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I find he following answer to the question What's the soonest Oumuamua could return? unsatisfying and oversimplifying and the last five words "Therefore it will never return." unsupported at a minimum.

Oumuamua as an object is remarkable, because it has a positive net energy, which means it is not bound to the gravitational well of our sun. Therefore it will never return.

So I'd like to ask: *What natural mechanisms could lead to the unlikely case of a rogue asteroid or planet passing through our solar system twice?"

  • Stick to accepted scientific facts and theories
  • Let's give an ample time limit of 10 billion years between visits.
  • For the purposes of this question, the solar system includes the Kuiper belt (50AU) but not the Oort cloud
  • Don't include intervention by artificial means or intervention (divine or otherwise)
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    $\begingroup$ Gravity from an extremely well-placed body? Actually, it'd probably take more than 1 body to get the direction right. $\endgroup$
    – PM 2Ring
    Commented Mar 18, 2019 at 9:50
  • $\begingroup$ @PM2Ring you've got me thinking: Could a trajectory around a large mass ever deflect by more than180 degrees due to general relativistic effects? $\endgroup$
    – uhoh
    Commented Mar 18, 2019 at 10:12
  • $\begingroup$ @PM2Ring there are 8 (or 9) planets $\endgroup$
    – Robert Frost
    Commented Mar 18, 2019 at 10:34
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    $\begingroup$ @user334732 'Oumuamua's trajectory isn't in the ecliptic plane . Even though its current distance from the Sun is roughly the same as Saturn's radial distance, 'Oumuamua is already well above the main plane of the solar system. So it's very unlikely to encounter anything of significant mass until it reaches the outer Oort cloud in several thousand years. And that cloud is rather sparse, Wikipedia says the distance between bodies there is on the order of tens of millions of kilometres. $\endgroup$
    – PM 2Ring
    Commented Mar 18, 2019 at 11:44
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    $\begingroup$ "It is unknown how long the object has been traveling among the stars.The Solar System is likely the first star system that ʻOumuamua has closely encountered since being ejected from its birth star system, potentially several billion years ago." I mean it surely is possible that the object has a close encounter with another star that alters its trajectory significantly, and then again with a third which throws it back at the original system, but that is so extremely contrived that you might as well call it magic or divine intervention. Saying it doesn't come back is imho justified. $\endgroup$
    – Polygnome
    Commented Mar 18, 2019 at 22:02

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Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is (Douglas Adams)

The wikipedia page notes that 'Oumuamua "has circulated the Milky Way several times" so it was unlikely to have formed around any nearby star. While it would take 'Oumuamua 600,000 years to travel here from a local star, the time before it can be expected to have another stellar encounter would be much greater, hundreds of millions of years. And the next star that it meets is unlikely to be a currently local star, it could be almost any star in the galaxy.

Once an interstellar object has left the solar system it will travel in a roughly straight line, (slowly curving with the general gravity of the galaxy). For it to return to the Earth it would have to have a gravitational encounter with another star (or more likely a series of encounters) that send it back towards the solar system. Even given a vast mind-boggling long period of time like 10 billion years the chance of this happening is minute.

Such an object will probably orbit the galaxy several times between stellar encounters, perhaps it has an encounter every 250 million years. And it is equally likely to encounter any of the disc stars (conservatively 50 billion stars). This means that the mean time between encounters of the sun is on the order of ten quintillion years.

It isn't impossible. It is quite easy to design a free-return flight to the moon. There is no reason why a free-return flight to another star can't be done. However, space is big "therefore ʻOumuamua will never return".

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  • $\begingroup$ How does one arrive at "250 million years" and "ten quintillion years"? If it's an estimate can you mention how it was done, if there's a source, a mention would be great. Thanks! $\endgroup$
    – uhoh
    Commented Mar 19, 2019 at 0:50
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    $\begingroup$ He was doing a back-of-the-envelope calculation. The point is that Oumuamua is really no more likely to come back around to the sun that it is to have an encounter with any other star. So he assumed an average number of years between stellar encounters, and approximately multiplied that by the number of stars. $\endgroup$
    – Mark Foskey
    Commented Mar 19, 2019 at 3:55
  • $\begingroup$ @MarkFoskey So where does the "average number of years between stellar encounters" come from? I'm asking how the values "250 million years" and "ten quintillion years" are obtained. If you can reproduce them that would be great, thanks. $\endgroup$
    – uhoh
    Commented Mar 19, 2019 at 8:27
  • $\begingroup$ The problem with back-of-the-envelope calculations is that because they are not precise, each estimate is different. The strength of back-of-the-envelope calculations is that they give you good order-of-magnitude results. If you estimate 1 billion years, it might be 100 million years and it might be five billion years -- but it isn't going to be 106 years next Tuesday. $\endgroup$
    – Mark Olson
    Commented Mar 19, 2019 at 11:21
  • $\begingroup$ Well, actually I think 250 million years is a gross underestimate. 'Oumuamua is no more likely to have a close encounter with a star than the Sun is to have such an encounter with another star, and any such encounter in the past few billion years would have disrupted the solar system. But, rounding, 250 million ~ 10^8, 50 billion ~ 10^11, and their product is 10^19. A quintillion is 10^18. Since 250 million is a clear underestimate, this makes the case that a double encounter would be vanishingly rare. $\endgroup$
    – Mark Foskey
    Commented Mar 19, 2019 at 14:45
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tl;dr: It's a hard problem. While this answer says

However, space is big "therefore ʻOumuamua will never return".

I say never say never!

What natural mechanisms could lead to the unlikely case of the same rogue asteroid or planet passing through our solar system twice?

I can think of three mechanisms.

1. Stellar and solar-system pinball

Under normal situations an asteroid or planet could not swing around a massive object, make a U-turn, and come straight back nearly along it's same path after a 180° deflection. However deflections by two or more large bodies can do it. Those two bodies could be

  • Two unrelated stars: This would make a triangular path; Sun, star-1, star-2, Sun again.
  • A double star: Three-body unbound orbits are crazy and chaotic and just about anything can happen including a 180°U-turn. This is my favorite since it only requires one encounter and star systems of two or higher are pretty common.
  • A star and its planet: Stars with Jupiter-size (and larger) planets are pretty common, and a hyperbolic orbit around a star doing much/most of the bend with the planet doing the rest of the 180°U-turn is possible.
  • other variations on this theme.

2. General Relativity (or Neutron Stars)

  • A black hole: As explained in answer(s) to Could a trajectory around a large mass ever deflect by more than 180 degrees due to general relativistic effects? "...there is no limit on how many "laps" a hyperbolic orbit can make before it returns to infinity." That's in the limiting case of a zero-mass object. For large mass ratios such as an asteroid or planet versus a black hole, the maxiumum number of turns an unbound orbit can make is about 0.41 times the mass ratio, which would be greater than a million for an Earth-sized planet and a vanilla stellar-collapse black hole.

  • A neutron star: I am not certain yet if the GR effects due to a neutron star are strong enough to produce a 180°U-turn without a destructively close approach; I'm currently working on this...

3. Galactic roller-coaster

The solar system orbits the galactic center at about 220 km/s with a period of about 250 million years. That speed is an order of magnitude or possibly two faster than the orbital speeds of asteroids and planets within solar systems. Jupiter orbits at 13 km/s and (486958) 2014 MU69 (Ultima Thule) moves at 4.6 km/s. So no mater which direction our rogue friend leaves the solar system, it's going to be moving around the galactic center with a nearly parallel orbit to that of the solar system, and it's going to be doing a complicated dance above and below the galactic plane as our solar system is doing, due to the restoring force of the galactic disks's gravity. This can bring us and our rogue friend back together in a way that would at first surprise us both until we remember how stars move in the disk.

Overall probability?

For all of these the probability is small, but the single collision options (double stars, solar systems, black holes, neutron stars) may make the probability better than the two unrelated star sequential scattering scenarios, and the galactic plane oscillations don't require a near approach to any other body at all.

So I think the statement:

...Therefore it will never return.

could be considered cavalier and possibly scientifically unsupported.

It's a hard problem, so never say never!

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  • $\begingroup$ Why the down vote? Is this not in fact a correct answer to What natural mechanisms could lead to the unlikely case of the same rogue asteroid or planet passing through our solar system twice? $\endgroup$
    – uhoh
    Commented May 10, 2020 at 2:01
  • $\begingroup$ It's rare for me to answer my own question and rarer-still for me to accept it, but as OP I feel this is the best answer. Please feel free to convince me otherwise! $\endgroup$
    – uhoh
    Commented Dec 9, 2020 at 23:14
  • $\begingroup$ "Under normal situations an asteroid or planet could not swing around a massive object, make a U-turn, and come straight back nearly along it's same path after a 180° deflection. " This is only true from the perspective of the reference frame where the object is stationary, from other reference frames the asteroid can appear to turn 180. A rouge gas gianst travelling through the oort cloud could give what is a 90 degree trun from it's reference frame but a 180 degree turn from the perspective of the sun's reference frame. $\endgroup$
    – blademan9999
    Commented May 17, 2023 at 11:35
  • $\begingroup$ @blademan9999 I think that a "rouge gas gianst travelling through the oort cloud" would not be considered "under normal situations". I think as written my statement is fine. Why not add a supplementary answer beginning with something like "While uhoh mentions 'normal situations' there are certainly other kinematic scenarios we can consider." I think that would be great! :-) $\endgroup$
    – uhoh
    Commented May 17, 2023 at 14:00
  • $\begingroup$ But my coment also applies to any passage near a star, Unless the star has a VERY low proper motion relative to the star, it will be possible for the asteroid to turn 180 degrees form the perspective of observers in the Sun's reference frame. $\endgroup$
    – blademan9999
    Commented May 18, 2023 at 10:04

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