Question 2

For the initial circular orbit of radius $r_0$ under gravitational force alone: $\frac{GMm}{r_0^2} = m\omega_0^2 r_0 \implies \omega_0^2 = \frac{GM}{r_0^3}$ The time period is $T_0 = \frac{2\pi}{\omega_0}$, so $T_0^2 = \frac{4\pi^2 r_0^3}{GM}$.

The additional central force is $F_c = -\frac{dV_c}{dr} = -\frac{d}{dr}\left(\frac{m\alpha}{r^3}\right) = \frac{3m\alpha}{r^4}$. Since $\alpha$ is positive, this force is directed away from the center.

In the combined potential, the net force towards the center for the same radius $r_0$ is: $F_{net} = \frac{GMm}{r_0^2} - \frac{3m\alpha}{r_0^4} = m\omega_1^2 r_0$ $\omega_1^2 = \frac{GM}{r_0^3} - \frac{3\alpha}{r_0^5}$ The new time period is $T_1 = \frac{2\pi}{\omega_1}$, so $T_1^2 = \frac{4\pi^2}{\omega_1^2}$.

We need to find $\frac{T_1^2 - T_0^2}{T_1^2} = 1 - \frac{T_0^2}{T_1^2} = 1 - \frac{\omega_1^2}{\omega_0^2}$. $1 - \frac{\omega_1^2}{\omega_0^2} = 1 - \frac{\frac{GM}{r_0^3} - \frac{3\alpha}{r_0^5}}{\frac{GM}{r_0^3}} = 1 - \left(1 - \frac{3\alpha}{GMr_0^2}\right) = \frac{3\alpha}{GMr_0^2}$