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This is kind of a weird one. Bare with me! :)

Background: (Skip to the math below if you just want to answer questions)

In my favorite game, Elite: Dangerous, I "own", or occupy I guess, a system named "59 Virginis", based on the real life counterpart inside the Virgo Constellation.

The game itself houses a unique, very close to life-like system called "Stellar Forge", which uses as close to real life accuracy with physics as possible.

In this system, my own group's "Home" is the Earth-Like planet "59 Virginis 4", which is in a binary (tidally locked) orbit with 59 Virginis 3 (a class III gas giant).

I am trying to create an "Alien Calendar" of sorts, and a key component of calendars is the year, or how long it takes to rotate the sun once. This is semi-important, seeing as the home planet itself is titled on it's axis, meaning the Earth-Like will have seasons! Meaning having a yearly cycle of seasons would be awesome, and making a calendar would be lots of fun!

This is where I'm incredibly stuck. The game doesn't outright tell us how long it takes for these two objects to circle the sun, but it does give a lot of extras on possibilities of finding it!

I've tried a few different methods, but I don't think I know enough math to be able to find such a thing.

Here is the information we have currently:

59 Virginis 3 (the Gas Giant)

  • "distanceToArrival": 964 light-seconds, (this value changes based on where the planets are in binary orbit at the time of information update),
  • "earthMasses": 2027.9021,
  • "radius": 72755.136 km,
  • "orbitalPeriod": 80.53994791666666 d,
  • "semiMajorAxis": 5.898593314670755e-5 au,
  • "orbitalEccentricity": 0.114867,
  • "orbitalInclination": 0.071319°,
  • "argOfPeriapsis": 246.659607°,
  • "rotationalPeriod": 1.7743948929398148 d,
  • "rotationalPeriodTidallyLocked": false,
  • "axialTilt": -1.236412°

59 Virginis 4

  • "distanceToArrival": 989 light-seconds,
  • "earthMasses": 1.795525,
  • "radius": 7228.5435 km,
  • "orbitalPeriod":80.53994791666666 d,
  • "semiMajorAxis": 0.0666149271869282 au,
  • "orbitalEccentricity":0.114867,
  • "orbitalInclination":0.071319°,
  • "argOfPeriapsis":66.659615°,
  • "rotationalPeriod":80.57560763888888 d,
  • "rotationalPeriodTidallyLocked":false, (it's pretty damn close tho)
  • "axialTilt":-0.293082°,

The above information can be taken from here which is the most accurate version that can be given. If you want it in an easier to read format, click on the bodies 3 and 4 here

I know that the orbital period of a binary object is this:

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

Which only describes the orbital period of the two objects around eachother, right?

Ideas that have been thrown around:

  • Using Kepler's Third law, which states "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."

  • Using the two object's "distance to arrival" to find the distance between the two objects, then finding where on that line the center of mass is, and then using that center of mass point to find the distance to the star again from that point, then using that distance to find the orbit. (This sounds like the most reasonable from my point of view, but possibly the hardest)

Here is a really bad paint drawing of what I think the one above could look like:

Here is a really bad paint drawing of what I think the one above could look like:

  • Possibly finding the sidereal period somehow, or synodic period in relation to the giant and the sun?

If you've gotten this far, you've already helped me more than most! Any advice in relation of where to go next would me MUCH appreciated! Thank you!!

The next step after finding how long it's sun's orbital period is to find how many years to reset the whole cycle, kind of like a leap year, but with the binary orbits. That should be slightly easier, but I'm stuck on this for now! :)

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  • $\begingroup$ So, roughly Earth orbiting Jupiter but just outside the Asteroid Belt. $\endgroup$
    – Carl Witthoft
    Commented Nov 8, 2019 at 16:10
  • $\begingroup$ What does “Arrival” mean here? $\endgroup$
    – Anton Sherwood
    Commented Feb 13, 2020 at 16:01

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I think that at the moment, all you have available for the pair's orbit around their star is a snapshot of their two distances; 964 and 989 seconds, or about 289 and 297 million kilometers. Your snapshot doesn't have any velocity information, so I think that there's no way to calculate the mass of the star from that data. Since the first planet is 2000 times heavier than the second, lets say that the pair's center of mass is 289 million km from the star. However your second link states explicitly that the star is 1.0195 solar masses. That means that the standard gravitational parameter or $GM$ is about 1.352E+20 m^3/s^2, so using

$$T= 2\pi\sqrt{\frac{a^3}{GM}}$$

I get a year of about 972 days for the center of mass of the pair. Since the lighter planet oscillates by +/- 10 million km (its semi-major axis) every 80 days, that's a 7% amplitude change in distance to the star or roughly a 14% amplitude modulation of the brightness of the sunlight every 80 days, so that may have some impact on the weather for the smaller planet.

Interestingly, the smaller planet goes "around the world in 80 days" if we name the larger planet "world".

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