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(C) The total energy of an artificial satellite in an orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.Initial energy at radius $r$ is $E_1 = -\f

Vedclass · Accepted Answer

$33.33$

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(C) The total energy of an artificial satellite in an orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.Initial energy at radius $r$ is $E_1 = -\frac{GMm}{2r}$.Final energy at radius $\frac{3r}{2}$ is $E_2 = -\frac{GMm}{2(\frac{3r}{2})} = -\frac{GMm}{3r}$.The energy increases as the satellite moves to a higher orbit. The change in energy is $\Delta E = E_2 - E_1 = -\frac{GMm}{3r} - (-\frac{GMm}{2r}) = \frac{GMm}{2r} - \frac{GMm}{3r} = \frac{GMm}{6r}$.The percentage increase is given by $\frac{\Delta E}{|E_1|} 	imes 100\%$.Percentage increase $= \frac{\frac{GMm}{6r}}{\frac{GMm}{2r}} 	imes 100\% = \frac{2}{6} 	imes 100\% = \frac{1}{3} 	imes 100\% = 33.33\%$.

The percentage increase in the energy for an artificial satellite to shift it from an orbit of radius $r$ to an orbit of radius $\frac{3r}{2}$ is

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