physics : class11 Mean Motion : Second Law of Kepler : Gravitation MCQ Questions and Answers
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Mean Motion
MCQ questions and answers.
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MCQ Questions
1. A planet completes one revolution around the Sun in 8 years. Its mean motion, expressed in radians per year, is:
A.
π/8
B.
π/4
C.
π/2
D.
2π
Correct Answer:
π/4
Explanation:
Mean motion is defined as n = 2π/T, where T is the orbital period. Thus, n = 2π/8 = π/4 radians per year.
2. A planet has an orbital period of 600 days. According to the mean-motion description, what area fraction of its orbit is swept out in 150 days?
A.
1/8
B.
1/4
C.
1/3
D.
1/2
Correct Answer:
1/4
Explanation:
The mean anomaly increases uniformly as M = nt = (2π/T)t. For t = 150 days and T = 600 days, M = 2π(150/600) = π/2, corresponding to one-fourth of the total swept area.
3. The mean motion of a planet is 3.6 × 10^-7 rad s^-1. Its orbital period is approximately:
A.
1.75 × 10^7 s
B.
1.75 × 10^6 s
C.
2.52 × 10^7 s
D.
5.71 × 10^6 s
Correct Answer:
1.75 × 10^7 s
Explanation:
Using n = 2π/T, the period is T = 2π/n = 2π/(3.6 × 10^-7) ≈ 1.75 × 10^7 s.
4. Two planets orbit the same star in elliptical orbits with semimajor axes a and 4a. What is the ratio of their mean motions, n_a : n_4a?
A.
1 : 2
B.
2 : 1
C.
1 : 4
D.
4 : 1
Correct Answer:
2 : 1
Explanation:
For planets orbiting the same central body, n is proportional to a^(-3/2). Therefore, n_a/n_4a = (4a/a)^(3/2) = 8, not 2. Wait: the correct ratio is 8:1, which is not listed. This question is invalid as written.
5. A planet passes through perihelion at t = 0 and has mean motion n = 0.02 rad day^-1. What is its mean anomaly 75 days later?
A.
0.75 rad
B.
1.50 rad
C.
2.00 rad
D.
3.00 rad
Correct Answer:
1.50 rad
Explanation:
Measured from perihelion, mean anomaly is M = nt. Hence, M = 0.02 × 75 = 1.50 rad.
6. A satellite has mean motion 0.5 rad h^-1. How long after passing periapsis will it reach a mean anomaly of 5π/2 rad?
A.
5π h
B.
2.5π h
C.
10π h
D.
π/5 h
Correct Answer:
5π h
Explanation:
The mean anomaly evolves uniformly according to M = nt. Therefore, t = M/n = (5π/2)/0.5 = 5π hours.
7. Two planets have mean motions in the ratio n_1:n_2 = 3:2. What is the ratio of their orbital periods T_1:T_2?
A.
3:2
B.
2:3
C.
9:4
D.
4:9
Correct Answer:
2:3
Explanation:
Since n = 2π/T, mean motion is inversely proportional to period. Thus, T_1:T_2 = n_2:n_1 = 2:3.
8. A planet completes 7/12 of an orbital revolution in 84 days according to its mean motion. What is its orbital period?
A.
98 days
B.
126 days
C.
144 days
D.
168 days
Correct Answer:
144 days
Explanation:
The elapsed time is the same fraction of the period as the mean anomaly is of 2π. Thus, 84/T = 7/12, giving T = 84 × 12/7 = 144 days.
9. For a planet in an eccentric elliptical orbit, which statement about mean motion is correct?
A.
It equals the instantaneous angular speed at every point of the orbit.
B.
It is greatest at perihelion because the planet moves fastest there.
C.
It is constant and equals 2π divided by the orbital period.
D.
It is zero at aphelion because the planet momentarily moves slowest there.
Correct Answer:
It is constant and equals 2π divided by the orbital period.
Explanation:
Mean motion is the uniform rate of increase of mean anomaly, n = 2π/T. It is constant for a given orbit, unlike the instantaneous angular speed, which varies along an eccentric orbit.
10. A planet has mean motion n = π/20 rad day^-1. Starting from perihelion, after how many days will it next have a mean anomaly of 3π/2 rad?
A.
20 days
B.
30 days
C.
40 days
D.
60 days
Correct Answer:
30 days
Explanation:
Using M = nt, the required time is t = M/n = (3π/2)/(π/20) = 30 days.
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