Gravitation JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Gravitation, free to read — no sign-in needed. The full chapter has 22 questions; sign in to attempt the remaining 17 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   A particle of mass $m$, and angular momentum $\ell$ is moving in a circular orbit of radius $r_0$ under the influence of an attractive force $\vec{F}(r) = -\dfrac{k}{r^2}\hat{r}$. Keeping its angular momentum unchanged, the particle is displaced radially by a small distance $\delta r \ll r_0$, due to which its radial distance varies periodically. The corresponding time period is:
   1. A.$\dfrac{2\pi\ell^3}{m k^2}$
   2. B.$2\pi\sqrt{\dfrac{m}{k}}$
   3. C.$\dfrac{2\pi\ell^3}{3 m k^2}$
   4. D.$\dfrac{2\pi\ell^3}{5 m k^2}$
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   Answer: (A)

   The equation of motion for the radial distance $r$ is given by: $m \dfrac{d^2r}{dt^2} = F(r) + \dfrac{\ell^2}{mr^3} = -\dfrac{k}{r^2} + \dfrac{\ell^2}{mr^3}$ For a circular orbit of radius $r_0$, the radial acceleration is zero: $-\dfrac{k}{r_0^2} + \dfrac{\ell^2}{mr_0^3} = 0 \Rightarrow r_0 = \dfrac{\ell^2}{mk}$ Let the particle be displaced by a small distance $x$ such that $r = r_0 + x$. The restoring force is: $m \dfrac{d^2x}{dt^2} = -\dfrac{k}{(r_0+x)^2} + \dfrac{\ell^2}{m(r_0+x)^3}$ Using the binomial expansion for $x \ll r_0$: $m \dfrac{d^2x}{dt^2} \approx -\dfrac{k}{r_0^2}\left(1 - \dfrac{2x}{r_0}\right) + \dfrac{\ell^2}{mr_0^3}\left(1 - \dfrac{3x}{r_0}\right)$ Since $\dfrac{k}{r_0^2} = \dfrac{\ell^2}{mr_0^3}$, the constant terms cancel out: $m \dfrac{d^2x}{dt^2} \approx \left( \dfrac{2k}{r_0^3} - \dfrac{3\ell^2}{mr_0^4} \right) x$ Substituting $\dfrac{\ell^2}{mr_0^4} = \dfrac{k}{r_0^3}$ into the equation: $m \dfrac{d^2x}{dt^2} = \left( \dfrac{2k}{r_0^3} - \dfrac{3k}{r_0^3} \right) x = -\dfrac{k}{r_0^3} x$ This represents simple harmonic motion with angular frequency $\omega = \sqrt{\dfrac{k}{mr_0^3}}$. The time period is $T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{mr_0^3}{k}}$. Substituting $r_0 = \dfrac{\ell^2}{mk}$: $T = 2\pi \sqrt{\dfrac{m}{k} \left( \dfrac{\ell^2}{mk} \right)^3} = 2\pi \sqrt{\dfrac{m \ell^6}{k m^3 k^3}} = \dfrac{2\pi \ell^3}{m k^2}$ Answer: $\dfrac{2\pi\ell^3}{m k^2}$
2. Q2JEE Advanced Adv 2023 (Paper 1)
   Two satellites $P$ and $Q$ are moving in different circular orbits around the Earth(radius $R$). The heights of $P$ and $Q$ from the Earth surface are ${h}_{P}$ and ${h}_{Q}$, respectively, where ${h}_{P}=\dfrac{R}{3}$. The accelerations of $P$ and $Q$ due to Earth’s gravity are ${g}_{P}$ and ${g}_{Q}$, respectively. If $\dfrac{{g}_{P}}{{g}_{Q}}=\dfrac{36}{25}$, what is the value of ${h}_{Q}$?
   1. A.$\dfrac{3R}{5}$
   2. B.$\dfrac{R}{6}$
   3. C.$\dfrac{6R}{5}$
   4. D.$\dfrac{5R}{5}$
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   Answer: (A)

   Acceleration due to gravity at any height from the Earth's surface is given by, $g=\dfrac{GM}{{r}^{2}}$, where $r=$distance from the centre of the Earth. Now, we can write $\dfrac{{g}_{P}}{{g}_{Q}}=\dfrac{\dfrac{GM}{{\left({r}_{P}\right)}^{2}}}{\dfrac{GM}{{\left({r}_{Q}\right)}^{2}}} \Rightarrow \dfrac{36}{25}={\left(\dfrac{{r}_{Q}}{{r}_{P}}\right)}^{2} \Rightarrow \dfrac{{r}_{Q}}{{r}_{P}}=\dfrac{6}{5} \Rightarrow {r}_{Q}=\dfrac{6}{5}\times \left(R+\dfrac{R}{3}\right) \Rightarrow R+{h}_{Q}=\dfrac{8}{5}R \Rightarrow {h}_{Q}=\dfrac{3}{5}R$
3. Q3JEE Advanced Adv 2022 (Paper 1)
   Two spherical stars $A$ and $B$ have densities ${\rho }_{A}$ and ${\rho }_{B}$, respectively. $A$ and $B$ have the same radius, and their masses ${M}_{A}$ and ${M}_{B}$ are related by ${M}_{B}=2{M}_{A}$. Due to an interaction process, star $A$ loses some of its mass, so that its radius is halved, while its spherical shape is retained, and its density remains ${\rho }_{A}$. The entire mass lost by $A$ is deposited as a thick spherical shell on $B$ with the density of the shell being ${\rho }_{A}$. If ${v}_{A}$ and ${v}_{B}$ are the escape velocities from $A$ and $B$ after the interaction process, the ratio $\dfrac{{v}_{B}}{{v}_{A}}=\sqrt{\dfrac{10n}{{15}^{\dfrac{1}{3}}}}$. The value of $n$ is
   Show answer & solution

   Answer: 2.3

   Given here: ${R}_{A}={R}_{B}=R$ and ${M}_{B}=2{M}_{A}$. Now, after interaction process, radius of remaining star $A$ is ${{R}_{A}}^{'}=\dfrac{R}{2}$ and its mass is ${{M}_{A}}^{'}={\rho }_{A}\dfrac{4}{3}\pi {\left(\dfrac{{R}_{A}}{2}\right)}^{3}=\dfrac{{M}_{A}}{8}$ Applying conservation of energy, $\dfrac{-G{{M}_{A}}^{'}m}{{{R}_{A}}^{'}}+\dfrac{1}{2}m{v}_{A}^{2}=0$ Escape velocity of star $A$ is ${v}_{A}=\sqrt{\dfrac{2G{M}_{A}}{8\times \left(\dfrac{R}{2}\right)}}=\dfrac{{v}_{0}}{2}$Now, for $B,$ mass collected over $B$ is ${{M}_{B}}^{'}={M}_{A}-\dfrac{{M}_{A}}{8}=\dfrac{7}{8}{M}_{A}$. Let the radius of star $B$ after interaction becomes $r$. Applying mass conservation, $\dfrac{4}{3}\pi \left({r}^{3}-{R}^{3}\right){\rho }_{A}=\dfrac{4}{3}\pi {R}^{3}\times \dfrac{7}{8}{\rho }_{A}$ $\Rightarrow r={\left(\dfrac{15}{8}\right)}^{\dfrac{1}{3}}R$ Escape velocity of star $B$ is $∴{v}_{B}=\sqrt{\dfrac{2G\times \left(2{M}_{A}+\dfrac{7}{8}{M}_{A}\right)}{{\left(15\right)}^{1/3}\left(\dfrac{R}{2}\right)}}$ $=\sqrt{\dfrac{2G{M}_{A}}{R}}\sqrt{\dfrac{2\left(\dfrac{16+7}{8}\right)}{{\left(15\right)}^{\dfrac{1}{3}}}}$ $={v}_{0}\times \sqrt{\dfrac{23\times 2}{8\times {\left(15\right)}^{\dfrac{1}{3}}}}=\dfrac{{v}_{0}}{2}\times \sqrt{\dfrac{23}{{\left(15\right)}^{\dfrac{1}{3}}}}$ Now, the ratio $\dfrac{{v}_{B}}{{v}_{A}}=\sqrt{\dfrac{23}{{\left(15\right)}^{\dfrac{1}{3}}}}=\sqrt{\dfrac{2.30\times 10}{{\left(15\right)}^{\dfrac{1}{3}}}}$ $∴n=2.30$
4. Q4JEE Advanced Adv 2021 (Paper 2)
   The distance between two stars of masses $3{M}_{S}$ and $6{M}_{S}$ is $9R$. Here $R$ is the mean distance between the centres of the Earth and the Sun, and ${M}_{S}$ is the mass of the Sun. Two stars orbit around their common centre of mass in circular orbits with period $nT$, where$T$ is the period of Earth's revolution around the Sun. The value of $n$ is _____.
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   Answer: 9

   $T=\dfrac{2\pi {R}^{3/2}}{\sqrt{G{M}_{S}}}$ For binary stars, $T'=\dfrac{2\pi {r}^{3/2}}{\sqrt{G\left({M}_{1}+{M}_{2}\right)}}$ Put ${M}_{1}=3{M}_{S}$,${M}_{2}=6{M}_{S}$ and $r=9R$ $T'=\dfrac{2\pi \times {9}^{3/2}\times {R}^{3/2}}{\sqrt{G\left(3{M}_{S}+6{M}_{S}\right)}}=\dfrac{2\pi \times {9}^{3/2}\times {R}^{3/2}}{{9}^{1/2}\sqrt{G{M}_{S}}}=9\left(\dfrac{2\pi {R}^{3/2}}{\sqrt{G{M}_{S}}}\right)$ $∴T'=9T$ Value of $n=9$.
5. Q5JEE Advanced Adv 2019 (Paper 1)
   Consider a spherical gaseous cloud of mass density $\rho \left(r\right)$ in free space where $r$ is the radial distance from its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common center with the same kinetic energy $K$ . The force acting on the particles is their mutual gravitational force. If $\rho \left(r\right)$ is constant in time, the particle number density $n\left(r\right)=\rho \left(r\right)/m$ is: [ $G$ is universal gravitational constant]
   1. A.$\dfrac{K}{\pi {r}^{2}{m}^{2}G}$
   2. B.$\dfrac{3K}{\pi {r}^{2}{m}^{2}G}$
   3. C.$\dfrac{K}{6\pi {r}^{2}{m}^{2}G}$
   4. D.$\dfrac{K}{2\pi {r}^{2}{m}^{2}G}$
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   Answer: (D)

   Given that all the particles of gaseous cloud has mass $‘m’$ and are moving due to mutual attraction with same kinetic energy $‘K’$ . Now in that cloud, let us consider a sphere of radius $‘r’$ containing a group of particles adding to a total mass $‘M’$ . Now, another particle of mass $‘m’$ is moving with speed $‘V’$ due to gravitational attraction of $M,$ in a circle of radius $‘r’$ . $∴$ Centripetal force is here given by $‘M’$ $\Rightarrow \dfrac{GM\cdot m}{{r}^{2}}=\dfrac{m{v}^{2}}{r}$ ....(1) But, kinetic energy, $K=\dfrac{1}{2}m{v}^{2}$ $\Rightarrow m{v}^{2}=2K$ Put it in $(1)$ $\Rightarrow \dfrac{GMm}{{r}^{2}}=\dfrac{2K}{r}$ $\Rightarrow M=\dfrac{2Kr}{Gm}$ Differentiating it we get, $dM=\dfrac{2K}{Gm}\cdot dr$ .....(2) Now, if density of cloud is $'{\rho }^{'},$ Then elementary mass, $dM=\left(4\pi {r}^{2}dr\right)\cdot \rho =\dfrac{2K}{Gm}\cdot dr$ $\Rightarrow \rho =\dfrac{K}{2\pi Gm{r}^{2}}$