A body of mass $m$ is placed on the earth's surface. It is taken from the earth's surface to a height $h=3 R…
A body of mass $m$ is placed on the earth's surface. It is taken from the earth's surface to a height $h=3 R,(R$ is radius of earth $)$. The change in gravitational potential energy of the body is
$\left(\frac{2}{3}\right) m g R$
$\left(\frac{3}{4}\right) m g R$
$\left(\frac{1}{2}\right) m g R$
$\left(\frac{1}{4}\right) m g R$
Solution
Gravitational potential energy on the surface of earth,
$
U_1=-\frac{G M m}{R}
$
where, $M=$ mass of earth.
and $\quad R=$ radius of earth.
Gravitational potential energy at height $h=3 R$ is given as
$
\begin{aligned}
U_2 & =-\frac{G M m}{R+h}=-\frac{G M m}{R+3 R} \\
U_2 & =-\frac{G M m}{4 R}
\end{aligned}
$
$\therefore$ Change in gravitational potential energy,
$
\Delta U=U_2-U_1=-\frac{G M m}{4 R}-\left(-\frac{G M m}{R}\right)
$
[from Eqs. (i) and (ii)]
$
\begin{aligned}
& =-\frac{G M m}{4 R}+\frac{G M m}{R}=\frac{3}{4} \cdot \frac{G M m}{R} \\
& =\frac{3}{4} \cdot \frac{g R^2 \cdot m}{R} \quad\left[\because G M=g R^2\right] \\
& =\frac{3}{4} m g R
\end{aligned}
$