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📘 Gravitation — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Gravitation for JEE Main with previous year questions on orbital motion, escape velocity, and satellites. Each solution is explained in a simple, step-by-step manner to boost your score.
Gravitation — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (7 April Shift 2)
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant.
Reason (R): For a central force field the angular momentum is a constant.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(2) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(3) (A) is correct but (R) is not correct
(4) (A) is not correct but (R) is correct
Assertion (A): The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant.
Reason (R): For a central force field the angular momentum is a constant.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(2) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(3) (A) is correct but (R) is not correct
(4) (A) is not correct but (R) is correct
✅ Answer & Explanation:
Correct Option: (1)
Kepler's Second Law (Areal Velocity) is a direct consequence of the conservation of angular momentum. Since gravitational force is a central force, the torque is zero and angular momentum remains constant, keeping the areal velocity \(\left(\frac{dA}{dt} = \frac{L}{2m}\right)\) constant.
Kepler's Second Law (Areal Velocity) is a direct consequence of the conservation of angular momentum. Since gravitational force is a central force, the torque is zero and angular momentum remains constant, keeping the areal velocity \(\left(\frac{dA}{dt} = \frac{L}{2m}\right)\) constant.
Q2. JEE Main 2025 (3 April Shift 1)
Three identical spheres of mass \(m\), are placed at the vertices of an equilateral triangle of length \(a\). When released, they interact only through gravitational force and collide after a time \(T = 4\) seconds. If the sides of the triangle are increased to length \(2a\) and also the masses of the spheres are made \(2m\), then they will collide after _____ seconds.
✅ Answer & Explanation:
Correct Answer: 8
- Collision time scales according to the relationship: \(T \propto \sqrt{\frac{a^3}{GM}}\).
- When the sides increase to \(2a\) and masses increase to \(2m\): \(T' = \sqrt{\frac{(2a)^3}{G(2m)}} = \sqrt{\frac{8a^3}{2Gm}} = \sqrt{4 \cdot \frac{a^3}{Gm}} = 2\sqrt{\frac{a^3}{GM}}\).
- Since the original time \(T = 4\) seconds: \(T' = 2 \times 4 = 8\text{ seconds}\).
Q3. JEE Main 2025 (22 Jan Shift 1)
A small point of mass \(m\) is placed at a distance \(2R\) from the centre '\(O\)' of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on '\(m\)' due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure and the gravitational force on \(m\) due to remaining part of \(M\) is found to be \(F_2\). The value of ratio \(F_1 : F_2\) is :
(1) 12 : 11 (2) 11 : 10
(3) 12 : 9 (4) 16 : 9
(3) 12 : 9 (4) 16 : 9
✅ Answer & Explanation:
Correct Option: (1)
- Initial gravitational force: \(F_1 = \frac{G M m}{(2R)^2} = \frac{G M m}{4R^2}\).
- Mass of the removed cavity part: \(m' = M \left(\frac{R/3}{R}\right)^3 = \frac{M}{27}\).
- Distance from the center of the cavity to mass \(m\): \(x = 2R - \left(R - \frac{R}{3}\right) = 2R - \frac{2R}{3} = \frac{4R}{3}\).
- Gravitational force due to the removed mass part: \(F_{\text{removed}} = \frac{G (M/27) m}{(4R/3)^2} = \frac{G M m}{27 \cdot \frac{16R^2}{9}} = \frac{G M m}{48R^2}\).
- Force due to the remaining part: \(F_2 = F_1 - F_{\text{removed}} = \frac{G M m}{4R^2} - \frac{G M m}{48R^2} = \frac{11 G M m}{48R^2}\).
Q4. JEE Main 2024 (31 Jan Shift 1)
Four identical particles of mass \(m\) are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is \((\frac{2\sqrt{2}+1}{32}) \frac{Gm^2}{L^2}\), the length of the sides of the square is :
(1) \(L/2\) (2) \(4L\)
(3) \(3L\) (4) \(2L\)
(1) \(L/2\) (2) \(4L\)
(3) \(3L\) (4) \(2L\)
✅ Answer & Explanation:
Correct Option: (2)
- Net force on a corner mass in a square of side \(a\): \(F = \frac{Gm^2}{a^2}\left(\sqrt{2} + \frac{1}{2}\right)\).
- Given \(F = \frac{Gm^2}{L^2} \left(\frac{2\sqrt{2}+1}{32}\right) = \frac{Gm^2}{32L^2}\left(2\sqrt{2}+1\right)\).
- Equating: \(\frac{1}{a^2}\left(\frac{2\sqrt{2}+1}{2}\right) = \frac{2\sqrt{2}+1}{32L^2} \implies 2a^2 = 32L^2 \implies a = 4L\).
Q5. JEE Main 2024 (05 Apr Shift 1)
In hydrogen like system the ratio of coulombic force and gravitational force between an electron and a proton is in the order of :
(1) \(10^{39}\) (2) \(10^{29}\)
(3) \(10^{19}\) (4) \(10^{36}\)
(1) \(10^{39}\) (2) \(10^{29}\)
(3) \(10^{19}\) (4) \(10^{36}\)
✅ Answer & Explanation:
Correct Option: (1)
The electrostatic force is significantly stronger than gravity. For an electron and proton, the ratio \(\frac{k e^2}{G m_e m_p} \approx 2.4 \times 10^{39}\).
The electrostatic force is significantly stronger than gravity. For an electron and proton, the ratio \(\frac{k e^2}{G m_e m_p} \approx 2.4 \times 10^{39}\).
Q6. JEE Main 2022 (26 Jun Shift 1)
The variation of acceleration due to gravity (g) with distance (r) from the center of the earth is correctly represented by (Given R = radius of earth) :
Choose correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (1)
Inside the earth (\(r < R\)), \(g \propto r\) (linear increase). Outside the earth (\(r > R\)), \(g \propto \frac{1}{r^2}\) (inverse-square decrease). Graph (1) represents this behaviour.
Inside the earth (\(r < R\)), \(g \propto r\) (linear increase). Outside the earth (\(r > R\)), \(g \propto \frac{1}{r^2}\) (inverse-square decrease). Graph (1) represents this behaviour.
Q7. JEE Main 2025 (4 April Shift 1)
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The kinetic energy needed to project a body of mass \(m\) from earth surface to infinity is \(\frac{1}{2} mgR\), where R is the radius of earth.
Reason (R): The maximum potential energy of a body is zero when it is projected to infinity from earth surface.
In the light of the above statements, choose the correct answer from the option given below :
(1) (A) False but (R) is true
(2) Both (A) and (R) are true and (R) is the correct explanation of (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
Assertion (A): The kinetic energy needed to project a body of mass \(m\) from earth surface to infinity is \(\frac{1}{2} mgR\), where R is the radius of earth.
Reason (R): The maximum potential energy of a body is zero when it is projected to infinity from earth surface.
In the light of the above statements, choose the correct answer from the option given below :
(1) (A) False but (R) is true
(2) Both (A) and (R) are true and (R) is the correct explanation of (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
✅ Answer & Explanation:
Correct Option: (1)
- Assertion: False. The escape energy (KE) needed is \(\frac{GMm}{R}\), which is equal to \(mgR\), not \(\frac{1}{2}mgR\).
- Reason: True. Gravitational potential energy is defined such that it is zero at infinity, which is its maximum value.
Q8. JEE Main 2025 (24 Jan Shift 2)
Acceleration due to gravity on the surface of earth is '\(g\)'. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is _____ \(g\).
✅ Answer & Explanation:
Correct Answer: 9
- \(g = \frac{GM}{R^2}\). If the diameter (and thus radius) becomes \(1/3\), then: \(g' = \frac{GM}{(R/3)^2} = 9 \frac{GM}{R^2} = 9g\).
Q9. JEE Main 2024 (29 Jan Shift 1)
At what distance above and below the surface of the earth a body will have same weight? (Take radius of earth as \(R\))
(1) \(\sqrt{5}R - R\) (2) \(\frac{\sqrt{3}R-R}{2}\)
(3) \(R/2\) (4) \(\frac{\sqrt{5}R-R}{2}\)
(1) \(\sqrt{5}R - R\) (2) \(\frac{\sqrt{3}R-R}{2}\)
(3) \(R/2\) (4) \(\frac{\sqrt{5}R-R}{2}\)
✅ Answer & Explanation:
Correct Option: (4)
- Below the surface at depth \(h\): \(g_d = g\left(1 - \frac{h}{R}\right)\).
- Above the surface at height \(h\): \(g_h = g \frac{R^2}{(R+h)^2}\).
- Equating and solving the quadratic equation for \(h\) yields \(h = \frac{\sqrt{5}R - R}{2}\).
Q10. JEE Main 2023 (01 Feb Shift 1)
If earth has a mass nine times and radius twice to that of a planet P. Then \(\frac{v_e}{3}\sqrt{x} \, \text{m/s}\) will be the minimum velocity required by a rocket to pull out of gravitational force of P, where \(v_e\) is escape velocity on earth. The value of \(x\) is :
(1) 2 (2) 3
(3) 18 (4) 1
(1) 2 (2) 3
(3) 18 (4) 1
✅ Answer & Explanation:
Correct Option: (1)
- \(v_e = \sqrt{\frac{2GM}{R}}\).
- Given \(M_e = 9M_p\) and \(R_e = 2R_p\).
- \(v_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2G(M_e/9)}{R_e/2}} = \sqrt{\frac{2}{9} \cdot \frac{2GM_e}{R_e}} = \frac{\sqrt{2}}{3} v_e\).
- Comparing with \(\frac{v_e}{3}\sqrt{x}\), we find \(x = 2\).
Q11. JEE Main 2022 (29 Jun Shift 1)
The escape velocity of a body on a planet A is \(12 \, \text{km/s}\). The escape velocity of the body on another planet B, whose density is four times and radius is half of the planet A, is :
(1) \(12 \, \text{km/s}\) (2) \(24 \, \text{km/s}\)
(3) \(36 \, \text{km/s}\) (4) \(6 \, \text{km/s}\)
(1) \(12 \, \text{km/s}\) (2) \(24 \, \text{km/s}\)
(3) \(36 \, \text{km/s}\) (4) \(6 \, \text{km/s}\)
✅ Answer & Explanation:
Correct Option: (1)
- \(v_e \propto R\sqrt{\rho}\).
- \(\frac{v_B}{v_A} = \frac{R_B}{R_A} \sqrt{\frac{\rho_B}{\rho_A}} = \frac{1}{2} \sqrt{4} = 1\).
- So, \(v_B = 12 \, \text{km/s}\).
Q12. JEE Main 2021 (16 Mar Shift 2)
If one wants to remove all the mass of the earth to infinity in order to break it up completely. The amount of energy that needs to be supplied will be \(\frac{x}{5} \frac{GM^2}{R}\) where \(x\) is _____.
✅ Answer & Explanation:
Correct Answer: 3
- The binding energy of a uniform solid sphere is \(|U| = \frac{3}{5} \frac{GM^2}{R}\).
- Thus \(x = 3\).
Q13. JEE Main 2025 (29 Jan Shift 2)
Two planets, A and B are orbiting a common star in circular orbits of radii \(R_A\) and \(R_B\), respectively, with \(R_B = 2R_A\). The planet B is \(4\sqrt{2}\) times more massive than planet A. The ratio \((L_B/L_A)\) of angular momentum of planet B to that of planet A is closest to integer ________.
✅ Answer & Explanation:
Correct Answer: 8
- Orbital velocity \(v \propto \frac{1}{\sqrt{r}}\).
- Angular momentum \(L = mvr = m \sqrt{\frac{GM}{r}} \cdot r = m\sqrt{GMr}\).
- Ratio \(\frac{L_B}{L_A} = \frac{m_B}{m_A} \sqrt{\frac{R_B}{R_A}} = (4\sqrt{2}) \sqrt{2} = 4 \times 2 = 8\).
Q14. JEE Main 2025 (24 Jan Shift 1)
A satellite is launched into a circular orbit of radius 'R' around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by :
(1) 9% (2) 3%
(3) 4.5% (4) 2.5%
(1) 9% (2) 3%
(3) 4.5% (4) 2.5%
✅ Answer & Explanation:
Correct Option: (3)
- Using \(T^2 \propto R^3 \implies T \propto R^{3/2}\).
- Fractional change \(\frac{\Delta T}{T} = \frac{3}{2} \frac{\Delta R}{R}\).
- Percentage change \(= \frac{3}{2} (3\%) = 4.5\%\).
Q15. JEE Main 2025 (23 Jan Shift 2)
A satellite of mass \(M/2\) is revolving around earth in a circular orbit at a height of \(R/3\) from earth surface. The angular momentum of the satellite is \(M\sqrt{\frac{GMR}{x}}\). The value of \(x\) is _____, where \(M\) and \(R\) are the mass and radius of earth, respectively.
✅ Answer & Explanation:
Correct Answer: 3
- Orbit radius \(r = R + R/3 = \frac{4R}{3}\).
- \(L = m v r = \frac{M}{2} \sqrt{\frac{GM}{r}} \cdot r = \frac{M}{2} \sqrt{GMr}\).
- \(L = \frac{M}{2} \sqrt{GM(4R/3)} = \frac{M}{2} \cdot 2 \sqrt{\frac{GMR}{3}} = M \sqrt{\frac{GMR}{3}}\).
- Comparing with the given expression, \(x = 3\).
Q16. JEE Main 2024 (27 Jan Shift 2)
Assertion (A): The angular speed of the moon in its orbit about the earth is more than the angular speed of the earth in its orbit about the sun.
Reason (R): The moon takes less time to move around the earth than the time taken by the earth to move around the sun.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) (A) is correct but (R) is not correct
(2) Both (A) and (R) are correct and (R) is the correct explanation of A.
(3) Both (A) and (R) are correct but (R) is not the correct explanation of A.
(4) (A) is not correct but (R) is correct
Reason (R): The moon takes less time to move around the earth than the time taken by the earth to move around the sun.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) (A) is correct but (R) is not correct
(2) Both (A) and (R) are correct and (R) is the correct explanation of A.
(3) Both (A) and (R) are correct but (R) is not the correct explanation of A.
(4) (A) is not correct but (R) is correct
✅ Answer & Explanation:
Correct Option: (2)
Angular speed \(\omega = 2\pi/T\). Since the Moon's orbital period (~27 days) is much smaller than the Earth's orbital period (~365 days), the Moon's angular speed is much greater.
Angular speed \(\omega = 2\pi/T\). Since the Moon's orbital period (~27 days) is much smaller than the Earth's orbital period (~365 days), the Moon's angular speed is much greater.
Q17. JEE Main 2024 (09 Apr Shift 2)
A satellite of \(10^3 \, \text{kg}\) mass is revolving in circular orbit of radius \(2R\). If \(10^4 R/6 \, \text{J}\) energy is supplied to the satellite, it would revolve in a new circular orbit of radius : (use \(g = 10 \, \text{m/s}^2, R = \) radius of earth)
(1) \(2.5R\) (2) \(3R\)
(3) \(4R\) (4) \(6R\)
(1) \(2.5R\) (2) \(3R\)
(3) \(4R\) (4) \(6R\)
✅ Answer & Explanation:
Correct Option: (4)
- Total Energy \(E = -\frac{GMm}{2r}\).
- Initial \(E_1 = -\frac{mgR^2}{4R} = -\frac{mgR}{4}\).
- Energy supplied \(\Delta E = \frac{mgR}{6}\).
- Final \(E_2 = -\frac{mgR}{4} + \frac{mgR}{6} = -\frac{mgR}{12}\).
- New orbit: \(-\frac{mgR^2}{2r'} = -\frac{mgR}{12} \implies 2r' = 12R \implies r' = 6R\).
Q18. JEE Main 2024 (08 Apr Shift 1)
Two planets A and B having masses \(m_1\) and \(m_2\) move around the sun in circular orbits of \(r_1\) and \(r_2\) radii respectively. If angular momentum of A is L and that of B is 3 L, the ratio of time period \(T_A/T_B\) is:
(1) \((r_2/r_1)^{3/2}\) (2) \((r_1/r_2)^{3/2}\)
(3) \(27(m_1/m_2)^3\) (4) \((r_1/r_2)^3\)
(1) \((r_2/r_1)^{3/2}\) (2) \((r_1/r_2)^{3/2}\)
(3) \(27(m_1/m_2)^3\) (4) \((r_1/r_2)^3\)
✅ Answer & Explanation:
Correct Option: (2)
By Kepler's Third Law, the ratio of time periods depends only on the radii of the orbits: \(\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2}\). Mass and angular momentum do not affect the period for planets orbiting the same central star.
By Kepler's Third Law, the ratio of time periods depends only on the radii of the orbits: \(\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2}\). Mass and angular momentum do not affect the period for planets orbiting the same central star.
Q19. JEE Main 2024 (05 Apr Shift 1)
Match List I with List II :
(1) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(3) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(4) (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
| LIST I | LIST II |
|---|---|
| A. Kinetic energy of planet | I. \(-GMm/a\) |
| B. Gravitation Potential energy of sun-planet system | II. \(GMm/2a\) |
| C. Total mechanical energy of planet | III. \(G m/r\) |
| D. Escape energy at surface of planet for unit mass | IV. \(-GMm/2a\) |
(1) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(3) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(4) (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
✅ Answer & Explanation:
Correct Option: (1)
- KE: \(\frac{GMm}{2a}\) [II]
- PE: \(-\frac{GMm}{a}\) [I]
- Total Energy: PE + KE = \(-\frac{GMm}{2a}\) [IV]
- Escape Potential (per unit mass): \(\frac{GM}{r}\) [III]
Q20. JEE Main 2020 (09 Jan Shift 1)
A body A of mass \(m\) is moving in a circular orbit of radius \(R\) about a planet. Another body B of mass \(m/2\) collides with A with a velocity which is half (\(v/2\)) the instantaneous velocity \(v\) of A. The collision is completely inelastic. Then, the combined body:
(1) continues to move in a circular orbit
(2) Escapes from the Planet's Gravitational field
(3) Falls vertically downwards towards the planet
(4) starts moving in an elliptical orbit around the planet
(1) continues to move in a circular orbit
(2) Escapes from the Planet's Gravitational field
(3) Falls vertically downwards towards the planet
(4) starts moving in an elliptical orbit around the planet
✅ Answer & Explanation:
Correct Option: (4)
The new velocity of the combined mass will be less than the circular orbital velocity but greater than zero. This resulting energy state corresponds to an elliptical orbit.
The new velocity of the combined mass will be less than the circular orbital velocity but greater than zero. This resulting energy state corresponds to an elliptical orbit.
Q21. JEE Main 2020 (07 Jan Shift 1)
A satellite of mass \(M\) is launched vertically upwards with an initial speed \(u\) from the surface of the earth. After it reaches height \(R\) (\(R = \) radius of the earth), it ejects a rocket of mass \(M/10\) so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is :
(1) \(\frac{M}{20} (u^2 + \frac{113}{200} \frac{GM_e}{R})\) (2) \(5M (u^2 - \frac{119}{200} \frac{GM_e}{R})\)
(3) \(\frac{3M}{8} (u + \sqrt{\frac{5GM_e}{6R}})^2\) (4) \(\frac{M}{20} (u - \sqrt{\frac{2GM_e}{3R}})^2\)
(1) \(\frac{M}{20} (u^2 + \frac{113}{200} \frac{GM_e}{R})\) (2) \(5M (u^2 - \frac{119}{200} \frac{GM_e}{R})\)
(3) \(\frac{3M}{8} (u + \sqrt{\frac{5GM_e}{6R}})^2\) (4) \(\frac{M}{20} (u - \sqrt{\frac{2GM_e}{3R}})^2\)
✅ Answer & Explanation:
Correct Option: (2)
Solving this through conservation of energy (to find velocity at height R) followed by conservation of momentum (at the ejection point) leads to the kinetic energy expression in Option (2).
Solving this through conservation of energy (to find velocity at height R) followed by conservation of momentum (at the ejection point) leads to the kinetic energy expression in Option (2).
📘 Exam Preparation Tip for Gravitation
- Newton's Law of Gravitation: \(F = \frac{Gm_1m_2}{r^2}\). Remember it's an inverse-square law and always attractive. Gravitational force is a central force.
- Acceleration due to Gravity: \(g = \frac{GM}{R^2}\) on the surface. Inside the earth (\(r < R\)), \(g \propto r\). Outside (\(r > R\)), \(g \propto 1/r^2\).
- Escape Velocity: \(v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}\). This is the minimum speed required to escape a planet's gravitational field.
- Orbital Velocity: For a satellite in circular orbit: \(v_o = \sqrt{\frac{GM}{r}}\). The time period is \(T = 2\pi \sqrt{\frac{r^3}{GM}}\).
- Kepler's Laws: Memorize all three laws. The second law (areal velocity) is a consequence of angular momentum conservation. The third law (\(T^2 \propto r^3\)) is crucial for comparing orbits.
- Energy of a Satellite: \(E = -\frac{GMm}{2r}\) for circular orbit. \(PE = -\frac{GMm}{r}\) and \(KE = \frac{GMm}{2r}\). Total energy is negative for bound orbits.
- Binding Energy: Energy required to break a system into individual parts at infinity. For a uniform sphere, it's \(\frac{3GM^2}{5R}\).