Stability of circular orbits in an arbitrary central force field

[pop_ianosd]

TL;DR Summary

I'm trying to understand why a stability criterion derived in a course for circular orbits in a general force field is valid.
I'm also trying to obtain a simple argument for why circular orbits in an inverse-cubed force field are unstable.

In this chapter, the stability of an object orbiting in a circular orbit of radius $r_c$ in an arbitrary force field $f$ is considered.
The author arrives at the equation of a harmonic oscillator, for small deviations $x$ from the circular orbit:
$$\ddot{x} + \left[-3\frac{f(r_c)}{r_c} - f'(r_c)\right]x= 0$$
From here follows that if the sign of the coefficient of $x$ is positive the orbit is stable.
My most specific question is about the case where $f = -\frac{c}{r^3}$. The author claims that in order to prove that it is unstable, we must keep the second order term when approximating the equation (308). That equation, with such a specific force would look like this:
$$\ddot{x} = \frac{h^2 - c}{(r_c + x)^3}$$
Since we are considering circular orbits, we have $-\frac{h^2}{r_c^3} = -\frac{c}{r_c^3}$ (eq. 306 in the link), so the right term in the equation above is always zero, and I therefore don't see how keeping the second order term would help.
My question is: what am I doing wrong, or do you have a better guess for what the author meant.
The way I would prove instability in this case is to derive the solutions to the unapproximated equation, and observe that the initial conditions that produce bounded solutions sit on a line, and not in a volume, in the initial conditions space. But I'm not too satisfied with this argument, as it deals with the concept of boundness rather than stability, and it just doesn't seem very elegant.

Returning to the criterion above, I'm also not convinced of it. I have the intuition that it would be a necessary condition for stability, but I wonder whether it is sufficient.

 

[BvU]

how keeping the second order term would help

it shows there is no restoring force: $\ \ddot x = 0$

 

[vanhees71]

As usual, it's much simpler using the Lagrange formalism. You have
$$L=\frac{m}{2} \dot{\vec{r}}^2 - V(r).$$
From rotational invariance you have angular-momentum conservation, and thus the orbit is in a plane perpendicular to it. This reduces the problem to a 1D effective motion for $r$:
$$L_{\text{eff}}=\frac{m}{2} \dot{r}^2-V_{\text{eff}}(r)$$
with
$$V_{\text{eff}}(r)=V(r)+\frac{l^2}{2m r^2}.$$
For the solution $r=r_c=\text{const}$ to be stable the effective potential must have a minimum at $r=r_c$. A sufficient condition is $V_{\text{eff}}'(r_c)=0$ and $m\omega^2:=V_{\text{eff}}''(r_c)>0$.

The setting $r=r_c+x$ the equation of motion for small $x$ can be approximated as the motion in the potential
$$\tilde{V}(x)=\frac{m}{2} \omega^2 x^2,$$
leading to the harmonic-oscillator equation
$$\ddot{x}=-\omega^2 x.$$
Since $\omega^2>0$, this admits solutions for $x$ staying small for all time, and that's what makes the circular orbit small.

If $V_{\text{eff}}$ has a maximum at $r=r_c$ the circular orbit is for sure unstable since then
$$\tilde{V}(x)=-\frac{m}{2} \lambda^2 x^2, \quad \lambda^2>0$$
and the EoM for $x$ is approximately (close to $x=0$)
$$\ddot{x}=+\lambda^2 x\; \Rightarrow \; x(t)=A \exp(\lambda t) +B \exp(-\lambda t),$$
i.e., you always find solutions exponentially growing, i.e., the circular orbit is not stable under arbitrary small perturbations, and the approximation breaks rapidly down.

For $V_{\text{eff}}''(r_c)=0$ you need to go to higher order in the series expansion of $V_{\text{eff}}$ in order to decide about the stability.

 

[pop_ianosd]

Thank you for the answers!
The image of the convex effective potential gives a nice intuitive reason for why the condition is sufficient for stability.
I'm not sure I understand what you mean by

The setting r=rc+x the equation of motion for small x can be approximated as the motion in the potential
V~(x)=m2ω2x2,
leading to the harmonic-oscillator equation
x¨=−ω2x.
Since ω2>0, this admits solutions for x staying small for all time, and that's what makes the circular orbit small.

Is this meant to be an argument for why the orbit is stable? Because if that's the case my problem with it is that I don't see how an approximation can prove something about the initial problem.

In the case of an inverse-cube force, you will then have a constant effective potential , if you set l such that Veff′(rc)=0. So then I guess the analysis is even simpler, in that the orbiting object is simply free to move in terms of r. Basically what @BvU said about there being no force.

Would it then be safe to conclude that the mention of using the second order terms is misleading?

The case

is special, since the first-order terms in the expansion of Equation (308) cancel out exactly, and it is necessary to retain the second-order terms. Doing this, it is easily demonstrated that circular orbits are also unstable for inverse-cube (

) forces.