A small mass $m$ is moved slowly from the surface of the earth to a height $h$ above the surface. The work done (by an external agent) in doing this is

m g h

, for all values of

h

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m g h

, for

h << R

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\frac{1}{2}

mgR, for h

=

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- \frac{1}{2}

mgR, for h

=

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Solution

The correct options are
B
$\frac{1}{2}$ mgR, for h $=$ R

C $m g h$ , for $h << R$

For, $h << R$ force is constant and is equal to mg

Therefore, work done is mgh in moving by distance $h$

For, $h = R$

gravitational potential at x is $- \frac{G M}{x}$

Work done in moving from $x = R$ to $x = 2 R$ is $- \frac{G M m}{2 R} + \frac{G M m}{R} = \frac{G M m}{2 R}$

which is equal to $\frac{1}{2} m g R$

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