I have two solar systems in my world that I'm not sure would be stable. I'm not concerned about whether they could evolve naturally (they didn't), just whether they would be stable if set running.

The first would be if there were two earth-like planets in the same orbit at each other's L4 and L5 Lagrange points. I know that most likely there was a planet in one of Earth's Lagrange points, Theia, that was pushed out of orbit by the gravity of the other planets and became the moon, but if there were no gravitational influence from other planets, would it be stable?

The second would be a binary star system with a planet orbiting the barycenter, but between the two stars. The stars are reasonably far apart, and the planet has a very small orbit in the center. If it only works with the planet not moving exactly in the center, that's fine too.

I've tried modeling these situations in Universe Sandbox with inconclusive results. Would these be stable in the long term?

  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Mar 13, 2023 at 3:08
  • $\begingroup$ Welcome Teagan. Please remember our one focused question per post requirement. Can you also tell us specifically what "long term" means here. $\endgroup$ Commented Mar 13, 2023 at 3:30
  • $\begingroup$ With the Solar System having a Lyapunov time in the range of somewhere between 2 million and 230 million years, we cannot predict planetary positions with any certainty beyond that time. We can estimate that the Earth orbit will remain stable enough for at least 1 or 2 billions of years (also about the same for objects in Earth's L4/L5 points). Anyone thinking in astronomical timescales will object to the word "stable". $\endgroup$
    – Klaws
    Commented Mar 13, 2023 at 15:17
  • 1
    $\begingroup$ a very quick search reveals that L4 and L5 are not stable $\endgroup$
    – njzk2
    Commented Mar 13, 2023 at 21:23
  • $\begingroup$ My understanding is that Theia didn't become the moon, it collided with proto-Earth causing a bunch of material to spray out and become the moon. Most of it is part of Earth now. $\endgroup$
    – Hearth
    Commented Mar 14, 2023 at 3:17

3 Answers 3


The first scenario is not stable. Lagrange points are only stable if you neglect the gravity the satellite exerts on the other two bodies, which requires that it be much less massive than them. For instance, a Trojan asteroid's gravitational pull on Jupiter or the Sun is practically nothing. If the satellite is similar in size to the planet, it will perturb that planet's orbit around the sun until you end up with double planets, similar to Pluto and Charon.

The second scenario is stable if and only if the planet and the two stars are the only bodies in the system, and if the planet is positioned with great care precisely between them. The slightest perturbation towards either star will be exaggerated over time by the stars' gravity - the further from dead center it is, the stronger the pull of the closer star, in a positive feedback loop that will eventually end in it being flung past that star and most likely ejected from the system altogether. Obviously this won't last any time at all (astronomically speaking) in nature, but if you want a sufficiently advanced alien art exhibit, it might be feasible, for awhile.

  • 19
    $\begingroup$ "The slightest perturbation towards either star will be exaggerated over time by the stars' gravity" this is precisely what it means for the position not to be stable $\endgroup$
    – Tristan
    Commented Mar 13, 2023 at 11:04
  • $\begingroup$ @Tristan True, stable probably isn't the word for it. What I mean is that the second configuration could exist temporarily, unlike the first configuration which will start to fall apart immediately under only those three bodies' gravity. $\endgroup$
    – Cadence
    Commented Mar 13, 2023 at 15:20
  • 13
    $\begingroup$ the phrase you're looking for is "unstable equilibrium" $\endgroup$
    – Tristan
    Commented Mar 13, 2023 at 15:23
  • $\begingroup$ I know little to nothing of astronomy, but am dubious about the second scenario. The two suns themselves would not be stable unless rotating around each other, in which case, the planets orbit would have to rotate around the same axis to match, but that results in a curved path, and no force to create that curve. That's not only unstable, that won't even last more than a few "days" by any measure. $\endgroup$ Commented Mar 14, 2023 at 18:10
  • $\begingroup$ @MooingDuck Of course the two stars will orbit around their common barycenter, that's always true of binary stars. A planet (that's not significantly closer to one star than the other) will also orbit their common barycenter - and in theory, it could do so at an arbitrarily close distance, i.e., between the stars. I'm not sure what you mean by "no force to create that curve"; like any orbit, it's maintained by the stars' gravity. $\endgroup$
    – Cadence
    Commented Mar 14, 2023 at 18:29

EDIT: Thanks to all the commenters for making me question statements I've made below. Both of shown graphs must be taken with a grain of salt - they do not account for Coriolis forces, and thus cannot describe complete picture.

Unfortunately, all three-body gravitational systems are unstable. Well, to be completely honest - there exist configurations, that are stable. But a tiny deviation from one (which will occur in real world) brings system back to chaos. If you are to dig more into that, look for "three body problem".

Now, onto exact systems you proposed:

two earth-like planets in the same orbit at each other's L4 and L5 Lagrange points.

At math perspective, stability is presented by local minimum of potential energy. So, to check if proposed system will be stable, we must check if planets are going to be in a local minimum of potential energy. This can be done with calculus, but it looks quite complicated, so here a nice graph it took from "Lagrange point" Wikipedia page:

Potential of a space around sun and earth

Here you can clearly see, that L4 and L5 points are at local maximum, not local minimum energetically. This implies, that an object is able to stay here, but space is going to feel "slippery" for it - any deviation from L4/L5 point will keep growing, if not negated by some sort of engine.

EDIT: This is exactly where Coriolis force comes into play, as it actually can negate some momentum deviation and thus make position stable. The key word here is "some momentum" - a planet is going to have lot more than that.

binary star system with a planet orbiting the barycenter, but between the two stars.

Same approach here, we need a graph of a potential energy. Here's what I found: source

Potential of a space around two massive objects

Despite it being an illustration for black holes, a main trend stays the same: there is no local minima, or "gravitational well" to orbit on between the stars. But if you look to the boundaries of a chart, you'll see a couple of co-centered rings - this is where a "gravitational well" exists, and this is where planets can have reasonably stable orbits.

EDIT: Same here. Can't really predict the way Coriolis will react here.

To sum up: unfortunately, these charming landscapes of several stars and planets in the sky, often imply that the system is totally unstable, and will soon face several catastrophic cosmic events.

  • 3
    $\begingroup$ if a tiny deviation from the arrangement makes it collapse, it is by definition unstable. There do also exist stable orbits. There is a figure-of-eight orbit for instance. These orbits have small domains of stability, but are still stable. Orbits based in lagrange point analyses solve the restricted three body problem, where one body is much lighter than the other two, and so cannot be used in the example with two planets each in the other's L4 & L5 $\endgroup$
    – Tristan
    Commented Mar 13, 2023 at 11:10
  • 4
    $\begingroup$ "Well, to be completely honest - there exist configurations, that are stable. But a tiny deviation from one (which will occur in real world) brings system back to chaos." <<< In physics we often make the distinction between "stable equilibrium" and "unstable equilibrium". I think in this case, you're not talking about a stable configuration, but about a configuration of unstable equilibrium. If a tiny deviation to the configuration brings it out of equilibrium, then by definition it's not "stable". $\endgroup$
    – Stef
    Commented Mar 13, 2023 at 12:08
  • 2
    $\begingroup$ Source: R. Montgomery [2001], A New Solution to the Three-Body Problem. Notices of the AMS vol 48:5. For "figure-of-eight", it's a "KAM stability. (...) the only kind of stability one can hope for in the N-body problem." KAM-stability is not really appliable to a real world, in my opinion. Espetially considering internal planet/star structure and other sources of uncontrolled deviations. $\endgroup$
    – Dzuchun
    Commented Mar 13, 2023 at 12:19
  • 3
    $\begingroup$ In your first plot (and second), L4 & L5 are stable equilibria, i.e. local minima, though you said the opposite. Of course, those plots neglect the gravitational effect of the two planets on each other. See also en.wikipedia.org/wiki/Lagrange_point $\endgroup$ Commented Mar 13, 2023 at 19:36
  • 2
    $\begingroup$ You're right for the wrong reasons here. L4 and L5 are unstable in the situation in the OP, but not because of any inherent limitations of Lagrange points. Stability of L4 and L5 requires that the mass of the third body is negligible relative to the mass of the first and second bodies, and a planet is very much not negligible in mass. $\endgroup$
    – Mark
    Commented Mar 14, 2023 at 4:18

You cannot have two equal-mass planets at each other's L4/L5 points. That configuration is only stable if the mass ratio between the larger and the smaller body is at least 25:1, and 1:1 is clearly less than that.

However, what you could have is two co-orbital planets of similar mass in stable horseshoe orbits with respect to each other, similar to the orbits of Saturn's moons Janus and Epimetheus.

I don't know what, if any, theoretical stability limits there are for horseshoe orbits, but the fact that we have an example of two similar-mass bodies in such a mutual orbit in our own solar system suggests that they cannot be too unstable.

Ps. The difference between a horseshoe orbit and a Lagrange point orbit is that the distance between the planets (or moons, as in the case of Janus and Epimetheus) in a horseshoe orbit doesn't remain constant.

Rather, one of the planets will orbit the sun slightly faster and slowly catch up to the other, at which point they will "switch orbits" so that the slower planet becomes the faster one and starts pulling away from the other one. Eventually the now faster planet will complete (almost) a full extra lap around the sun and again catch up to the other one, at which point they switch orbits again and the cycle repeats.

(This is super weird unless you have a good intuitive grasp of orbital mechanics, and specifically the counterintuitive fact that a force pulling an orbiting body forward along its orbit will lift it to a higher orbit and thus cause it to orbit slower, not faster.)

Viewed from the surface of one of the planets, the other one would appear as a "morning / evening star" depending on the mutual position of the planets in their orbits, a bit like Venus appear from Earth. However, whereas Venus moves from ahead of the Earth (i.e. visible at dawn, since Earth rotates prograde) to behind the Earth (i.e. visible at dusk) and back every 1.6 years, the corresponding period for a horseshoe orbit would be measured in decades, centuries or even millennia depending on the difference in the orbital periods. (For comparison, a quick back-of-the-envelope calculation based on this illustration of Janus and Epimetheus indicates that they take over 4000 orbits around Saturn to complete a full horseshoe cycle.)

The two planets would be closest to each other (and thus appear brightest from each other's surface) at the midpoint of the orbital switch. How close they'd get to each other, how long the closest approach would last and how frequently they'd occur all depend on the difference in the orbital radii. Qualitatively, the larger the difference, the faster the horseshoe cycle would be, the closer the planets would get to each other during closest approach, and the quicker that approach would be. But too large a difference will make the system unstable and likely lead to a quick collision or ejection of one of the planets. (AIUI, too small a difference is only unstable in the sense that random perturbations will tend to increase it.) You'd probably have to run some orbital simulations if you'd like some actual numbers.

Depending on the length of the cycle and the variation in orbital radii (and thus period), the change in year length between the "slow" and the "fast" portion of the cycle would likely be measurable by even prehistoric astronomers, at least if they had a change to observe an orbit swap (which should also be a conspicious celestial event in its own right, at least to someone carefully observing the sky and comparing it to old records) or several. Of course, the change in the position of the "morning / evening star" in the sky relative to the sun would also be observable, although a correct interpretation of it might be tricky without a good understanding of the solar system and its heliocentric nature.

Actually explaining the physics behind the orbit swap, of course, also requires at least a Newtonian understanding of gravity and some insight and calculation on top of that. (I suspect Isaac Newton could probably have figured it out, at least qualitatively, had he lived on a planet where the question was relevant. At least he had most of the mathematical tools available.)

The change in that planets' orbital radii throughout the horseshoe cycle would also likely drive a climate cycle of some kind, perhaps similar to the Earth's Milankovitch cycles, and associating these cycles with the appearance and disappearance of the morning / evening star would be a fairly obvious conclusion. Even if the relative difference in orbital radii between the "slow" (further from the sun) and "fast" (closer to the sun) parts of the cycle was small (say, around 0.01% to 0.1%, as they are for Janus and Epimetheus), even a tiny change in insolation over a sufficiently long time is likely to have noticeable climate effects. And the smaller the difference was, the longer the cycle would take.

  • $\begingroup$ Wow, this might be really inspiring for a world builders! You may come up with a story in the world, where two planets are almost colliding time to time, and IT WILL ACTUALLY BE REALISTIC. $\endgroup$
    – Dzuchun
    Commented Mar 15, 2023 at 8:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .