Question

A body of mass 'm' is raised to a height ${}^{\prime }10R^{\prime }$ from the surface of earh, where 'R' is the radius of earth. The increase in potential energy is (G$=$ universal constant of gravitation, M$=$mass of earth and g$=$ acceleration due to gravity).

• A. $\frac{GMm}{11R}$
• B. $\frac{GMm}{10R}$
• C. $\frac{mgR}{11G}$
• D. $\frac{10GMm}{11R}$

Answer 1

The correct option is D $\frac{10GMm}{11R}$

Potential energy at point $U=-\frac{GMm}{r}$

where $r$ is the distance of the point from the earth surface.

Thus potential energy at earth surface i.e. $r=R$, $U_1=\frac{-GMm}{R}$

Potential energy at a height $10R$ above earth surface i.e. $r=11R$, $U_2=\frac{-GMm}{11R}$

Increase in potential energy $\Delta U=U_2-U_1$

Or $\Delta U=-\frac{GMm}{11R}-(-\frac{GMm}{R})$

Or $\Delta U=\frac{GMm}{R}[-\frac{1}{11}+1]$

$⟹$ $\Delta U=\frac{10GMm}{11R}$