physics : class11 Distance-Dependent Orbital Speed : Second Law of Kepler : Gravitation MCQ Questions and Answers
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Distance-Dependent Orbital Speed
MCQ questions and answers.
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MCQ Questions
1. A planet moves in an elliptical orbit with perihelion distance r_p and aphelion distance r_a = 4r_p. What is the ratio of its speeds at perihelion and aphelion, v_p/v_a?
A.
1/4
B.
1/2
C.
2
D.
4
Correct Answer:
4
Explanation:
By Kepler's second law, the areal velocity is constant. At perihelion and aphelion, the velocity is perpendicular to the radius, so rv is constant. Hence v_p/v_a = r_a/r_p = 4.
2. A planet follows an elliptical orbit of eccentricity e. If its speed at perihelion is four times its speed at aphelion, what is the eccentricity of the orbit?
A.
0.20
B.
0.40
C.
0.60
D.
0.75
Correct Answer:
0.60
Explanation:
At perihelion and aphelion, rv is constant, so v_p/v_a = r_a/r_p. Since r_p = a(1 - e) and r_a = a(1 + e), the ratio is (1 + e)/(1 - e) = 4. Solving gives e = 3/5 = 0.60.
3. At two points on an orbit, a planet is at distances r_1 and r_2 from the central body, where r_1 = 2r_2. If the velocity is perpendicular to the radius at both points, what is v_1/v_2?
A.
1/4
B.
1/2
C.
2
D.
4
Correct Answer:
1/2
Explanation:
Kepler's second law gives constant areal velocity, so r_1v_1 = r_2v_2 for perpendicular velocities. Therefore v_1/v_2 = r_2/r_1 = 1/2.
4. At point A, a planet is at distance 3R from the focus, moving with speed 4V at an angle of 30 degrees to the radius. At point B, it is at distance 2R, moving with speed v at an angle of 90 degrees to the radius. If the orbit obeys Kepler's second law, what is v?
A.
V
B.
2V
C.
3V
D.
6V
Correct Answer:
3V
Explanation:
The areal velocity is (1/2)rv sin(theta). Equating the values at A and B gives (1/2)(3R)(4V)sin30 degrees = (1/2)(2R)v sin90 degrees. Thus 3RV = Rv, so v = 3V.
5. A planet moves in an elliptical orbit with perihelion distance 0.8 AU and aphelion distance 1.2 AU. Its speed at perihelion is 30 km/s. What is its speed when it is at a distance of 1.0 AU from the focus?
A.
18.0 km/s
B.
20.0 km/s
C.
24.5 km/s
D.
30.0 km/s
Correct Answer:
24.5 km/s
Explanation:
The semimajor axis is a = (0.8 + 1.2)/2 = 1.0 AU. Using the distance-dependent orbital-speed relation v^2 = mu(2/r - 1/a), the perihelion speed satisfies 30^2 = mu(2/0.8 - 1/1.0) = 1.5mu. At r = a = 1.0 AU, v^2 = mu, so v = 30/sqrt(1.5) approximately 24.5 km/s.
6. A planet moves through two small orbital arcs having the same angular width. The distances from the focus at the two arcs are r and 2r. What is the ratio of the times taken to traverse the arcs, t_2/t_1?
A.
1/4
B.
1/2
C.
2
D.
4
Correct Answer:
4
Explanation:
For a small angular displacement dtheta, Kepler's second law gives dt proportional to r^2 dtheta. Since the angular widths are equal, t_2/t_1 = (2r/r)^2 = 4.
7. At one point in an orbit, a planet is at distance R and has tangential speed 6V. At another point, it is at distance 3R. If the tangential component of its velocity at the second point is required by Kepler's second law, what is its value?
A.
V
B.
2V
C.
3V
D.
18V
Correct Answer:
2V
Explanation:
Kepler's second law requires r v_t to remain constant, where v_t is the transverse or tangential velocity component. Thus R(6V) = 3R(v_t), giving v_t = 2V.
8. For a planet at perihelion and aphelion, the ratio of its speeds is v_p/v_a = 3. What is the eccentricity of its elliptical orbit?
A.
0.25
B.
0.33
C.
0.50
D.
0.67
Correct Answer:
0.50
Explanation:
At the apsides, rv is constant, so v_p/v_a = r_a/r_p = (1 + e)/(1 - e). Setting this equal to 3 gives 1 + e = 3 - 3e, hence e = 1/2.
9. At point A, a planet is at distance R from the focus, and its velocity makes an angle of 30 degrees with the radius. At point B, it is at distance 2R, and its velocity makes an angle of 60 degrees with the radius. What is the ratio v_A/v_B?
A.
sqrt(3)/2
B.
1
C.
sqrt(3)
D.
2sqrt(3)
Correct Answer:
2sqrt(3)
Explanation:
Kepler's second law requires rv sin(theta) to be constant. Therefore Rv_A sin30 degrees = 2Rv_B sin60 degrees. Hence v_A/v_B = 2sin60 degrees/sin30 degrees = 2sqrt(3).
10. A planet travels through equal small angular intervals near two points of its orbit. Its distances from the focus at those points are R and 3R. What is the ratio of the times required, t_{3R}/t_R, to traverse the respective intervals?
A.
1/3
B.
1
C.
3
D.
9
Correct Answer:
9
Explanation:
For equal angular intervals, Kepler's second law gives dt proportional to r^2. Therefore t_{3R}/t_R = (3R/R)^2 = 9, showing that orbital motion is slower at the greater distance.
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