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Question: Answered & Verified by Expert
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 :
PhysicsRotational MotionTS EAMCETTS EAMCET 2006
Options:
  • 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 $\frac{2}{5} M R^2 \omega_0^2\left[1+\left(\frac{\omega}{\omega_0}\right)^2+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]$
Solution:
2962 Upvotes Verified Answer
The correct answer is: $\frac{M R^2 \omega_0^2}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]$
$(\mathrm{KE})_{\text {total }}=\frac{1}{2} \times \frac{2}{5} M R^2 \omega_0^2+\frac{1}{2} M R^2\left(\frac{D^2}{R^2}\right) \omega^2$
$=\frac{M R^2 \omega_0^2}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]$

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