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A body of mass $2 m$ is placed on earth's surface. Calculate the change in gravitational potential energy, if this body is taken from earth's surface to a height of $h$, where $h=4 R$.
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The correct answer is:
$\frac{8}{5} m g R$
Potential energy, $U=-\frac{G M m}{r}=\frac{-2 G M m}{r} \quad[\because m=2 m]$
At earth's surface, $r=R$
$\therefore \quad U_e=-\frac{2 G M m}{R}$
Now, if a body is taken to height, $h=4 R$, then the potential energy is given by
$\begin{aligned} U_h & =-\frac{2 G M m}{R+h} \quad[\because r=h+R] \\ & =\frac{-2 G M m}{5 R}\end{aligned}$
Thus, change in gravitational potential energy,
$\begin{aligned} \Delta U & =U_h-U_e \\ & =-\frac{2 G M m}{5 R}+\frac{2 G M m}{R}=\frac{8}{5} \frac{G M m}{R}\end{aligned}$
$\begin{aligned} \therefore \quad \Delta U & =\frac{8}{5} \frac{g R^2 m}{R} \quad\left[\because G M=g R^2\right] \\ & =\frac{8}{5} m g R\end{aligned}$
At earth's surface, $r=R$
$\therefore \quad U_e=-\frac{2 G M m}{R}$
Now, if a body is taken to height, $h=4 R$, then the potential energy is given by
$\begin{aligned} U_h & =-\frac{2 G M m}{R+h} \quad[\because r=h+R] \\ & =\frac{-2 G M m}{5 R}\end{aligned}$
Thus, change in gravitational potential energy,
$\begin{aligned} \Delta U & =U_h-U_e \\ & =-\frac{2 G M m}{5 R}+\frac{2 G M m}{R}=\frac{8}{5} \frac{G M m}{R}\end{aligned}$
$\begin{aligned} \therefore \quad \Delta U & =\frac{8}{5} \frac{g R^2 m}{R} \quad\left[\because G M=g R^2\right] \\ & =\frac{8}{5} m g R\end{aligned}$
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