physics : class11 Elliptical Orbit Geometry : Second Law of Kepler : Gravitation MCQ Questions and Answers
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Elliptical Orbit Geometry
MCQ questions and answers.
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MCQ Questions
1. A planet moves in an elliptical orbit of eccentricity 0.60 around the Sun, which lies at one focus. What is the ratio of its orbital speed at perihelion to that at aphelion, using Kepler's second law?
A.
1.5
B.
2.5
C.
4.0
D.
6.25
Correct Answer:
4.0
Explanation:
Equal areas are swept in equal times, so the areal velocity (1/2)rv is constant. At perihelion and aphelion, the velocity is perpendicular to the radius, giving 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) = 1.6/0.4 = 4.0.
2. A planet completes one revolution in period T while moving in an elliptical orbit of semimajor axis a and semiminor axis b. What area is swept by the line joining the planet to the Sun during a time interval T/4?
A.
πab/8
B.
πab/4
C.
πab/2
D.
πab
Correct Answer:
πab/4
Explanation:
The total area of the ellipse is πab. By Kepler's second law, equal areas are swept in equal times, so in one-fourth of the period the swept area is one-fourth of the total elliptical area: πab/4.
3. The polar equation of an elliptical orbit, with the Sun at the focus and perihelion as the reference direction, is r = a(1 − e²)/(1 + e cos θ). If e = 0.50, what is the ratio r(60°)/r(120°)?
A.
1/2
B.
3/5
C.
5/3
D.
2
Correct Answer:
3/5
Explanation:
Using the polar equation, r(60°) is proportional to 1/(1 + 0.5 × 0.5) = 1/1.25, while r(120°) is proportional to 1/(1 + 0.5 × (−0.5)) = 1/0.75. Therefore, r(60°)/r(120°) = 0.75/1.25 = 3/5.
4. For an elliptical orbit with semimajor axis a and eccentricity e, which expression gives the ratio of the aphelion distance to the perihelion distance?
A.
(1 − e)/(1 + e)
B.
1 − e²
C.
(1 + e)/(1 − e)
D.
1/(1 − e²)
Correct Answer:
(1 + e)/(1 − e)
Explanation:
The perihelion and aphelion distances are r_p = a(1 − e) and r_a = a(1 + e). Dividing gives r_a/r_p = (1 + e)/(1 − e), which directly reflects the geometry of the ellipse relative to its focus.
5. A planet has perihelion distance 2 × 10^10 m and aphelion distance 18 × 10^10 m. What is the eccentricity of its elliptical orbit?
A.
0.40
B.
0.60
C.
0.80
D.
0.90
Correct Answer:
0.80
Explanation:
For an ellipse, r_p = a(1 − e) and r_a = a(1 + e). Eliminating a gives e = (r_a − r_p)/(r_a + r_p). Thus e = (18 − 2)/(18 + 2) = 16/20 = 0.80.
6. A planet takes time Δt to move from perihelion to aphelion along one half of its elliptical orbit. What time does it take to move from aphelion back to perihelion along the other half?
A.
Δt/2
B.
Δt
C.
2Δt
D.
πΔt
Correct Answer:
Δt
Explanation:
The two halves of the ellipse have equal areas. Since Kepler's second law states that equal areas are swept in equal times, the travel times along the two halves are equal.
7. An elliptical orbit has semimajor axis 10 m and eccentricity 0.8. What are the perihelion and aphelion distances from the focus?
A.
2 m and 18 m
B.
8 m and 12 m
C.
0.8 m and 10.8 m
D.
4 m and 16 m
Correct Answer:
2 m and 18 m
Explanation:
The focal distances at perihelion and aphelion are r_p = a(1 − e) and r_a = a(1 + e). Therefore, r_p = 10(0.2) = 2 m and r_a = 10(1.8) = 18 m.
8. Two points on an elliptical orbit have the same distance from the Sun and lie symmetrically on opposite sides of the major axis. Which statement about the time required to travel from perihelion to each point is correct?
A.
The times are always unequal because the orbital speeds differ.
B.
The times are equal because the corresponding swept areas are equal.
C.
The time to the point above the major axis is twice the other time.
D.
The times are equal only if the ellipse is a circle.
Correct Answer:
The times are equal because the corresponding swept areas are equal.
Explanation:
Points with equal distances from the focus and symmetric positions about the major axis subtend equal geometric sectors from the perihelion direction. Hence the areas swept from perihelion to those points are equal, and Kepler's second law makes the corresponding travel times equal.
9. The area of an elliptical orbit is 6.0 × 10^20 m² and the orbital period is 12 years. What is the areal velocity of the planet with respect to the Sun?
A.
5.0 × 10^19 m²/year
B.
1.0 × 10^20 m²/year
C.
2.0 × 10^20 m²/year
D.
7.2 × 10^21 m²/year
Correct Answer:
5.0 × 10^19 m²/year
Explanation:
The areal velocity is the total area swept divided by the orbital period. Therefore, dA/dt = (6.0 × 10^20)/12 = 5.0 × 10^19 m²/year. Kepler's second law states that this value remains constant throughout the orbit.
10. For an elliptical orbit, the semimajor axis is a and the eccentricity is e. Which expression correctly relates the perihelion distance r_p, aphelion distance r_a, and semimajor axis?
A.
r_p + r_a = a
B.
r_p + r_a = 2ae
C.
r_p + r_a = 2a
D.
r_p r_a = 2a
Correct Answer:
r_p + r_a = 2a
Explanation:
The focus-to-vertex distances are r_p = a(1 − e) and r_a = a(1 + e). Adding them gives r_p + r_a = a(1 − e + 1 + e) = 2a, a fundamental geometric property of the elliptical orbit.
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