Moon revolves around the earth in an orbit of radius R with time period of revolution T. It also rotates about its own axis with a time period T. If mass of the moon is M and its radius is 'r', the total kinetic energy of the moon is

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$\frac{{Mπ}^{2} R^{2}}{2 T^{2}}$

$\frac{{Mπ}^{2} R^{2}}{2 T^{2}} + \frac{4 {Mr}^{2} π^{2}}{5 T^{2}}$

$\frac{2 {Mπ}^{2} R^{2}}{T^{2}} + \frac{4 {Mr}^{2} π^{2}}{5 T^{2}}$

$\frac{4 {Mr}^{2} π^{2}}{5 T^{2}}$

answer is A.

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Detailed Solution

$K E = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

$K E = \frac{2 {Mπ}^{2} R^{2}}{T^{2}} + \frac{4 {Mr}^{2} π^{2}}{5 T^{2}}$

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