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To understand how gravity influence objects, time and space, I have been thinking of how a planets shape would change the orbits of its moons.

More specifically: can I design a planet whose moon move in a square orbit?

Below is a diagram of my first intuitive try. For simplicity I imagine a two-dimensional shape extruded with the moon moving around it.

1. Is there in theory a shape that would create a square orbit for objects moving around it?

2. If yes, what is that shape?

enter image description here

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    $\begingroup$ Notice that at large enough distances (compared to the radius-like characteristic length of the star), all orbits would be elliptical (either circles or non-circular ellipses) regardless of the shape and composition of the star. You can see it as a consequence of the multipole expansion or simply as the intuitive heuristic that any object is a point-like object from a large enough distance. $\endgroup$
    – ACat
    Feb 2 at 17:43
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    $\begingroup$ The Binet equation doesn't like sharp turns in orbits, but everywhere-smooth, approximately square orbits may be OK with a suitable force law, which in turn can result from a suitable spherically asymmetric mass distribution in the planet (which gravity will only allow if the planet is small). By "smooth", I mean $\frac{d^2}{d\theta^2}\frac1r$ is finite. $\endgroup$
    – J.G.
    Feb 2 at 17:50
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    $\begingroup$ different but potentially of interest in Space SE: Could you stably orbit around a square (cubic) body? Would the orbit destabilize automatically if not corrected by input? $\endgroup$
    – uhoh
    Feb 3 at 23:38
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    $\begingroup$ "A planet is a celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit." $\endgroup$
    – Mazura
    Feb 4 at 4:25
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    $\begingroup$ Maybe the universe is square and this planet is rolling around the outside edges of this square universe? $\endgroup$
    – niemiro
    Feb 4 at 23:58

7 Answers 7

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Let us consider what forces are needed for a square orbit. As Newton pointed out, as long as there are no forces it will move in a straight line... so there must be no gravity along the sides. Then suddenly the moon turns 90 degrees, which implies a lot of force accelerating it. So there must be an enormous force just near the corners, and not along the sides.

This is awkward to achieve with real gravity. The gravitational force is a central force: each particle with mass exerts a $GMm/r^2$ force directed towards it, and you cannot shield from this by putting over masses in front: all the contributions from different mass particles sum together. So you cannot have just gravity bending the trajectory 90 degrees at the corner, since the gravitation from there will also affect the trajectory along the edge.

A general thing is that a complex shape of a planet produces a gravitational field that can be expressed using spherical harmonics. These tend to decay fast with distance if they are high frequence/sharp ("higher order"): weird planet shapes only affect very nearby orbits.

A four planet trick

If you allow for four fixed pointlike "planets" in a square I think one can prove there exists a nearly square orbit. Think of the moon approaching one of them, with the impact parameter $b$ (how far from the straight collision course it starts) a free variable. If $b$ is too large, the trajectory will just bend slightly and sweep past. If $b$ is too small you get more than a 90 degree turn. By continuity there is some $b$ that gives an exact 90 degree turn. That means, by conservation of energy, that it will move with exactly the same speed as it started when it gets far away from the planet. So, we can arrange that it does the same with the next, next and next planets and return to the starting point. The result is an orbit that is like a smoothed square. But it is not so much an orbit around a planet.

Somewhat square orbit around four fixed mass points, found via iterative refinement of the initial conditions

There is a subtlety above: the influence of the other three mass points will be felt at all places, so the gravitational bending is not going to be the perfect 2-body encounter I am assuming. Inside the Hill sphere of the planet its gravity dominates over all others and 2-body dynamics is a good approximation. However, proving the existence of the closed orbit requires more. Fortunately this is a continuous situation: if we color the starting points of the moon by how close they get to the second planet at closest approach, there will be some point that reaches the right distance to do a perfect 90 degree turn. Near that point, if we instead color by distance to the third planet, there will be an optimal point that causes three near-90 degree turns. Using the same method for the last planet and the starting point I think one can convince oneself that such an orbit must exist. Things are slightly trickier, since we should also do the same for how velocities change: we want to find a fixed point of the mapping $f:(x_{start},v_{start})\rightarrow (x_{end},v_{end})$ created by the planets so that $f(x,v)=f(x,v)$ (technically, finding a fixed point of the Poincare map). Doing this analytically is likely a nightmare, but one can use software optimization methods.

My plot above was made by starting with a region of initial values selected by hand, finding the one trajectory that got closest to its initial condition (in terms of position and velocity), zooming in on that to find even better starting values, and so on.

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    $\begingroup$ That squarish orbit is unstable, right? If the moon moves off the correct path by even a micrometer, it will quickly go farther and farther off the correct path until the orbit is no longer anything at all like a square. $\endgroup$ Feb 3 at 1:21
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    $\begingroup$ @TannerSwett - Yes, it is unstable. The local Poincare map is expansive around the fixed point. It is not as unstable as one would expect, but I need about four digits of precision to make it close for one orbit. So were the distances measured in megameters, a micrometer error would mess up the orbit after 3 orbits. $\endgroup$ Feb 3 at 12:06
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    $\begingroup$ @TannerSwett - Addendum: I actually did plot the Poincare map, and the square orbit is a saddle point rather than wholly unstable. There are hence starting conditions along a stable manifold that will cause a displaced moon to approach the orbit, although overall stretching in the unstable direction is vastly stronger and what matters "in practice". $\endgroup$ Feb 3 at 12:54
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    $\begingroup$ @J... in principle we could have the four planets connected to each other by light rigid braces. If the planets are far enough apart the forces between them ought to be small enough that making the braces light enough is possible whilst maintaining sufficient rigidity. The actual limitation here would be getting enough material to produce the braces, as well as actually constructing them. It would also require that the planets have no individual rotation, but instead the entire assembly rotates as a whole (which may mess up the orbit, I'm not sure) $\endgroup$
    – Tristan
    Feb 3 at 13:41
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    $\begingroup$ space is vast, but the forces between the bodies drop off fairly quickly. If the planets are far enough apart to barely be felt then the necessary station-keeping force from the brace is trivial. The braces could therefore have a tapering cross-section, being wide at the base mostly to support its own weight. This weight could also be used to provide some of the centripetal force needed for a rotating assemblage, allowing you to also cut the cross-section needed (similar to space elevators using the weight of the cable to keep the top in place) $\endgroup$
    – Tristan
    Feb 3 at 13:55
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If you just want a body to have a stable-ish square path relative to a parent system's barycenter, this can easily be done with a retrograde orbit. Illustration of concept:

enter image description here

I drew this as a planet in a binary star system, but it doesn't necessarily have to be— Be warned that the body the green planet or moon is orbiting does need to be big enough to have a mass and sphere of influence sufficient to capture the green planet at an appropriate orbital distance and period, though.

Run a Fourier transform or something like that to get the orbital parameters you'll need, and then solve for the mass of each body.

In fact, you may even be able to create significantly more complicated stable-ish orbits than a simple square:

Homer Simpson's orbit: A fun application of the Discrete Fourier Transform

(Not everything will be possible due to the constraints on mass and SOI/Hill Sphere. But there's potential.)

So… To frame it as in the question: You can't really have a single shape that will produce a square orbit for a moon around it, but you can probably have a simple orbital system where a moon can be added such that it will take on a squarish trajectory.

UPDATE: The radius and period required may not be very stable, or on the edge of stability. See the comments by @rob.

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    $\begingroup$ Fantastic answer. I like that "discrete Fourier transform with two terms" sounds much more modern than "retrograde epicycle." Is your visual just an illustration, or is it a simulation? (And if it's a simulation, what are the mass ratios?) I'm skeptical about stability under any circumstances, but I could possibly be convinced. $\endgroup$
    – rob
    Feb 4 at 21:37
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    $\begingroup$ Consider a three-body system where a primary object with mass $M$ is orbited by a secondary with mass $m$ at radius and frequency $R,\Omega$; a low-mass particle orbits the secondary with radius and frequency $r,\omega$. The distance from a square of side $a$ to its center varies between $a/2$ and $a/\sqrt 2$, which requires $r/R = 3-2\sqrt2 \approx 1/6$; to get four corners, require $\omega = 4\Omega$. Kepler's law fixes the mass ratio: $\frac{m}{M+m} = \frac{\omega^2 r^3}{\Omega^2 R^3} \approx \frac{1}{12}$. ... $\endgroup$
    – rob
    Feb 5 at 6:31
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    $\begingroup$ ... (continued) ... A plot of the generalized potential in the rotating frame (see e.g. this pdf) suggests this orbit is within the Hill sphere of the secondary, but far enough out that the equipotential surfaces aren't terribly spherical. My doubts about stability are unchanged: the Hill sphere isn't a hard-and-fast boundary. $\endgroup$
    – rob
    Feb 5 at 6:31
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    $\begingroup$ How stable is this orbit? Is attraction from the inner star included in your simulation? $\endgroup$
    – gerrit
    Feb 5 at 11:34
  • $\begingroup$ @gerrit It is just an illustration, rather than a simulation, unfortunately. @rob finds exact mass ratios to estimate the stability in his comments. I will update the answer if I look into this more; Could be interesting to see how close to square-ish you can get. (BTW ω=3Ω, not I think? Each corner is only 270°. And that would just make it more unstable… Empirically real moons do seem to bottom out around .) $\endgroup$
    – Will Chen
    Feb 5 at 18:01
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The 1st thing to look at would be the velocity. Non-zero velocity would imply a smooth orbit; therefore your moon would have to slow down to complete stop at the corner before accelerating into an orthogonal direction.

The 2nd thing to look at would be the acceleration. As the moon approaches, say, upper right corner the acceleration ought to be downward to slow it from the strictly vertical velocity before the corner to the complete stop. And right after the upper right corner the acceleration has to be horizontal to accelerate the moon strictly horizontally.

Moreover, acceleration cannot be zero at the corner: Newtonian's law are 2nd degree, so if an object has zero velocity and acceleration it's not going anywhere. So the moon's acceleration is discontinuous at the corner, from non-zero vertical to non-zero horizontal.

The next question is, can acceleration caused by Newtonian gravity be discontinuous? Well, if your Planet X has finite density then certainly no. Moreover, if your arrangement doesn't include infinite density singularities placed at the corners of the moon's orbit the answer is still no, the reason's being that Laplace potential would have a smooth local solution outside singularities.

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    $\begingroup$ “so if an object has zero velocity and acceleration it's not going anywhere” – en.wikipedia.org/wiki/Norton%27s_dome $\endgroup$ Feb 3 at 9:42
  • $\begingroup$ @leftaroundabout, very cool! I didn't consider a non-differentiable potential. $\endgroup$
    – Michael
    Feb 3 at 21:22
  • $\begingroup$ @leftaroundabout It's a good demonstration of why divide-by-zero and divide-by-infinitesimal aren't the same thing. :) $\endgroup$
    – Graham
    Feb 4 at 17:12
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A sharp corner indicates infinite curvature, so the ratio of the acceleration to the velocity would have to be infinite, which implies that the velocity is zero. Besides the instability of that requirement, that means that as it's approaching a corner, there must be an acceleration away from that corner (that is, the speed is decreasing, so the acceleration vector is opposite the velocity vector). However, that would mean that the net force is tangential to the its displacement vector relative to the planet, but gravity is always in the same direction as the displacement vector.

I don't think there's any way to do it without bodies outside the orbit supplying a force. But if the bodies are outside the orbit, then the force would presumably increase as the object gets away from the center of the orbit. This conflicts with the need for the force pushing the object away from the corner to go to zero as it reaches the corner.

And of course any planet-sized object with the shape you depict would quickly collapse under its own gravity; the force on the corners would be tremendous.

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  • $\begingroup$ "so the ratio of the acceleration to the velocity would have to be infinite, which implies that the velocity is zero" - or that the acceleration is infinite. :-P $\endgroup$
    – Vikki
    Feb 4 at 8:00
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Just throwing ideas out here but what about this:

enter image description here

If your planet looks a shell of sufficiently high mass then we have that inside the shell there is no gravity and outside there is high gravity. In the region without gravity the path of the moon will be straight which would allow it to form a nice square. If we punch a couple holes in your planet then the moon could occasionally peek outside where the high gravity causes its path to redirect inside.

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  • $\begingroup$ The only way to make such a shell is to reduce its mass as much as physically possible. Using steel you could create a structure about the size of the moon but with only microgravity (E-5m/s). Even with candidate supermaterials that we may one day be able to create this would be completely impossible. Any rocky moon/planet/body you would mean to redirect in this way would be by far the more massive body so it would, in fact, be the shell that would move. $\endgroup$
    – J...
    Feb 4 at 14:56
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Unfortunately such an orbit is not possible. Consider Newton's second law: $\vec F = m \mathbf{\vec a}$ . Along the side of the square, the acceleration $\mathbf{\vec a}$ is a vector pointing in the direction of the orbit, but the force must be pointing somewhere in the direction of the planet, so you can't have an equality.

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  • $\begingroup$ Thanks for your answer. So would it be correct to say that a moon in orbit around my 'planet X' would eventually crash into the side of one of the corner points? $\endgroup$
    – erik m
    Feb 2 at 17:18
  • $\begingroup$ If the orbital radius is large enough you can probably have close-to-regular elliptical orbits, since the planet will be seen to the moon as roughly a point mass. $\endgroup$
    – user341440
    Feb 2 at 17:39
  • $\begingroup$ This answer assumes the speed would have to be constant. But nothing in the question requires that. It's easy to construct a perfectly square trajectory whose acceleration is bounded everywhere, you just need to slow down before each corner. $\endgroup$ Feb 3 at 9:25
  • $\begingroup$ @leftaroundabout where do you see this assumption in the answer ? if the speed is constant then the acceleration would be zero. This is not what is considered here $\endgroup$
    – J. Delaney
    Feb 3 at 17:11
  • $\begingroup$ @J.Delaney ah, I confused this answer with the first paragraph of Anders Sandberg's answer, which this comment was addressing. Sorry. $\endgroup$ Feb 3 at 17:15
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In reality you can't have a spiky planet as you have drawn. The peaks will collapse under the gravity of the planet. And there's no known way that a "four planet trick" as described by Anders Sandberg could remain stable in reality. A planet's shape can have only very minor influence on it's satellites orbit. The bulge at the earth's equator (which is caused by the centrifugal effect of the earth's rotation) does however influence the precession of orbits of some artificial satellites. This is used to advantage in the "sun synchronous" orbit, where an artifical satellite's polar orbit precesses once a year.

There are however five "Lagrange points" which orbit together with a planet around its star, where the gravity of the planet and star have equal influence. This enables an object to orbit with the planet while being distant from it. Three of these, L1 L2 L3 are in line with the planet and star. Artificial satellites have been put into such orbits, but require station keeping as they are unstable. The other two, L4 and L5, lead and lag the planet by 60 degrees and objects can remain stable in these. Thus there are two groups of asteroids, the "greeks" and the "trojans" leading and lagging Jupiter in its orbit around the sun. These asteroids "librate" about the Lagrange points.

Then there are the "Hildas", a group of asteroids which orbit the sun 3 times in the time that Jupiter orbits twice. The orbits of these asteroids are really elliptical. But from the point of view of an observer at Jupiter, they appear to cover a triangular path, once for every 2 orbits that Jupiter makes of the sun. The orbits remain synchronous as the asteroids interact with the three Sun-Jupiter lagrange points L3 L4 L5 which are mutually 120 degrees apart.

In principle an artificial satellite in a 2:3 resonance could be put in place between Earth and the moon. It would appear to an observer on the moon to follow a triangular path.

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