# Do two objects with intersecting orbits have to collide given enough time?

If I have two objects orbiting a central body, both in elliptical orbits, with an intersection at their shared periapsis, is it inevitable that they collide?

I think the answer is no, but I'm wondering if there's a clear way to express that or constrain the problem so the answer is yes.

I'm thinking of "collide" as in they come very close to each other, so within some small distance $$\epsilon$$.

• Almost the same question here: astronomy.stackexchange.com/questions/47513/31410 Apr 22 at 12:37
• @fasterthanlight I was just about to mention trojans, but the question you linked covers that scenario. However, since this question doesn't rule out trojans, that's one obvious example of objects with intersecting orbits (in fact nearly identical orbits, just offset ahead or behind on the same path) that won't collide. Apr 22 at 17:20
• There are two reasons that they might not collide 1) resonance, or 2) their influence on each other causes their orbits to change enough that they are no longer intersecting. Apr 23 at 15:57
• If I had to guess, I suspect that they'd probably develop a resonance or eject each other before colliding. Of course, given that this is a question about a massive family of analytical problems with no generalized closed-form expression, I doubt you'll find a proof or even the start of a proof any time soon. Your best insights might come from doing a LOT of modeling & starting to piece together the Poincaré map of a limited 3-body family of systems. See this guy's channels for inspiration re: analyzing chaotic systems via Poincaré mapping: youtube.com/c/Zymplectic/videos Apr 25 at 17:38

I can't offer analysis of your hypothetical ideal case, but in real cases of small or random objects in the solar system, it seems that the trajectory becomes sufficiently indeterminate with increasing intervals from the present, that collisions and other events become really unpredictable. The most likely next event is usually a somewhat close but non-collisionary encounter that perturbs the orbit enough and uncertainly enough to render the subsequent course unpredictable.

For example 2060 Chiron, a minor planet and/or giant comet discovered in 1977 with a period presently of roughly 50 years, has been studied by both forwards and backwards integrations of its orbit -- see the report "Rapid dynamical evolution of giant comet Chiron", G. Hahn & M. E. Bailey, Nature 348 (1990) 132–136 and a study by Horner et al. (2004).

The indications for Chiron are that it has had and will again have perturbing encounters, especially with Saturn, not very close to actual collision, but enough to change its orbital axis and period. At intervals beyond such events, the course is increasingly uncertain. In effect, the closer any perturbing encounter, the more the departing trajectory and its estimation are affected by uncertainties both of real trajectory, due to arbitrary tiny influences, and of estimation, caused by any errors in the account. So the medium-long-term future of Chiron and other comparable objects can only be described in terms of statistical chance. Hahn and Bailey reported: "Simulations extending ±100,000 years from the present suggest that on this time-scale, Chiron is about twice as likely to have been a short-period comet at some time in the past as to become one in the future. The mean half-life for such evolution is ∼0.2 Myr, much less than the ∼l-Myr lifetime for ejection from the Solar System, implying that Chiron may have been a short-period comet in the past, and will probably become one in the future."

So the conclusion seems to be that in real cases, possible collisions and other orbital changes can't be identified with any certainty -- unless they happen to be imminent along the present orbital track and due to occur soon enough that there will no appreciable non-collisionary perturbing encounters to cause uncertainties in the meantime.

One way to look at it, is in terms of the orbital periods. The gray dot will be at the intersection point on every multiple of its period, minus any initial offset: $$k_gT_g - \theta_g$$

If the purple dot starts at the intersection point (i.e. initial offset is 0), then it will be back at the intersection point every multiple of its period: $$k_pT_p$$

So if we modulo the multiple of the gray period by the purple period, whenever that is 0, they'll both be at the intersection point.

$$k_gT_g - \theta_g \mod T_p = 0$$

So if I plot that number over $$T_p$$, I can see how close the two are in terms of a fraction of the purple orbit's period: $$k_gT_g - \theta_g \mod T_p \over T_p$$

Picking a random example, the red circle shows when they will intersect:

However, if you make the orbits periodic with each other, if there's any offset, they never intersect:

To summarize:

• In the general case: no it is not inevitable that they collide. The counter examplse is when the orbits are periodic with each other, and there's an offset, they won't ever intersect.
• If they've "collided" once, then it seems inevitable that they will collide again based on playing around with some examples. (There's likely a more rigorous way to show this).
• I'm not sure if there's a general way to tell if they will ever collide in the future without just calculating out to some limit.

(The code for the charts is available here: https://observablehq.com/@pcarleton/do-two-objects-with-intersecting-orbits-have-to-collide)

• Both objects will return to their same respective positions after the least common multiple of their periods. So if they do collide, they will again after that time. And if they don't collide in that span, they never will.
– MJD
Apr 22 at 18:19
• The term you're looking for here is orbital resonance.
– Mark
Apr 22 at 22:33
• "The counter examplse is when the orbits are periodic with each other, and there's an offset, they won't ever intersect." But does perfect periodicity have non-zero probability measure? Apr 23 at 22:34
• "If they've "collided" once, then it seems inevitable that they will collide again" - this is false, at least for massless point-particles with perfectly stable orbits. After the first collision, if their periods are P and Q, then they'll collide again whenever nP = kQ for some integers n,k. In other words, when P = (k/n) Q. If P is rational and Q is not, then no k/n will satisfy that. Apr 24 at 7:04
• @BlueRaja-DannyPflughoeft According to the OP, collision is defined as "coming within some small distance $\epsilon$." So it should be quite possible for orbital periods with an irrational ratio to collide multiple times in this sense, even if they will only ever coincide at most once. Apr 24 at 14:16

The answer depends on the depth of modelling and the nature of the bodies.

In order to predict a collision at all, your modelling has to be accurate at least down to the objects' sizes. Point objects never collide, do they?

And then again, what should be considered a collision? Does a tidal disintegration of one of the bodies count? Are they themselves bound by gravity or by other forces? Are they rigid or can tidally deform on approach to each other?

All 3-body systems are in theory unstable. There are configurations that can be stable for a great deal of time (e.g. orbital resonances), but they are not stable forever.

For the usual celestial bodies the orbital modelling down to the object sizes accuracy can get increasingly hard for long timespans even if you completely skip the non-gravity interactions, but two marginal cases are clear:

• If the sizes of the objects are minor in regard to the distances between them, chances are that one of the bodies will be at some point ejected on an escape trajectory and the other one will get a lower orbit. How much minor is a good question, but approaches to it exist in the other answers.

• If bodies are bigger, chances are that two of them will collide before one of them gets the chance to escape. One of the collision participants may or may not be the "central" body.

And, on the top of everything that, if we model the problem even deeper, all orbits decay because of the gravitational waves emission and/or tides. Depending on the masses and distances involved, this effect alone can bring a collision of both orbiting objects with the central body before other scenarios get their chance.

• While point objects would not collide they'll no longer be in their orbits if they pass close enough. If they're big enough and far enough out (Jupiter) one may even be ejected. It's even possible (but quite unlikely) to have one ejected and one immolated. Apr 24 at 6:42
• No orbit in a 3-body system is exactly Keplerian so they will never stay "in their orbits". The only possible outcome of a 3-body system with point objects is one of them being ejected. Apr 24 at 8:53
• Always ejected? What about a system with two Mercurys. They don't have the power to eject each other. Apr 24 at 19:25
• Two point Mercurys around a point Sun? Trivial. One of them sinks halfway to the Sun, done. Apr 24 at 19:43
• That's them separating themselves, not one being ejected. I do agree they will eventually end up separating themselves unless it's a resonance situation. Apr 24 at 23:12

In a pure three-body system, I think the Lagrange points L4 and L5 are examples. I think a small object close to L4 or L5 will stay close, but, viewed in the frame of the largest mass, the orbit of the object at L4 or L5 will cross the orbit of the larger object, rather than staying on the same orbit. So that will be an example of orbits that cross but the two bodies never collide.

Horseshoe orbits may also be examples in some cases.

Assuming their orbits are never perturbed and ignoring object sizes so we can assume arriving at the same point in orbit at the same time is a "collision" the answer is clearly yes.

Assume you have two bodies that intersect at the same point in their orbits but have different orbital periods. Let's call their periods A and B. To determine if they intersect at the same time we need to find out if there are multiples of each orbital period that produce the same product (ie end at the same length of time). Mathematically does there exist an X and Y such that X * A = Y * B.

To prove that there is always an intersection you need to find an X and Y that satisfy the equation and there is an obvious answer that does.

Y = A X = B

B * A = A * B