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An object of mass $m$ is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance $R$ to $2 R$ from the centre of the earth. What is the gain in its potential energy?
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Let us consider an object of mass $m$ is lifted from the surface of the earth to a height equal to the radius of the earth $(R)$.
Potential energy of the body on the surface of the earth
$$
\left(E_{P_i}\right)=-\frac{G M m}{R}
$$
PE of the object at a height equal to the radius of the earth
$$
\left(E_{P g}\right)=-\frac{G M m}{(2 R)}
$$
So, gain in P.E. of the object,
$\begin{aligned}\left(E_{P f}\right)-\left(E_{P i}\right) &=\frac{-G M m}{2 R}-\left(-\frac{G M m}{R}\right) \\ &=\frac{-G M m+2 G M m}{2 R}=+\frac{G M m}{2 R} \\ \text { Gain in P.E. }=& \frac{g R^2 \times m}{2 R}=\frac{1}{2} m g R \\ \quad\left(\because G M=g R^2\right) \end{aligned}$
Potential energy of the body on the surface of the earth
$$
\left(E_{P_i}\right)=-\frac{G M m}{R}
$$
PE of the object at a height equal to the radius of the earth
$$
\left(E_{P g}\right)=-\frac{G M m}{(2 R)}
$$
So, gain in P.E. of the object,
$\begin{aligned}\left(E_{P f}\right)-\left(E_{P i}\right) &=\frac{-G M m}{2 R}-\left(-\frac{G M m}{R}\right) \\ &=\frac{-G M m+2 G M m}{2 R}=+\frac{G M m}{2 R} \\ \text { Gain in P.E. }=& \frac{g R^2 \times m}{2 R}=\frac{1}{2} m g R \\ \quad\left(\because G M=g R^2\right) \end{aligned}$
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