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. The relative velocity of approach at a separation 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{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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Solution

The correct option is B ${[\frac{2 G}{r} (m_{1} + m_{2})]}_{1}^{1 / 2}$
$∵ C h a n g e i n P . E . = C h a n g e i n K . E .$
$\frac{1}{2} μ v^{2} = \frac{G m_{1} m_{2}}{r}$
where, $μ$ is reduced mass
$μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}$
$∴ \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} v^{2} = \frac{G m_{1} m_{2}}{r}$
$v = \sqrt{\frac{2 G}{r} (m_{1} + m_{2})}$

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