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(D) The total energy $E$ of a satellite is the sum of its potential energy $PE$ and kinetic energy $KE$.$E = PE + KE = -\frac{GMm}{R + h} + \frac{1}{2

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$-\frac{m g_0 R^2}{2(R + h)}$

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(D) The total energy $E$ of a satellite is the sum of its potential energy $PE$ and kinetic energy $KE$.$E = PE + KE = -\frac{GMm}{R + h} + \frac{1}{2}mv^2$For a satellite in a circular orbit,the gravitational force provides the necessary centripetal force:$\frac{mv^2}{R + h} = \frac{GMm}{(R + h)^2} \implies v^2 = \frac{GM}{R + h}$Substituting $v^2$ into the energy equation:$E = -\frac{GMm}{R + h} + \frac{1}{2}m\left(\frac{GM}{R + h}\right) = -\frac{GMm}{2(R + h)}$Since the acceleration due to gravity at the surface is $g_0 = \frac{GM}{R^2}$,we have $GM = g_0 R^2$.Substituting this into the expression for $E$:$E = -\frac{m g_0 R^2}{2(R + h)}$

$A$ satellite of mass $m$ is orbiting the Earth (of radius $R$) at a height $h$ from its surface. The total energy of the satellite in terms of $g_0$,the value of acceleration due to gravity at the Earth's surface,is

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