Question

If ${\int }_{x=\infty }^{x=R}\frac{GMm}{x^2}dx=\frac{xGMm}{R}$ , Find value of x where G, M & m are constant.

Answer 1

${\int }_{\infty }^R\frac{GMm}{x^2}dx=GMm{\int }_{\infty }^R\frac{1}{x^2}dx$
$=GMm{\int }_{\infty }^R{x}^{-2}dx=GMm{\left[\frac{x^{-1}}{-1}\right]}_{\infty }^R$
$=-GMm{\left[\frac{1}{x}\right]}_{\infty }^R=-Gmm{[\frac{1}{R}-\frac{1}{\infty }]}_{\infty }^R$
$=\frac{GMm}{R}(\because \frac{1}{\infty }=0)$