Practicing Success
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 |
$\left(\frac{Gr}{M+m}\right)^{\frac{1}{2}}$ $\left[\frac{2 G(M+m)}{r}\right]^{\frac{1}{2}}$ $2 G(M+m)^{\frac{1}{2}}$ $\frac{2 Gr}{M+m}$ |
$\left[\frac{2 G(M+m)}{r}\right]^{\frac{1}{2}}$ |
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 $\int\limits_0^v v d v=-G(M+m) \int\limits_{\infty}^h x^{1 / 2} d x$ $\frac{v^2}{v}=-G(M+m)\left[\frac{x^{-2+1}}{-2+1}\right]_{\infty}^{r}=-G(M+m)\left[\frac{-1}{r}\right]_{\infty}^{-r}$ ∴ $v=\sqrt{\frac{2 G(M+m)}{r}}$ |