Question

Solve the integral $I={\int }_{\infty }^R\frac{GMm}{x^2}dx$

• A.
• B.
• C.
• D. - GMm ln(R)

Answer 1

The correct option is A

$I={\int }_{\infty }^R\frac{GMm}{x^2}dx=GMm{\int }_{\infty }^R\frac{dx}{x^2}=GMm[-\frac{1}{x}{]}_{\infty }^R$

=$GMm|-\frac{1}{R}+\frac{1}{\infty }|=\frac{-GMm}{R}$

This expression represents the gravitational potential energy which is obtained by integrating $GMm/x^2$ between the limits infinity $\left(\infty \right)$ and radius of the earth (R).