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)

$(KE{)}_{\text{total }}=\frac{1}{2}\times \frac{2}{5}M{R}^2{\omega }_0^2+\frac{1}{2}M{R}^2\left(\frac{D^2}{R^2}\right){\omega }^2$
$=\frac{M{R}^2{\omega }_0^2}{5}\left[1+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2\right]$