Answer
a) $KE_R=2.57\times10^{29}J$
b) $KE_R=2.68\times10^{33}J$
Work Step by Step
Rotational kinetic energy: $$KE_R=\frac{1}{2}I\omega^2$$
a) The earth can be considered a solid sphere with the axis through its center, so $I=\frac{2}{5}MR^2$
We have Earth's mass $M=5.97\times10^{24}kg$ and radius $R=6.38\times10^6m$
The Earth turns an angle of $\theta=2\pi rad$ (1 revolution) in $t=24h=8.64\times10^4s$, so $$\omega=\frac{\theta}{t}=7.27\times10^{-5}rad/s$$
Therefore, $$KE_R=\frac{1}{2}\Big(\frac{2}{5}MR^2\omega^2\Big)=2.57\times10^{29}J$$
b) The earth is the only particle rotating around the Sun in this exercise, so $I=M_ER_{orbit}^2$
The radius of orbit $R_{orbit}=1.5\times10^{11}m$. Therefore, $I=1.34\times10^{47}kg.m^2$
The Earth finishes an angle of $\theta=2\pi rad$ (1 revolution) in $t=365.25\text{ days}=3.16\times10^7s$, so $$\omega=\frac{\theta}{t}=2\times10^{-7}rad/s$$
Therefore, $$KE_R=\frac{1}{2}I\omega^2=2.68\times10^{33}J$$
