Question

Two stars of masses m & 3m are released. The initial distance between them is r. Find the speed of approach when distance between them decreases to $\frac{r}{2}$.

• A. $\sqrt{\frac{8Gm}{r}}$
• B. $\sqrt{\frac{3Gm}{r}}$
• C. $\sqrt{\frac{2Gm}{r}}$
• D. $\sqrt{\frac{16Gm}{r}}$

Answer 1

The correct option is A $\sqrt{\frac{8Gm}{r}}$

By conservation of momentum, when distance between them is $\frac{r}{2}$, if speed of m is v , speed of 3m will be $\frac{v}{3}$ in opposite direction.

By energy conservation,

$KE+PE=KE+PE0-\frac{G\times m\times 3m}{r}=\frac{1}{2}m{v}^2+\frac{1}{2}3m\frac{v^2}{9}-\frac{Gm.3m}{\frac{r}{2}}-\frac{3G{m}^2}{r}=\frac{1}{2}m{v}^2(1+\frac{1}{3})-\frac{6G{m}^2}{r}\frac{3G{m}^2}{r}=\frac{1}{2}m{v}^2\left(\frac{4}{3}\right)\frac{3G{m}^2}{r}=\frac{2m{v}^2}{3}$ $v=\sqrt{\frac{9Gm}{2r}}and{v}_{approach}=v+\frac{v}{3}=\frac{4}{3}\sqrt{\frac{9Gm}{2r}}=\sqrt{\frac{16}{9}.\frac{9Gm}{2r}}=\sqrt{\frac{8Gm}{r}}$