Gravitation NEET PYQs (2005–2026) with Solutions

10 solved Gravitation PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyGravitational Potential & Energy

   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:

   1. AOption A: 2mgR
   2. BOption B: mgR/4
   3. COption C: mgR
   4. DOption D: mgR/2

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   Correct answer: (D) mgR/2

   Explanation

   Work done equals change in gravitational potential energy: W = U₂ − U₁ Initial position: at Earth's surface (r = R) U₁ = −GMm/R Final position: height = R above surface Thus final distance from Earth's center: r = 2R U₂ = −GMm/(2R) Therefore: W = (−GMm/2R) − (−GMm/R) W = GMm/2R Using: GM/R² = g ⇒ GM = gR² Substitute: W = (gR²m)/(2R) W = mgR/2 Hence option (D) is correct.

2. Question 2 (NEET 2026)

   NEET 2026EasyEscape speed

   Two planets $P_1$ and $P_2$ with equal mass have radii $R_1$ and $R_2$, respectively, where $R_2 = \frac{R_1}{2}$. The escape speeds of $P_1$ and $P_2$ are $v_1$ and $v_2$, respectively. Then $\frac{v_2}{v_1}$ is:

   1. AOption A: $\frac{1}{\sqrt{2}}$
   2. BOption B: $1$
   3. COption C: $\sqrt{2}$
   4. DOption D: $2$

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   Correct answer: (C) $\sqrt{2}$

   Explanation

   Escape speed $v_e = \sqrt{\frac{2GM}{R}}$, so for equal masses $v_e \propto \frac{1}{\sqrt{R}}$. $\frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} = \sqrt{\frac{R_1}{R_1/2}} = \sqrt{2}$, option (C).

3. Question 3 (NEET 2026)

   NEET 2026EasyKepler's law of periods

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

   1. AOption A: $R^{1/2}$
   2. BOption B: $R^{3/2}$
   3. COption C: $R^2$
   4. DOption D: $R^3$

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   Correct answer: (B) $R^{3/2}$

   Explanation

   For a circular orbit, gravity provides the centripetal force: $\frac{GMm}{R^2} = \frac{mv^2}{R} \Rightarrow v = \sqrt{\frac{GM}{R}}$. $T = \frac{2\pi R}{v} = 2\pi\sqrt{\frac{R^3}{GM}}$, so $T^2 \propto R^3$ (Kepler's third law). Hence $T \propto R^{3/2}$, option (B).

4. Question 4 (NEET 2025)

   NEET 2025MediumMotion of Satellite, Orbital Velocity and Time Period

   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.

   1. AOption A: 100 days
   2. BOption B: 105 days
   3. COption C: 115 days
   4. DOption D: 108 days

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   Correct answer: (D) 108 days

   Explanation

   Using conservation of angular momentum: Iω = constant. For a uniform sphere, I ∝ R². If radius doubles, angular velocity becomes one-fourth. Since T = 2π/ω, the time period becomes four times. Therefore, T = 4 × 27 = 108 days.

5. Question 5 (NEET 2025)

   NEET 2025EasyKepler's Laws

   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?

   1. AOption A: 88 earth days
   2. BOption B: 225 earth days
   3. COption C: 172 earth days
   4. DOption D: 124 earth days

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   Correct answer: (A) 88 earth days

   Explanation

   Using Kepler’s third law: $$T^2 \propto r^3$$ Given: $$r_M = 4r_m$$ Therefore: $$\left(\frac{T_M}{T_m}\right)^2 = 4^3 = 64$$ $$\frac{T_M}{T_m} = 8$$ $$T_m = \frac{687}{8} \approx 86 \text{ days}$$ Closest option is 88 Earth days.

6. Question 6 (NEET 2025)

   NEET 2025EasyAcceleration due to Gravity

   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 :

   1. AOption A: 16 N
   2. BOption B: 27 N
   3. COption C: 32 N
   4. DOption D: 36 N

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   Correct answer: (B) 27 N

   Explanation

   Weight varies inversely as square of distance from Earth’s centre. Height: $$h = \frac{R}{3}$$ Distance from centre: $$r = R + h = \frac{4R}{3}$$ Therefore: $$W' = 48 \times \left(\frac{R}{4R/3}\right)^2$$ $$W' = 48 \times \frac{9}{16} = 27 \text{ N}$$

7. Question 7 (NEET 2024)

   NEET 2024EasyAcceleration Due to Gravity

   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:

   1. AOption A: 4.9 m s⁻²
   2. BOption B: 3.92 m s⁻²
   3. COption C: 19.6 m s⁻²
   4. DOption D: 9.8 m s⁻²

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   Correct answer: (B) 3.92 m s⁻²

   Explanation

   Acceleration due to gravity is: $$g = \frac{GM}{R^2}$$ Given: $$M_p = \frac{M_e}{10}, \quad R_p = \frac{R_e}{2}$$ Therefore, $$g_p = \frac{G(M_e/10)}{(R_e/2)^2}$$ $$g_p = \frac{GM_e}{10} \times \frac{4}{R_e^2}$$ $$g_p = \frac{4}{10}g_e = 0.4 \times 9.8$$ $$g_p = 3.92\,m\,s^{-2}$$ Hence, option (B) is correct.

8. Question 8 (NEET 2024)

   NEET 2024HardSatellites & Orbital Motion

   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:

   1. AOption A: $\dfrac{GmM}{2R}$
   2. BOption B: $\dfrac{GmM}{3R}$
   3. COption C: $\dfrac{5GmM}{6R}$
   4. DOption D: $\dfrac{2GmM}{3R}$

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   Correct answer: (C) $\dfrac{5GmM}{6R}$

   Explanation

   The satellite is launched into a circular orbit at altitude: $$h = 2R$$ Hence orbital radius: $$r = R + 2R = 3R$$ Initial energy on Earth's surface: $$E_i = -\frac{GMm}{R}$$ Final total energy in circular orbit: $$E_f = -\frac{GMm}{2r}$$ Substituting $r=3R$:$$E_f = -\frac{GMm}{6R}$$Minimum energy required:$$\Delta E = E_f - E_i$$$$\Delta E = -\frac{GMm}{6R}+\frac{GMm}{R}$$$$\Delta E = \frac{5GMm}{6R}$$Thus the required energy is:$$\boxed{\frac{5GMm}{6R}}$$Hence, the correct answer is: **(3)$\dfrac{5GMm}{6R}$**.

9. Question 9 (NEET 2022)

   NEET 2022MediumGravitational Constant, Potential and Intensity

   Match List-I with List-II:

   1. AOption A: (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
   2. BOption B: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
   3. COption C: (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
   4. DOption D: (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)

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   Correct answer: (C) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

   Explanation

   Dimensions: G → [M⁻¹L³T⁻²] Potential energy → [ML²T⁻²] Potential → [L²T⁻²] Intensity → [LT⁻²] Thus matching: (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii) Hence option (C) is correct.

10. Question 10 (NEET 2021)

    NEET 2021EasyEscape Velocity

    The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius four times that of Earth and same density is

    1. AOption A: 4v
    2. BOption B: v
    3. COption C: 2v
    4. DOption D: 3v

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    Correct answer: (A) 4v

    Explanation

    Escape velocity is given by: $$v_e = \sqrt{\frac{2GM}{R}}$$ For a planet of uniform density: $$M = \frac{4}{3}\pi R^3 \rho$$ Substituting: $$v_e = \sqrt{\frac{2G\left(\frac{4}{3}\pi R^3\rho\right)}{R}}$$ $$v_e = \sqrt{\frac{8\pi G\rho R^2}{3}}$$ Since density remains the same: $$v_e \propto R$$ New planet radius = 4R $$\frac{v_1}{v} = \frac{4R}{R} = 4$$ $$v_1 = 4v$$ Hence, option (A) is correct.