Approximation of Stable Orbits as Harmonic Oscillators

Question

A textbook on classical mechanics I am currently reading considers the stable orbit (at $r_0$) of a body subject to the power law:

$$\mathbf{F}(r)=-Kr^n\mathbf{\hat{r}},\quad n\in\mathbb{Z}$$ $$\mathbf{F}(r_0)=-Kr_0^n\mathbf{\hat{r}}\tag{1}$$

such that, in polar coordinates:

$$L=mr^2\dot\theta, \quad F(r)=m\ddot{r}-r\dot\theta^2$$ $$F(r)= m\ddot{r}-\frac{L^2}{mr^3}.\tag{2}$$

It suggests that for small displacements around $r_0$ in the form:

$$r=r_0+x, \quad x \ll r_0,\tag{3}$$

both sides can be approximated with a Taylor expansion, resulting in:

$$m\ddot x = -x\left(nKr_0^{n-1}+3\frac{L^2}{mr_0^4}\right)\tag{4}$$ $$m\ddot x=-xKr_0^{n-1}\left(n+3\right),\tag{5}$$

using the fact that, at equilibrium,

$$Kr_0^{n-1}=\frac{L^2}{mr_0^4}.$$

Equation (5) should then indicate that the orbit of the body behaves like a harmonic oscillator for $n > -3$. However, I don't understand how the Taylor approximation was applied in equation (4), and to what order. I would be very thankful if someone could give an explanation, without using Lagrangian mechanics, of how the approximation was taken.

Answer 1

The strategy is to rewrite OP's Eq. $(2)$ leaving on one side the terms that depend only on $r$ and on the other side the part that depends on derivatives of $r$, namely

$$m\ddot{r}=F(r)+\frac{L^2}{mr^3}.\tag{1}$$

The stable circular orbit at $r=r_0$ satisfies

$$Kr_0^n=\frac{L^2}{mr_0^3}.\tag{2}$$

For small deviations $r=r_0+x$ with $x\ll r_0$, the LHS of $(1)$ is

$$m\ddot{r}=m(\ddot{r_0}+\ddot{x})=m\ddot{x},$$

and the RHS can be approximated to first order in $x$ in the following way: let's call $h(r)$ the RHS of Eq. $(1)$

$$h(r)=F(r)+\frac{L^2}{mr^3}=-Kr^n+\frac{L^2}{mr^3}.\tag{3}$$

Now we write the Taylor expansion of $h(r)$ around the equilibrium value $r_0$, up to first order

\begin{align} h(r)&\approx h(r_0)+h'(r_0)(r-r_0)\\ &=h(r_0)+h'(r_0)x. \end{align}

By Eq. $(2)$, we have $h(r_0)=0$. It's easy to check, then, that

$$m\ddot{x}=h'(r_0)x$$

is OP's Eq. $(4)$.