Two particles of masses $m_{1}$ and $m_{2}$ , initially at rest at infinite distance from each other, move under the action of mutual gravitational pull. Show that at any instant their relative velocity of approach is $\sqrt{2 G (m_{1} + m_{2}) / R}$ , where R is their separation at that instant.

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Solution

The gravitational force of attraction on $m_{1}$ due to $m_{2}$ at a separation r is
$F_{1} = \frac{G m_{1} m_{2}}{r^{2}}$
Therefore, the acceleration of $m_{1}$ is
$a_{1} = \frac{F_{1}}{m_{1}} = \frac{G m_{2}}{r^{2}}$
Similarly, the acceleration of $m_{2}$ due to $m_{1}$ is
$a_{2} = - \frac{G m_{1}}{r^{2}}$
The negative sign is put as $a_{2}$ is directed opposite to $a_{1}$ . The relative acceleration of approach is
$a = a_{1} - a_{2} = \frac{G (m_{1} + m_{2})}{r^{2}}$ (i)
If v is the relative velocity, then
$a = \frac{d v}{d t} = \frac{d v}{d r} \frac{d r}{d t}$
But $- d r / d t = v$ (negative sign shows that r decreases with increasing t).
$∴ a = - \frac{d v}{d r} v$ (ii)
$v d v = - \frac{G (m_{1} + m_{2})}{r^{2}} d r$
Integrating, we get
$\frac{v^{2}}{2} = \frac{G (m_{1} + m_{2})}{r} + C$
At $r = \infty, v = 0$ (given), and so $C = 0$ . Therefore,
$v^{2} = \frac{2 G (m_{1} + m_{2})}{r}$
Let $v = v_{R}$ when $r = R$ . Then
$v_{R} = \sqrt{\frac{2 (G m_{1} + m_{2})}{R}}$

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Two particles of masses m1 and m2, initially at rest at infinite distance from each other, move under the action of mutual gravitational pull. Show that at any instant their relative velocity of approach is √2G(m1+m2)/R, where R is their separation at that instant.

Two particles of masses $m_{1}$ and $m_{2}$ , initially at rest at infinite distance from each other, move under the action of mutual gravitational pull. Show that at any instant their relative velocity of approach is $\sqrt{2 G (m_{1} + m_{2}) / R}$ , where R is their separation at that instant.