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(D) The gravitational potential energy at the surface of the Earth is $U_{i} = -\frac{GMm}{R}$.The gravitational potential energy at a height $h = \fr

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$\frac{mgR}{3}$

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(D) The gravitational potential energy at the surface of the Earth is $U_{i} = -\frac{GMm}{R}$.The gravitational potential energy at a height $h = \frac{R}{2}$ is $U_{f} = -\frac{GMm}{R + R/2} = -\frac{GMm}{3R/2} = -\frac{2GMm}{3R}$.The increase in potential energy is $\Delta U = U_{f} - U_{i}$.$\Delta U = -\frac{2GMm}{3R} - (-\frac{GMm}{R}) = -\frac{2GMm}{3R} + \frac{GMm}{R} = \frac{GMm}{3R}$.Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this into the expression for $\Delta U$,we get $\Delta U = \frac{(gR^2)m}{3R} = \frac{mgR}{3}$.

If the gravitational acceleration at the surface of the Earth is $g$,then the increase in potential energy in lifting an object of mass $m$ to a height equal to half of the radius of the Earth from the surface will be:

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