NEET Gravitation PYQs & Practice Questions

10 free NEET Physics questions on Gravitation, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Escape Velocity · Angular Momentum Conservation · Work done to lift a mass · Kepler's Third Law · Variation of g with Height · Acceleration due to Gravity · Kepler's Laws · Satellites · Inertial and Gravitational Mass · Acceleration Due to Gravity on Another Planet · Minimum Energy to Launch Satellite into Orbit · Gravitational Potential at Null Point

1. Q1NEET 2026Escape Speed

   Two planets P1 and P2 with equal mass have radii R1 and R2, respectively, where R2 = R1/2. The escape speeds of P1 and P2 are v1 and v2, respectively. Then v2/v1 is:

   • A. 1
   • B. sqrt(2)
   • C. 2
   • D. 1/sqrt(2)
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   Answer: A — 1

   Escape speed v = sqrt(2GM/R). Since both planets have equal mass M, v2/v1 = sqrt(R1/R2) = sqrt(R1/(R1/2)) = sqrt(2).

2. Q2NEET 2026Kepler's Third Law

   In the solar system, the time-period of revolution of a planet tracing a circular orbit of radius R is proportional to:

   • A. R^(3/2)
   • B. R²
   • C. R³
   • D. R^(1/2)
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   Answer: —

   By Kepler's third law of planetary motion, T² ∝ R³, so T ∝ R^(3/2).

3. Q3NEET 2026Work done to lift a mass

   The amount of work done to raise a mass 'm' from the surface of the Earth to a height equal to the radius of the Earth 'R', will be:

   • A. mgR/2
   • B. 2mgR
   • C. mgR/4
   • D. 2mgR
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   Answer: —

   W=GMm/R-GMm/2R=GMm/2R=mgR/2.

4. Q4NEET 2025Angular Momentum Conservation

   The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

   • A. 100 days
   • B. 105 days
   • C. 115 days
   • D. 108 days
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   Answer: C — 115 days

   Conservation of angular momentum: I1ω1 = I2ω2 ⟹ R1²/T1 = R2²/T2. With R2 = 2R1: T2 = 4T1 = 4×27 = 108 days.

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5. Q5NEET 2025Kepler's Third Law

   The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth Days. Then which of the following is the length of 1 year on Mercury?

   • A. 88 earth days
   • B. 225 earth days
   • C. 172 earth days
   • D. 124 earth days
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   Answer: —

   By Kepler's third law T²∝r³: (T1/T2)² = (r1/r2)³ = (4r/r)³ = 64, so T1/T2 = 8. T2 = T1/8 = 687/8 ≈ 86.1 ≈ 88 earth days.

6. Q6NEET 2025Variation of g with Height

   A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is :

   • A. 16 N
   • B. 27 N
   • C. 32 N
   • D. 36 N
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   Answer: A — 16 N

   g' = g/(1+h/Re)². With h = Re/3: g' = g/(1+1/3)² = g/(4/3)² = 9g/16. Weight at height = 48×(9/16) = 27 N.

7. Q7NEET 2024Acceleration Due to Gravity on Another Planet

   The mass of a planet is 1/10th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:

   • A. 4.9 ms⁻²
   • B. 3.92 m s⁻²
   • C. 19.6 m s⁻²
   • D. 9.8 m s⁻²
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   Answer: A — 4.9 ms⁻²

   g' = G(M/10)/(R/2)² = G·M/(10) × 4/R² = (4/10) × GM/R² = (4/10) × g = 0.4 × 9.8 = 3.92 m/s².

8. Q8NEET 2024Minimum Energy to Launch Satellite into Orbit

   The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is:

   • A. GmM/2R
   • B. GmM/3R
   • C. 5GmM/6R
   • D. 2GmM/3R
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   Answer: B — GmM/3R

   Total energy on surface = -GMm/R. Total energy in orbit at altitude 2R (orbital radius 3R) = -GMm/(2×3R) = -GMm/6R. Minimum energy required E = (-GMm/6R) - (-GMm/R) = GMm/R - GMm/6R = 5GMm/6R.

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9. Q9NEET 2023Gravitational Potential at Null Point

   Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (G = gravitational constant):

   • A. -12Gm/R
   • B. -16Gm/R
   • C. -20Gm/R
   • D. -8Gm/R
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   Answer: A — -12Gm/R

   At the null point where field is zero, Gm/x² = G(9m)/(R-x)², giving x = R/4. The potential at this point VP = -Gm/x - G(9m)/(R-x) = -Gm/(R/4) - 9Gm/(3R/4) = -4Gm/R - 12Gm/R = -16Gm/R.

10. Q10NEET 2023Satellite Orbital Period and Earth Density

    A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity 3π/(Gd) represents:

    • A. T²
    • B. T³
    • C. √T
    • D. T
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    Answer: —

    For a satellite orbiting just above Earth's surface, T² = 4π²R³/(GM). Substituting M = (4/3)πR³d: T² = 4π²R³/(G×(4/3)πR³d) = 3π/(Gd).