Question

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 $\omega_{0}$ with respect to its centre and $\omega$ with respect to the centre of the sun, the total kinetic energy of earth is :

• A. $\frac{M R^{2} \omega_{0}^{2}}{5}\left[1+\left(\frac{\omega}{\omega_{0}}\right)^{2}+\frac{5}{2}\left(\frac{D \omega}{R \omega_{0}}\right)^{2}\right]$
• B. $\frac{M R^{2} \omega_{0}^{2}}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_{0}}\right)^{2}\right]$
• C. $\frac{2}{5} M R^{2} \omega_{0}^{2}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_{0}}\right)^{2}\right]$
• D. $\frac{2}{5} M R^{2} \omega_{0}^{2}\left[1+\left(\frac{\omega}{\omega_{0}}\right)^{2}+\frac{5}{2}\left(\frac{D \omega}{R \omega_{0}}\right)^{2}\right]$

Answer 1

Answer: (B)

$( KE )_{\text {total }} =\frac{1}{2} \times \frac{2}{5} M R^{2} \omega_{0}^{2}+\frac{1}{2} M R^{2}\left(\frac{D^{2}}{R^{2}}\right) \omega^{2} $
$=\frac{M R^{2} \omega_{0}^{2}}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_{0}}\right)^{2}\right]$