Question 705023

The effective potential is $V_{eff} = \frac{l^2}{2mr^2} + \frac{1}{2}kr^2$. For a circular orbit, $\frac{d V_{eff}}{d r} = 0$, so $-\frac{l^2}{mr^3} + kr = 0$, which gives $r_0^4 = \frac{l^2}{mk}$ or $r_0^2 = \sqrt{\frac{l^2}{mk}} = \frac{l}{\sqrt{mk}}$. The total energy $E = V_{eff}(r_0) = \frac{l^2}{2m r_0^2} + \frac{1}{2} k r_0^2 = \frac{l^2}{2m} (\frac{\sqrt{mk}}{l}) + \frac{1}{2} k (\frac{l}{\sqrt{mk}}) = \frac{l}{2} \sqrt{\frac{k}{m}} + \frac{l}{2} \sqrt{\frac{k}{m}} = l \sqrt{\frac{k}{m}} = \sqrt{\frac{k l^2}{m}}$.