Time Period Variation in Combined Gravitation and Central Force Field

To find the value of $\frac{T_1^2 - T_0^2}{T_1^2}$, we analyze the central force acting on the particle of mass $m$ in both cases.

1. Motion in the initial gravitational field:

The gravitational force providing the required centripetal force for the circular orbit of radius $r_0$ with time period $T_0$ is given by: $F_g = \frac{G M m}{r_0^2}$

For a circular orbit with angular speed $\omega_0 = \frac{2\pi}{T_0}$: $F_g = m \omega_0^2 r_0$ $\frac{G M m}{r_0^2} = m \left(\frac{2\pi}{T_0}\right)^2 r_0$ $\frac{G M}{r_0^3} = \frac{4\pi^2}{T_0^2} \quad \implies \quad \frac{1}{T_0^2} = \frac{G M}{4\pi^2 r_0^3} \quad \text{--- (1)}$

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2. Motion in the combined potential field:

The additional central potential energy is: $V_c(r) = \frac{m \alpha}{r^3}$

The additional central force $F_c(r)$ corresponding to this potential energy is: $F_c(r) = -\frac{d V_c(r)}{d r} = -\frac{d}{d r} \left(\frac{m \alpha}{r^3}\right) = \frac{3 m \alpha}{r^4}$

Since $F_c(r) > 0$, this force acts radially outwards (repulsive force). Therefore, the net attractive radial force acting on mass $m$ at distance $r_0$ is: $F_{\text{net}} = F_g(r_0) - F_c(r_0) = \frac{G M m}{r_0^2} - \frac{3 m \alpha}{r_0^4}$

For the new circular orbit of radius $r_0$ with new time period $T_1$ and angular speed $\omega_1 = \frac{2\pi}{T_1}$: $F_{\text{net}} = m \omega_1^2 r_0$ $\frac{G M m}{r_0^2} - \frac{3 m \alpha}{r_0^4} = m \left(\frac{2\pi}{T_1}\right)^2 r_0$

Dividing by $m r_0$: $\frac{4\pi^2}{T_1^2} = \frac{G M}{r_0^3} - \frac{3 \alpha}{r_0^5}$

Dividing by $4\pi^2$: $\frac{1}{T_1^2} = \frac{G M}{4\pi^2 r_0^3} - \frac{3 \alpha}{4\pi^2 r_0^5} \quad \text{--- (2)}$

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3. Expressing in terms of $T_0$ and $T_1$:

Substitute equation (1) into equation (2): $\frac{1}{T_1^2} = \frac{1}{T_0^2} - \frac{3 \alpha}{r_0^2} \cdot \left(\frac{1}{4\pi^2 r_0^3}\right)$

From equation (1), $\frac{1}{4\pi^2 r_0^3} = \frac{1}{G M T_0^2}$, so: $\frac{1}{T_1^2} = \frac{1}{T_0^2} - \frac{3 \alpha}{r_0^2} \cdot \frac{1}{G M T_0^2}$ $\frac{1}{T_1^2} = \frac{1}{T_0^2} \left( 1 - \frac{3 \alpha}{G M r_0^2} \right)$

Multiplying both sides by $T_1^2$: $1 = \frac{T_1^2}{T_0^2} \left( 1 - \frac{3 \alpha}{G M r_0^2} \right)$

Rearranging terms: $\frac{T_0^2}{T_1^2} = 1 - \frac{3 \alpha}{G M r_0^2}$ $1 - \frac{T_0^2}{T_1^2} = \frac{3 \alpha}{G M r_0^2}$ $\frac{T_1^2 - T_0^2}{T_1^2} = \frac{3 \alpha}{G M r_0^2}$

Hence, the correct option is (A).