Gravitation JEE Main previous year questions with solutions

5 solved JEE Main questions on Gravitation, free to read — no sign-in needed. The full chapter has 221 questions; sign in to attempt the remaining 216 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 1)Acceleration Due to Gravity
   When one moves from a point $16$ km below the earth's surface to a point $16$ km above the earth's surface. The change in $g$ is approximately $\alpha$ %. The value of $\alpha$ is _____. (Take radius of the earth $= 6400$ km.)
   1. A.$0.12$
   2. B.$0.25$
   3. C.$0.50$
   4. D.$0.75$
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   Answer: (B)

   The acceleration due to gravity at a depth $d$ below the earth's surface is given by: $g_d = g\left(1 - \dfrac{d}{R}\right)$ The acceleration due to gravity at a height $h$ above the earth's surface (for $h \ll R$) is given by: $g_h = g\left(1 - \dfrac{2h}{R}\right)$ Given $d = 16$ km and $h = 16$ km, we can find the values of $g_d$ and $g_h$: $g_d = g\left(1 - \dfrac{16}{6400}\right) = g\left(1 - \dfrac{1}{400}\right)$ $g_h = g\left(1 - \dfrac{2 \times 16}{6400}\right) = g\left(1 - \dfrac{2}{400}\right)$ The magnitude of the change in acceleration due to gravity when moving from the depth $d$ to the height $h$ is: $\Delta g = g_d - g_h = g\left(1 - \dfrac{1}{400}\right) - g\left(1 - \dfrac{2}{400}\right)$ $\Delta g = g\left(\dfrac{2}{400} - \dfrac{1}{400}\right) = \dfrac{g}{400}$ The percentage change in $g$ is approximately: $\dfrac{\Delta g}{g} \times 100 = \dfrac{\dfrac{g}{400}}{g} \times 100 = \dfrac{100}{400} = 0.25\%$ Thus, the value of $\alpha$ is $0.25$. Answer: $0.25$
2. Q2JEE Main 2025 (24 Jan, Shift 2)Gravitational field
   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 $\ldots\ldots$ g.
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   Answer: 9

   $\because$ acceleration due to gravity on surface is given by $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^2}$ Now since diameter is reduced to $1 / 3^{\text {rd }}$, radius also reduces to $1 / 3^{\text {rd }}$, keeping mass constant New value of acceleration due to gravity on Earth's surface is $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\left(\frac{\mathrm{R}_{\mathrm{e}}}{3}\right)^2}=9 \frac{\mathrm{GMe}}{\mathrm{R}_{\mathrm{e}}^2}=9 \mathrm{~g}$
3. Q3JEE Main 2024 (09 Apr, Shift 1)Gravitational potential and Potential Energy
   An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $318.5 \mathrm{~km}$. From earth's surface to the orbit, the change in total mechanical energy of the ball is $x \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}$. The value of $x$ is (take $\left.\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km}\right)$ :
   1. A.10
   2. B.12
   3. C.9
   4. D.11
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   Answer: (D)

   $\begin{aligned} & \mathrm{h}=318.5 \approx\left(\frac{\mathrm{R}_{\mathrm{e}}}{20}\right) \\ & \mathrm{T} \cdot \mathrm{E}_{\mathrm{i}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}} \\ & \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\mathrm{h}\right)}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)} \\ & \Rightarrow \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-10 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}\end{aligned}$ Change in total mechanical energy $\begin{aligned} & =\mathrm{TE}_{\mathrm{f}}-\mathrm{TE}_{\mathrm{i}} \\ & =\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\operatorname{Re}}\left[1-\frac{10}{21}\right]=\frac{11 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \operatorname{Re}}\end{aligned}$
4. Q4JEE Main 2023 (08 Apr, Shift 2)Kepler laws
   The orbital angular momentum of a satellite is $L$, when it is revolving in a circular orbit at height $h$ from earth surface. If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be
   1. A.$8L$
   2. B.$9L$
   3. C.$4L$
   4. D.$3L$
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   Answer: (D)

   For a satellite revolving around the Earth in a circular orbit, it can be written that $\dfrac{m{v}^{2}}{r}=\dfrac{GMm}{{r}^{2}}...\left(1\right)$ ($r=R+h$ is the distance from the centre of the Earth) Simplify equation (1) to obtain the linear speed of the satellite. $v=\sqrt{\dfrac{GM}{r}}$ Thus, the angular momentum of the satellite is given by $\begin{matrix}L & = & mvr \\ & = & mr\sqrt{\dfrac{GM}{r}} \\ & = & m\sqrt{GMr}...\left(2\right)\end{matrix}$ Similarly, for distance $r'$ from the centre of the Earth, the angular momentum is given by $L'=m\sqrt{GMr'}...\left(3\right)$ Divide equation (3) by equation (2) to obtain the required angular momentum. $\begin{matrix}\dfrac{L'}{L} & = & \dfrac{m\sqrt{GMr'}}{m\sqrt{GMr}} \\ & = & \sqrt{\dfrac{r'}{r}}...\left(4\right)\end{matrix}$ Now, ${r}^{'}=r+8r=9r$ Substitute the known values of the parameters into equation (4) to obtain the required angular momentum. $\begin{matrix}\dfrac{L'}{L} & = & \sqrt{9} \\ & \Rightarrow & L'=3L\end{matrix}$
5. Q5JEE Main 2022 (27 Jul, Shift 1)Motion of satellite
   Two satellites $A$ and $B$ having masses in the ratio $4:3$ are revolving in circular orbits of radii $3r$ and $4r$ respectively around the earth. The ratio of total mechanical energy of $A$ to $B$ is
   1. A.$9:16$
   2. B.$16:9$
   3. C.$1:1$
   4. D.$4:3$
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   Answer: (B)

   Given that $\dfrac{{m}_{1}}{{m}_{2}}=\dfrac{4}{3},\dfrac{{r}_{1}}{{r}_{2}}=\dfrac{3}{4}$ Now the total energy of a satellite $TE=\dfrac{1}{2}m{{v}_{o}}^{2}+\left(\dfrac{-GMm}{r}\right)$ Since the orbital speed of a satellite ${v}_{o}=\sqrt{\dfrac{GM}{r}}$ $\Rightarrow TE=-\dfrac{GMm}{2r}\propto \dfrac{m}{r}$ $\Rightarrow \dfrac{T{E}_{1}}{T{E}_{2}}=\dfrac{{m}_{1}}{{m}_{2}}\cdot \dfrac{{r}_{2}}{{r}_{1}}=\dfrac{4}{3}\times \dfrac{4}{3}=\dfrac{16}{9}$