Q. A satellite of mass m is orbiting the Earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of $g_0$, the value of acceleration due to gravity at the Earth's surface, is
A) $\frac{mg_0R^2}{2(R + h)}$
B) $-\frac{mg_0R^2}{2(R + h)}$
C) $\frac{2mg_0R^2}{R + h}$
D) $-\frac{2mg_0R^2}{R + h}$
Explanation
[Solution] $E = -\frac{GMm}{2(R + h)}$ with $GM = g_0R^2$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.