Two masses $M_{1}$ & $M_{2}$ are initially at rest and are separated by a very large distance. If the masses approach each other subsequently due to gravitational attraction between them, their relative velocity of approach at a separation $d$ is

\sqrt{\frac{2 G (M_{1} + M_{2})}{d}}

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\sqrt{\frac{4 G (M_{1} + M_{2})}{d}}

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\sqrt{\frac{4 G (M_{1} M_{2})}{d}}

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\sqrt{\frac{G (M_{1} + M_{2})}{d}}

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Solution

The correct option is A $\sqrt{\frac{2 G (M_{1} + M_{2})}{d}}$

Relative velocity of $M_{1}$ w.r.t. $M_{2}$ or vice-versais,
$v_{r e l} = v_{1} + v_{2}$ ........(1)
Initial energy of system,
$E_{i} = 0$ ,
Final energy of system,
$E_{f} = \frac{- G M_{1} M_{2}}{d} + \frac{M_{1} v_{1}^{2}}{2} + \frac{M_{2} v_{2}^{2}}{2}$
From conservation of energy,
$E_{i} = E_{f}$
$\frac{- G M_{1} M_{2}}{d} + \frac{M_{1} v_{1}^{2}}{2} + \frac{M_{2} v_{2}^{2}}{2} = 0$ ..........(2)
By momentum conservation,
$M_{1} v_{1} - M_{2} v_{2} = 0$
$v_{2} = \frac{M_{1}}{M_{2}} v_{1}$
Substituting the value of $v_{2}$ in (2)
$\frac{(M_{1})^{2} (v_{1})^{2}}{2 M_{2}} + \frac{M_{1} (v_{1})^{2}}{2} = \frac{G M_{1} M_{2}}{d}$
$v_{1} = \sqrt{\frac{2 G (M_{2})^{2}}{d (M_{1} + M_{2})}}$
$v_{2} = \sqrt{\frac{2 G (M_{1})^{2}}{d (M_{1} + M_{2})}}$
Substituting the value of $v_{1}, v_{2}$ in (1) we get
$v_{r e l} = \sqrt{\frac{2 G (M_{1} + M_{2})}{d}}$

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