Question

A solid spherical planet of mass $3m$ and radius 'R' has a very small tunnel along its diameter. A small cosmic particle of mass $2m$ is at a distance $2R$ from the center of the planet as shown. Both are initially at rest, and due to gravitational attraction, both start moving towards each other. After sometime, the cosmic particle passes through the centre of the planet. (Assume the planet and the cosmic particle are isolated from other planets)

• A. Displacement of the cosmic particle till that instant is $1.2R$.
• B. Acceleration of the cosmic particle at that instant is zero.
• C. Velocity of the cosmic particle at that instant is $\sqrt{\frac{8Gm}{3R}}$.
• D. Total work done by the gravitational force on both the particle is $+\frac{6G{m}^2}{R}$.

Answer 1

Answer: (A), (B), (D)

Applying momentum conservation.

$0=2m{v}_1-3m{v}_2$

$\Rightarrow v_2=\frac{2v_1}{3}............\left(i\right)$

From energy conservation,

$k_i+U_i=k_f+U_r$

$0+(-\frac{G3m}{2R}2m)=\frac{1}{2}2m{v}_1^2+\frac{1}{2}\left(3m\right)v_2^2$

$+(-\frac{3}{2}\frac{G\left(3m\right)}{R})\left(2m\right).........\left(ii\right)$

Solving eqn.(i) & (ii) get,

$v_1=\sqrt{\frac{18Gm}{5R}}$

(A) COM will be fixed and assume that x is displacement of cosmic particle

$S_{cm}=\frac{m_1{s}_1+m_2{s}_2}{m_1+m_2}\Rightarrow x=1.2R$

$\left(B\right)F_{net}=0\Rightarrow a=0$

$\left(D\right)W_{gr}=U\downarrow$

$\Rightarrow W_{gr}=(-\frac{G\left(3m\right)}{2R})2m-\left(\frac{3}{2}\frac{G\left(3m^2\right)}{R}\right)2m=\frac{6G{m}^2}{R}$