Is it possible for two stars to exist close to each other?

"Close" is relative; let's assume that two stars are close to each other if they are at the center of the same solar system.

It's possible for this to happen with 3 stars. But what about more? How many stars can be the center of exactly one solar system without collapsing because of gravity?

  • $\begingroup$ Sure, if you could get them to orbit eachother. $\endgroup$
    – user19
    Commented Sep 24, 2013 at 22:30
  • $\begingroup$ @Undo Well, if you could get them to orbit each other, one of them will eventually fly off and never come back. The probability of it staying in is well less than 1%, and under specific conditions of mass and orbital parameters. $\endgroup$
    – Cheeku
    Commented Sep 24, 2013 at 22:31
  • 2
    $\begingroup$ @Cheeku in an infinite universe, likelihoods are more significant ;) $\endgroup$ Commented Sep 24, 2013 at 22:33
  • $\begingroup$ @ZoltánSchmidt yeah! I think I have read that such a system was found. I am searching for the paper to give an elaborate answer. $\endgroup$
    – Cheeku
    Commented Sep 24, 2013 at 22:35
  • $\begingroup$ CHECK YT for N-Body Stable Orbits, video sims youtube.com/results?search_query=n+body+stable+orbits $\endgroup$ Commented Jan 3, 2018 at 4:24

2 Answers 2


Well, if they have an angular momentum with respect to their center of mass, or in other words , if they are in orbit, gravitational collapse is not the issue.

This is what happens when three bodies are in orbit around each other. Don't get disappointed so soon. There is indeed a way to avoid this. If the three bodies are of comparable mass, and one of the bodies is situated at L4 or L5, tuning the orbital parameters may work just fine, but then you gotta worry about tidal forces.

Lagrange points

So, if the size of our third star exceeds the area given to it at L4/L5, it will experience tidal forces and will start ejecting mass to either of the two other stars, or both.

But as Zolan commented above, astronomy with an infinite universe has a possibility for everything. What are listed above are the most common problems which arise, that is the reasons which are mainly responsible for making such systems highly unlikely of existence.

But here's a system which works the way you described :

A binary system exists, and the third star is far enough that it experiences the gravity as that of a single star. And the nearest planet(as you want it to be a "solar system"), is so far that it experiences the gravity of the three stars as that of one. Now, since "close" is a relative term, for the planet, the three stars are appreciably close.

So, the answer to your question, YES and NO

  • $\begingroup$ We have found as many as 7 stars in a single star system. $\endgroup$
    – Caters
    Commented Oct 10, 2014 at 3:51
  • $\begingroup$ @caters Post your report link here. I want to see it. $\endgroup$
    – Cheeku
    Commented Oct 10, 2014 at 7:56
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    $\begingroup$ en.wikipedia.org/wiki/Star_system#Septuple $\endgroup$
    – Caters
    Commented Oct 10, 2014 at 9:08

At least 7.

We currently know of two star systems with 7 stars: Nu Scorpii and AR Cassiopeiae. The two have different structures, both of which are complicated but appear to be stable on stellar timescales.

Nu Scorpii

This system has two components, Nu Scorpii AB and Nu Scorpii CD.

  • Nu Scorpii A is a triple star system itself, with a binary system in the center. The inner two stars (Aa and Ab) cannot be resolved, in part because the fainter of the two is quite dim. The third star, Nu Scorpii Ac, is also faint. Nu Scorpii B is a single star orbiting Nu Scorpii A.
  • Nu Scorpii C is also an individual star and is orbited by Nu Scorpii D, which is likely, like Nu Scorpii Aa and Ab, a binary system that cannot be resolved.

Nu Scorpii AB and Nu Scorpii CD have a long orbital period (compared to the days-long period of Nu Scorpii Aa and Ab), likely in the hundreds of thousands of years.

AR Cassiopeiae

This system has a slightly different arrangement. It consists of a central triple-star system (AR Cassiopeiae AB) orbited by two binary systems (AR Cassiopeiae CD and FG).

  • AR Cassiopeiae A is an eclipsing binary with a period on the order of days, just like with Nu Scorpii A. AR Cassiopeiae B orbits these two at a large distance.
  • AR Cassiopeiae C and D are similarly sized stars, although they are faint and hard to distinguish.
  • AR Cassiopeiae F and G form a system similar to AR Cassiopeiae CD: two stars of similar masses, orbiting the primary triple-star system at the same distance (to within an order of magnitude).

As with Nu Scorpii CD, AR Cassiopeia CD and FG orbit AR Cassiopeiae AB on a timescale of hundreds of thousands of years.

The key thing to note about these high-multiplicity multiple-star systems is that their components are often binary systems themselves. In other words, you don't have a case where there's one or two central stars, with the rest independently orbiting them, or where you have a bunch of stars independently orbiting the same point. It's much stabler to create a few tight binaries and then set them in motion - much like a normal triple-star system, except each component is actually two stars.

So, why haven't we found star systems with more than 7 stars in them? There are a few possible reasons:

  • We can't resolve certain components that we think are only single stars. We only know that Nu Scorpii Aab is a binary system by looking at its spectral lines and seeing how they shift over time as the components move in their orbits around each other. It's possible that we can't even detect these spectroscopic binaries in some systems.
  • The initial protostellar cloud may not be massive enough for these systems to form easily. To form a system of 7 stars, unless one or more components were captured by the others (which seems unlikely, given that the arrangements are delicate), you need either one very, very massive cloud (if the fission hypothesis somehow holds for the entirety of the system - which is probably not the case) or several massive clouds collapsing around the same time as each other, which also may not be plausible.
  • It is likely that such systems are unstable to either complete collapse (i.e. ejecting members from the system) or losing members to passing stars. Given the high separations necessary for the components to orbit stably, it's not far-fetched that a passing star could perturb them enough to leave the system.
  • $\begingroup$ Incredible! My head is spinning. It looks like each step in the hierarchy is one or two orders of magnitude in separation. Although I've currently accepted the answer to the question State vectors of “interesting” multiple stars there is always room for a better one. $\endgroup$
    – uhoh
    Commented Jan 3, 2018 at 3:37
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    $\begingroup$ @uhoh You should check out Cris Moore's insane braided orbits. See tuvalu.santafe.edu/~moore/gallery.html for links to animated diagrams and papers. Of course, these orbits are extremely unlikely to occur naturally. They're reasonably stable to small perturbations of position and velocity, but not to variations in the masses, which need to be almost identical. $\endgroup$
    – PM 2Ring
    Commented Jan 4, 2018 at 21:58

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