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

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

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

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

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

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Solution

The correct option is B
$[\frac{2 G}{r} (m_{1} + m_{2})]^{\frac{1}{2}}$

Let velocities of these masses at r distance from each other be $v_{1}$ and $v_{2}$ respectively.
By conservation of momentum
$m_{1} v_{1} - m_{2} v_{2} = 0$
$\Rightarrow m_{1} v_{1} = m_{2} v_{2} . . . . (i)$
By conservation of energy
Change in P.E. = change in K.E.
$\frac{G m_{1} m_{2}}{r} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
$\Rightarrow \frac{m_{1}^{2} v_{1}^{2}}{m_{1}} + \frac{m_{2}^{2} v_{2}^{2}}{m_{2}} = \frac{2 G m_{1} m_{2}}{r} . . . . (i i)$
On solving equation (i) and (ii)
$v_{1} = \sqrt{\frac{2 G m_{2}^{2}}{r (m_{1} + m_{2})}} a n d v_{2} = \sqrt{\frac{2 G m_{1}^{2}}{r (m_{1} + m_{2})}}$
$∴ v_{a p p} = | v_{1} | + | v_{2} | = \sqrt{\frac{2 G}{r} (m_{1} + m_{2})}$

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Two bodies of masses M1 and M2 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

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