Question

A body of mass $m$ is taken from earth's surface to the height equal to the radius of earth, the change in potential energy will be of

• A. $mg{R}_e$
• B. $mg{R}_e/2$
• C. $2mg{R}_e$
• D. $mg{R}_e/4$

Answer 1

Answer: (B)

$mg{R}_e/2$

The gravitational potential energy at a height $h$ is
$U_h=\frac{-G{M}_{e}m}{R_e+h}$
If $h=R_e\Rightarrow U_h=\frac{-G{M}_{e}m}{+2R_e}=\frac{-mg{R}_e}{2}$
& the gravitational potential energy on the surface of earth is
$U_s=\frac{-G{M}_{e}m}{R_e}$
So change in potential energy $△U=U_h-U_s$
$=\frac{-G{M}_{e}m}{2R_e}+\frac{G{M}_{e}m}{R_e}=\frac{G{M}_{e}m}{2R_e}$
$△U=\frac{mg{R}_e}{2}$
It means that to move a body of a mass $m$ from the surface of earth to a height $h=R_e$, the energy required is $\frac{G{M}_{e}m}{2R_e}=\frac{mg{R}_e}{2}$