Dual Potential - Circular Orbit

Consider the equations of motion of the system, which can be derived using any standard methods available:

$r^2\dot{\theta} = K$ $\ddot{r} = r\dot{\theta}^2 -\frac{1}{r^2} + \frac{3}{r^4}$

In the above equation, $K$ is a constant. The speed of the particle is:

$V = \sqrt{\dot{r}^2 + r^2\dot{\theta}^2}$

Now, it is given that the solution is:

$r = 3$ $\implies \dot{r} = 0$ $\implies \ddot{r} = 0$

and that the speed of the particle is constant. This means:

$V = r\dot{\theta} = 3\dot{\theta}$

Since the speed is constant, this implies that $\dot{\theta}$ is also constant. Consider the equation of motion:

$\ddot{r} = r\dot{\theta}^2 -\frac{1}{r^2} + \frac{3}{r^4}$

Plugging in the solution gives:

$0 = 3\dot{\theta}^2 -\frac{1}{9} + \frac{3}{81}$ $\implies \dot{\theta}^2 = \frac{2}{81}$ $\implies \boxed{\lvert \dot{\theta} \rvert = \frac{\sqrt{2}}{9}}$

The force is central. It is given by $F(r)=-\dfrac{dV(r)}{dr}=\dfrac{3-r^2}{r^4}$ . This must equal $-m\dot {\theta}^2r=-\dot {\theta}^2r$ (since we are dealing with constant speed). Hence, $\dot {\theta}=\sqrt {\dfrac {r^2-3}{r^5}}=\dfrac{\sqrt 2}{9}\approx \boxed {0.1571348}$ . For the condition of uniform circular motion to prevail, $\dot {\theta}$ must be real, which holds true for $r^2-3\geq 0\implies r\geq \boxed {\sqrt 3}$ .

@Steven Chase Sir, for the Bonus part, I think that constant-speed circular orbits are possible for $r\ge\sqrt{3}$ only...