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From an interesting ScienceDaily article, I read this

Before the groups of stars disperse, binary stars move through their birth sites and the group studied how they interact with other stars gravitationally. "In many cases the pairs are torn apart into two single stars, in the same way that a pair of dancers might be separated after colliding with another couple on a crowded dance floor," explains Michael Marks, a PhD student and member of the International Max-Planck Research School for Astronomy and Astrophysics. The population of binaries is therefore diminished before the stars spread out into the wider Galaxy.

The stellar nurseries do not all look the same and are crowded to different extents, something described by the density of the group. The more binaries form within the same space (higher density groups), the more interaction will take place between them and the more binary systems will be split up into single stars. This means that every group has a different composition of single and binary stars when the group disperses, depending on the initial density of stars.

Here's the question: what are the types of gravitational interactions that make binary systems split up into single stars?

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All of them? I don't understand the question. Depending on mass and distance of the interloper, any close encounter has the potential to "pull" one of the stars out of the dance. How do you define differences in the "types" of interactions? –  ghoppe Sep 19 '11 at 12:04
    
Hm, well, what physically happens when a star gets pulled out? Is it analogous to the situation where a spaceprobe gets accelerated by Jupiter's gravity? –  InquilineKea Sep 19 '11 at 18:22
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@InquilineKea It's complicated, but yes, overall the process is somewhat similar. It requires 3 or more stars, usually a close binary and a visitor, or 2 pairs of close binaries, etc. When they get close to each other they start a complex "dance" around the common barycenter. Rarely, it so happens that one of the stars, already possessing a momentarily high speed due to the "dance", gets pulled close to a close group of stars moving fast in the opposite direction. As the sole star orbits the group, it steals kinetic energy, so its speed gets even higher. It's spit out at high velocity. –  Florin Andrei Sep 21 '11 at 18:03

2 Answers 2

up vote 4 down vote accepted

2-body situations alone (pairs of stars) cannot eject anything, usually. What happens a lot is that a close binary is visited by another star, or by another close binary. The outcome is hard to visualize, but sometimes it ejects one star from the group at high velocity. It could eject one of the binary components, or the visitor - all bets are off.

I used to do simulations like that, way before, and I was always amazed by the speedy rogues that were ejected from populous groups. You could do that yourself with the Universe Sandbox:

http://universesandbox.com/

If you can't setup a binary + singlet collision, just select the galaxy collision and watch the rogues spewed out.

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Wow - amazing game! –  InquilineKea Sep 21 '11 at 18:27

It is just one interaction - gravity:

As you mention in your comment, it is exactly analogous to the situation where a spaceprobe gets accelerated by Jupiter's gravity.

In a cluster of stars, any time a pair of stars comes close to another pair or more (close here just means close enough for one of the pair to be affected more than the other) the difference in forces can accelerate them in very different directions.

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Two-body interactions rarely eject stars out of a cluster. It's usually 3-body or more (close binaries plus a visitor, or two binary pairs) that project high-velocity rogues out of the group. –  Florin Andrei Sep 20 '11 at 20:52
    
You are absolutely right @Florin - amended appropriately. –  Rory Alsop Sep 21 '11 at 0:16
    
Oh I see - so since it's 3-body, it can't be analytically solved in most cases, right? –  InquilineKea Sep 21 '11 at 18:27

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