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Suppose two stars orbit each other and the planets orbit around the two stars' barycenter. I'm thinking of having at least one as a main sequence G or K class. For the sake of argument, let the other star be a red dwarf. How do I determine the habitable zone of this or any binary system?

Thanks!

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    $\begingroup$ How far away is the red dwarf? Binary stars can be extremely far away from each other such that the 2nd smaller star is irrelevant. $\endgroup$
    – Schwern
    Dec 27, 2022 at 2:44
  • $\begingroup$ Why, please? Is this about real astrophysics, or your own fictional world? $\endgroup$ Dec 27, 2022 at 22:03
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    $\begingroup$ @RobbieGoodwin I want my fictional worlds to possess enough science to balance the artistry whereby suspension of disbelief is maintained. $\endgroup$
    – Ylahris
    Dec 28, 2022 at 0:11

3 Answers 3

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Treat the two stars as a single star by summing their effects

This is not a perfect solution, but I'd avoid a perfect solution. I'll explain that momentarily.

Just as we reasonably simplify calculating the effects of gravity by representing any mass as a single, infinitely small point in space, treat your binary stars as a single star by summing up those effects (solar wind, luminosity, etc.) that contribute to determining the habitable zone. Most can be directly summed (e.g., the power output of the stars can be summed). Some will likely need to be normalized before summing (like dealing with the temperature of each star, if they're particularly disparate).

Once you've created your "representational" star, the calculations should be straightforward.

Why not try to make it perfect?

In a word: orbits.

Not only is the orbit of the two stars a significantly complicating factor (at times during the planet's orbit the lower mass star will be on the near side of the larger mass star or on the far side of the larger mass star), but the orbit of the planet itself is a significantly complicating factor.

Remember when I said that gravity calculations can be reasonably simplified by reducing the dimensions of a mass to a single infinitely small point in space? You can do that with your binary stars... reduce the source of the gravitational effect to a single infinitely small point in space representing both stars... but that point moves! And as it moves it throws the planet around with it.

Perfectly calculating the "habitable zone" of a binary star system doesn't result in a lovely ring of distance X from the star and width Y, as it does with a single star system. It results in a function f(x) from the star and a function f(y) for its width. That's messy. That's really messy. And due to the orbital complications I mentioned earlier, it might be true that given any binary star system no planet will consistently remain within the habitable zone.

But you're looking for suspension of disbelief, not specific reality. (Please tell me you're not looking for specific reality....) Which is why I'm advocating simplification and not trying to be perfect.

OK, so exactly how do I do what you're asking?

  • The simplest case: I upvoted @user98816's answer (you should, too) because he/she is right, if the primary star is much, much more massive than the secondary star, then you can simply ignore the effects of the secondary star and use the primary star as the single-star reference for your calculation.

  • The slightly more complex case: If the two stars are basically equal in size and type, then you can simply "double" the star (find the star on the star type charts that represents 2X the mass with the same temperature) and use that as your reference star.

  • And we're starting to get complex: If your stars are spread far apart, you'll discover that it's really hard to have any habitable zone due to how far away your planet must be. The distance between the pair matters a lot. Keep it as short as you can stand.

After that it gets more complex. Grab yourself a copy of Circumstellar Habitable Zones by Jimmy Newland (I think, it comes from his site, but the PDF itself doesn't identify an author) and work through the details. You'll need to find things like ambient luminosity, etc. But it shouldn't be difficult so long as you work through what needs to be normalized before summation. If you need additional help, asking for specific details in our chat, The Factory Floor, will work great.

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    $\begingroup$ Additional challenges are posed by orbit stability. Close-orbiting planets (i.e. orbits like those in our S.S.) in close binary systems will likely result in ejected planets within no more than a few GY. Stable orbits are possible, but difficult. A solution is to simply place your second sun far away, a few hundred to a few thousand AU or more. $\endgroup$
    – Izzy
    Dec 27, 2022 at 5:32
  • $\begingroup$ If the point you use for calculating the combined gravitational effect of the stars on the planet is the barycenter of the two stars, that point actually does not move. Its lack of movement is a major part of why it is useful to calculate with. $\endgroup$
    – Douglas
    Dec 27, 2022 at 7:24
  • $\begingroup$ @Douglas But that's an oversimplification that has consequences for calculating the orbit of a planet. I might not state it clearly enough in my answer, but there are negative consequences to oversimplifying anything. $\endgroup$
    – JBH
    Dec 27, 2022 at 7:28
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Ylharis! While not an expert, I have researched circumstellar habitable zones in the past, and I reckon that the answer you’re looking for is pretty much the same as a non-binary system, if the other star is a red dwarf. Such stars have a habitable zone so close that even mercury would be considered chilly if it orbited one, so while you may get some interesting lighting effects, assume a standard habitable zone of 1 Earth-distance if the primary is a G-class star, and about the same or maybe mars’ distance if it is a K-type star due to more intense radiation.

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The red dwarf has few orders of magnitude lower luminosity than a G or K main sequence star. This is why it can be completely skipped when calculating the habitable zone.

E.g. Our Solar system will look exactly like it is if any of our planets (except Earth, of course) is replaced with a red dwarf.

It will, of course, somewhat affect the stability of orbits, so Venus and Mars are probably best left alone if you want Earth for anything important (but see below).

In regard to replacing Jupiter or any other planet beyond the ice line with a red dwarf, one will get a tiny island of habitability around it, mostly unaffected by the far away Sun.

And then, the bordering option - e.g. a red dwarf at the Mars orbit. In this case you may arrange your planet to get most of its heat from its host red dwarf and its light for photosynthesis from the big, but far away bigger star.

The planet will be tidally locked to the red dwarf, so its host-star side will be a hot and maybe lifeless desert (or an ocean where most of the storms form) and its dark side will be something that has whatever moderate-ish climate you want and a day-night cycle.

p.s. be aware that the red dwarfs are rather nervous with their solar storms way stronger than whatever we have. Either get a very strong magnetic field for your borderline planet, or prepare to gradually lose its atmosphere and water.

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