Answer
The KE increases by a factor of $6\times10^5$. The energy comes from the initial gravitational PE.
Work Step by Step
The rotational kinetic energy is $\frac{1}{2}I\omega^2$. The final angular velocity, from Problem 39, is $=(5.625\times10^5 \frac{rev}{month})$.
$$\frac{KE_f}{KE_i}=\frac{\frac{1}{2}I_f\omega_f^2}{\frac{1}{2}I_i\omega_i^2}$$
$$\frac{KE_f}{KE_i}=\frac{\frac{2}{5}MR_f^2 \omega_f^2}{\frac{2}{5}MR_i^2\omega_i^2}$$
$$\frac{KE_f}{KE_i}=\frac{R_f^2 \omega_f^2}{R_i^2\omega_i^2}$$
$$\frac{KE_f}{KE_i}=\frac{(8000m)^2 (5.625\times10^5rev/month)^2}{(6\times10^6m)^2(1rev/month)^2}\approx6\times10^5$$The KE increases by a factor of $6\times10^5$. The energy comes from the initial gravitational PE.
