Two bodies of masses $m_{1}$ and $m_{2}$ are initially at rest at an 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 B ${[\frac{2 G}{r} (m_{1} + m_{2})]}_{1}^{1 / 2}$

(When allowed to move towards each other)
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
Decrease in P.E.=Increase 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)
On solving equation (i) and (ii)
$v_{1} = \sqrt{\frac{2 G m_{2}^{2}}{r (m_{1} + m_{2})}}$ and $v_{2} = \sqrt{\frac{2 G m_{1}^{2}}{r (m_{1} + m_{2})}}$
$∴ v_{app} = | v_{1} | + | v_{2} | = \sqrt{\frac{2 G}{r} (m_{1} + m_{2})}$

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Two bodies of masses m1 and m2 are initially at rest at an 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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