Question 4

In a central force field, the radial equation of motion for a particle is given by:

$m \frac{d^2r}{dt^2} = -\frac{k}{r^2} + \frac{\ell^2}{mr^3}$

For a stable circular orbit at $r = r_0$, the net radial acceleration is zero, which gives the equilibrium condition:

$-\frac{k}{r_0^2} + \frac{\ell^2}{mr_0^3} = 0 \implies \frac{\ell^2}{m} = k r_0 \quad \text{--- (i)}$

Let the particle be displaced radially by a small distance $\delta r$, such that $r = r_0 + \delta r$. Substituting this directly into the equation of motion, we get:

$m \frac{d^2(r_0 + \delta r)}{dt^2} = -\frac{k}{(r_0 + \delta r)^2} + \frac{\ell^2}{m(r_0 + \delta r)^3}$

Since $r_0$ is a constant, its second derivative is zero $\left(\frac{d^2 r_0}{dt^2} = 0\right)$. Factoring out $r_0$ from the denominators on the right-hand side:

$m \frac{d^2(\delta r)}{dt^2} = -\frac{k}{r_0^2} \left(1 + \frac{\delta r}{r_0}\right)^{-2} + \frac{\ell^2}{m r_0^3} \left(1 + \frac{\delta r}{r_0}\right)^{-3}$

Since the displacement is extremely small ($\delta r \ll r_0$), we can apply the binomial approximation, $(1 + x)^n \approx 1 + nx$:

$m \frac{d^2(\delta r)}{dt^2} \approx -\frac{k}{r_0^2} \left(1 - 2\frac{\delta r}{r_0}\right) + \frac{\ell^2}{m r_0^3} \left(1 - 3\frac{\delta r}{r_0}\right)$

Expanding the terms and regrouping the constant parts and $\delta r$ parts together:

$m \frac{d^2(\delta r)}{dt^2} = \left(-\frac{k}{r_0^2} + \frac{\ell^2}{m r_0^3}\right) + \left(\frac{2k}{r_0^3} - \frac{3\ell^2}{m r_0^4}\right)\delta r$

From the equilibrium condition (i), the first term in the brackets is exactly zero. Now, substituting $\frac{\ell^2}{m} = k r_0$ into the second term:

$m \frac{d^2(\delta r)}{dt^2} = 0 + \left(\frac{2k}{r_0^3} - \frac{3(k r_0)}{r_0^4}\right)\delta r$

$m \frac{d^2(\delta r)}{dt^2} = \left(\frac{2k}{r_0^3} - \frac{3k}{r_0^3}\right)\delta r$

$m \frac{d^2(\delta r)}{dt^2} = -\frac{k}{r_0^3}\delta r$

Rearranging this into the standard differential equation format for Simple Harmonic Motion (SHM):

$\frac{d^2(\delta r)}{dt^2} + \left(\frac{k}{m r_0^3}\right)\delta r = 0$

Comparing this with the standard SHM equation $\frac{d^2(\delta r)}{dt^2} + \omega^2 \delta r = 0$, the angular frequency $\omega$ is:

$\omega = \sqrt{\frac{k}{m r_0^3}}$

Substituting the value of the stable radius $r_0 = \frac{\ell^2}{mk}$:

$\omega = \sqrt{\frac{k}{m\left(\frac{\ell^2}{mk}\right)^3}} = \sqrt{\frac{k \cdot m^3 k^3}{m \ell^6}} = \frac{m k^2}{\ell^3}$

The time period $T$ of these small periodic oscillations is:

$T = \frac{2\pi}{\omega} = \frac{2\pi \ell^3}{m k^2}$

Hence, the correct option is (A).