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

  $\frac{M R^2 \omega_0^2}{5}\left[1+\left(\frac{\omega}{\omega_0}\right)^2+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]$
• Correct answer
  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

  $\frac{2}{5} M R^2 \omega_0^2\left[1+\left(\frac{\omega}{\omega_0}\right)^2+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]$

Hence, the correct answer is option: 'B'.