Given the following planetary conditions, what would be the exact length of a year?

Question

List of conditions:

• it's an earth-size planet
• it's located in a binary star system, where both suns have the same size and mass
• the summed up mass of both suns is equivalent to that of our sun
• yes, the orbit is stable
• there is a moon the same mass as earth's moon, only less dense and therefore, bigger
• the distance between the stars is about 5 million km
• the distance between the planet in question and the suns is the same as our earth The masses for both suns are in my notes, I'll add them the minute I find them.

These are elements I'm not willing to change unless you provide solid arguments of them not making any sense. I've already done some rough calculations to make these things possible, but only in terms of making the orbit stable and for making a planet capable of supporting life. I don't know how to calculate how long would a year be in such a setting and I need this to create a calendar. To be honest, the moon here is a whole different question I'll address later so for your answer please just stick to what I wrote here for the time being. I'll edit later. Please do ask for more details if you feel the need. I have lots of information but I can't fit it all here without making this post seem messy and overloaded.

Answer 1

Frame challenge

Here's a problem for you:

• it's located in a binary star system, where both suns have the same size and mass
• the summed up mass of both suns is equivalent to that of our sun

does not square with

• the distance between the planet in question and the suns is the same as our earth

The problem is that the luminosity of a star doesn't scale linearly with its mass. A useful approximation is the mass-luminosity relation, which for stars between ~0.43x and 2x the mass of the sun looks like this: $$\frac{L}{L_\odot} \approx \left(\frac{M}{M_\odot}\right)^4$$ where $L$ and $L_\odot$ are the luminosities of your star and the Sun respectively, and $M$ and $M_\odot$ are the masses of your star and the Sun. A star with half the mass of the Sun has approximately 6% of its luminosity. A pair of such stars give you ~12% of the Sun's luminosity, but that's going to leave any world that's a whole AU out feeling very chilly. For specifics of what your stars might look like, have a read up on red dwarfs... a star with half the mass of the Sun would have a classification like M1V.

There are various habitable zone calculators out there (here's one) which you can use to work out how close your world would need to be. A radius of 0.33 to 0.58 AU seems like a reasonable estimate. The orbital period $T$ of a planet around a much larger body of mass $M$ is $T = 2\pi \sqrt{\frac{a^3}{GM}}$ where $a$ is the orbital semi-major axis and $G$ is the gravitational constant. You can handwave your binary stars as a point mass which isn't quite correct here, but it is close enough for your needs. It'll come out as something like 110.26 days at a radius of .45 AU.

Now, since I started typing this you made a comment on L.Dutch's answer suggesting that what you really meant was that the star's combined luminosity was the same as the Sun, even though what you said was that their combined mass was the same.

Now, you can rearrange the mass-luminosity relation to give you the ratio of your star's mass to the Sun... $\sqrt[4]{0.5}$ or about .84 (making them somewhere around a G9V or K0V star). This should let you keep the planet's orbital radius as 1AU, but feeding the new mass into the orbital period equation gives you a year length of more like 281.8 days.

Answer 2

Considering that the star mass is the same as our Sun and the distance of the planet is the same as our Earth, in first approximation we can conclude that the orbital period will be equally the same: 1 year.

This because the orbital period is determined by the mass of the primary and the distance from it of the orbiting body.

Answer 3

I did not calculate it too deeply, but probably there is no stable orbit so close to use with this configuration. My reasoning is that if the stars are perpendicular to the planet, it will be attracted with an about 1/3600 smaller gravity as if they are parallel. That causes a periodic excitation to its orbit, practically there is no way for this to have a zero effect on the long-term. The orbit of the planet will slowly deviate from the stable nearly-circular orbit, only God can say, exactly how, but the most likely outcome is a stable orbit much more far away (maybe Jupiter distance or so).

What can help, if you have luck and the planets orbit is somehow in a synchronous with the stars.

Answer 4

About the same as the our Earth, but with more variability.

The length of a sidereal year varies over time due to many factors. See this question. The changes you're making, specifically the binary stars, will likely increase the variability of this value.

All that being said, the general orbital mechanics don't distinguish between orbiting a star, or a barycenter between stars. It's factors other than general gravitation that will make the difference (such as tides, solar wind, the ever changing mass of stars, etc.).