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(A) The total energy $E$ of a satellite of mass $m$ orbiting the Earth of mass $M_e$ at a distance $r$ is given by:$E = -\frac{G M_e m}{2r}$This shows

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(A) The total energy $E$ of a satellite of mass $m$ orbiting the Earth of mass $M_e$ at a distance $r$ is given by:$E = -\frac{G M_e m}{2r}$This shows that $E \propto \frac{1}{r}$.Let the initial radius be $r_1$ and the initial energy be $E_1 = -\frac{G M_e m}{2r_1}$.When the radius is reduced to half,$r_2 = \frac{r_1}{2}$.The new energy is $E_2 = -\frac{G M_e m}{2r_2} = -\frac{G M_e m}{2(r_1/2)} = -\frac{G M_e m}{r_1} = 2 E_1$.The change in total energy is $\Delta E = E_2 - E_1 = 2E_1 - E_1 = E_1$.The percentage change is given by $\frac{|\Delta E|}{|E_1|} 	imes 100\% = \frac{|E_1|}{|E_1|} 	imes 100\% = 100\%$.

$A$ satellite is orbiting the Earth. If its orbital radius is reduced to half of its initial value,then the percentage change in its total energy is: (in $\%$)

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