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Question
If the acceleration due to gravity at the surface of the earth is g, the work done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is
Options
\[\frac{1}{2}mgR\]
\[2mgR\]
\[mgR\]
\[\frac{1}{4}mgR\]
Solution
\[\frac{1}{2}mgR\]
Work done = \[-\](final potential energy\[-\]initial potential energy)
\[\Rightarrow W = - \left( \frac{GMm}{2R} - \frac{GMm}{R} \right)\]
\[ \Rightarrow W = \frac{1}{2}\frac{GMm}{R} = \frac{1}{2}mR \times \left( \frac{GM}{R^2} \right)\]
\[ \Rightarrow W = \frac{1}{2}mRg \left[ \because g = \frac{GM}{R^2} \right]\]
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