[QUE/ME-12009] ME-PROBLEM

A particle moves in a spherically symmetric potential \[V(r)=\frac{1}{2}\mu \omega^2 + \frac{\lambda^2}{2\mu r^2}\] where \(\mu\) is the mass of the particle, and \(\omega,\lambda\) are real constants.

• Does there exist a circular orbit for zero orbital angular momentum?
• Assume \(L=0, E= \frac{25}{2}\mu\omega^2a^2\), \(\lambda=12\mu\omega a^2\) Use initial conditions \[ r(t)\big|_{t=0} = 4a;\quad \dot{r}(t)\big|_{t=0} = 0 \text{ and } \dot{\theta}(t)\big|_{t=0}=0 \] solve the equations of motion and obtain \(r\), \(\theta\) as function of time. Describe the motion that takes place under conditions specified here.

 If angular momentum is zero, we have \(\mu r^2\dot{\theta}=0\) which implies that \(\dot{\theta}=0\). Thus \(\theta\) is constant and motion will be in a straight line. This can also seen by setting \(\vec{r}\times \vec{p}=0\) implies that position and momentum are always parallel (follows from conservation of angular momentum). Therefore the motion is not in a circle. Now we assume that \(\vec{L}=0\). The effective potential in this case is equal to the potential. Since angular momentum is conserved the orbits will lie in a plane. Taking \(r,\theta\) as coordinates, the total energy can be written as \begin{eqnarray} E &=& \frac{\mu}{2} \Big(\dd[r]{t}\Big)^2 + V(r)\\ &=& \frac{\mu}{2} \Big(\dd[r]{t}\Big)^2 +\frac{1}{2}\mu \omega^2r^2 +\frac{\lambda^2}{2\mu r^2} \end{eqnarray} Using the initial conditions at time \(t=0\), the energy is \(E=\frac{25}{2}\mu\omega^2 a^2\), and we get \begin{eqnarray} \frac{\mu}{2}\Big(\dd[r]{t}\Big)^2 = E - \frac{1}{2}\mu\omega^2 r^2 -\frac{\lambda^2}{2\mu\omega^2 r^2}\\ \Big(\dd[r]{t}\Big)^2=25\omega^2 a^2 -\omega^2 r^2 -\frac{\lambda^2}{\mu^2r^2}. \end{eqnarray} Taking square root we get \begin{equation} dt = \frac{dr}{\sqrt{25\omega^2a^2-\omega^2r^2- \frac{\lambda^2}{\mu^2r^2}}}. \end{equation} Substituting the given value \(\lambda=12\mu\omega a^2\), and defining the dimensionless variable \(y=r/a\), after some straightforward algebra we get \begin{eqnarray} dt &=& \int \frac{a dy}{\sqrt{25 \omega^2 a^2 - \omega^2 a^2 y^2 -\frac{144 a^2}{\omega^2 y^2}}}\\ \omega t &=& \int\frac{dy}{\sqrt{25-y^2 - 144/y^2}}\\ &=& \int \frac{ydy}{\sqrt{25y^2-y^4 -144}} \end{eqnarray} Changing the integration variable to \(\sigma\) where \(\sigma=y^2 (=r^2/a^2)\) and with \(d\sigma=2y dy\), \begin{eqnarray} \omega t&=&\int \frac{d\sigma }{2\sqrt{(25\sigma-y^2- 144)}}\\ 2\omega t &=& \int \frac{d\sigma }{\sqrt{49/4 - (\sigma-\sigma_0)^2}},\qquad \text{where } \sigma_0 = 25/2. \end{eqnarray} Therefore \begin{eqnarray} \int\frac{d\sigma}{\sqrt{(7/2)^2 - (\sigma -\sigma_0)^2}} &=& 2\omega t +\phi. \end{eqnarray} Integrating over \(\sigma\) gives \begin{eqnarray} \sin^{-1}\frac{(\sigma -\sigma_0)}{7/2} &=&2\omega t+\phi \quad \mbox{\HighLight{used \(\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(x/a)\)}} \end{eqnarray} Thus we get \begin{equation} \sigma =\sigma_0+\frac{7}{2}\sin(2\omega t +\phi) = \frac{25}{2} + \frac{7}{2}\sin(2\omega t+\phi). \end{equation} Substituting \(\sigma=y^2=r^2/a^2\), and applying initial conditions gives the final answer for the orbit. \begin{equation} r^2=\frac{25}{2} a^2 + \frac{7}{2}a^2\cos(2\omega t). \end{equation} The particle oscillates between \(3a\) and \(4a\) along the line \(\theta=0\).