The shape of a fountain on a small planet

I left off the previous post about particles with a fixed initial speed with a conjecture about the inverse-square gravitational field case. After some more exploration, I’ve come up with a reasonably satisfactory explanation for why the final shape in attractive gravity is an ellipse.

I’ll refer to the center of the gravitational field as the “planet”, matching the mental image of running a fountain on a planet that’s so small that we can’t pretend gravity is uniform. The planet will always be directly below the origin, so that, for example, “horizontal” means “perpendicular to the line through the origin and the planet”.

Once again, here’s a visualization of the dynamics of the problem.

An elliptical family

After some thinking, I remembered that I know a little about orbital mechanics¹ and wondered if that could apply here, since the inverse-square gravity means that the trajectories are Kepler orbits. In those terms, all possible paths for the particle have the same total orbital energy; I figured it would be really nice for that to, in turn, mean that all the trajectories have the same semimajor axis, which I went ahead and proved to be the case.² (That’s not at all a novel result, of course; I just didn’t know it offhand.)

So the set of possible trajectories for the particle is the family of ellipses with a focus at one fixed point (the planet), passing through another fixed point (the origin), with a fixed semimajor axis.

For any point on an ellipse, the sum of the distances from that point to the two foci equals twice the semimajor axis of the ellipse. Since the distance from the origin (a point on every ellipse) to the planet (one focus of every ellipse) is fixed, the distance from the origin to the other focus of each ellipse in the family is fixed (i.e., the other foci lie on a circle centered at the origin (i.e., that circle is the focus locus)). I’ll refer to this non-planet focus as the free focus.

A fortuitous drawing

While tinkering with visualizations of particle paths, I noticed an interesting phenomenon, shown here:

That is, with evenly spaced initial velocities, certain pairs of paths seemed to intersect right on the horizontal line through the origin; a little counting showed that they were paths whose starting velocities made complementary angles³ to that line. At first I thought this must be a red herring, an artifact of the gravitational field being approximately uniform in that area: that’s exactly what happens in uniform gravity, which seemed like far too much of a coincidence. But fiddling with parameters convinced me that those intersections stay the same even when the gravity is definitely not approximately uniform (as in the image above, where the particles have a chance to reverse horizontal direction before crossing the line).

If two starting velocities are at complementary angles, what does that mean for the free foci of the corresponding trajectories? Consider the reflection property of ellipses: any ray of light emitted from one focus of an ellipse will hit the other focus after reflecting off the ellipse once. In this case, one such ray comes up the negative $y$-axis and reflects off each ellipse at the origin, where the tangent to an ellipse is the direction of its initial velocity. So the direction from the origin to the free focus is the reflection of the upward (positive $y$) vector across the initial velocity.

A velocity that is complementary to itself makes an angle of an eighth of a turn with the horizontal, and the focus in that case is on the horizontal line through the origin. From there, rotating the velocity by some amount either clockwise or counterclockwise produces two complementary velocities, and the corresponding foci make the same angle as each other to the horizontal (twice the change in angle of velocity), so they are symmetric across the horizontal line.

So complementary trajectories have free foci that are reflections of each other across the line. Knowing that, we can consider the following diagram to confirm that they really do hit the line at the same point:

The sum of the length of the solid blue line and that of either dashed blue line equals the common major axis of the ellipses; the same is true of the solid orange line and either dashed orange line. Thus, the dashed blue lines have the same length, as do the dashed orange lines, so the origin and other intersection point both lie on the perpendicular bisector of the segment between the two foci. Since the foci were originally taken to be symmetric about the horizontal red line, that line is the perpendicular bisector.

And in fact the same logic works for any line, not just this special horizontal one! We didn’t actually use the direction of the line at all. So the non-origin intersection of any two of the ellipses lies on the line that contains the origin and passes halfway between their free foci.

From there, we can see that the elliptical orbit that hits any ray starting at the origin the farthest from the origin is the one with its focus on that ray: given a ray and two different foci that are symmetric across it, if we push the foci closer to the ray, that would shorten the dashed orange lines if the intersection didn’t move; thus, the intersection actually does move farther from the origin in order to keep the total orange length the same.⁴

Conclusion

With that last statement in hand, we can draw one last diagram like the previous one, but with just one focus placed onto an arbitrary ray:

The two orange segments connect the two foci of one of the ellipses of the family to a point on that ellipse and the blue one is a radius of the circular focus locus, so the total length of all of them together is fixed. The blue segment and one orange segment both lie on the ray, so that total length is the sum of the distances from the farthest point to the origin and the planet. Thus the farthest points for all the rays therefore do indeed lie on an ellipse whose foci are the origin and the planet!

So, does this satisfy my desire for an intuitive explanation of the result? Partially, I’d say. I like that this is uses mostly geometric-rather-than-algebraic logic, which is much of what I was hoping for, but the initial facts about Kepler orbits ultimately bottom out in algebra, as far as I know. Given how elegant the final outcome is, I would still really like there to be a trick of coordinate transformation or something that gets to the end even more directly.

There’s also still the question of the repulsive gravity case. I imagine it’s possible to either (1) redo the geometry based on hyperbolae having a constant difference (rather than sum) of distances to foci or (2) turn all the geometry into algebra that lets you then do the repulsive case by just flipping the appropriate signs. I’m not terribly inclined to actually go through it; doing the attractive case makes me entirely certain that it all works out.

Footnotes