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(A) For the initial circular orbit,the gravitational force provides the centripetal force: $F_0 = \frac{GMm}{r_0^2} = m \omega_0^2 r_0$,where $\omega_

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$\frac{3 \alpha}{G M r_0^2}$

Vedclass · Answer

(A) For the initial circular orbit,the gravitational force provides the centripetal force: $F_0 = \frac{GMm}{r_0^2} = m \omega_0^2 r_0$,where $\omega_0 = \frac{2\pi}{T_0}$.Thus,$\omega_0^2 = \frac{GM}{r_0^3}$.When the additional potential $V_c(r) = \frac{m \alpha}{r^3}$ is introduced,the additional force is $F_c = -\frac{dV_c}{dr} = -\frac{d}{dr} (m \alpha r^{-3}) = 3m \alpha r^{-4}$.The total force acting on the particle is $F_{total} = \frac{GMm}{r_0^2} + \frac{3m \alpha}{r_0^4} = m \omega_1^2 r_0$,where $\omega_1 = \frac{2\pi}{T_1}$.Dividing by $m r_0$,we get $\omega_1^2 = \frac{GM}{r_0^3} + \frac{3 \alpha}{r_0^5}$.Since $\omega^2 = \frac{4\pi^2}{T^2}$,we have $\frac{4\pi^2}{T_1^2} = \frac{GM}{r_0^3} + \frac{3 \alpha}{r_0^5} = \frac{GM}{r_0^3} \left( 1 + \frac{3 \alpha}{G M r_0^2} \right)$.Substituting $\frac{4\pi^2}{T_0^2} = \frac{GM}{r_0^3}$,we get $\frac{1}{T_1^2} = \frac{1}{T_0^2} \left( 1 + \frac{3 \alpha}{G M r_0^2} \right)$.Rearranging,$\frac{T_0^2}{T_1^2} = 1 + \frac{3 \alpha}{G M r_0^2}$.Then,$1 - \frac{T_0^2}{T_1^2} = -\frac{3 \alpha}{G M r_0^2}$,which implies $\frac{T_1^2 - T_0^2}{T_1^2} = \frac{3 \alpha}{G M r_0^2}$.

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