A particle of mass 𝑚 is under the influence of the gravitational field of a body of mass 𝑀 (≫ 𝑚). The particle is moving in a circular orbit of radius 𝑟0 with time period 𝑇0 around the mass 𝑀. Then, the particle is subjected to an additional central force, corresponding to the potential energy 𝑉c(𝑟) = 𝑚𝛼/𝑟 3 , where 𝛼 is a positive constant of suitable dimensions and 𝑟 is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius 𝑟0 in the combined gravitational potential due to 𝑀 and 𝑉c(𝑟), but with a new time period 𝑇1, then (𝑇1 2 − 𝑇0 2 )/𝑇1 2 is given by [𝐺 is the gravitational constant.]

To solve the problem, we need to analyze the motion of a particle of mass $m$ in a circular orbit of radius $r_0$ around a much larger mass $M$, under the influence of both gravitational force and an additional central force with potential energy $V_c(r)=\frac{m\alpha }{r^3}$.

Step 1: Gravitational Force and Centripetal Force

Initially, the particle is moving in a circular orbit of radius $r_0$ with time period $T_0$ under the gravitational force alone. The gravitational force provides the centripetal force required for circular motion:

$$\frac{GMm}{r_0^2}=m{\omega }_0^2{r}_0$$

where ${\omega }_0=\frac{2\pi }{T_0}$ is the angular velocity. Solving for ${\omega }_0^2$:

$${\omega }_0^2=\frac{GM}{r_0^3}$$

Step 2: Combined Potential and New Angular Velocity

When the additional central force with potential $V_c(r)=\frac{m\alpha }{r^3}$ is introduced, the total potential energy becomes:

$$V_{\text{total}}(r)=-\frac{GMm}{r}+\frac{m\alpha }{r^3}$$

The effective force is the negative gradient of the potential:

$$F_{\text{total}}(r)=-\frac{d{V}_{\text{total}}}{dr}=-\frac{GMm}{r^2}+\frac{3m\alpha }{r^4}$$

For the particle to move in the same circular orbit of radius $r_0$, the total force must provide the centripetal force:

$$-\frac{GMm}{r_0^2}+\frac{3m\alpha }{r_0^4}=m{\omega }_1^2{r}_0$$

where ${\omega }_1=\frac{2\pi }{T_1}$ is the new angular velocity. Solving for ${\omega }_1^2$:

$${\omega }_1^2=\frac{GM}{r_0^3}-\frac{3\alpha }{r_0^5}$$

Step 3: Relating $T_1$ and $T_0$

The time periods $T_0$ and $T_1$ are related to the angular velocities by:

$${\omega }_0=\frac{2\pi }{T_0},{\omega }_1=\frac{2\pi }{T_1}$$

Substituting ${\omega }_0^2$ and ${\omega }_1^2$:

$${\left(\frac{2\pi }{T_0}\right)}^2=\frac{GM}{r_0^3},{\left(\frac{2\pi }{T_1}\right)}^2=\frac{GM}{r_0^3}-\frac{3\alpha }{r_0^5}$$

Dividing the second equation by the first:

$$\frac{T_0^2}{T_1^2}=1-\frac{3\alpha }{GM{r}_0^2}$$

Rearranging:

$$\frac{T_1^2-T_0^2}{T_1^2}=\frac{3\alpha }{GM{r}_0^2}$$

Final Answer

The correct answer is:

$$\boxed{{\displaystyle \text{(A) }\frac{3\alpha }{GM{r}_0^2}}}$$