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)

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velocity of the cosmic particle at that instant is $\sqrt{\frac{8 Gm}{3 R}}$

Total work done by the gravitational force on both the particle is $- \frac{2 {Gm}^{2}}{R}$

Acceleration of the cosmic particle at that instant is zero

Displacement of the cosmic particle till that instant is $\frac{4 R}{3}$

answer is C.

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Detailed Solution

Applying momentum conservation,
0 =mv₁ - 2mv₂
$\Rightarrow v_{2} = \frac{v_{1}}{2} . . . . (i)$
From energy conservation,
$\begin{array}{l} k_{i} + U_{i} = k_{f} + U_{f} \\ 0 + (- \frac{G (2 m)}{2 R}) m = \frac{1}{2} {mv}_{1}^{2} + \frac{1}{2} (2 m) v_{2}^{2} + (- \frac{3}{2} \frac{G (2 m)}{R}) (m) . . . . (ii) \end{array}$
Solving eqn.(i) & (ii) get,
$v_{1} = \sqrt{\frac{8 Gm}{3 R}}$
(A) COM will be fixed so,
$s_{cm} = \frac{m_{1} s_{1} + m_{2} s_{2}}{m_{1} + m_{2}}$
$\begin{array}{l} 0 = \frac{(m) (x) + (2 m) (- (2 R - x))}{m + 2 m} \Rightarrow x = \frac{4 R}{3} \\ (B) F_{net} = 0 \Rightarrow a = 0 \\ (D) W_{gr} = U ↓\Rightarrow W_{gr} = (- \frac{G (2 m)}{2 R}) m - (- \frac{3}{2} \frac{G (2 m)}{R}) m \end{array}$