A particle of mass $m$ moves under an attractive central force $Kr^4$ with angular momentum $L.$ For what energy will the motion be circular and what is the radius of the circle?

In order to find the radius I have been taught to use $F=ma$, multiply by $\dot r$ and then integrate. Then set $\frac{dV}{dr}=0.$ I have done what I can of this below.

Set $r^2\dot\theta=h.$

$$Kr^4=m(\ddot r-r\ddot\theta^2)$$

$$\Rightarrow Kr^4=m\bigg(\ddot r-\frac{h^2}{r^3}\bigg)$$

$$\Rightarrow Kr^4\dot r=mr\dot r -\frac{h^2}{r^3}\dot r$$

$$\frac{Kr^5}{5}=\frac{1}{2}\dot r^2+\frac{h^2}{2r^2}$$

$$\frac{1}{2}m\dot r^2+\frac{h^2}{2r^2}-\frac{kr^5}{5}=0$$

Then $\frac{dV}{dr}=0$ or $\frac{h^2}{r^3}=-Kr^4.$

And so, $$r=\sqrt[7]{\frac{-h^2}{K}}$$

I'm not convinced that this is correct. Could someone verify this and also explain how to go about finding the energy required for circular orbit?

  • $\begingroup$ I assume that K is $-ve$? If it's a circular orbit ($r$ = constant) why do you have $\ddot r, \dot r$ terms? $\endgroup$
    – jim
    Mar 17, 2016 at 20:16
  • $\begingroup$ @jim I thought I needed the $\ddot r $ and $\dot r$ terms because I don't know that the motion is circular, I'm trying to find the conditions for which it is. $\endgroup$
    – MHW
    Mar 17, 2016 at 20:42
  • $\begingroup$ Sorry, it sounded as if you were asking for circular motion. $\endgroup$
    – jim
    Mar 17, 2016 at 21:17
  • $\begingroup$ @jim Sorry about that. Maybe it's me not fully understanding the question. The way I've interpreted it is that I need to find the conditions that make the motion circular, in which case I guess you can't make that assumption in finding the energy. If you think I'm reading it wrong perhaps you could explain what I'm really being asked to do? $\endgroup$
    – MHW
    Mar 18, 2016 at 0:35
  • $\begingroup$ OK, perhaps it is my mis-interpretation, I'll remove my answer. $\endgroup$
    – jim
    Mar 18, 2016 at 9:51

1 Answer 1


I was worried about stability: I think the simplest approach is to assume a circular orbit (central forces admit circular orbits as solutions) and see where this takes you.

So for a circular orbit, you can set time derivatives for $r$ to zero. Start by equating the centripetal force to the attractive central force (it appears from your question that you have been given the angular momentum $L$ and for a central force $L$ is conserved): $-\frac{m v^2}{r} = Kr^4$ (taking $K -ve$ which is standard for an attractive force), then use $L = mvr$ so that you can write $\frac{m v^2}{r} = \frac{L^2}{m r^3}$ giving $\frac{m v^2}{r} = \frac{L^2}{m r^3}$ and the result for $r$ follows (this seems similar to what you have).

The next thing to check is stability, i.e. what happens when you perturb the orbit by a small amount. The condition for stability. For this have a look at http://www2.ph.ed.ac.uk/~egardi/MfP3-Dynamics/Dynamics_lecture_19.pdf (if you are unhappy with setting the time derivatives of $r$ to zero this may help you for the general case).

When you say "what energy" if you mean the kinetic energy, then this is easily determined from $L$ (which appears to be given) and $r$ (now determined): The (kinetic) energy is given as $T = \frac{1}{2}mv^2 = \frac{L^2}{2mr^2}$ with $L$ given and $r$ given in terms of the constant $K$ and the angular momentum $L$.


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