Considering a motion of a body under an attractive inverse cube central force,

$\textbf{F}(\textbf{r}) = -\frac{k}{r^3} \hspace{1mm}\hat{\textbf{r}}$ with $k>0$.

Is it possible for a body to move in an stable circular orbit? Since the derivation of the effective potential

$U_{eff}(r) = \frac{l^2}{2mr^2}+U(r)$

(where $l$ is the angular momentum)

has to be $0$ for a circular orbit, the only solution would be that $k = \frac{l^2}{m}$. But that would lead to an effective potential $U_{eff}(r) = 0$ for any $r$ (except $r = 0$). Is this a valid solution?

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    $\begingroup$ The derivative of the potential has to be zero I think. This way a test particle will stay inside the 'potential well' and so it can be a stable orbit. $\endgroup$
    – WarreG
    Commented Jan 19, 2019 at 13:56
  • $\begingroup$ After a small calculation the derivative seems to be zero at $k=l^2/m$, so I guess it is a valid solution. $\endgroup$
    – WarreG
    Commented Jan 19, 2019 at 14:01
  • $\begingroup$ You should clarify what $l$ is. For $L$ being the angular momentum, the solution is $|k|=L^2/(mr)$ $\endgroup$
    – FGSUZ
    Commented Jan 19, 2019 at 14:03
  • $\begingroup$ So for small deviations from a circular orbit with radius $r_{0}$ we just get some other circular orbits with radii $r_{0}+dr$ (Because the derivative of the effective potential is still 0)? But as a whole the initial circular orbit is not stable? $\endgroup$ Commented Jan 19, 2019 at 14:58
  • $\begingroup$ @Peter Hoferf to have a circle motion, for your force ansatz i get for k, $k=-2\,L^2/m$. if the force sign is positive we get for k the same solution. so way your force sign is negative ? $\endgroup$
    – Eli
    Commented Jan 19, 2019 at 21:16

4 Answers 4


The possible trajectories of a particle subject to an inverse-cube force $F = - k/r^3$ can actually be derived exactly; they are known as Cotes's spirals. Depending on the relative values of the particle's angular momentum $\ell$, its mass $m$, and the constant of proportionality $k$ in the force law, they take on the form $$ r(\theta) = \begin{cases} (A \cos C \theta + B \sin C \theta)^{-1} & km < \ell^2 \\ (A \cosh C \theta + B \sinh C \theta)^{-1} & km > \ell^2 \\ (A + B \theta)^{-1} & km = \ell^2 \end{cases} $$ where $$ C = \sqrt{\left| \frac{ m k}{\ell^2} - 1 \right|} $$ and $A$ and $B$ are determined by the initial conditions of the trajectory.

It is not hard to see that almost all of these functions will either have $r \to \infty$ for some finite value of $\theta$, or $r \to 0$ as $\theta \to\infty$, or both. The only case in which $r$ is bounded and does not reach $r = 0$ is when $km = \ell^2$ and $B = 0$, which corresponds to circular motion. Most perturbations from this trajectory will either involve changing $\ell$ (if the particle is given an extra tangential "push"), or changing $B$ (if the particle is given a purely radial push, since now we must have $dr/d\theta = 0$.) Thus, most perturbations will lead to the particle either spiraling in to $r = 0$ or flying out to $r \to \infty$.

Thinking about this in terms of the effective potential: to have a circular orbit in an inverse-cube field, you must have $\ell = \sqrt{km}$. A perturbation will either change $\ell$ or leave $\ell$ the same. If $\ell$ is changed from its initial value, then the effective potential becomes $U_\text{eff}(r) = Q/r^2$ for some value of $Q$, which has no maxima or minima; the radial motion must either go to $0$ or $\infty$. If $\ell$ is unchanged, then we must have $dr/dt = 0$, and the effective 1-D problem is that of a particle moving with some initial velocity in a potential $U_\text{eff} = 0$. Again, this radial motion must either go to $0$ or $\infty$.


For stable orbit, we need to have $d^2V_{eff}/dr^2>0$ at $r=r_0$ and we need to find $r_0$ from the solution of $dV_{eff}/dr=0$

We can find the potential by $$V=-\int Fdr$$

Hence $$V=-\frac {k} {2r^2}$$ so $$V_{eff}=\frac {l^2} {2mr^2}-\frac {k} {2r^2}$$

and at $r=r_0$, $dV_{eff}/dr=0$ hence

$$dV_{eff}/dr=\frac {-l^2} {mr_0^{3}}+\frac {k} {r_0^3}=0$$

so we have,


Now we need to find $d^2V_{eff}/dr^2$ at $r=r_0$

$$d^2V_{eff}/dr^2=\frac {3l^2} {mr_0^{4}}-\frac {3k} {r_0^4}$$ Using the above relationship we find that,

$$d^2V_{eff}/dr^2=\frac {3k} {r_0^{4}}-\frac {3k} {r_0^4}=0$$ which is exactly zero. So there cannot be any stable circular orbit.

  • $\begingroup$ So what descriptively happens if there are small deviations from the circular orbit with radius $r_0$? Would the body just describe other unstable circular orbits since $V_{eff} = 0$ (due to our chosen value for $k$)? $\endgroup$ Commented Jan 19, 2019 at 16:34
  • $\begingroup$ There cannot be any stable circular motion as I shown. The criteria for circular orbit is not satisfied. To have a stable orbit for perturbation we definitly need $V^2_{eff}/dr^2>0$ but since we have $V^2_{eff}/dr^2=0$ any perturbation from the orbit cannot lead any stable circular motion. $\endgroup$
    – seVenVo1d
    Commented Jan 19, 2019 at 17:43
  • $\begingroup$ Well yes as you said the body would do unstable circular motion according. If you draw the $V-r$ graph you ll also see that there cannot be any stable circular motion. $\endgroup$
    – seVenVo1d
    Commented Jan 19, 2019 at 17:50
  • $\begingroup$ I find the same question that you asked in my textbook, Marion-Thornton Classical Dynamic Question 8.22. If you can find the solution manuel you ll see a more comprehensive answer, but the basic idea is the same as I explained here. $\endgroup$
    – seVenVo1d
    Commented Jan 19, 2019 at 17:53

for a circle motion die circle radius $r(t)$ must be constant.

we can calculate the EOM's with Euler-Lagrange method.

kinetic energy:

$$ T=\frac{1}{2}\,m\left(\dot{r}^2+r^2\,\dot{\phi}^2\right)$$

potential energy

$$U=\frac{L^2}{2\,m\,r^2}+ U(r)$$

where $L$ is the angular momentum and $U(r)$ the unknown potential for a circle motion

The equation of motions are:

$$\ddot{r}=r\,\dot{\phi}^2+\frac{L^2}{m^2\,r^3}-\frac{1}{m}\frac{d\,U}{d\,r}\tag 1$$

$$m\,r\left(\ddot{\phi}\,r+2\,\dot{r}\,\dot{\phi}\right)=0\quad \Rightarrow$$

$$m\,\frac{d}{dt}\left(r^2\,\dot{\phi}\right)=0\tag 2$$

we obtain from equation (2) that $\dot{\phi}=h/r^2\quad$ with $h=L/m$ we put this result in equation (1) and get:

$$\ddot{r}=\frac{2\,L^2}{m^2\,r^3}\,-\frac{1}{m}\,\frac{d\,U}{d\,r}\tag 3$$

for a circle motion $r(t)=const\quad \Rightarrow\quad \ddot{r}=0$ ; we solve equation (3) for $\frac{d\,U}{d\,r}$


  • $\begingroup$ Unless m is dimensionless and equal to 1, something is very wrong here! $\endgroup$
    – Bill N
    Commented Jan 19, 2019 at 19:02
  • $\begingroup$ Yes I also think so, I Have to check $\endgroup$
    – Eli
    Commented Jan 19, 2019 at 19:33
  • $\begingroup$ @Bill N I found my error i forgot the mass in the kinetic energy, I think this solution is o.k ? but if i compare force sign in the quotation i got deference sing, and anther value for $k$. $\endgroup$
    – Eli
    Commented Jan 19, 2019 at 20:45

In case this helps anyone in the future. Defining Veff essentially turns this into a 1d problem with only r as the dynamical variable. If we want circular orbit you can either (do explicit calculation as Layla's answer does) or just see that Veff = const × 1/r^2 for inverse-cubic force law. Now, this function can be either monotonically increasing (const<0) or decreasing (const>0) except when the const =0. For monotonic functions you don't get any stationary points of Veff so no circular orbit is possible. Only when const = 0 => Veff = 0, you can get a circular orbit.

Talking about stability of circular orbit: As Veff is just zero for all r, so it's like a 1d question of a ball on a plane surface. If you give it a push it will just keep going. If the push is inwards it will collide with force center, or if the push is outward then it will move put to infinity. In 2D, our particle will just trace out a spiral inwards or outwards respectively.


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