Gravitation

Concept:For a small radial perturbation from a stable circular orbit under a central inverse‑square force, the radial motion is simple harmonic. The period is found by linearising the radial equation of motion.Explanation:For a particle of mass $m$ in a central force $\vec{F}(r) = -\frac{k}{r^2}\hat{r}$, conservation of angular momentum $\ell$ gives $\dot{\theta} = \ell/(mr^2)$.The radial equation of motion is:$m\ddot{r} - \frac{\ell^2}{m r^3} = -\frac{k}{r^2}$.For a circular orbit of radius $r_0$, $\ddot{r}=0$ and $\dot{r}=0$, so:$\frac{\ell^2}{m r_0^3} = \frac{k}{r_0^2} \;\Rightarrow\; r_0 = \frac{\ell^2}{mk}$.Let $r = r_0 + \delta r$ with $|\delta r| \ll r_0$. Expand the equation to first order in $\delta r$:$m \ddot{\delta r} = \frac{\ell^2}{m r_0^3}\left(1 - 3\frac{\delta r}{r_0}\right) - \frac{k}{r_0^2}\left(1 - 2\frac{\delta r}{r_0}\right)$.Using the equilibrium condition, the constant term cancels, leaving:$m\ddot{\delta r} = -\left(\frac{3\ell^2}{m r_0^4} - \frac{2k}{r_0^3}\right)\delta r$.Substitute $\ell^2 = mk r_0$ to get:$\frac{3\ell^2}{m r_0^4} - \frac{2k}{r_0^3} = \frac{3k}{r_0^3} - \frac{2k}{r_0^3} = \frac{k}{r_0^3}$.Hence $\ddot{\delta r} = -\frac{k}{m r_0^3}\,\delta r$, which is SHM with angular frequency $\omega = \sqrt{\frac{k}{m r_0^3}}$.Time period $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m r_0^3}{k}}$.Replace $r_0 = \ell^2/(mk)$:$T = 2\pi\sqrt{\frac{m}{k}\left(\frac{\ell^2}{mk}\right)^3} = 2\pi\sqrt{\frac{\ell^6}{m^2 k^4}} = \frac{2\pi \ell^3}{m k^2}$.Answer:Option A: $\frac{2 \pi \ell^3}{m k^2}$.