6
$\begingroup$

Here is a link to an online animation of stars' movements in a trinary system.

Basically, the three stars are orbiting around a barycenter; one (yellow in the animation, mass = 1) closer to it and the other two (red and blue; mass = 0.25) orbiting around each other as well.

How far apart would the planets and stars have to be for there to be a stable orbit in a habitable zone around either the yellow star or in a circumbinary orbit around the red and blue stars?

The mass of the yellow star can vary as needed, as long as it is roughly 4 times the mass of red and blue as in the animation.

$\endgroup$
  • 1
    $\begingroup$ I'm going to edit this question, because it is interesting mathematically, to make it less broad and more mathematically rigorous. I'm going to add the hard-science tag, which is not something I normally would do. Feel free to remove it if you want. $\endgroup$ – kingledion May 13 '17 at 2:43
  • $\begingroup$ This question won't get answered right away, so you aught to be patient, but I know that I, at least, will come back to it eventually. $\endgroup$ – kingledion May 13 '17 at 2:48
  • $\begingroup$ @kingledion I agree with everything, except is there a scientific reason why you wrote that the mass of the yellow star should be roughly sun-like? I don't necessarily feel that way. $\endgroup$ – CHEESE May 13 '17 at 2:54
  • 2
    $\begingroup$ Can you even get a stable system with four bodies? $\endgroup$ – Joe Kissling May 13 '17 at 3:32
  • 1
    $\begingroup$ @CHEESE Stars much larger than Sol (our sun) generally have much shorter life-spans. In order to develop life, which is what you want to know about, the star can't be too big. An estimate of stellar lifespan is 10$^10$ years, times the -2.5 power the star's mass in Sol-equivalents. The sun's life is 10 billion years, but a star 1.5 times the size of Sol has a life of only 3.6 billion years, not long enough for life. Hence, the star must be Sol-like or else it will turn to a red giant and die, and in its death probably exterminate any life developing around the other stars as well. $\endgroup$ – kingledion May 13 '17 at 11:12
3
$\begingroup$

This kind of system could definitely host a planet given certain constraints.

A useful rule of thumb for long-term stability in a gravitational orbit is that all orbits must be separated at least by a factor of 7 in radius (see, e. g. Publications of the Astronomical Society of the Pacific, 115:825–836, 2003 July). For example, a planet could orbit at 1 AU from a star, and both can orbit a barycenter with another star which is at 7 AU a safety margin, use a factor of 10.

Having the planet orbit the single big star is easy. You can put it at 1 AU, and place the binary at least 10 AU away from the big star.

If you want to have the planet orbit a single star in the binary, you need a bigger system. You could have the planet at 0.25 AU from one star, have the binary stars orbit eachother at 2.5 AU, and have the third star and the binary orbit eachother at 25 AU.

Having the planet orbit the both stars in the binary is unfortunately very difficult. The problem is that the habitable zone of a 0.5 solar mass star is only 0.25 AU from the star (L~M^4). This means that to maintain the appropriate separation you would need the stars to orbit eachother at only 0.025 AU or around 5 solar radii. I think the stars can escape collision, but a system forming this way seems unlikely due to the tight constraints.

If you scaled all of the masses up to avoid this problem the big star would die relatively quickly (within a few billion years), which might have a disastrous effect on life developing on the planet.

If you want a circumbinary planet, you should probably have all the stars have the same mass as the Sun - then the binary could orbit at a more comfortable 0.1 AU and the planet could be placed at around 1 AU, with the third star much farther away.

Edit: Added a reference

$\endgroup$
  • $\begingroup$ Welcome to Worldbuilding! For your first answer you have tried a hard-science question which has specific requirements for answers. Please read the tag description and the yellow post notice! If you’re unfamiliar with SE in generel, start with the tour. $\endgroup$ – JDługosz May 14 '17 at 15:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.