physics : class11 Semi-Major Axis Effects : Second Law of Kepler : Gravitation MCQ Questions and Answers
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Semi-Major Axis Effects
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MCQ Questions
1. Two planets move around the same star in elliptic orbits having the same eccentricity. The semi-major axis of planet Q is four times that of planet P. According to Kepler's second law and the dependence of orbital period on semi-major axis, what is the ratio of their areal velocities, (dA/dt)Q : (dA/dt)P?
A.
1 : 4
B.
1 : 8
C.
2 : 1
D.
4 : 1
Correct Answer:
2 : 1
Explanation:
For an orbit, the areal velocity is total orbital area divided by period. With the same eccentricity, orbital area is proportional to a^2, while the period is proportional to a^(3/2). Hence areal velocity is proportional to a^(1/2). Since aQ/aP = 4, the ratio is sqrt(4) : 1 = 2 : 1.
2. A planet travels in an elliptical orbit of semi-major axis a and eccentricity e. During a time interval equal to one-fourth of its orbital period, what fraction of the total area of the ellipse is swept out by the radius vector, regardless of the planet's position at the beginning of the interval?
A.
1/16
B.
1/4
C.
sqrt(1 - e^2)/4
D.
It depends on whether the interval begins at perihelion or aphelion
Correct Answer:
1/4
Explanation:
Kepler's second law states that the areal velocity is constant. Therefore, the swept area is proportional only to elapsed time. In one-fourth of the orbital period, one-fourth of the complete elliptical area is swept, independent of the starting position and eccentricity.
3. Two orbits around the same central body have semi-major axes a and 9a and the same eccentricity. If the time taken to sweep a particular fixed fraction of each orbit's total area is compared, what is the ratio of the times for the larger and smaller orbits?
A.
1 : 3
B.
1 : 9
C.
3 : 1
D.
27 : 1
Correct Answer:
27 : 1
Explanation:
A fixed fraction of the total area is swept in the same fraction of the orbital period. Kepler's third-law dependence gives T proportional to a^(3/2). Thus Tlarge/Tsmall = (9a/a)^(3/2) = 9^(3/2) = 27.
4. Two planets orbit the same star in ellipses with equal semi-major axes but different eccentricities. Which statement about the area swept by the radius vector in one-half of each orbital period is correct?
A.
The more eccentric orbit sweeps twice as much area
B.
The less eccentric orbit sweeps twice as much area
C.
Both sweep half of their respective total elliptical areas
D.
The swept areas are equal because the semi-major axes are equal
Correct Answer:
Both sweep half of their respective total elliptical areas
Explanation:
Kepler's second law makes swept area proportional to elapsed time as a fraction of the orbital period. In half a period, each planet sweeps half of its own ellipse. The actual total areas may differ because the eccentricities differ, even though the semi-major axes are equal.
5. For a family of elliptic orbits around the same central mass, the eccentricity is kept constant while the semi-major axis is increased by a factor of 16. By what factor does the areal velocity increase?
A.
2
B.
4
C.
8
D.
64
Correct Answer:
4
Explanation:
The total area of an ellipse is proportional to a^2 when eccentricity is fixed, and the orbital period is proportional to a^(3/2). Therefore, areal velocity is proportional to a^2/T, or a^(1/2). Increasing a by 16 increases the areal velocity by sqrt(16) = 4.
6. A planet moves in an ellipse with semi-major axis a and eccentricity e. The area swept from perihelion to a point is exactly one-eighth of the total area of the ellipse. What fraction of the orbital period has elapsed since perihelion?
A.
1/16
B.
1/8
C.
1/4
D.
The answer depends on e
Correct Answer:
1/8
Explanation:
By Kepler's second law, equal areas are swept in equal times. Since the swept area is one-eighth of the complete elliptical area, the elapsed time is one-eighth of the orbital period. The eccentricity affects the angular position corresponding to that point, but not the area-time fraction.
7. Two planets orbit the same star in orbits with the same eccentricity. Planet B has twice the semi-major axis of planet A. If both sweep the same fraction of their respective orbital areas, what is the ratio of the swept physical areas, AB : AA?
A.
1 : 2
B.
1 : 4
C.
2 : 1
D.
4 : 1
Correct Answer:
4 : 1
Explanation:
For equal eccentricity, the area of an elliptical orbit is proportional to a^2. Since the same fraction is swept in each orbit and aB = 2aA, the swept-area ratio is (2aA)^2 : aA^2 = 4 : 1.
8. A planet is transferred from an elliptical orbit of semi-major axis a to another orbit around the same star with semi-major axis 4a, while the eccentricity remains unchanged. Compare the time required to sweep one-quarter of the total area in the two orbits.
A.
The new time is one-half the old time
B.
The new time is twice the old time
C.
The new time is four times the old time
D.
The new time is eight times the old time
Correct Answer:
The new time is eight times the old time
Explanation:
One-quarter of the total area is swept in one-quarter of the orbital period. For the same central mass, T is proportional to a^(3/2). Thus the period, and therefore the time for one-quarter of it, changes by 4^(3/2) = 8.
9. For two elliptic orbits around the same central body, the ratio of their semi-major axes is 25, and their eccentricities are equal. What is the ratio of the angular momentum magnitudes per unit mass, h2/h1, if the orbit with the larger semi-major axis is labeled 2?
A.
1/5
B.
1/25
C.
5
D.
25
Correct Answer:
5
Explanation:
The areal velocity is h/2. For an elliptic orbit, h is proportional to sqrt[mu a(1 - e^2)]. With the same central mass and equal eccentricity, h is proportional to sqrt(a). Therefore h2/h1 = sqrt(25) = 5. This is the semi-major-axis dependence underlying the second law.
10. Two planets orbit the same star in elliptic orbits with equal eccentricities. Their areal velocities are in the ratio 3 : 2. What is the ratio of their semi-major axes, with the first planet listed first?
A.
3 : 2
B.
9 : 4
C.
27 : 8
D.
4 : 9
Correct Answer:
9 : 4
Explanation:
For equal eccentricity around the same central mass, areal velocity is proportional to sqrt(a). Hence (dA/dt)1/(dA/dt)2 = sqrt(a1/a2). Squaring the ratio 3/2 gives a1/a2 = 9/4.
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