Uniqueness and existence of polygonal orbits through a spherical shell

Say we have a spherical wire mesh raised to a negative voltage. Then let's say we release a proton from near the surface, and away from the surface, at some angle and speed. Also, imagine that the proton will always miss the mesh wires and pass through the surface.

The field in the center of this sphere is zero, so the proton travels in straight lines there, and outside of the sphere the proton travels in elliptical orbits with one focus located at the center of the sphere.

If the proton exits 3 times and winds up back at the same point it started, I imagine its trajectory would look something like this, where the dotted lines are the other part of the elliptical orbit that did not travel.

• Is the above idea possible in the first place?
• would it be possible to have a 2-exit orbit in the same sense?
• Is there only one combination of angle and velocity you can release it at to get a N-gon orbit?
• Alternatively, is there a relationship between angle and speed that fits such an N-gon orbit?

The above picture is imagining a "rounded" triangle orbit. If the last case is true (that there isn't a single unique N-gon orbit), then I wonder if there are varying degrees of roundness of the corners you could produce. If not, I imagine the orbital geometry would force all N-gon orbits to be similar to each other. I can't answer this easily to myself, and it seems such an interesting question.

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For the 2-exit orbit, I will cheat a bit and launch the proton on a purely radial orbit: it will go back and forth on its straight line all the time. For the other cases, let me think a bit :) –  JJ Fleck May 1 '13 at 12:59
@JJFleck yep, and that will even work for any given speed. For some reason, I had never even considered a non-trivial 2-gon (pardon the abuse of terminology) until I made that image. As two conjoined ellipses it's probably the most simple periodic path (that's neither a line nor a circle). So questions about degrees of freedom are probably best answered looking at that shape. –  AlanSE May 1 '13 at 14:05

A few thoughts about this. Let's first take it the other way around: determine if it would work with a given set of ellipses and then find for which mesh radius it would work.

The general polar equation for a newtonian ellipse is

$$r(\theta) = \frac{p}{1 + e\cos(\theta-\theta_0)}$$

where $p$ is the ellipse parameter, $e$ the excentricity and $\theta_0$ the angle of which the ellipse has been turned compared to the $x$-axis. Thus, we need the equation of the three ellipses to be

$$r_1 = \frac{p}{1 + e\cos(\theta)} \qquad r_2 = \frac{p}{1 + e\cos(\theta - 2\pi/3)} \qquad\text{and}\qquad r_3 = \frac{p}{1 + e\cos(\theta-4\pi/3)}$$

Now, to find the radius of the mesh sphere, you just have to find a common tangent to two ellipses (which is always possible) and by symmetry, you will have your $3$-gon. See the next picture where $e=0.8$:

Indeed, a non trivial $2$-gon is also possible for the same eccentricity:

And we can find $N$-gon all the same using tangent. You may notice that for a given ellipse size, the radius of the mesh is growing, so for a given mesh, you just have to change the value of parameter $p$ to find your answer.

Note that for the time being, we did not play with $e$, so for a given mesh, you can have as much 3-gon as you like if you only change the value of the eccentricity. In short, when launching your proton, the initial velocity (and position but it's starting on the mesh, isn't it ?) determine the half-big axis $a=p(1-e^2)$ (just as for gravitationnal dynamics: $E_m=-\frac{GMm}{2a}$) and the initial angle will completely determine your ellipse. Here is an example of 2 such $3$-gon for a given mesh (I didn't took the time to turn the second in order to make the two start from a same point but I think we get the idea)

• There are plenty of (speed,angle) couples that could achieve an $N$-gon, but there are certainly limits (for the least, the orbit has to be bound)
Just to add some more fun to the answer, once $N$ get big enough, you can imagine some other funny orbits as seen in the second image below.