Question

Two particles of masses $m$ and $M$ are at infinity and at rest. Both particles move towards each other due to mutual interaction. When the distance between them is $d$, the relative velocity of approach will be

• A. ${\left[\frac{2Gd}{M+m}\right]}^{1/2}$
• B. ${\left[\frac{2Gd(M+m)}{d}\right]}^{1/2}$
• C. $2G{\left[\frac{M+m}{d}\right]}^{1/2}$
• D. $\frac{2Gd}{M+m}$

Answer 1

Answer: (B)

${\left[\frac{2Gd(M+m)}{d}\right]}^{1/2}$

Since only internal forces act, center of mass must be at rest.

Thus, $MV=mv$

Also, from energy conservation,

$\frac{P^2}{2M}+\frac{P^2}{2m}-\frac{GMm}{d}=0$

$P=Mm\sqrt{\frac{2G}{(M+m)d}}$

Thus,

$v=\frac{P}{m}=M\sqrt{\frac{2G}{(M+m)d}}$

$V=\frac{P}{M}=m\sqrt{\frac{2G}{(M+m)d}}$

Relative velocity is

$v_{rel}=v+V=\sqrt{\frac{2G(M+m)}{d}}$

Answer is option B.