Question

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

• A. $\frac{M{R}^2{\omega }_0^2}{5}\left[1+\left(\frac{\omega }{{\omega }_0}\right)+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2\right]$
• 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. $\frac{2}{5}M{R}^2{\omega }_0^2\left[1+\left(\frac{\omega }{{\omega }_0}\right)+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2\right]$

Answer 1

Answer: (A)

Total kinetic energy = $\frac{1}{2}I{\omega }_0^2+\frac{1}{2}m{\nu }^2+\frac{1}{2}I{\omega }^2$ $\frac{1}{2}\left[\frac{2}{5}M{R}^2\right]{\omega }_0^2+\frac{1}{2}M{D}^2{\omega }^2+\frac{1}{2}\left[\frac{2}{5}M{R}^2\right]{\omega }^2$ = $\frac{M{R}^2{\omega }_0^2}{5}\left[1+\frac{{\omega }^2}{{\omega }_0^2}+\frac{5}{2}\frac{D^2{\omega }^2}{R^2{\omega }_0^2}\right]$ (or) $\frac{M{R}^2{\omega }_0^2}{5}\left[1+\frac{5}{2}\left(\frac{D^2{\omega }^2}{R^2{\omega }_0^2}\right)\right]$