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A mass ' \( m \) ' on the surface of the Earth is shifted to a target equal to the radius of the Earth. If
' \( R \) ' is the radius and ' \( M \) ' is the mass of the Earth, then work done in this process is
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' \( R \) ' is the radius and ' \( M \) ' is the mass of the Earth, then work done in this process is
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Verified Answer
The correct answer is:
\( \frac{m g R}{2} \)
Gravitational potential energy on surface on Earth is
\[
U_{E}=\frac{-G M m}{R}
\]
where \( M \) is mass of Earth; \( m \) is mass on surface of Earth; \( R \) is radius; \( G \) is gravitational constant
When mass \( m \) is shifted to a target then potential energy is
\[
U=\frac{-G M m}{(R+R)}=\frac{-G M m}{2 R}
\]
Work done in the process is
\[
\begin{array}{l}
W=U-U_{E}=-\frac{G M m}{2 R}-\left(-\frac{G M m}{R}\right)=-\frac{G M m}{2 R}+\frac{G M m}{R} \\
\Rightarrow W=\frac{G M m}{2 R}
\end{array}
\]
Now substitute \( G M=g R^{2} \), we get
\[
W=\frac{g R^{2} m}{2 R}=\frac{g R m}{2}
\]
Therefore, work done in process \( =\frac{m g R}{2} \)
\[
U_{E}=\frac{-G M m}{R}
\]
where \( M \) is mass of Earth; \( m \) is mass on surface of Earth; \( R \) is radius; \( G \) is gravitational constant
When mass \( m \) is shifted to a target then potential energy is
\[
U=\frac{-G M m}{(R+R)}=\frac{-G M m}{2 R}
\]
Work done in the process is
\[
\begin{array}{l}
W=U-U_{E}=-\frac{G M m}{2 R}-\left(-\frac{G M m}{R}\right)=-\frac{G M m}{2 R}+\frac{G M m}{R} \\
\Rightarrow W=\frac{G M m}{2 R}
\end{array}
\]
Now substitute \( G M=g R^{2} \), we get
\[
W=\frac{g R^{2} m}{2 R}=\frac{g R m}{2}
\]
Therefore, work done in process \( =\frac{m g R}{2} \)
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