A particle of mass $m$, and angular momentum $\ell$ is moving in a circular orbit of radius $r_0$ under the influence of an attractive force $\vec{F}(r) = -\frac{k}{r^2}\hat{r}$. Keeping its angular momentum unchanged, the particle is displaced radially by a small distance $\delta r \ll r_0$, due to which its radial distance varies periodically. The corresponding time period is:

Step 1: Analyze the initial circular orbit.

For a circular orbit of radius $r_0$, the centripetal force is provided by the attractive central force.

$$\frac{m v^2}{r_0} = \frac{k}{r_0^2}$$

The angular momentum $\ell$ is given by:

$$\ell = m v r_0$$

Step 2: Relate the angular momentum to the orbit radius.

Eliminating the velocity $v$ from the circular orbit equations yields the relationship between $\ell$ and $r_0$.

$$\ell^2 = m k r_0$$ $$r_0 = \frac{\ell^2}{mk}$$

Step 3: Define the effective potential.

The radial motion of the particle is described using the one-dimensional effective potential $U_{\text{eff}}(r)$.

$$U_{\text{eff}}(r) = \frac{\ell^2}{2mr^2} - \frac{k}{r}$$

Step 4: Find the first derivative of the effective potential.

The equilibrium radius $r_0$ corresponds to the minimum of the effective potential.

$$\frac{dU_{\text{eff}}}{dr} = -\frac{\ell^2}{mr^3} + \frac{k}{r^2}$$

Setting this derivative to zero at $r = r_0$ confirms the circular orbit condition.

$$-\frac{\ell^2}{mr_0^3} + \frac{k}{r_0^2} = 0$$ $$\ell^2 = m k r_0$$

Step 5: Set up the equation of motion for small radial oscillations.

For a small radial displacement $r = r_0 + \delta r$ where $\delta r \ll r_0$, the equation of motion is obtained by Taylor expanding about the minimum.

$$m \delta\ddot{r} = -U_{\text{eff}}''(r_0) \delta r$$

Step 6: Calculate the second derivative of the effective potential.

Differentiate the first derivative with respect to $r$.

$$U_{\text{eff}}''(r) = \frac{3\ell^2}{mr^4} - \frac{2k}{r^3}$$

Evaluate this second derivative at $r = r_0$ using the relation $\ell^2 = m k r_0$.

$$U_{\text{eff}}''(r_0) = \frac{3k}{r_0^3} - \frac{2k}{r_0^3}$$ $$U_{\text{eff}}''(r_0) = \frac{k}{r_0^3}$$

Step 7: Determine the angular frequency of the radial oscillations.

The angular frequency $\omega_r$ is determined by the effective spring constant.

$$\omega_{r}^2 = \frac{U_{\text{eff}}''(r_0)}{m}$$ $$\omega_{r}^2 = \frac{k}{mr_0^3}$$

Step 8: Express the angular frequency in terms of angular momentum.

Substitute $r_0 = \frac{\ell^2}{mk}$ into the expression for $\omega_r^2$.

$$\omega_{r}^2 = \frac{k}{m} \left(\frac{mk}{\ell^2}\right)^3$$ $$\omega_{r}^2 = \frac{m^2 k^4}{\ell^6}$$ $$\omega_{r} = \frac{mk^2}{\ell^3}$$

Step 9: Calculate the time period of the radial oscillations.

The time period $T$ is related to the angular frequency by $T = \frac{2\pi}{\omega_r}$.

$$T = \frac{2\pi\ell^3}{mk^2}$$

Answer: The corresponding time period is $T = \dfrac{2\pi\ell^3}{mk^2}$.