Question

Assume the Earth's orbit is circular around the Sun and the distance between their centers is $D$, mass of the earth is $M$ and its radius is $R$. If earth has an angular velocity ${\omega }_0$ with respect to its center and $\omega$ with respect to the center of the sun, the total kinetic energy of the earth is :

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

Answer 1

The correct option is A $\frac{M{R}^2{\omega }_o^2}{5}[1+(\frac{\omega }{{\omega }_o}{)}^2+\frac{5}{2}(\frac{D\omega }{R{\omega }_o}{)}^2]$

Total kinetic energy
$K{E}_T=\frac{1}{2}I{\omega }_0^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}M{D}^2{\omega }^2$

(We have used Parallel Axis Theorem for the Moment of Inertia for revolution about the sun )
or
$\frac{1}{2}[\frac{2}{5}M{R}^2]{\omega }_0^2+\frac{1}{2}[\frac{2}{5}M{R}^2]{\omega }^2+\frac{1}{2}M{D}^2{\omega }^2$
or
$\frac{M{R}^2{\omega }_0^2}{5}[1+\frac{{\omega }^2}{{\omega }_0^2}+\frac{5}{2}\frac{D^2{\omega }^2}{R^2{\omega }_0^2}]$