KCET · Physics · Work Power Energy
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
- A \( \frac{m g R}{2} \)
- B mgR
- C \( 2 \) mgR
- D \( \frac{m g R}{4} \)
Answer & Solution
Correct Answer
(A) \( \frac{m g R}{2} \)
Step-by-step Solution
Detailed explanation
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
\(W=U-U_{E}=-\frac{G M m}{2 R}-\left(-\frac{G M m}{R}\right)=\) \(-~\frac{G M m}{2 R}+\frac{G M m}{R} \)
\(\Rightarrow W=\frac{G M m}{2 R}\)
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
\(W=U-U_{E}=-\frac{G M m}{2 R}-\left(-\frac{G M m}{R}\right)=\) \(-~\frac{G M m}{2 R}+\frac{G M m}{R} \)
\(\Rightarrow W=\frac{G M m}{2 R}\)
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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