Two bodies of masses $m_{1}$ and $m_{2}$ are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance $r$ between them is

{[2 G \frac{(m_{1} - m_{2})}{r}]}^{1 / 2}

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{[\frac{2 G}{r} (m_{1} + m_{2})]}_{1}^{1 / 2}

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{[\frac{r}{2 G (m_{1} m_{2})}]}^{1 / 2}

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{[\frac{2 G}{r} m_{1} m_{2}]}_{1}^{1 / 2}

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Solution

The correct option is D ${[\frac{2 G}{r} (m_{1} + m_{2})]}_{1}^{1 / 2}$
By applying law of conservation of momentum,
$m_{1} v_{1} - m_{2} v_{2} = 0 \Rightarrow m_{1} v_{1} = m_{2} v_{2}$ ........ (i)
Where $v_{1}$ and $v_{2}$ are the velocities of masses $m_{1}$ and $m_{2}$ at a distance $r$ from each other.
By conservation of energy,
Change in $P . E$ = change in $K . E .$
$\frac{G m_{1} m_{2}}{r} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$ ........ (ii)
Solving eqn. (i) and (ii) we get
$v_{1} = \sqrt{\frac{2 G m_{2}^{2}}{r (m_{1} + m_{2})}}$ and $v_{2} = \sqrt{\frac{2 G m_{1}^{2}}{r (m_{1} + m_{2})}}$
Relative velocity of approach, $v_{R}$
$= | v_{1} | + | v_{2} | = \sqrt{\frac{2 G}{r} (m_{1} + m_{2})}$

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