Q. The gravitational potential energy of a body of mass $'m'$ at the earth’s surface is $- mgR_{e}$. Its gravitational potential energy at a height $R_{e}$ from the earth’s surface will be (Here $R_{e}$ is the radius of the earth)
A) $- 2mgR_{e}$
B) $2mgR_{e}$
C) $\frac{1}{2}mgR_{e}$
D) $- \frac{1}{2}mgR_{e}$
Explanation
(d) $\mathrm{\Delta}U = U_{2} - U_{1} = \frac{mgh}{1 + \frac{h}{R_{e}}} = \frac{mgR_{e}}{1 + \frac{R_{e}}{R_{e}}} = \frac{mgR_{e}}{2}$
$\Rightarrow U_{2} - \left( - mgR_{4} \right) = \frac{mgR_{e}}{2} \Rightarrow U_{2} = - \frac{1}{2}mgR_{e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.