# Circular orbit problem

A particle moves under the action of the central force $Kr^4$ with angular momentum $l$. Find the energy for which the motion is circular and find the radius of that circular orbit.

From a previous question I know that $l=mr^2\dot \theta.$ I also know that for motion in polar coordinates the acceleration is given by $a=(\ddot r-r\dot \theta^2)\hat r+(2\dot r \dot \theta + r \ddot \theta) \hat \theta.$

Then using $F=ma$ I get the following:

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

$$\Rightarrow Kr^4=m\ddot r -mr \dot \theta^2$$

Then using my formula for angular momentum I get $$m \ddot r -\frac{l^2}{mr^3}-Kr^4=0$$

Then multiplying by $\dot r$ gives

$$m\dot r \ddot r -\frac{l^2 \dot r}{mr^3}-Kr^4 \dot r$$

$$\Rightarrow \frac{d}{dt}\bigg(\frac{1}{2}m\dot r^2\bigg)-\frac{l^2}{mr^3}\dot r-Kr^4 \dot r=0$$

Then integrating with respect to $t$ gives

$$\frac{1}{2}m \dot r^2 + \frac{l^2}{2mr^2}-\frac{Kr^5}{5}=E$$

So the effective potential $U(r)$ is given by $$U(r)=\frac{l^2}{2mr^2}-\frac{Kr^5}{5}$$

The orbit will be circular when $U(r)$ is a minimum, so when $$Kr^4=\frac{-l^2}{mr^3}$$ or $$r=\bigg(\frac{-l^2}{Km}\bigg)^{\frac{1}{7}}$$

Is this correct for the radius? This answer really doesn't look right to me.

• Have a look at the following math.stackexchange.com/questions/1702131/… A central potential always accommodates circular orbits (so you can set time derivatives of $r$ to zero), though you need to check that the orbit is stable.
– jim
Apr 4, 2016 at 12:05
• @jim This seems to give back the same as me but with $r^7$ positive instead of negative. Do you know why this is? I feel like my method is along the right lines, I'm just missing something.
– MHW
Apr 4, 2016 at 16:58
• Both get $r^7 = \frac{-L^2}{m \, K}$
– jim
Apr 4, 2016 at 17:28
• @jim So then would I just compute E at this value of $r$ in order to find the energy for which the motion is circular?
– MHW
Apr 4, 2016 at 19:32
• By E do you mean the kinetic energy, $\frac{1}{2} mv^2$ or the total energy? What value are you using for the angular momentum, $L$?
– jim
Apr 4, 2016 at 19:43

I think the simplest approach is to assume a circular orbit (central forces admit circular orbits as solutions) and see where this takes you. In your answer, remember $K$ is $-ve$.
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).