Q. The Earth is assumed to be a sphere of radius $R.$ A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is $fv$, where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is
A) $\frac{1}{3}$
B) $\frac{1}{2}$
C) $\sqrt{2}$
D) $\frac{1}{\sqrt{2}}$
Explanation
(d) Escape velocity of the body from the surface of earth is $v = \sqrt{2gR}$
Escape velocity of the body from the platform
Potential energy + Kinetic energy = 0
$\Rightarrow - \frac{GMm}{2R} + \frac{1}{2}mv_{p}^{2} = 0 \Rightarrow v_{p} = \sqrt{\frac{GM}{R^{2}}.R} = \sqrt{gR}$
$= \frac{1}{\sqrt{2}}\sqrt{2gR} = \frac{1}{\sqrt{2}};\ \therefore f = \frac{1}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.