Q. The escape velocity of a body from earth’s surface is $v_{e}$. The escape velocity of the same body from a height equal to $7R$ from earth’s surface will be
A) $\frac{v_{e}}{\sqrt{2}}$
B) $\frac{v_{e}}{2}$
C) $\frac{v_{e}}{2\sqrt{2}}$
D) $\frac{v_{e}}{4}$
Explanation
(c) $v_{e} \propto \frac{1}{\sqrt{r}}$ where $r$ is a position of body from the surface
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{1}}{r_{2}}} = \sqrt{\frac{R + 7R}{R}} \Rightarrow v_{2} = \frac{v_{1}}{2\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.