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Assume the earth’s orbit around the sun as circular and the distance between their centers as D. Mass of the earth is M and its radius is R. If earth has an angular velocity $ω_{0}$ with respect to its center and $ω$ with respect to the center of the sun, the total kinetic energy of the earth is

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$\frac{M R^{2} ω_{0}^{2}}{5} [1 + {(\frac{ω}{ω_{0}})}^{2} + \frac{5}{2} {(\frac{D ω}{R ω_{0}})}^{2}]$

$\frac{M R^{2} ω_{0}^{2}}{5} [1 + \frac{5}{2} {(\frac{D ω}{R ω_{0}})}^{2}]$

$\frac{2}{5} M R^{2} ω_{0}^{2} [1 + \frac{5}{2} {(\frac{D ω}{R ω_{0}})}^{2}]$

$\frac{2}{5} M R^{2} ω_{0}^{2} [1 + {(\frac{ω}{ω_{0}})}^{2} + \frac{5}{2} {(\frac{D ω}{R ω_{0}})}^{2}]$

answer is D.

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Detailed Solution

$K E = \frac{2}{5} M R^{2} ω_{0}^{2} [1 + {(\frac{ω}{ω_{0}})}^{2} + \frac{5}{2} {(\frac{D ω}{R ω_{0}})}^{2}]$

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