Two bodies of masses m-1 and m-2 are initially rest infinite distance apart- They are then allowed to move towards each other under mutual gravitational attraction- Their relative velocity of approach a separation distance r between them is displaystyleleft-frac-r-2G-m-1m-2-right-1-2-displaystyleleft-frac-2G-r-m-1m-2right-1-2-displaystyleleft-2Gfrac-m-1-m-2-r-right-1-2-displaystyleleft-frac-2G-r-m-1-m-2-right-1-2-

Question

Toppr Admin · Accepted Answer

By applying law of conservation of momentum,
$m_{1} v_{1} - m_{2} v_{2} = 0 \Rightarrow m_{1} v_{1} = m_{2} v_{2}$ ........ (i)
Where $v_{1}$ and $v_{2}$ are the velocities of masses $m_{1}$ and $m_{2}$ at a distance $r$ from each other.
By conservation of energy,
Change in $P . E$ = change in $K . E .$
$\frac{G m_{1} m_{2}}{r} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$ ........ (ii)
Solving eqn. (i) and (ii) we get
$v_{1} = \sqrt{\frac{2 G m_{2}^{2}}{r (m_{1} + m_{2})}}$ and $v_{2} = \sqrt{\frac{2 G m_{1}^{2}}{r (m_{1} + m_{2})}}$
Relative velocity of approach, $v_{R}$
$= | v_{1} | + | v_{2} | = \sqrt{\frac{2 G}{r} (m_{1} + m_{2})}$