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(D) The gravitational potential energy at the surface of the Earth is $U_i = -\frac{GMm}{R}$.At a height $h = \frac{R}{2}$,the distance from the cente

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$\frac{mgR}{3}$

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(D) The gravitational potential energy at the surface of the Earth is $U_i = -\frac{GMm}{R}$.At a height $h = \frac{R}{2}$,the distance from the center of the Earth is $r = R + \frac{R}{2} = \frac{3R}{2}$.The potential energy at this height is $U_f = -\frac{GMm}{r} = -\frac{GMm}{3R/2} = -\frac{2GMm}{3R}$.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 gravitational acceleration 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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