Two particles of mass m and M are initially at rest at infinite distance. Find their relative velocity of approach due to gravitational attraction when d is their separation at any instant

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$\sqrt{\frac{2 G (M + m)}{d}}$

$\sqrt{\frac{G (M + m)}{2 d}}$

$\sqrt{\frac{G (M + m)}{d}}$

$\sqrt{\frac{G (M + m)}{4 d}}$

answer is A.

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Detailed Solution

$\frac{GMm}{d} = \frac{1}{2} {mv}_{1}^{2} + \frac{1}{2} {mv}_{2}^{2} \frac{2 GMm}{d} = {mv}_{1}^{2} + {mv}_{2}^{2}$
Velocity of centre of mass of the two objects is zero

${\vec{v}}_{CM} = \frac{{mv}_{1} - {Mv}_{2}}{m + M} {mv}_{1} = {Mv}_{2} \frac{m}{M} = \frac{v_{2}}{v_{1}} \frac{m + M}{M} = \frac{v_{2} + v_{1}}{v_{1}}$
Relative velocity of approach
$v_{2} + v_{1} = (\frac{m + M}{M}) v_{1}$
Initially the objects are at infinite distance and at rest
So, $K . E + P . E = 0 \to (1)$
When the separation between the masses is d
$K . E + P . E = \frac{1}{2} {mv}_{1}^{2} + \frac{1}{2} {Mv}_{2}^{2} - \frac{GMm}{d} \to (2)$
From (1) & (2)
$\frac{1}{2} {mv}_{1}^{2} + \frac{1}{2} {Mv}_{2}^{2} = \frac{GMm}{d} {mv}_{1}^{2} + \frac{1}{2} {Mv}_{2}^{2} = \frac{2 GMm}{d} {mv}_{1}^{2} + M {(\frac{{Mv}_{1}}{M})}^{2} = \frac{2 GMm}{d} {mv}_{1}^{2} + M \frac{m^{2} v_{1}^{2}}{M^{2}} = \frac{2 GMm}{d} v_{1}^{2} \frac{(M + m)}{M} = \frac{2 GM}{d} v_{1}^{2} = \frac{2 {GM}^{2}}{d (M + m)} v_{1} = M \sqrt{\frac{2 G}{d (M + m)}} v_{2} + v_{1} = \frac{m + M}{M} \times M \sqrt{\frac{2 G}{d (M + m)}} v_{2} + v_{1} = \sqrt{\frac{2 G (M + m)}{d}}$