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One of the cultures in my story lives on the dark side of a tidally locked planet. I had an idea that they would mark longer spans of time by the appearance of the three outer planets in the system in the sky. For example, every time the closest planet appears in the sky marks a "week", the next one a "month", etc.

An idea I had was that this culture would call a general assembly of chieftains (like the early Icelandic Althing) every time the three outer planets meet at the same point in the sky.

My planet orbits a red dwarf star and has an orbital period of approximately 48 hours. The system is loosely based on the Trappist 1 system, but you don't have to strictly stick to that in your answers. For context, the outer three planets of the Trappist 1 system have respective orbital periods of roughly 9, 12, and 19 days.

What would the ratios of the planets' orbits have to be for this meeting to happen roughly every 1-2 Earth years?

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  • $\begingroup$ Is your planet orbiting its star in a week? And are the other planets also quite close (though not as close) to the star? If the other planets are "outer planets" similar to those in our system, they simply won't move fast enough along their orbital arc to line up anywhere near to often. $\endgroup$ Commented Mar 28, 2023 at 20:01
  • $\begingroup$ Sorry, I just clarified some details in an edit to the question $\endgroup$
    – DMacc1917
    Commented Mar 28, 2023 at 20:10

1 Answer 1

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Easy Peasy Mathematics

What you're looking for is the lowest common denominator.

Step #1

Convert (for fun and giggles) all four planets' orbital periods into hours.

48, 216, 288, 456

Step #2

Convert them all to fractions:

1/48, 1/216, 1/288, 1/456

Step #3

Use a (thank goodness!) online lowest common denominator calculator to quickly find that all too critical lowest common denominator, then flip it right-side-up.

16,416 hours or 1.874 years

What this tells us is that all four planets will line up in the sky once for...

  • every 342 orbits of your primary world.
  • every 76 orbits of Trappist 1 Outer A
  • every 57 orbits of Trappist 1 Outer B
  • every 36 orbits of Trappist 1 Outer C

I've made an assumption about what you mean by "meet in the night sky." If this isn't what you were expecting, you need to provide a clear definition of exactly what you mean by "meet in the night sky.

However, it likely doesn't matter. In other words, it doesn't matter where the three outer planets are in the night sky. If you're looking for that to happen exactly, then it'll happen only once per 16,416 hours.

Except

This assumes an alignment with your primary planet, which must also be at one specific point of its orbit around the sun for this simple calculation to work. If you're looking for an arbitrary alignment of those three outer planets regardless of where your primary planet is in its own orbit... this gets a lot more complicated.

I'd vote to stick with the simple math.

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  • $\begingroup$ Perfect answer, thanks! $\endgroup$
    – DMacc1917
    Commented Mar 29, 2023 at 17:04

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