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 𝑉_c(𝑟) = 𝑚𝛼/𝑟³ , 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 𝑉_c(𝑟), but with a new time period 𝑇₁, then $(T_1^2-<!--\; -\; -->T_0^2)/T_1^2$ is given by

[𝐺 is the gravitational constant.]

To find the value of $(T_1^2-T_0^2)/T_1^2$, we analyze the forces acting on the particle of mass m in both circular orbits of radius $r_0$.

Step 1: Initial Circular Orbit (Only Gravitational Field)
Initially, the particle is in a circular orbit of radius $r_0$ around a body of mass M. The centripetal force is provided solely by gravity:
$\frac{GMm}{r_0^2}=m{\omega }_0^2{r}_0$
where ${\omega }_0$ is the initial angular velocity. This simplifies to:
${\omega }_0^2=\frac{GM}{r_0^3}$

Since the time period is related to angular velocity by $T_0=\frac{2\pi }{{\omega }_0}$, we have:
$T_0^2=\frac{4{\pi }^2}{{\omega }_0^2}$

Step 2: Combined Potential
The additional central force corresponds to the potential energy:
$V_c\left(r\right)=\frac{m\alpha }{r^3}$
The force associated with a potential energy V(r) is given by $F=-\frac{dV}{dr}$. Thus, the additional central force is:
$F_c\left(r\right)=-\frac{d}{dr}\left(\frac{m\alpha }{r^3}\right)=\frac{3m\alpha }{r^4}$

The total radial force acting on the particle is the sum of the gravitational force (which is attractive, directed toward the center, hence negative) and this additional force:
$F_{\text{net}}=-\frac{GMm}{r^2}+\frac{3m\alpha }{r^4}$

For the particle to continue moving in a circular orbit of the same radius $r_0$ with a new angular velocity ${\omega }_1$, the net force must equal the centripetal force:
$-m{\omega }_1^2{r}_0=-\frac{GMm}{r_0^2}+\frac{3m\alpha }{r_0^4}$

Dividing both sides by $-m{r}_0$ gives:
${\omega }_1^2=\frac{GM}{r_0^3}-\frac{3\alpha }{r_0^5}$

Substitute ${\omega }_0^2=\frac{GM}{r_0^3}$ into the equation:
${\omega }_1^2={\omega }_0^2-\frac{3\alpha }{r_0^5}$

Step 3: Calculating the Ratio
Using the relationship between the time period and angular velocity ($T^2\propto \frac{1}{{\omega }^2}$), we can express the required ratio as:
$\frac{T_1^2-T_0^2}{T_1^2}=1-\frac{T_0^2}{T_1^2}=1-\frac{{\omega }_1^2}{{\omega }_0^2}$

Substituting the expression for ${\omega }_1^2$:
$1-\frac{{\omega }_1^2}{{\omega }_0^2}=1-(1-\frac{3\alpha }{r_0^5{\omega }_0^2})=\frac{3\alpha }{r_0^5{\omega }_0^2}$

Substitute ${\omega }_0^2=\frac{GM}{r_0^3}$ back into the expression:
$\frac{T_1^2-T_0^2}{T_1^2}=\frac{3\alpha }{r_0^5(\frac{GM}{r_0^3})}=\frac{3\alpha }{GM{r}_0^2}$