$\text{A particle of mass}m\text{is under the influence of the gravitational field of a body of mass}M(M\gg m).$

$\text{The particle is moving in a circular orbit of radius}{\mathrm{ r}}_0\text{with time period}{\mathrm{ T}}_0\text{ around the mass}\mathrm{ \; M}.$

$\text{Then, the particle is subjected to an additional central force, corresponding to the potential energy}{\mathrm{ V}}_c\left(r\right)=\frac{m{a}^3}{r^3},$

$\text{where}\alpha \text{ α is a positive constant of suitable dimensions and}\mathrm{ r }\text{is the distance from the center of the orbit.}$

$\text{If the particle moves in the same circular orbit of radius }{r}_0\text{ in the combined gravitational potential}$

$\text{due to}\mathrm{ M}\text{ and}{\mathrm{ V}}_c\left(r\right),\text{ but with a new time period}{\mathrm{ T}}_1,\text{then}\frac{{\mathrm{ T}}_1^2-T_0^2}{T_1^2}\text{ is given by}$

$\text{[G is the gravitational constant.]}$

The correct option is:

$\frac{3\alpha }{GM{r}_0^2}$

Step 1: Motion under pure gravitational potential
For a particle of mass _m moving in a circular orbit of radius _r0 around a massive body _M, the centripetal force is provided by the gravitational force:

$F_g=\frac{GMm}{r_0^2}$

The equation of motion for the circular orbit gives:

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

Simplifying for angular frequency _ω0:

${\omega }_0^2={\left(\frac{2\pi }{T_0}\right)}^2=\frac{GM}{r_0^3}$

Therefore, we get:

$\frac{1}{T_0^2}=\frac{GM}{4{\pi }^2{r}_0^3}$

Step 2: Motion under the combined potential
When an additional potential _{Vc(r) = m α / r³} is introduced, the additional central force _Fc(r) acting on the particle is given by:

$F_c\left(r\right)=-\frac{d{V}_c}{dr}=-\frac{d}{dr}\left(\frac{m\alpha }{r^3}\right)=\frac{3m\alpha }{r^4}$

The net inward force required for the same circular radius _r0 with new angular frequency _ω1 is:

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

Dividing both sides by _{m r0} gives:

${\omega }_1^2={\left(\frac{2\pi }{T_1}\right)}^2=\frac{GM}{r_0^3}+\frac{3\alpha }{r_0^5}$

This can be rewritten in terms of _T1 as:

$\frac{1}{T_1^2}=\frac{GM}{4{\pi }^2{r}_0^3}(1+\frac{3\alpha }{GM{r}_0^2})$

Step 3: Calculating the required ratio
Substituting _{1 / T0²} into the equation for _{1 / T1²}:

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

Multiplying both sides by _T1²:

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

Rearranging to find the fractional change _{(T1² − T0²) / T1²}:

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