Gravitational Potential and Gravitational Potential Energy: NEET MCQs with Answers

15 free questions on Gravitational Potential and Gravitational Potential Energy for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
The work done in bringing a mass from infinity to the surface of Earth is: [NEET 2020]
  1. GMm
  2. -GMm/R
  3. GMm/R²
  4. -GMm/R²
Show answer
Answer: (B) -GMm/R
Explanation: Work done = ΔU = U_final - U_initial = (-GMm/R) - 0 = -GMm/R. The negative sign indicates that gravity does positive work (the mass falls), or equivalently, external agent does negative work (gravity assists). The magnitude GMm/R is the gravitational binding energy at Earth's surface.

Question 2

Previous-year question
A particle is dropped from rest at a height h above Earth's surface. Neglecting air resistance, the velocity with which it strikes the ground is: [NEET 2023]
  1. v = √(2gh)
  2. v = √(gh)
  3. v = 2√(gh)
  4. v = √(gRh/(R+h))
Show answer
Answer: (A) v = √(2gh)
Explanation: Energy conservation gives ½mv² = GMm/R − GMm/(R + h), so v = √(2gRh/(R + h)). For h much smaller than R this reduces to v = √(2gh).

Question 3

Previous-year question
Inside a uniform spherical shell, the gravitational potential: [NEET 2024]
  1. Increases linearly from center to surface
  2. Remains constant throughout
  3. Decreases from center to surface
  4. Becomes zero at all points
Show answer
Answer: (B) Remains constant throughout
Explanation: Inside a uniform spherical shell, the gravitational field is zero at all points (shell theorem). Therefore, the potential is constant everywhere inside, with value V = -GM/R, where R is the shell's radius. This remarkable result shows that a spherically symmetric mass distribution produces no field in its interior.

Question 4

Previous-year question
Work done bringing mass from infinity to Earth's surface: [NEET 2021]
  1. GMm
  2. -GMm/R
  3. GMm/R²
  4. -GMm/R²
Show answer
Answer: (B) -GMm/R
Explanation: W = ΔU = (-GMm/R) - 0 = -GMm/R. Negative means gravity does positive work. Magnitude = binding energy at surface.

Question 5

Previous-year question
Gravitational potential energy U at distance r: [NEET 2024]
  1. U = GMm/r
  2. U = -GMm/r
  3. U = GMm/r²
  4. U = mg(R-r)
Show answer
Answer: (B) U = -GMm/r
Explanation: U = -GMm/r with zero at infinity. Negative means bound (work needed to separate to infinity). Near Earth: ΔU ≈ mgh for small h.

Question 6

Previous-year question
Inside uniform spherical shell: field and potential are: [NEET 2023]
  1. Field = 0, Potential = 0
  2. Field = 0, Potential = constant
  3. Field = GM/r², Potential = constant
  4. Both vary linearly
Show answer
Answer: (B) Field = 0, Potential = constant
Explanation: Shell theorem: inside shell, gravitational field is zero everywhere, but potential is constant = -GM/R throughout. Key result in gravitation.

Question 7

Previous-year question
The work done to raise a mass m from the surface of the earth to a height h equal to the radius of the earth is [NEET 2019]
  1. mgR
  2. 2mgR
  3. ½mgR
  4. (3/2)mgR
Show answer
Answer: (C) ½mgR
Explanation: W = ΔU = GMm/R – GMm/2R = ½mgR.
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Question 8

Previous-year question
At what height from the surface of the earth are the gravitational potential and the value of g equal to –5.4 × 10⁷ J kg⁻¹ and 6.0 m s⁻² respectively? (radius of the earth = 6400 km) [NEET 2016]
  1. 2600 km
  2. 1600 km
  3. 1400 km
  4. 2000 km
Show answer
Answer: (A) 2600 km
Explanation: V = –GM/(R + h) and g = GM/(R + h)², so R + h = |V|/g = 5.4 × 10⁷/6.0 = 9 × 10⁶ m. Hence h = 9000 – 6400 = 2600 km.

Question 9

Previous-year question
In a gravitational field, at a point where the gravitational potential is zero, [AIPMT 1990]
  1. the gravitational field is necessarily zero
  2. the gravitational field is not necessarily zero
  3. nothing can be said definitely about the gravitational field
  4. None of the above
Show answer
Answer: (A) the gravitational field is necessarily zero
Explanation: Gravitational potential due to any mass is V = −Gm/r, which is negative everywhere at a finite distance, and masses are always positive so the contributions can never cancel. Hence the total potential of any mass distribution is strictly negative at every finite point, and V = 0 can only occur infinitely far from all the masses — where the field is also zero. So in a gravitational field V = 0 does necessarily imply E = 0.

Question 10

Previous-year question
A planet is moving in an elliptical orbit around the Sun. If T, U, E and L stand respectively for its kinetic energy, gravitational potential energy, total energy and the magnitude of the angular momentum about the centre of force, which of the following is correct? [AIPMT 1990]
  1. T is conserved
  2. U is always positive
  3. E is always negative
  4. L is conserved but the direction of vector L changes continuously
Show answer
Answer: (C) E is always negative
Explanation: For a bound orbit the total energy E = −GMm/2a is negative, which keeps the planet from escaping. Kinetic energy varies along the orbit, U = −GMm/r is always negative, and since gravity is a central force L is conserved in both magnitude and direction.

Question 11

Previous-year question
A body of mass m rises to a height h = R/5 from the Earth's surface, where R is the Earth's radius. If g is the acceleration due to gravity at the Earth's surface, the increase in potential energy is [AIPMT 1991]
  1. (5/6)mgh
  2. (4/5)mgh
  3. mgh
  4. (6/7)mgh
Show answer
Answer: (A) (5/6)mgh
Explanation: The exact change in potential energy is ΔU = mgh/(1 + h/R). With h = R/5, h/R = 1/5 : ΔU = mgh/(1 + 1/5) = mgh/(6/5) = (5/6)mgh.

Question 12

Previous-year question
A body attains a height equal to the radius of the Earth. The velocity of the body with which it was projected is [AIPMT 2001]
  1. √(GM/R)
  2. √(2GM/R)
  3. √(5GM/4R)
  4. √(3GM/R)
Show answer
Answer: (A) √(GM/R)
Explanation: Energy conservation between the surface and the top of the flight, where the speed is zero: ½mu² − GMm/R = −GMm/2R. This gives ½u² = GM/2R, so u = √(GM/R).

Question 13

Previous-year question
With what velocity should a particle be projected so that the height it attains becomes equal to the radius of the Earth? [AIPMT 2001]
  1. (GM/R)^(1/2)
  2. (8GM/R)^(1/2)
  3. (2GM/R)^(1/2)
  4. (4GM/R)^(1/2)
Show answer
Answer: (A) (GM/R)^(1/2)
Explanation: Conservation of energy between the surface and the highest point (where the speed is zero): ½mu² − GMm/R = 0 − GMm/(R + R) ½u² = GM/R − GM/2R = GM/2R u² = GM/R u = √(GM/R).

Question 14

Previous-year question
A body of mass m is placed on the Earth's surface. It is then taken from the Earth's surface to a height h = 3R. The change in gravitational potential energy is [AIPMT 2002]
  1. mgh/R
  2. (2/3)mgR
  3. (3/4)mgR
  4. mgR/2
Show answer
Answer: (C) (3/4)mgR
Explanation: U_i = −GMm/R and U_f = −GMm/(R + 3R) = −GMm/4R. ΔU = −GMm/4R + GMm/R = 3GMm/4R. Using GM = gR² : ΔU = 3gR²m/4R = (3/4)mgR.

Question 15

Previous-year question
The height at which the weight of a body becomes 1/16th of its weight on the surface of the Earth (radius R) is [AIPMT 2012]
  1. 5R
  2. 15R
  3. 3R
  4. 4R
Show answer
Answer: (C) 3R
Explanation: w_h/w = [R/(R + h)]² = 1/16 R/(R + h) = 1/4 ⟹ R + h = 4R ⟹ h = 3R.

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