Question

Assume the earth's orbit around the sun as circular and the distance between their centre as $\mathrm{D}$. Mass of the earth is $\mathrm{M}$ and its radius is $\mathrm{R}$. 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

Accepted Answer

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

Answer

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

Answer

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

Answer

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

JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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