Q. A satellite of mass $m$ is orbiting close to the surface of the earth (Radius R $= 6400\ $km) has a kinetic energy $k$. The corresponding kinetic energy of the satellite to escape from the earth’s gravitational field is
A) $K$
B) $2K$
C) $mgR$
D) $mK$
Explanation
(b) $K = \frac{GMm}{2R}$
Escape velocity $V_{e} = \sqrt{\frac{2GM}{R}}$
Kinetic energy to escape $\left( K^{'} \right) = \frac{1}{2}m \times 2\frac{GM}{R}$
$K^{'} = 2K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.