physics : class11 Eccentricity and Areal Motion : Second Law of Kepler : Gravitation MCQ Questions and Answers
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Eccentricity and Areal Motion
MCQ questions and answers.
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MCQ Questions
1. A planet moves in an elliptical orbit with perihelion distance 2.0 × 10^10 m and aphelion distance 8.0 × 10^10 m. What is the eccentricity of its orbit?
A.
0.40
B.
0.50
C.
0.60
D.
0.75
Correct Answer:
0.60
Explanation:
For an ellipse, r_p = a(1 − e) and r_a = a(1 + e). Hence e = (r_a − r_p)/(r_a + r_p) = (8 − 2)/(8 + 2) = 0.60.
2. A planet has eccentricity 0.60. According to Kepler's second law, what is the ratio of its speed at perihelion to its speed at aphelion?
3. A planet moves around the Sun in an elliptical orbit of eccentricity 0.50. What is the ratio of its angular speed about the Sun at perihelion to that at aphelion?
A.
3
B.
6
C.
9
D.
12
Correct Answer:
9
Explanation:
Kepler's second law gives r^2 dθ/dt = constant. Thus the angular-speed ratio is (r_a/r_p)^2 = [(1 + e)/(1 − e)]^2 = (1.5/0.5)^2 = 9.
4. A planet sweeps out an area of 1.20 × 10^20 m^2 in 10 days while orbiting its star. How much area will it sweep in 25 days?
A.
2.40 × 10^20 m^2
B.
3.00 × 10^20 m^2
C.
3.60 × 10^20 m^2
D.
4.80 × 10^20 m^2
Correct Answer:
3.00 × 10^20 m^2
Explanation:
By Kepler's second law, equal areas are swept in equal times, so swept area is proportional to time. The required area is 1.20 × 10^20 × 25/10 = 3.00 × 10^20 m^2.
5. A planet's areal velocity is 2.0 × 10^15 m^2 s^−1. What area does it sweep during one-fourth of its orbital period T?
A.
2.0 × 10^15 T m^2
B.
5.0 × 10^14 T m^2
C.
5.0 × 10^15 T m^2
D.
8.0 × 10^15 T m^2
Correct Answer:
5.0 × 10^14 T m^2
Explanation:
Areal velocity is dA/dt = 2.0 × 10^15 m^2 s^−1. In time T/4, the swept area is (2.0 × 10^15)(T/4) = 5.0 × 10^14 T m^2.
6. A planet's distance from its star is 3.0 × 10^10 m at perihelion and 9.0 × 10^10 m at aphelion. If its speed at aphelion is 20 km s^−1, what is its speed at perihelion?
7. For an elliptical orbit, the areal velocity is 1.5 × 10^15 m^2 s^−1 and the orbital period is 2.0 × 10^7 s. What is the area of the ellipse?
A.
1.5 × 10^22 m^2
B.
3.0 × 10^22 m^2
C.
6.0 × 10^22 m^2
D.
9.0 × 10^22 m^2
Correct Answer:
3.0 × 10^22 m^2
Explanation:
Over one complete orbit, the swept area equals the entire ellipse area. Thus A = (dA/dt)T = (1.5 × 10^15)(2.0 × 10^7) = 3.0 × 10^22 m^2.
8. A planet moves in an ellipse with semimajor axis a and eccentricity e. What fraction of the total elliptical area lies between the star and the planet when the planet is at a point where its radius vector makes a small angular displacement dθ?
A.
[(1 − e^2)/(2(1 + e cos θ))] dθ
B.
[a^2(1 − e^2)^2/(2(1 + e cos θ)^2)] dθ
C.
[a^2(1 − e^2)/(2(1 + e cos θ))] dθ
D.
[a^2(1 + e cos θ)/(2(1 − e^2))] dθ
Correct Answer:
[a^2(1 − e^2)^2/(2(1 + e cos θ)^2)] dθ
Explanation:
The small swept area is dA = (1/2)r^2 dθ. Using r = a(1 − e^2)/(1 + e cos θ), the expression becomes dA = [a^2(1 − e^2)^2/(2(1 + e cos θ)^2)]dθ.
9. Two planets orbit the same star in elliptical orbits. At corresponding points, their distances from the star are in the ratio 2:3 and their speeds are in the ratio 3:2. What is the ratio of their areal velocities?
A.
1:1
B.
2:3
C.
3:2
D.
4:9
Correct Answer:
1:1
Explanation:
Areal velocity is (1/2)rv for transverse orbital motion. The given products are (2)(3) and (3)(2), which are equal, so the areal velocities have ratio 1:1.
10. A planet in an elliptical orbit has perihelion and aphelion speeds in the ratio 5:3. What is the eccentricity of its orbit?
A.
0.20
B.
0.25
C.
0.33
D.
0.40
Correct Answer:
0.25
Explanation:
By constant areal velocity, v_p/v_a = r_a/r_p = (1 + e)/(1 − e). Thus 5/3 = (1 + e)/(1 − e), giving 5 − 5e = 3 + 3e and e = 1/4 = 0.25.
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