13
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ Maybe not equidistant, but for example: Alpha Centuari A/B + Proxima Centuari $\endgroup$
    – CrazyPyro
    Sep 2, 2019 at 16:58

2 Answers 2

15
$\begingroup$

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.

$\endgroup$
11
  • 5
    $\begingroup$ 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? $\endgroup$
    – MSalters
    Sep 1, 2019 at 0:15
  • 2
    $\begingroup$ @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. $\endgroup$
    – peterh
    Sep 1, 2019 at 7:24
  • 1
    $\begingroup$ @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. $\endgroup$
    – peterh
    Sep 1, 2019 at 7:57
  • 2
    $\begingroup$ @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. $\endgroup$
    – PM 2Ring
    Sep 1, 2019 at 9:33
  • 1
    $\begingroup$ @PM2Ring You can put a planet at the other Lagrange point, or to a different orbit around the main star. $\endgroup$ Sep 1, 2019 at 10:05
9
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Why? If there is a huge star, then two other stars could orbit it. They would be each others' Trojans. $\endgroup$
    – peterh
    Aug 31, 2019 at 16:34
  • 3
    $\begingroup$ 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 $\endgroup$
    – PM 2Ring
    Aug 31, 2019 at 16:44
  • 2
    $\begingroup$ 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. cloudynights.com/topic/91088-colors-of-beta-monocerotis & en.wikipedia.org/wiki/Beta_Monocerotis $\endgroup$ Sep 1, 2019 at 11:12
  • $\begingroup$ @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. $\endgroup$
    – James K
    Sep 1, 2019 at 18:46
  • $\begingroup$ @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. $\endgroup$
    – PM 2Ring
    Sep 1, 2019 at 19:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .