physics : class11 Gravitational Force in Orbital Motion : Gravitational Force : Gravitation MCQ Questions and Answers
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Gravitational Force in Orbital Motion
MCQ questions and answers.
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MCQ Questions
1. A satellite moves in a circular orbit of radius r around a planet of mass M. If the orbital radius is increased to 4r while M remains unchanged, the gravitational force on the satellite becomes what fraction of its original value?
A.
1/2
B.
1/4
C.
1/8
D.
1/16
Correct Answer:
1/16
Explanation:
The gravitational force is F = GMm/r². Replacing r by 4r gives F' = GMm/(4r)² = F/16.
2. A satellite of mass m revolves in a circular orbit of radius r around a planet of mass M. If its mass is doubled without changing the orbit, which pair correctly describes the changes in gravitational force and orbital speed?
A.
Force doubles; speed doubles
B.
Force doubles; speed remains unchanged
C.
Force remains unchanged; speed doubles
D.
Force remains unchanged; speed remains unchanged
Correct Answer:
Force doubles; speed remains unchanged
Explanation:
The force F = GMm/r² is proportional to satellite mass, so it doubles. The orbital speed v = √(GM/r) is independent of the satellite's mass, so it remains unchanged.
3. Two satellites orbit the same planet in circular orbits of radii r and 9r. What is the ratio of their orbital speeds, v_r : v_9r?
A.
1 : 3
B.
3 : 1
C.
9 : 1
D.
1 : 9
Correct Answer:
3 : 1
Explanation:
For a circular orbit, v = √(GM/r), so v_r/v_9r = √(9r/r) = 3. Therefore, the speed ratio is 3:1.
4. A satellite is transferred from a circular orbit of radius r to another circular orbit of radius 4r around the same planet. What is the ratio of its orbital periods in the larger and smaller orbits?
A.
2
B.
4
C.
8
D.
16
Correct Answer:
8
Explanation:
Kepler's third law gives T proportional to r^(3/2). Thus, T_2/T_1 = (4r/r)^(3/2) = 4^(3/2) = 8.
5. A satellite moves in an elliptical orbit around a planet. At the nearest point its distance from the planet's center is r, and at the farthest point it is 4r. What is the ratio of its speeds at the nearest and farthest points?
A.
1 : 4
B.
1 : 2
C.
2 : 1
D.
4 : 1
Correct Answer:
4 : 1
Explanation:
At the nearest and farthest points, the velocity is perpendicular to the radius. Conservation of angular momentum gives m v_near r = m v_far(4r), so v_near/v_far = 4.
6. A satellite of mass m is in a circular orbit of radius r around a planet of mass M. Its total mechanical energy is:
A.
-GMm/r
B.
-GMm/(2r)
C.
GMm/(2r)
D.
0
Correct Answer:
-GMm/(2r)
Explanation:
For a circular orbit, kinetic energy is GMm/(2r), while gravitational potential energy is -GMm/r. Their sum is E = -GMm/(2r).
7. A satellite in a circular orbit has orbital speed v. At the same orbital radius, the escape speed from the planet is related to v by:
A.
v_escape = v/2
B.
v_escape = v
C.
v_escape = √2v
D.
v_escape = 2v
Correct Answer:
v_escape = √2v
Explanation:
The circular orbital speed is v = √(GM/r), whereas escape speed is v_escape = √(2GM/r). Therefore, v_escape = √2v.
8. A satellite is orbiting a planet in a circular orbit. If the planet's mass is increased by a factor of 4 while the orbital radius remains fixed, how does the gravitational force and orbital period change?
A.
Force becomes 4 times; period becomes 2 times
B.
Force becomes 4 times; period becomes half
C.
Force becomes 2 times; period becomes half
D.
Force remains unchanged; period becomes half
Correct Answer:
Force becomes 4 times; period becomes half
Explanation:
The gravitational force F = GMm/r² is proportional to M, so it becomes four times. Since T = 2π√(r³/GM), increasing M by 4 reduces T by a factor of 2.
9. A satellite is placed in a circular orbit at radius r and then moved to a circular orbit at radius 2r. What happens to its orbital angular momentum about the planet's center?
A.
It becomes half its original value
B.
It remains unchanged
C.
It becomes √2 times its original value
D.
It becomes 2 times its original value
Correct Answer:
It becomes √2 times its original value
Explanation:
For a circular orbit, angular momentum L = mvr. Since v = √(GM/r), L = m√(GMr), which is proportional to √r. Increasing r to 2r makes L become √2 times larger.
10. A satellite moves in a circular orbit under the gravitational force of a planet. If the orbital radius is doubled, what fraction of the original gravitational potential energy remains, assuming the satellite mass and planet mass are unchanged?
A.
1/4
B.
1/2
C.
2
D.
4
Correct Answer:
1/2
Explanation:
Gravitational potential energy is U = -GMm/r. At radius 2r, U' = -GMm/(2r) = U/2, so its algebraic value is half the original value.
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