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NEET Gravitation MCQs with answers

308 questions on Gravitation for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easyNEET 2020

1. Escape velocity from Earth's surface:

  1. 8 km/s
  2. 11.2 km/s
  3. 15 km/s
  4. 3 km/s
Show answer and explanation

Answer: (B) 11.2 km/s

ve = √(2gR) ≈ 11.2 km/s. ve = √2 × orbital velocity.

Quick tip: ve = 11.2 km/s. Orbital vo = 7.9 km/s. ve = √2 × vo.

NCERT Class 11, Chapter 8, page 188

easy

2. According to Kepler's second law, the areal velocity of a planet revolving around the Sun is constant because the gravitational force exerted by the Sun on the planet:

  1. is independent of the distance of the planet from the Sun
  2. acts along the line joining the planet to the Sun, producing zero torque
  3. does positive work on the planet throughout its orbit
  4. is conservative, keeping total mechanical energy zero
Show answer and explanation

Answer: (B) acts along the line joining the planet to the Sun, producing zero torque

The gravitational force between the Sun and the planet is purely radial (a central force). Because the position vector r and the force F are collinear, the net torque τ = r × F = 0. Zero torque implies the angular momentum L is conserved, leading directly to constant areal velocity dA/dt = L / (2m).

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3. The orbital period of a planet around the Sun is T, and its orbit encloses a total area A. If the mass of the planet is m, what is the magnitude of its orbital angular momentum about the Sun?

  1. 2mA / T
  2. mA / (2T)
  3. 4mA / T
  4. mA / T
Show answer and explanation

Answer: (A) 2mA / T

According to Kepler's second law, areal velocity is constant: dA/dt = L/(2m). Integrating over one complete revolution of period T, the total area enclosed is A = (L/(2m)) × T. Rearranging for angular momentum gives L = 2mA/T.

hard

4. A satellite moves in an elliptical orbit around Earth such that its closest distance to Earth's center is R and its farthest distance is 3R. If its speed at the closest point is v₀, what is its speed at the farthest point?

  1. v₀ / 3
  2. v₀ / 9
  3. v₀ / √3
  4. 3v₀
Show answer and explanation

Answer: (A) v₀ / 3

Since the gravitational pull is a central force, the angular momentum of the satellite about the center of Earth is conserved: L = m × r_min × v_max = m × r_max × v_min. Therefore, R × v₀ = (3R) × v_far, which yields v_far = v₀/3.

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5. If the acceleration due to gravity at a depth d₁ is equal to that at a height h = R/3, then the value of depth d₁ in terms of the Earth's radius R is:

  1. d₁ = 7R/16
  2. d₁ = 9R/16
  3. d₁ = 3R/4
  4. d₁ = 5R/8
Show answer and explanation

Answer: (A) d₁ = 7R/16

At height h = R/3, g_h = g / (1 + 1/3)² = 9g/16. Setting this equal to g_d = g(1 - d₁/R) gives 1 - d₁/R = 9/16, which yields d₁/R = 1 - 9/16 = 7/16, so d₁ = 7R/16.

easy

6. Which of the following physical quantities has the same SI base unit representation as that of the Universal Gravitational Constant G?

  1. Force × (Velocity)² / (Mass)³
  2. (Acceleration) × (Area) / (Mass)
  3. Work / (Momentum)²
  4. Power × (Length) / (Force)²
Show answer and explanation

Answer: (B) (Acceleration) × (Area) / (Mass)

The dimensional formula of G is [M⁻¹ L³ T⁻²]. Checking “(Acceleration) × (Area) / (Mass)”: Acceleration [L T⁻²] × Area [L²] / Mass [M] = [M⁻¹ L³ T⁻²], which exactly matches the dimensions of G.

hard

7. A hypothetical spherical planet has uniform density ρ and radius R. A narrow tunnel is drilled through its center. If a particle of mass m is released from the surface into the tunnel, what is the magnitude of the gravitational acceleration experienced by the particle when it is at a distance r (r < R) from the center?

  1. 4πGρr
  2. (4/3)πGρ(R - r)
  3. Gρr² / R
  4. (4/3)πGρr
Show answer and explanation

Answer: (D) (4/3)πGρr

Inside a uniform solid sphere at a distance r from the center, the gravitational field is produced solely by the mass inside the sphere of radius r: M_r = (4/3)πr³ρ. Thus, g(r) = GM_r / r² = G[(4/3)πr³ρ] / r² = (4/3)πGρr.

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8. The escape velocity from the Earth is 11.2 km/s. For a planet whose mass is 10 times and radius 2 times that of the Earth, the escape velocity is:

  1. 25.04 km/s
  2. 56 km/s
  3. 50.09 km/s
  4. 5.01 km/s
Show answer and explanation

Answer: (A) 25.04 km/s

v_e = √(2GM/R) ∝ √(M/R). So v = 11.2 × √(10/2) = 25.04 km/s.

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

NCERT Class 11, Chapter 7

easy

9. According to Kepler's third law, the ratio of the square of the orbital period T to the cube of the semi-major axis a (i.e., T²/a³) for planets orbiting the Sun depends on:

  1. The mass of the orbiting planet
  2. The mass of the Sun only
  3. Both the mass of the Sun and the mass of the planet equally
  4. The eccentricity of the elliptical orbit
Show answer and explanation

Answer: (B) The mass of the Sun only

By equating gravitational attraction to centripetal acceleration, T²/a³ = 4π²/(G M_Sun), assuming the planet's mass is negligible compared to the Sun. Hence, this constant ratio depends only on the mass of the Sun.

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10. The gravitational potential in a region is given by V = -(5x + 12y) J kg⁻¹, where x and y are in metres. The magnitude of the gravitational field intensity at the point (2 m, 1 m) is:

  1. 13 N kg⁻¹
  2. 17 N kg⁻¹
  3. 7 N kg⁻¹
  4. √119 N kg⁻¹
Show answer and explanation

Answer: (A) 13 N kg⁻¹

The gravitational field components are E_x = -∂V/∂x = -(-5) = 5 N kg⁻¹ and E_y = -∂V/∂y = -(-12) = 12 N kg⁻¹. Thus, E = √(5² + 12²) = 13 N kg⁻¹.

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11. Taking g = 9.8 m/s² on the surface and the Earth’s radius as 6400 km, the acceleration due to gravity at a height of 2560 km is:

  1. 19.21 m/s²
  2. 1.96 m/s²
  3. 5 m/s²
  4. 7 m/s²
Show answer and explanation

Answer: (C) 5 m/s²

g_h = g(R/(R+h))² = 9.8 × (6400/8960)² = 5 m/s². (The approximation g(1 − 2h/R) holds only when h ≪ R.)

Quick tip: For a large height use g(R/(R+h))², not g(1 − 2h/R).

NCERT Class 11, Chapter 7

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12. A mass M is distributed uniformly along a semi-infinite straight line along the positive x-axis extending from x = a to x = ∞, with a constant linear mass density λ. A point particle of mass m is placed at the origin x = 0. The magnitude of the gravitational attraction on the point particle is:

  1. G m λ / a
  2. G m λ / a²
  3. 2 G m λ / a
  4. Zero
Show answer and explanation

Answer: (A) G m λ / a

Consider a small element of the line mass dx located at coordinate x. Its mass is dm = λ dx. The gravitational force on mass m at the origin due to this element is dF = G·m·(λ dx)/x². Integrating from x = a to x = ∞ gives F = ∫_a^∞ (G m λ / x²) dx = G m λ [-1/x]_a^∞ = G m λ / a.

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13. A geostationary satellite orbits at a distance of 7R from the centre of the Earth. What will be the orbital time period of another satellite whose orbital distance from the centre of the Earth is 1.75R?

  1. 6 hours
  2. 3 hours
  3. 1.5 hours
  4. 12 hours
Show answer and explanation

Answer: (B) 3 hours

According to Kepler's third law, T² ∝ r³. Therefore, (T₂/T₁)² = (r₂/r₁)³ = (1.75R / 7R)³ = (1/4)³ = 1/64. Taking square root on both sides gives T₂/T₁ = 1/8. For a geostationary satellite, T₁ = 24 hours, so T₂ = 24 / 8 = 3 hours.

easy

14. The acceleration due to gravity on the surface of the Earth is g. If a body of mass m is taken to a height h = 4R above the surface of the Earth, where R is the radius of the Earth, its weight becomes:

  1. mg / 5
  2. mg / 16
  3. mg / 9
  4. mg / 25
Show answer and explanation

Answer: (D) mg / 25

The acceleration due to gravity at height h is given by g_h = g / (1 + h/R)². For h = 4R, g_h = g / (1 + 4)² = g / 25. Thus, the weight becomes W' = m g_h = mg / 25.

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15. The acceleration due to gravity at a height equal to R/4 (R = radius of Earth) is what percentage of its value at the surface?

  1. 75 %
  2. 80 %
  3. 64 %
  4. 50 %
Show answer and explanation

Answer: (C) 64 %

g_h = g/(1 + h/R)² = g/(1 + 1/4)² = 64% of g (exact form, since h is not small).

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