Two particles of masses m and M are initially at rest and infinitely separated from each other. Due to gravitational attraction the particles approach each other. Their relative velocity of approach at a separation r between them is
Answer & explanation
Correct answer: option 2
Suppose at an instant the masses m and M are at x distance apart then gravitational force of attraction between them is
F = $\frac{GMm}{x^2}$
$a_{m}=\frac{GM}{x^2} ; a_{m}=\frac{GM}{x^2}$
Net acceleration of approach is
$a=a_{m}+a_{m}=\frac{G(M+m)}{x^2}$
As v increases, when x decreases, we have
$a=\frac{-v d v}{d x}=\frac{G(M+m)}{x^2}$
$v d v=\frac{-G(M+m)}{x^2} d x$
Integrating
$\int\limits_0^v v d v=-G(M+m) \int\limits_{\infty}^h x^{1 / 2} d x$
$\frac{v^2}{v}=-G(M+m)\left[\frac{x^{-2+1}}{-2+1}\right]_{\infty}^{r}=-G(M+m)\left[\frac{-1}{r}\right]_{\infty}^{-r}$
∴ $v=\sqrt{\frac{2 G(M+m)}{r}}$