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Kepler’s laws of planetary motion: NEET MCQs with answers

57 questions on this topic of Gravitation (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The ratio of kinetic energy of a planet at perihelion to that at aphelion is 9 : 4. What is the ratio of its distance from the Sun at aphelion to that at perihelion?

  1. 27 : 8
  2. 9 : 4
  3. √3 : √2
  4. 3 : 2
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Answer: (D) 3 : 2

By conservation of angular momentum, L = m·v·r = constant, so v_p·r_p = v_a·r_a. The ratio of kinetic energies is K_p / K_a = (v_p / v_a)² = (r_a / r_p)². Since K_p / K_a = 9/4, we have r_a / r_p = √(9/4) = 3/2.

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2. Given below are two statements: Statement I: A planet moves fastest when it is closest to the Sun and slowest when it is farthest from the Sun. Statement II: The areal velocity of the radius vector joining the planet to the Sun is inversely proportional to its distance from the Sun. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because angular momentum L = mvr is constant, so v is inversely proportional to r (fastest at perihelion, slowest at aphelion). Statement II is incorrect because Kepler's second law explicitly states that areal velocity dA/dt = L/(2m) is strictly constant, independent of distance.

hard

3. A planet of mass m moves around the Sun of mass M in an elliptical orbit of semi-major axis a. If L is the magnitude of its angular momentum about the Sun, the total area of the elliptical orbit is:

  1. (2 L T) / m
  2. (L T) / m
  3. (L T) / (2 m)
  4. (L T) / (4 m)
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Answer: (C) (L T) / (2 m)

Kepler's second law gives areal velocity dA/dt = L / (2m). Since areal velocity is constant, the total area A swept in one complete revolution of period T is A = (dA/dt) × T = (L / (2m)) × T = (L T) / (2m).

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4. Given below are two statements: Statement I: A planet moves along an elliptical orbit with the Sun at one focus; its speed is maximum when it is closest to the Sun. Statement II: The total mechanical energy of the planet-Sun system varies periodically as the planet moves from perihelion to aphelion. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because by conservation of angular momentum, speed is inversely proportional to distance, making it maximum at perihelion. Statement II is incorrect because gravity is a conservative force and no external non-conservative forces act, so the total mechanical energy remains strictly constant.

easy

5. A planet orbits the Sun in an elliptical path where the perihelion distance is r_p and the aphelion distance is r_a. What is the semi-major axis 'a' of this orbit?

  1. √(r_p × r_a)
  2. 2r_p r_a / (r_p + r_a)
  3. (r_p + r_a) / 2
  4. r_a - r_p
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Answer: (C) (r_p + r_a) / 2

The major axis of an elliptical orbit is the total distance across the ellipse passing through both foci, which equals the sum of perihelion distance and aphelion distance: 2a = r_p + r_a. Thus, the semi-major axis is a = (r_p + r_a)/2.

hard

6. A planet of mass m moves around the Sun of mass M in an elliptical orbit. If the total mechanical energy of the planet is E = -GMm / (2a), the work done by the gravitational force on the planet as it moves from perihelion (distance r₁) to aphelion (distance r₂) is:

  1. GMm / (2a)
  2. GMm (1/r₁ - 1/r₂)
  3. Zero
  4. GMm (1/r₂ - 1/r₁)
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Answer: (D) GMm (1/r₂ - 1/r₁)

By the work-energy theorem, Work done by gravity = -ΔU = -(U_aphelion - U_perihelion) = -[(-GMm/r₂) - (-GMm/r₁)] = GMm (1/r₂ - 1/r₁). Since r₂ > r₁, this work is negative, corresponding to the planet slowing down from perihelion to aphelion.

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7. A satellite orbiting a planet has a period of 1 day. Another satellite orbits the same planet at 4 times the orbital radius. Its period is:

  1. 16 days
  2. 4 days
  3. 2 days
  4. 8 days
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Answer: (D) 8 days

Kepler's third law: T² ∝ r³, so T = 1 × 4^(3/2) = 8 days.

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8. The time period of revolution of a planet around the Sun in an elliptical orbit of semi-major axis a is T. If the mass of the Sun were suddenly increased to 4 times its original value while the orbit's semi-major axis remained a, the new time period would be:

  1. T/2
  2. 2T
  3. 4T
  4. T/4
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Answer: (A) T/2

Kepler's third law states T² = 4π²a³/(GM). Thus, T is inversely proportional to √M. When M becomes 4M, the new period becomes T' = T/√4 = T/2.

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9. A satellite orbiting a planet has a period of 2 days. Another satellite orbits the same planet at 9 times the orbital radius. Its period is:

  1. 18 days
  2. 6 days
  3. 162 days
  4. 54 days
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Answer: (D) 54 days

Kepler's third law: T² ∝ r³, so T = 2 × 9^(3/2) = 54 days.

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10. A satellite orbiting a planet has a period of 1 day. Another satellite orbits the same planet at 16 times the orbital radius. Its period is:

  1. 16 days
  2. 256 days
  3. 4 days
  4. 64 days
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Answer: (D) 64 days

Kepler's third law: T² ∝ r³, so T = 1 × 16^(3/2) = 64 days.

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11. If the mean orbital radius of Earth around the Sun were to double, what would be the new length of an Earth year (taking the current year as 365 days)?

  1. 730 days
  2. 516 days
  3. 1032 days
  4. 1460 days
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Answer: (C) 1032 days

According to Kepler's third law, T² ∝ R³. Therefore, T₂ = T₁ × (R₂/R₁)^(3/2) = 365 × 2^(3/2) = 365 × 2√2 ≈ 365 × 2.828 ≈ 1032 days.

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12. A satellite orbiting a planet has a period of 5 days. Another satellite orbits the same planet at 9 times the orbital radius. Its period is:

  1. 45 days
  2. 135 days
  3. 405 days
  4. 15 days
Show answer and explanation

Answer: (B) 135 days

Kepler's third law: T² ∝ r³, so T = 5 × 9^(3/2) = 135 days.

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