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So is it 'possible' for 3 stars to maintain at least a somewhat stable triangular configuration?

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- #1

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So is it 'possible' for 3 stars to maintain at least a somewhat stable triangular configuration?

- #2

.Scott

Homework Helper

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If three stars are placed into such an orbit, then the orbit of the smallest one will always be unstable - a slight change in its orbit which moves it towards either of the other two stars will result in a continued acceleration in that direction. That will break the "triangular configuration" you are looking for. But a figure 8 configuration might be possible.

Here is a link to a page that describes some of these orbits.

http://www.atlasoftheuniverse.com/orbits.html

- #3

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Can you essentially place 3 stars on equidistant points on a circle?

As a result I don't think there are 'smallest' orbits like you said.

- #4

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You really should look at the link he provided. That shows the possible stable orbits of 3 stars. It answers your question, which is why he provided it.

Can you essentially place 3 stars on equidistant points on a circle?

As a result I don't think there are 'smallest' orbits like you said.

- #5

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But anyway no, I didn't understand if it would answer my question. I was unclear if he understood my question properly. I hadn't understood if the page considered my question.

Additionally of course this doesn't explain what the issue with my triangular orbit.

- #6

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But he DID explain why your proposal is not a stable orbit although I suppose if you were to insist on the staggeringly unlikely existence of 3 stars of identical mass, his explanation would not hold.Additionally of course this doesn't explain what the issue with my triangular orbit.

- #7

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These orbital ideas were shown and mathematically proven by Kempler. They ARE special cases, and would tend to be either extreme coincidence, or an artificially made structure. Has shown up in various Science Fiction at times, but is based on real math. (Larry Niven does a good example of the form).

- #8

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Lagrange points are stable - if the mass of two orbiting bodies combined does not exceed about 3,8 % of primary mass.

Therefore, it should be possible to have an equilateral triangle of a B dwarf of 5,0 solar masses and two red dwarfs of 0,09 solar masses each.

- #9

.Scott

Homework Helper

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By smallest one, I meant the smallest star, not the smallest orbit.

Can you essentially place 3 stars on equidistant points on a circle?

As a result I don't think there are 'smallest' orbits like you said.

If you put three stars of the same mass into an orbit about a common center of gravity where they start out as points on an equilateral triangle with circular orbits, ideally they would stay in those orbits and remain equidistant from each other. But that is not stable. Any slight variation will become amplified and the paths will no longer be circles or ellipses.

The only reason I picked on the smallest one is that, if it is small enough, you basically have a two-body system (which is stable) with a small visitor. So there is always a problem with nudging the smallest one.

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Alright well that answers the post. Thanks folks, not necessarily that I understood all words and explanations, but got the idea.

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No. The stability of Lagrange points depends on mass. For suitable masses, they are stable.If you put three stars of the same mass into an orbit about a common center of gravity where they start out as points on an equilateral triangle with circular orbits, ideally they would stay in those orbits and remain equidistant from each other. But that is not stable. Any slight variation will become amplified and the paths will no longer be circles or ellipses.

- #12

.Scott

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With a triangle, none are at the Lagrange point of any of the others. I believe I read that with six, there is more stability because they do line up with Lagrange points.No. The stability of Lagrange points depends on mass. For suitable masses, they are stable.

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With a triangle, none are at the Lagrange point of any of the others.

Um, no. Lagrange points L4 and L5 each form an equilateral triangle with the other two masses.

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Normal Lagrangian Points L3, L4 and L5, along with the 'prime' mass, about a central point, you would need to add points between L3 and L4, and Between L3 and L5 at the same 60 degrees between masses and orbital distance/speed. As far as I know this is the only basic known theoretical stable Kempler Rosette, with 7 masses being the most stable, but it is of a special set of very unlikely configurations for it to happen very often naturally.

3 masses would be subject to outside distortion of orbits with no or diminishing stabilization from the other masses, the old 3-body problem rears it's head. The 6 or preferably 7 mass solution gets around that as it is a self-reinforcing system of reverberations of orbits that is able to maintain a reasonably long-term stability.

However, Space and Time are a very large playground within to work, and all of that energy matters!

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No. The stability of Lagrange points depends on mass. For suitable masses, they are stable.

And it seems to work with main sequence stars (until the O-type star goes supernova).

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Um? And how stable is that if the masses are slightly different, and distances slightly deviate?.Scott, I believe you are correct that with 6 masses of same size and orbital speed and distance, within a plane, then they should be able to hold their form,

- #17

Tom.G

Science Advisor

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Use these searches and try it yourself:Um? And how stable is that if the masses are slightly different, and distances slightly deviate?

https://www.google.com/search?&q=planetary+orbit+program

https://www.google.com/search?&q=n+body+orbit+simulator

You can probably find other online interactive or downloadable ones too.

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Like I said before, they would be an extremely unlikely thing to form in Nature, but Not Impossible.

But yes, take the time to play with some of the orbital dynamics programs and have then extrapolate for a few millions of orbits with different patterns and setups. There are literally endless possibilities of combinations

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Otherwise, you have to pack binaries in orbital hierarchies. Two stars orbit each other. Another two stars orbit each other. The

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