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I know next to nothing about astrophysics, but I've read a few Wikipedia articles, and it seems like a "normal" kind of binary system includes two stars of similar mass orbiting a common center. I know that there are a couple types of planet; the kind that orbits one of the stars, and the kind that orbits both (orbiting, I believe, around the common center of the stars.) I don't know how fast any of these things actually happen, though. Is it feasible that a planet of the second type (in the habitable zone) orbit at a speed such that it and the star complete one revolution in the same time?

Bonus: What effects would this have, if possible, on the planet's physiology/seasons?

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The second kind is called a circumbinary system. The short answer is no, the planet will take much longer.

The formula for the orbital period $T$ in such a system is

$$T = 2\pi \sqrt{\frac{a^3}{G(M_1+M_2)}}$$

where $a$ is the orbital distance (technically the semi-major axis of the motion of a body in the frame of reference of one of the stars), and $M_i$ are the masses of the stars (in such a system, the mass of the planet is negligible), And $G=6.67\times10^{-11}$ is the gravitational constant.

The only variable that determines the orbital period in such a system is the orbital distance $a$, and since in a circumbinary system, the orbital distance between the stars is much less than the distance from the stars to the planet, it follows that the orbital period of the stars must be much less than that of the planet.

The effect is that at most times, two "suns" would be visible, They would appear close together in the sky but sometimes one would be eclipsed by the other. The eclipses would be quick, probably less than an hour.

A few numbers: if the two stars were of a similar mass to the sun, and orbited each other at 20 solar-diameters (so they would appear about 1 "hand" apart in the sky) they would have an orbital period of 2.3 Earth days. Whereas the planet would have a period of about 400 days (depending on how you calculate "habitable".

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  • $\begingroup$ What is G in the formula you used? $\endgroup$
    – CHEESE
    May 12, 2017 at 18:38
  • $\begingroup$ G=6.76e-11, it is the graviational constant $\endgroup$
    – James K
    May 12, 2017 at 20:37
  • $\begingroup$ Your link isn't working ... $\endgroup$
    – Glorfindel
    Feb 24, 2022 at 21:59
  • $\begingroup$ Perhaps that's better $\endgroup$
    – James K
    Feb 24, 2022 at 22:10

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