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Assuming an identical Solar system except for the strange sun(s) and that

  • the binary star system's barycentre coincides with the Sun's centre of mass
  • mass of the combined star system is equal to the mass of the Sun
  • the stars are identical with a radius 2-1/3 times that of the Sun (assuming density same as the sun: is this possible, given the other constraints?)
  • combined surface luminosity when the stars are equidistant from the earth (as felt from the earth) is equal to the surface luminosity of the Sun
  • rotational period of the binary star system is of the order of a few days, say a week

how would things like day and night, seasons, eclipses etc. change?

I hope the orbit of the binary star system being coplanar with the planets would be enough for stability. Please correct me if I'm wrong.

There is a similar question on WBSE, but with different premises. I can't seem to find my answer there anyway.

EDIT: I did the calculations for relations between sequence star size, mass, temperature and luminosity and it turns out that selecting any one parameter fixes the rest as well. So such a system cannot be identical to the solar system and even if it was in terms of geometry, the suns would be too cold to allow any life on Earth.

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closed as too broad by Aify, sphennings, Thucydides, Ash, adaliabooks Oct 8 '17 at 10:23

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ There are so many variables which you have left unspecified that the only possible answer is that all the things you mentioned would change in accordance with the new conditions. Please edit your question and specify the masses of the stars, their radii, the distance between them, their surface luminosities, and the size, rotation speed and orbital radius of the planet in question; if necessary, also a description of its orbit. A hint as to why the system is stable would also be appreciated. $\endgroup$ – AlexP Oct 7 '17 at 12:23
  • $\begingroup$ This refernce might be of some interest arxiv.org/abs/0705.3444 $\endgroup$ – Slarty Oct 7 '17 at 22:00
  • $\begingroup$ Such a system is not possible. Two stars with a combined mass equal to the sun, such that each is half as massive as the sun, cannot have a combined luminosity equal to the sun. A star half the mass of the sun will have a luminosity a bit less than 9% that of the sun, so two such stars will have a combined luminosity of less than 1/5th that of the sun. If everything else about the system remains the same, the Earth will simply freeze. $\endgroup$ – Logan R. Kearsley Oct 8 '17 at 2:09
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A close binary with identical or near-identical members seems to me a bit unrealistic, but it's not impossible. For two half-solar mass stars to orbit their barycenter with a period of one week, they would have to be about 10 million km from it on average. That's not very far away; tidal effects of each star on the other one will be considerable, though I haven't the faintest idea how that would play out. Also, you didn't mention other planets, but if there's a Mercury there, I don't know if its orbit will be stable.

But you're interested in Earth. If the orbits of the star pair and the Earth are approximately coplanar, you'll expect to have an eclipse twice on every orbit. As one star blocks the light of the other one, the total energy received on Earth will drop by half. Since each star has a diameter of 1.1 million km and their average orbital speed is 55.6 km/s, a very crude calculation says that the eclipse should last the time that one star takes to move 550000 km at that speed (this is half the 1.1M km because both stars will be moving in apparently opposite directions); that's about 2 hours 45 minutes. This is a grossly simplified calculation and it won't hold if the stars and Earth are not exactly coplanar, or if their orbits are not exactly circular, and many other factors.

At maximum separation, the stars will be about 20 million km from each other; if the line that joins them at that time is precisely perpendicular to the line that joins Earth and the stars' barycenter, then they'll appear in Earth's sky about 7.6° apart, or about 14 times the average apparent diameter of the Moon. If this happens at sunrise, the second star will rise about half an hour after the first one (that's ${7.6 \over 360} \times 24$ hours), and likewise at sunset.

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    $\begingroup$ Mercury gets as close as 46 Gm to our Sun at perihelion, and close to 70 Gm at aphelion. It's probably a pretty safe bet that closest approach being at about 35 Gm from half a solar mass, and up to 55 Gm from another half solar mass, can't be stable; the difference in gravitational attraction from the two stars would be considerable. It's not quite bad enough to cause black hole-style spaghettification (thankfully!), but the difference is going to cause distinctly different gravitational attraction in different directions, which certainly doesn't lend itself to orbital stability. $\endgroup$ – a CVn♦ Oct 7 '17 at 21:34

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