I've been building a world, partly for test reasons (making sure my spreadsheet works, and so i know what i need to add to it) and partly for fun, but for calender reasons i need to know the orbital period of my binary star system.

More specifically, as i am putting this in a spreadsheet, i need the calculations to figure out the orbital period - preferably dumbed down a little as, when searching myself, i couldn't make heads or tails of it.

Star 1:

Mass: 1.07 Solar Masses

Distance from Barycentre: 0.0709724389 AU

Star 2

Mass: 0.853 Solar Masses

Distance from Barycentre: 0.0890275611 AU

Orbital Eccentricity: 0.41

  • $\begingroup$ You need the masses of both bodies and the distance between their centers to calculate the orbital period (you can express the mass in terms of solar masses if it's more convenient). $\endgroup$
    – pablodf76
    Jul 20, 2017 at 20:12
  • $\begingroup$ I've added those to the question. $\endgroup$
    – Hannah
    Jul 20, 2017 at 20:31
  • $\begingroup$ Your distances from the barycenter of the system only apply if the orbits are circular, otherwise you'll have to qualify them. I assume these are the maximum distances from each other? $\endgroup$
    – Pak
    Jul 20, 2017 at 21:18
  • $\begingroup$ No. Their absolute max separation is 0.4512... i'm getting the sinking feeling that i've been worldbuilding wrong $\endgroup$
    – Hannah
    Jul 20, 2017 at 21:23
  • $\begingroup$ Barycenter, from ancient Greek barys (heavy); from the same Greek word we have barometer (measures the weight of the atmosphere) and baritone (deep voice). $\endgroup$
    – AlexP
    Jul 20, 2017 at 21:45

1 Answer 1


You seem to have it pretty figured out already. You just need to apply Kepler's Third Law. For two bodies orbiting a barycenter, the square of the orbital period is proportional to the cube of their mean distance and inversely proportional to the sum of their masses.

For these things I always use WolframAlpha to avoid doing calculations by hand. For your parameters (with a semimajor axis of 0.08 AU, which is the average distance from each star to the barycenter) the result is 143 hours.

  • $\begingroup$ Ok, when finding the mean of my two distances, i come out with either 0.08 (with (Distance1+Distance2)/2 ) or 0.12 with (Distance1+Distance2/2). Also, What do you mean by inversly proportional? $\endgroup$
    – Hannah
    Jul 20, 2017 at 21:20
  • $\begingroup$ @Hannah I've just corrected the calculation (it should be 0.08 AU as you said, I think). The "inversely proportional" bit refers to Kepler's Third Law. The more massive the objects, the shorter their orbital period. $\endgroup$
    – pablodf76
    Jul 20, 2017 at 22:10
  • $\begingroup$ Ok. ta. I'm still a little confused over the exact equations (that i'll need to put into the spreadsheet) but you've been very helpful $\endgroup$
    – Hannah
    Jul 20, 2017 at 22:39
  • $\begingroup$ ... I think i've found the equation i'm looking for - as it seems to have given the most accurate result (According to Wolfram Alpha) of all the ones i've tried, which is P = SQRT(A^3/(M1+M2)... where P is the orbital period and A is the Max separation... a second opinion would be helpful though, as my result is 1dp off. $\endgroup$
    – Hannah
    Jul 21, 2017 at 9:27
  • $\begingroup$ What you found is what I found, which is basically a form of Kepler's 3rd, but I'm still doubtful what A means in this case (since Kepler's laws are usually explained with reference to very different masses). $\endgroup$
    – pablodf76
    Jul 21, 2017 at 10:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .