# Nearly circular orbits (and their angular frequency)

In 'nearly' circular motion, the radius is not constant. If the force is central the and the angular momentum is still conserved, we have a central force:

where $r_0$ is equilibrium position.

For this motion, the orbital radius will 'oscillate' due to the centripetal and centrifugal force. However, in such motion, I am not sure how the angular frequency is given by:

whereby $Ω$ is the angular frequency due to the 'oscillating' radius, and ω is the angular frequency of the orbit itself.

Any help in explaining this concept and explaining how the angular frequency is derived will be appreciated.

Conservation of energy $$E = \frac{1}{2}m(\dot{r}^2 + r^2\omega^2) + U(r) = \frac{1}{2}m\dot{r}^2 + \frac{L^2}{2mr^2} + U(r)$$ yields the equation of motion $$m\ddot{r} = \frac{L^2}{mr^3} - \frac{\partial U}{\partial r} = - \frac{\partial U_\text{eff}}{\partial r}.$$ For a circular orbit with radius $r_0$, we have $\ddot{r}\equiv 0$, so that $$\left.\frac{\partial U_\text{eff}}{\partial r}\right|_{r=r_0}=0\qquad \text{and}\qquad \frac{L^2}{mr_0^3} = \left.\frac{\partial U}{\partial r}\right|_{r=r_0}\quad.$$ For nearly circular stable orbits, we can write $r = r_0 +x$, with $x\ll r_0$, and the equation of motion becomes $$m\ddot{x} = -\left.\frac{\partial U_\text{eff}}{\partial r}\right|_{r=r_0+x}\quad.$$ We can Taylor expand the right-hand side up to first order in $x$: $$\left.\frac{\partial U_\text{eff}}{\partial r}\right|_{r=r_0+x} = \left.\frac{\partial U_\text{eff}}{\partial r}\right|_{r=r_0} + x\left.\frac{\partial^2 U_\text{eff}}{\partial r^2}\right|_{r=r_0} + \mathcal{O}(x^2) = x\left.\frac{\partial^2 U_\text{eff}}{\partial r^2}\right|_{r=r_0} + \mathcal{O}(x^2),$$ so we obtain $$m\ddot{x} + x\left.\frac{\partial^2 U_\text{eff}}{\partial r^2}\right|_{r=r_0} = 0.$$ This is the equation of a harmonic oscillator, with frequency $$\Omega = \sqrt{\frac{1}{m}\left.\frac{\partial^2 U_\text{eff}}{\partial r^2}\right|_{r=r_0}}\quad.$$ Your example involves a central force $F(r) = -Kr^n$. The rest is a straightforward calculation.