Question

A body of mass $m$ is lifted up from the surface of earth to a height three times the radius of the earth $R$.The change in potential energy of the body is

• A. $3mgR$
• B. $\frac{5}{4}mgR$
• C. $\frac{3}{4}mgR$
• D. $2mgR$

Answer 1

Answer: (C)

$\frac{3}{4}mgR$

$\Delta U=\frac{mgh}{1+\frac{h}{R}}$
$h=3R$

hence ,
$\Delta U=\frac{mg\left(3R\right)}{1+\frac{3R}{R}}=\frac{3mgR}{4}$