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Escape velocity and satellite motion, period and energy: NEET MCQs with answers

90 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. If a body is projected from the surface of the Earth at an angle of 30° to the horizontal, its escape velocity will be:

  1. Equal to the escape velocity when projected vertically
  2. √3/2 times the escape velocity when projected vertically
  3. Half of the escape velocity when projected vertically
  4. Double the escape velocity when projected vertically
Show answer and explanation

Answer: (A) Equal to the escape velocity when projected vertically

Escape velocity depends only on the mass and radius of the planet (ve = √(2GM/R)) and does not depend on the angle of projection relative to the horizontal.

medium

2. The escape velocity from Earth is 11.2 km/s. A planet has 2 times Earth's mass and 4 times Earth's radius. The escape velocity from that planet is:

  1. 31.7 km/s
  2. 15.8 km/s
  3. 7.92 km/s
  4. 5.6 km/s
Show answer and explanation

Answer: (C) 7.92 km/s

vₑ = √(2GM/R) ∝ √(M/R). So vₑ = 11.2 × √(2/4) = 7.92 km/s.

hard

3. The binding energy of a satellite of mass m revolving in a circular orbit of radius r around a planet of mass M is E_b. What is its angular momentum L about the center of the planet?

  1. √(m r² E_b)
  2. √(2 m r² E_b)
  3. √(4 m r² E_b)
  4. 2 √(m r² E_b)
Show answer and explanation

Answer: (B) √(2 m r² E_b)

The total energy of the satellite is E = -GMm/(2r), so its binding energy is E_b = -E = GMm/(2r). The kinetic energy of the satellite is K = (1/2)m v² = GMm/(2r) = E_b. The angular momentum is L = m v r. We know K = L² / (2 I) = L² / (2 m r²). Setting K = E_b gives E_b = L² / (2 m r²), which leads to L = √(2 m r² E_b).

medium

4. The escape velocity from Earth is 11.2 km/s. A planet has 8 times Earth's mass and 2 times Earth's radius. The escape velocity from that planet is:

  1. 44.8 km/s
  2. 5.6 km/s
  3. 11.2 km/s
  4. 22.4 km/s
Show answer and explanation

Answer: (D) 22.4 km/s

vₑ = √(2GM/R) ∝ √(M/R). So vₑ = 11.2 × √(8/2) = 22.4 km/s.

easy

5. A satellite of mass m is in a circular orbit of radius r around a planet of mass M. What is the total mechanical energy of the satellite?

  1. - GMm / r
  2. - GMm / (2r)
  3. + GMm / (2r)
  4. + GMm / r
Show answer and explanation

Answer: (B) - GMm / (2r)

For a circular orbit of radius r, the kinetic energy is K = GMm / (2r) and the potential energy is U = - GMm / r. The total mechanical energy E = K + U = GMm/(2r) - GMm/r = - GMm / (2r).

hard

6. A satellite is in a circular orbit of radius r around a planet of mass M. If a space collision reduces its orbital speed by 20%, what is the kinetic energy of the satellite immediately after the collision in terms of G, M, m, and r?

  1. 0.32 GMm/r
  2. 0.18 GMm/r
  3. 0.40 GMm/r
  4. 0.50 GMm/r
Show answer and explanation

Answer: (A) 0.32 GMm/r

The initial orbital speed is v₀ = √(GM/r). After collision, its speed becomes v = 0.8v₀. The kinetic energy is K = (1/2)mv² = (1/2)m(0.8v₀)² = 0.32 m v₀² = 0.32 GMm/r.

medium

7. The escape velocity from the Earth is 11.2 km/s. For a planet whose mass is 6 times and radius 4 times that of the Earth, the escape velocity is:

  1. 13.72 km/s
  2. 16.8 km/s
  3. 9.14 km/s
  4. 54.87 km/s
Show answer and explanation

Answer: (A) 13.72 km/s

v_e = √(2GM/R) ∝ √(M/R). So v = 11.2 × √(6/4) = 13.72 km/s.

Quick tip: v_e ∝ √(M/R): take the square root of the mass-to-radius ratio.

NCERT Class 11, Chapter 7

medium

8. The escape velocity from the Earth is 11.2 km/s. For a planet whose mass is 12 times and radius 2 times that of the Earth, the escape velocity is:

  1. 27.43 km/s
  2. 54.87 km/s
  3. 67.2 km/s
  4. 4.57 km/s
Show answer and explanation

Answer: (A) 27.43 km/s

v_e = √(2GM/R) ∝ √(M/R). So v = 11.2 × √(12/2) = 27.43 km/s.

Quick tip: v_e ∝ √(M/R): take the square root of the mass-to-radius ratio.

NCERT Class 11, Chapter 7

medium

9. A planet has 12 times the mass and 3 times the radius of the Earth. If the escape velocity from the Earth is 11.2 km/s, the escape velocity from the planet is:

  1. 67.2 km/s
  2. 5.6 km/s
  3. 44.8 km/s
  4. 22.4 km/s
Show answer and explanation

Answer: (D) 22.4 km/s

v_e = √(2GM/R) ∝ √(M/R). So v = 11.2 × √(12/3) = 22.4 km/s.

Quick tip: v_e ∝ √(M/R): take the square root of the mass-to-radius ratio.

NCERT Class 11, Chapter 7

easy

10. A satellite of mass m is orbiting close to the surface of a planet of mass M and radius R. The angular momentum of the satellite about the centre of the planet is:

  1. m√(2GMR)
  2. m√(GMR)
  3. √(GMm²R³)
  4. m√(GM/R)
Show answer and explanation

Answer: (B) m√(GMR)

The orbital speed close to the surface is v₀ = √(GM/R). The angular momentum is L = m v₀ R = m √(GM/R) × R = m√(GMR).

easy

11. Given below are two statements: Statement I: The total mechanical energy of a satellite orbiting in an elliptical path about the Earth is negative. Statement II: For a satellite moving in an elliptical orbit, the speed of the satellite is greatest when it is farthest from the Earth (apogee). 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 for any bound gravitational system, the total mechanical energy is negative (E = -GMm/2a). Statement II is incorrect because by conservation of angular momentum (m v r = constant), the speed is maximum at closest approach (perigee) and minimum at apogee.

medium

12. The escape velocity from Earth is 11.2 km/s. A planet has 2 times Earth's mass and 2 times Earth's radius. The escape velocity from that planet is:

  1. 16.8 km/s
  2. 5.6 km/s
  3. 11.2 km/s
  4. 22.4 km/s
Show answer and explanation

Answer: (C) 11.2 km/s

vₑ = √(2GM/R) ∝ √(M/R). So vₑ = 11.2 × √(2/2) = 11.2 km/s.

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