• # question_answer Two particles of mass m & M are initially at rest and infinitely separated from each other. Due to mutual interaction (gravitational) they approach each other. Their relative velocity of approach when they are at a distance d is A) $\sqrt{\frac{2Gd}{M+m}}$                        B) $\sqrt{\frac{2G(M+m)}{d}}$ C) $\sqrt{\frac{2GMm}{(M+m)d}}$                            D) $\sqrt{\frac{2Gm}{d}}$

1st method- Let us say the spheres are moving with velocities ${{v}_{1}}\And {{v}_{2}}$ when they are at a separation of d. Then from momentum conservation, $m{{v}_{1}}=M{{v}_{2}}$ From energy conservation, $\left( \frac{mv_{1}^{2}}{2}+\frac{Mv_{1}^{2}}{2} \right)-0=\left[ -\frac{GMm}{d}-0 \right]$ After solving above equation we get ${{v}_{1}}+{{v}_{2}}=\sqrt{\frac{2G(M+m)}{d}}$ II2nd  Method : Using C frame or reduced mass concept. Let ${{v}_{r}}$ be the relative velocity when they are at a separation of d. then,$\left( \frac{mM}{m+M} \right)\frac{v_{r}^{2}}{2}=-\left[ -\frac{GMm}{d}-0 \right]$ $\Rightarrow$${{v}_{r}}=\sqrt{\frac{2G(M+m)}{d}}$