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Just like binary stars can a system of three stars mutually equidistant from each other? What I mean is three stars at the vertices of a equilateral If yes than what will be the orbit of a planet out there If no what will be the problem for it's nonexistence?

Heth Ghetiya
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2 Answers2

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It is possible in a Trojan configuration:

enter image description here

In the place of the "Planet" on the image, also a small star could exist. The third star would be at $L_4$ or at $L_5$. This configuration could be made stable.

However, as this link shows,

In unnormalized units, this criterion becomes

$$\frac{m_2}{m_1+ m_2} < 0.0385$$

We thus conclude that the $L_4$ and $L_5$ Lagrange points are stable equilibrium points, in the co-rotating frame, provided that mass $m_2$ is less than about $4\%$ of mass $m_1$.

Thus, the mass of the second star should be at most 3.85% of the central star.

As far I know, no such known star system exists, but if it would, it would be stable.

Stable planetary orbit is possible to

  • either very close to one of the stars (compared to the size of the triangle)
  • in the other Lagrange point
  • or very far from all of them.

If the triangle is big, then a planet even in the habitable zone is possible.

peterh
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    3.85% is not a problematic percentage. Stellar mass can vary by about 3 magnitudes (0.1 solar mass to 100 solar masses). However, it appears that the third star can be only 10% of the mass of the second start, so 0.385% of the central star. And at this point, that means short-lived O/B stars with a pair of red dwarfs. And at that point, we get into evolutionary questions. Are these dwarfs captured? If not, how did they evolve in the proximity of the central star? – MSalters Sep 01 '19 at 00:15
  • @MSalters that's simple. It's part of a Type-IV civilization's signature/watermark they insert into any artificial galaxy they create. Or the entire signature of a Type-III civilization's artificial star systems. – John Dvorak Sep 01 '19 at 04:27
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    @MSalters The smallest known star has 0.085 solar mass. Thus, the central star should have at least 2 solar mass. It could live around 2billion years before going to nova. Such a star system is not impossible, only very unlikely. – peterh Sep 01 '19 at 07:24
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    @MSalters If the third star could be at most 10% of the second, then the smallest possible star masses are 0.085, 0.85, $\approx$ 20 solar masses. It makes the life of the central star to below 10million years. – peterh Sep 01 '19 at 07:57
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    @John I'd be more impressed with one of Cris Moore's n-body systems, even just the basic 3-body figure 8. It's reasonably stable to small perturbations of position and velocity, but the masses do need to be fairly close to identical, IIRC. – PM 2Ring Sep 01 '19 at 09:33
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    @PM2Ring You can put a planet at the other Lagrange point, or to a different orbit around the main star. – John Dvorak Sep 01 '19 at 10:05
  • @John Indeed! I was just prompting Peter to complete his answer. – PM 2Ring Sep 01 '19 at 10:10
  • If the planet is the second star, with no more than 3.85% the mass of the first/central star, then what's the mass limitation of the third star at the Lagrange point? – candied_orange Sep 02 '19 at 12:07
  • @candied_orange I did not find a reference for that, but the first comment refers here another answer, and estimates that it can be at most 10% of the second. In this case, we have star masses 0.085, 0.85 and around 20 solar mass. It makes the lifetime of the central star below 6million years. – peterh Sep 02 '19 at 12:49
  • @peterh what if the central star is a black hole whose accretion disk is just small enough to not fry all life orbiting it, but big enough to still be visible? – John Dvorak Sep 02 '19 at 12:57
  • @JohnDvorak The question asks for stars. Black holes are not stars. Actually, they radiate a lot, because a large part of the mass of the infalling matter is radiated away as photons. Their radiation is hard to differentiate from the neutron stars, but it is possible. It could work - the gravitational field of a black hole (except close to the event horizon) is the same as of a star with the same mass. – peterh Sep 02 '19 at 13:06
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Systems of three stars can exist, but a system of three stars in a triangle is unstable and won't exist in reality. There are configurations of three stars that are stable, for example, two stars in a close orbit about their common centre of gravity, and a third star in a distant orbit.

Planets can exist in such a system, they could orbit around the distant third star (like a moon orbits a planet), or they could be circumbinary, around the two close stars. Such complex systems are more likely to be unstable on the scale of billions of years. The key to stability is having each body in approximately an inverse-square gravitational field, so its orbit can be approximated by a Keplerian ellipse. That is not the case if three equal mass bodies are in an equilateral triangle.

James K
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  • Why? If there is a huge star, then two other stars could orbit it. They would be each others' Trojans. – peterh Aug 31 '19 at 16:34
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    There are stable equilateral systems, but not with equal masses. There's some info here & anims here. For some interesting stable n-body systems of equal mass see Cris Moore's gallery. Eg, the figure-8 orbit – PM 2Ring Aug 31 '19 at 16:44
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    Beta monocerotis is an example of "two stars in a close orbit and a third star in a distant orbit". As a bonus, the 3 stars are very similar in size, mass and temperature. The triple looks wonderful in an amateur telescope. https://www.cloudynights.com/topic/91088-colors-of-beta-monocerotis/ & https://en.wikipedia.org/wiki/Beta_Monocerotis – Eric Duminil Sep 01 '19 at 11:12
  • @PM2Ring, The figure 8 orbit etc are only stable in the sense that a pencil balanced on its point is stable. Perturb the orbits and they very rapidly and catastrophically stop being stable. With three stars of roughly equal mass the thes orbits are not stable over the medium to long term. – James K Sep 01 '19 at 18:46
  • @James As I said here IIRC the 3 body figure 8 is reasonably stable, but the masses do need to be almost identical. But it's over a decade since I read the paper, and my phone won't let me read the article linked from Moore's site. – PM 2Ring Sep 01 '19 at 19:50