the gravitational potential energy
measure of:
between these two objects.
(a) The gravitational field strength at that point.
(a) 6.67  10–11 J
(b) The amount of potential energy possessed by
(b) –6.67  10–10 J an object at that point.
(c) –3.67  10–10 J (c) The mass of the object at that point.
(d) 6.47  10–10 J (d) The acceleration is due to gravity at that point.
16. Calculate the escape velocity for a rocket launched 20. Which of the following is the correct relation for the
from the surface of the Earth. Given that the mass of gravitational potential energy at the surface of a
the Earth is 5.972  1024 kg and its radius is 6.371  hollow sphere?
106 m. GM GM
(a) 11234 m/s (a) U = − 2 (b) U = −
r r
(b) 23423 m/s
GM
(c) 11186 m/s (c) U = − (d) None of the above
(d) 13244 m/s 2r
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Answer Key
1. (b) 11. (c)
2. (c) 12. (a)
3. (b) 13. (c)
4. (c) 14. (b)
5. (c) 15. (b)
6. (c) 16. (c)
7. (b) 17. (a)
8. (b) 18. (a)
9. (a) 19. (b)
10. (c) 20. (b)
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Hints and Solutions

1. (b) 4. (c)
Explanation: Explanation:
• A body weighs more at the poles than at the • According to the universal law of gravitation,
equator, because the earth is not a perfect the gravitational force between two objects is
square, but it is flattened at the poles. The inversely proportional to the square of the
distance between the equator and the center of distance between their centers.
the earth is less than at the poles; therefore, the • This means that as the distance between the
force of gravitation is more at the poles than at objects increases, the gravitational force
the equator; and So, it weighs more at the poles between them decreases.
than equator.
5. (c)
2. (c) Explanation:
Explanation: • Newton’s law of gravitation, statement that any
• Inside a hollow sphere, if the object is located particle of matter in the universe attracts any
at a distance (r) from the center that is less than other with a force varying directly as the
the radius of the sphere (R), the gravitational product of the masses and inversely as the
field strength (E) at that point is zero. This is square of the distance between them
because the gravitational forces from all the
mass distributions within the hollow sphere 6. (c)
cancel each other out. Explanation:
• Due to the symmetrical distribution of mass in • The escape velocity from a planet is the
the hollow sphere, the net gravitational field minimum velocity an object needs to escape the
inside becomes zero. This means that there is gravitational pull of the planet and move away
no gravitational force acting on an object inside indefinitely. It depends on both the mass and
the hollow sphere. the radius of the planet.
• The formula for escape velocity (v) is given by:
3. (b) 2GM
Explanation: v=
R
• This statement represents the universal law of
gravitation, which was formulated by Sir Isaac • Where v is the escape velocity, G is the
Newton. gravitational constant, M is the mass of the
• The law states that any two objects in the planet, and R is the radius of the planet.
universe attract each other with a force that is • A planet with a larger mass or a larger radius
directly proportional to the product of their will have a higher escape velocity.
masses and inversely proportional to the square
of the distance between their centers. 7. (b)
• Mathematically, the universal law of Explanation:
gravitation can be expressed as • The gravitational force is indeed a long-range
mm force. It acts between two objects with mass,
F = G 12 2 regardless of their distance apart.
r • Unlike contact forces, such as friction or
Where normal force, which require physical contact,
• F represents the gravitational force between the gravitational force can act over large distances,
two objects, even astronomical distances. This is one of the
• G is the gravitational constant (a fundamental fundamental characteristics of gravity.
constant of nature),
• m1 and m2 are the masses of the two objects, 8. (b)
and is the distance between the centers of the Explanation:
two objects. • When an object is in free fall near the surface
of the Earth, the acceleration remains constant.
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This is known as the acceleration due to 11. (c)

gravity, denoted by the symbol "g." Given:
• The acceleration due to gravity near the surface Mass of sphere = 10 kg, the radius of sphere R=5m,
of the Earth is approximately 9.8 ms–2 and is the distance of point = 3m.
directed downward. Calculation:
• Regardless of the mass or size of the object, if As the r < R, So, the point is enclosed inside the
it is in free fall (not experiencing significant air solid sphere.
resistance), it will accelerate at this rate. GMr
By using the equation, E = 3 we will find the
R
9. (a) gravitational field inside the sphere.
Explanation: 6.67  10−11  10  3 200.1 10−11
• The mass of the satellite does not affect the E= 
53 125
period of its orbit around a planet.
= 1.6  10 N/kg
–11
• According to Kepler's third law of planetary
motion, the period of an orbiting object only
12. (a)
depends on the mass of the planet being orbited
Given:
and the radius of the orbit. The mass of the
satellite cancels out when calculating the Height = 500 km, the mass of earth = 5.97  1024 kg,
period. radius of the earth = 6400 km
• The equation for the period of an orbiting Calculation:
satellite is g
By using the equation, g' = 2
r3  h 
T = 2  1 + 
GM  Re 
• Where T is the period, r is the radius of the 9.8 9.8 9.8
 g =  = = 8.43 m/s2
orbit, G is the gravitational constant, and M is  500 
2
 69 
2 1.145
the mass of the planet being orbited. As you 1 +   
can see, the mass of the satellite does not  6400   64 
appear in the equation.
13. (c)
Explanation:
10. (c)
• Since the object is moving downward, the
Explanation:
acceleration due to gravity will be positive, as it
• The correct relation for the gravitational field is in the same direction as the object's motion.
inside a solid sphere is given by:
• Therefore, the acceleration due to gravity
GMr
E= 3 experienced by the object as it falls downward
R is 9.8 m/s².
• where E represents the gravitational field 14. (b)
strength, G is the gravitational constant, m is Given:
the mass of the sphere enclosed by the radius r, The radius of the earth = 6400 km, depth = 200 km
and R is the radius of the sphere. Calculation:
• This formula shows that the gravitational field  d 
strength (E) inside a solid sphere varies linearly By using the equation, g = g 1 − 
 Re 
with the distance from the center (r) and
 200  62
inversely with the cube of the radius of the  g = 9.8 1 −   9.8  = 9.49 m/s2
sphere (R³).  6400  64
• Inside a solid sphere, the gravitational field is
directly proportional to the mass enclosed within 15. (b)
a given radius. As you move closer to the center, Given:
the mass enclosed within that radius increases, m1 = 5 kg, m2 = 10 kg, r =5 m
leading to a stronger gravitational field. Calculation:
Gm1m 2
By using the equation, U P = −
r
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6.67 10−11  5 10 19. (b)

 UP = − Explanation:
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• Gravitational potential at a point in space is a
 –6.67  10–10 J/kg
measure of the potential energy possessed by
an object placed at that point within a
16. (c)
gravitational field.
Given:
• It is directly related to the work done in
Mass of the earth = 5.9721024 kg,
bringing a unit mass from infinity to that point
radius = 6.371106m
in the gravitational field.
Calculation:
• The gravitational potential energy is defined as
2GM
By using the equation, Ve = the work done per unit mass, and the
R gravitational potential is a scalar quantity that
2  6.67  10−11  5.972  1024 represents this potential energy per unit mass.
 Ve =
6.371  106 20. (b)
= 11186 m/s Explanation:
• The correct relation for the gravitational
17. (a) potential energy at the surface of a hollow
Explanation: sphere is:
• According to Kepler's first law, also known as GM
the law of elliptical orbits. It states that the path U=−
R
of each planet around the Sun is an ellipse, with
the Sun located at one of the two foci of the • Where G is the gravitational constant, U is the
ellipse. gravitational potential energy, R is the radius of
the hollow sphere, and M is the mass of the
• This means that the distance between the planet
hollow sphere.
and the Sun varies throughout its orbit.
• This formula represents the gravitational
18. (a) potential energy between an object of mass M
Given: and a point on the surface of the hollow sphere.
The negative sign indicates that the
T1 = 5 hr, r1 = 1.2  106 km, r2 = 3.6  106 km, T2 =?
gravitational potential energy is a negative
Calculation:
2 3 quantity, indicating an attractive force.
T  r 
By using the equation,  2  =  2 
 T1   r1 
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 T   3.6  10 km 
2 6
 2  = 
 5   1.2  106 km 
T
 2 = 27  T2 = 5  5.19  26hr.
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