Question

A solid spherical planet of mass $2m$ and radius 'R' has a very small tunnel along its diameter. A small cosmic particle of mass $m$ is at a distance $2R$ from the centre of the planet as shown. Both are initially at rest, and due to gravitational attraction, both start moving toward each other. After some time, 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 $\frac{4R}{3}$
• 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 planet and particle is $-\frac{2{Gm}^2}{R}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Displacement of the cosmic particle till that instant is $\frac{4R}{3}$
B Acceleration of the cosmic particle at that instant is zero
C velocity of the cosmic particle at that instant is $\sqrt{\frac{8Gm}{3R}}$


Applying momentum conservation,
$$0={mv}_1-2{mv}_2$$
$\Rightarrow v_2=\frac{v_1}{2}$ -(i)
From energy conservation,
$$k_i+U_i=k_f+U_f$$
$0+(-\frac{G\left(2m\right)}{2R})m=\frac{1}{2}{mv}_1^2+\frac{1}{2}\left(2m\right)v_2^2+(-\frac{3}{2}\frac{G\left(2m\right)}{R})\left(m\right)$ --(ii)

Solving eqn. (i) $\&$ (ii) get,
$v_1=\sqrt{\frac{8Gm}{3R}}$
(A) COM will be fixed so,
$$\begin{array}{c}{S}_{cm}=\frac{m_1{s}_1+m_2{s}_2}{m_1+m_2}\\ 0=\frac{\left(m\right)\left(x\right)+\left(2m\right)(-(2R-x\left)\right)}{m+2m}\Rightarrow x=\frac{4R}{3}\end{array}$$
(B) $F_{\text{net}}=0\Rightarrow a=0$
(D) $W_{gr}=-\Delta U\downarrow \Rightarrow W_{gr}=(-\frac{G\left(2m\right)}{2R})m-(-\frac{3}{2}\frac{G\left(2m\right)}{R})m$