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Assume the earth’s orbit around the sun as circular and the distance between their centres as ‘D’. Mass of the earth is ‘M’ and its radius is ‘R’. If earth has an angular velocity of rotation $ω_{0}$ and angular velocity of revolution $ω$ with respect to the centre of the sun, the total kinetic energy of the earth when observed from sun is: [Neglect rotation of sun and $ω_{0}$ || $ω$ ]

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

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

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

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

answer is B.

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Detailed Solution

$K = \frac{1}{2} M V_{C}^{2} + \frac{1}{2} I_{C} ω_{0}^{2} W h e r e V_{C} = D ω$

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