Q. A body is released from a point distance $r$ from the centre of earth. If $R$ is the earth and $r > R$, then the velocity of the body at the time of striking the earth will be
A) $\sqrt{gR}$
B) $\sqrt{2gR}$
C) $\sqrt{\frac{2gR}{r - R}}$
D) $\sqrt{\frac{2gR(r - R)}{r}}$
Explanation
(d) Using law of conservation of energy
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac{GMm}{r} = \frac{1}{2}{mv}^{2} - \frac{GMm}{R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{v^{2}}{2} = \frac{GM}{R} - \frac{GM}{r}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = GM\left( \frac{r - R}{rR} \right) = gR\left( \frac{r - R}{r} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \sqrt{2gR(r - R)/r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.