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I have looked around for some time attempting to find a post that could help me answer the question I have, but I've come up short. I'm attempting to visualize in greater detail what my planet would look like, but I can't quite get the details just right.

My planet is in a trinary star system. Here are the details:

Star X:

  1. Type F Star
  2. Mass= 2.72813E+30 (kg)
  3. Diameter= 1,757,799 (km)
  4. Luminosity= 1.16E+027 (W)
  5. Surface Temp= 6,778.6 (K)

Star y:

  1. Type M Star
  2. Mass= 8.74911E+29 (kg)
  3. Diameter= 757,674 (km)
  4. Luminosity= 2.17E+025 (W)
  5. Surface Temp= 3,817 (K)

Star z:

  1. Type M Star
  2. Mass= 8.9082E+29
  3. Diameter= 767,844 (km)
  4. Luminosity= 2.31E+025 (W)
  5. Surface Temp= 3,851.9 (K)

Stars "y" and "z" are a binary pair that orbit around Star "X." My planet orbits the binary. Currently, the length of one of my planet's days is equivalent to 1.253 Earth days (30 hours). With this in mind, it takes approximately 1,222.334 days for the binaries (and the planet) to orbit Star "X." "y" and "z" has an average distance of 56,847,191 (km) from the planet.

Any commentary on what the sky would look like, assuming that the planet is not tidally locked and will have a day/night cycle? Sunrise, sunset, midday, etc? I'm mostly trying to figure out when each star would be visible throughout a single day and a single year. I'm not sure how much it will matter, but the eccentricity will be large enough to where Star X will appear slightly larger in the sky for roughly half of the orbital period. I feel like I'm missing other pertinent details here. Any feedback on this would be most helpful!

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    $\begingroup$ brings up the classic three body problem. this would not be a stable system $\endgroup$ – Topcode Apr 18 '20 at 4:08
  • $\begingroup$ Do you care about physics or just want a simple answer on what would be visible? $\endgroup$ – Galactic Apr 18 '20 at 4:41
  • $\begingroup$ @galactic_analyzer I like as much detail as possible, so both? But, as much as you're willing to help out with. I know it can be time-consuming and probably headache-inducing trying to cover all bases. $\endgroup$ – Rauri Apr 18 '20 at 5:06
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    $\begingroup$ @Rauri, I know that you're new, but here at Worldbuilding Stack Exchange, we are ready to answer as much as possible. A question that I asked today: worldbuilding.stackexchange.com/questions/174486/… got a really good response only 3.5 hours later (the one by @cyber101). It's time-consuming and headache-inducing for sure, but we're willing to do it as a hobby. $\endgroup$ – Galactic Apr 18 '20 at 5:10
  • $\begingroup$ I would really like to see images of this situation done in a graphics program with adjustable light sources. $\endgroup$ – Willk Apr 18 '20 at 19:59
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Key assumptions: all the stars and the planet orbit in the same plane. The planet has zero axial tilt. The orbits are circular or nearly so. If you change these assumptions then it could look very different indeed.

For a person standing on the equator

Season phase 1: Star y appears over the horizon at “dawn” followed a short while later by star z (or vice versa). Both increase in elevation until each passes directly overhead and then slowly drop down over the opposite horizon. During this period stars y and z will get closer together or move further apart according to their orbital period around one another.

There will be two solar eclipses every binary rotation period, once by star y in front of star z and once by star z in front of star y. Each will be visible from an entire hemisphere. As stars y and z are setting star X is rising over the opposite horizon (no true night). Star X will then also pass directly overhead before setting over the opposite horizon as stars y and z are rising again.

Season phase 2: During the 1222 day “year” Star X will rise earlier and earlier compared to stars y and z and at mid “summer/winter” all three stars will appear close together in the sky. There will be a true night followed by 3 stars rising in variable sequence and tracking across the sky. Stars y and z will both eclipse star X every day. Star X will never eclipse either star y or z. Once every 1222 days there will be a double eclipse visible from one entire hemisphere with Star X being eclipsed by both star y and star z. The sequence would alternate nearest to furthest y ,z, X and z, y, X.

Additional case - effects of an eccentric planetary orbit:

Assumption: the orbits of the stars are all circular or near circular around each other and only the planet has an eccentric orbit around y/z.

Firstly the greater the degree of eccentricity the greater the instability of the system as the planet will be subject to much decreased gravitational attraction to y/z and an increased attraction to X on a periodic basis. This will tend to shift the system from an approximate planet + y/z orbiting X into a 3 body problem planet – y/z – X and resultant chaotic behaviour. So some eccentricity might be OK but the planet should never stray too far from y/z compared to X – an order of magnitude I would suggest so always 10 time closer to y/z than X.

With an eccentric orbit the pattern of the movement of the stars would be very similar, however the stars would vary in size and luminosity a little bigger and brighter at each type of perihelion and dimmer and smaller at each type of aphelion, but not that much (with a low eccentricity).

The relative positions of the stars would also change at a variable rate and this would make the calculation of the timing of eclipses a lot more complex (for primitive people) as the planet would be moving more slowly at aphelion and much faster at perihelion rather than be moving at a uniform rate as would be the case in a circular orbit.

One further case for a person not standing at the equator:

As an observer moved north or south of the equator the arc of the stars passing across the sky would occur at a decreasing elevation until at the poles all of the stars would appear to move around the horizon in permanent although probably fairly bright twilight.

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  • $\begingroup$ Thank you so much! Do you have any feedback on how this would look with a more eccentric orbit? This would be a key point. You were correct in assuming zero axial tilt, but the elliptical orbit is important since it helps the species on my planet keep up with the year more easily and it functions as a "seasonal"/temperature change as the planet/binary pair move closer to the primary star. The eccentricity would be absolutely less than a parabola but comfortably in the ellipse range. I haven't gotten the math just right for this detail- obviously an important detail. 😔 $\endgroup$ – Rauri Apr 18 '20 at 17:45
  • $\begingroup$ I will have a think and attempt to extend my answer later. There are two major issues firstly the whole set up is complex and additionally the more bodies you have the more variables that need to be specified such as size, orbital period, axial tilt, orbital eccentricity and so on. Having zero axial tilt helps and assuming all orbits are in the plane of the ecliptic also helps - otherwise we might have to deal with suns spinning around each other in the sky. $\endgroup$ – Slarty Apr 19 '20 at 9:01
  • $\begingroup$ Thank you, thank you! You're the only person who has really answered. It's been a few days now, so I doubt I'll be receiving much more feedback than what I've had so far. If you're able to expand on your answer regarding the eccentric orbit, I'd really appreciate it! $\endgroup$ – Rauri Apr 20 '20 at 17:38
  • $\begingroup$ Well I hope I'm right! It can be difficult to visualise these things as I'm sure you will understand, but I think it should be roughly right. Not sure exactly what you want me to expand on concerning the eccentric orbit. But basically the view is the same as that from the non-eccentric orbit except that the stars would have a greater change in relative size and luminosity and the speed of their movement in the sky would not be as uniform due to speed changes in the planetary orbit. Any specific questions? $\endgroup$ – Slarty Apr 20 '20 at 19:58
  • $\begingroup$ I see what you mean. Fairly simple answer but logical and very helpful. Thanks so much, Slarty! ☺️ $\endgroup$ – Rauri Apr 20 '20 at 20:00

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