Q. The escape velocity of a body from the earth is $v_{e}$. If the radius of earth contracts to 1/4th of its value, keeping the mass of the earth constant, the escape velocity will be
A) Doubled
B) Halved
C) Tripled
D) Unaltered
Explanation
(a) ${Escape\ velocity\ v}_{e} = \sqrt{\frac{2GM}{R}}$
$If\ R^{'} = \frac{R}{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {v'}_{e} = 2\sqrt{\frac{2GM}{R}}$
Since, G and M are constant hence,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {v'}_{e} = {2v}_{e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.