Q. The radius of a planet is 1/4 of earth’s radius and its acceleration due to gravity is double that of earth’s acceleration due to gravity. How many times will the escape velocity at the planet’s surface be as compared to its value on earth’s surface
A) $\frac{1}{\sqrt{2}}$
B) $\sqrt{2}$
C) $2\sqrt{2}$
D) 2
Explanation
(a) $v = \sqrt{2gR} \Rightarrow \frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{e}}{R_{p}}} = \sqrt{2 \times \frac{1}{4}} = \frac{1}{\sqrt{2}}$
$\therefore v_{p} = \frac{v_{e}}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.