<p>If the outer planet orbit is 18.77 d and for every 2 orbits of this planet, the inner planets executed exactly 3, 4, 6, 9, 15 and 24 orbits respectively (roughly the case for Trappist-1), then all the planets would return to their &quot;starting positions&quot; every 37.54 days.</p>

CC BY-SA 4.0
2023-06-25T06:25:01.540Z
2023-06-30T07:59:41.843Z
2023-06-30T07:59:41.843Z
ProfRob
<p>If the resonances are 2:3:4:6:9:15:24, we need to find the smallest number that is a multiple of all those numbers, so each planet has made an integer number of revolutions, and is back where it started.</p>
<p>The most obvious way to do this is to work out the prime factors of those numbers. They are:</p>
<pre><code>2: 2
3: 3
4: 2x2
6: 2x3
9: 3x3
15: 3x5
24: 2x2x2x3
</code></pre>
<p>For each prime, we take the largest number of copies needed for any number: so three 2s from 24, two 3s from 9 and one 5 from 15 and multiply them together to get the least common multiple, or 360.</p>
<p>So, it would take 360 units of time for the planets to be in the same place relative to the stars, where the orbit of the closest planet is two units of time.</p>
<p><a href="https://iopscience.iop.org/article/10.3847/PSJ/abd022" rel="nofollow noreferrer">The orbital period of the closest planet is 1.5 days</a>, so our unit is 0.75 earth days, so it takes 0.75 x 360 earth days (about 9 months) for the planets to return to their initial configuration.</p>

CC BY-SA 4.0
2022-10-17T18:42:25.203Z
I don't understand this math at all. You're making some unwritten jumps.
skout
CC BY-SA 4.0
2022-10-18T11:19:36.377Z
Edited to try to make it more clear. What specifically are you struggling with?
David McKee
CC BY-SA 4.0
2022-10-18T12:16:01.363Z
@skout, after the edit (didn't see it before), I can follow it OK.  You might want to ask follow-up questions if you don't understand parts of it.
Greg Miller
CC BY-SA 4.0
2022-10-18T20:52:02.723Z
any thoughts on [Just how "locked" are resonant-chains of exoplanets thought to be? (e.g. K2-138 and TOI-178)](https://astronomy.stackexchange.com/q/23980/7982) (just bountied)
uhoh
CC BY-SA 4.0
2022-10-19T14:11:40.323Z
Nothing beyond https://en.wikipedia.org/wiki/Orbital_resonance#Mean-motion_resonances_among_extrasolar_planets
David McKee
CC BY-SA 4.0
2022-10-23T03:22:23.600Z
Okay, so what I am not getting is the cart nature of resonance. I would think that if this resonance is how I am thinking, every time the outer planet completes 2 full orbits, the inner planet would complete 24 full orbits, and every planet is completing their orbits in the same amount of time, so everyone would align in the time it took the inner planet to do 24 or the outer to do 2 or the middle to do 6, etc.
skout
CC BY-SA 4.0
2022-10-24T12:19:26.557Z
Whilst 24 divides by 2 nicely, it doesn't divide so nicely by 15. The 15 planet comes back to its starting position at times 15, 30, 45, 60, 75, 90, 105, 120. The 24 planet comes back to its starting position at times 24, 48, 72, 96 and 120. Only 120 divides neatly, so you have to wait until the 24 planet has been around 5 times for those two planets to match up. But this doesn't divide by 9 (so not all the planets are back to their starting positions) nor does 240, but 360 does...
David McKee
CC BY-SA 4.0
2023-06-24T00:27:10.337Z
@David McKee Ah, yeah, you're wrong. The orbital period of planet 1 is the unit, each planet gets back to it's starting position every orbit, which means planet one gets back to it's starting position at the end of every period of time
skout
CC BY-SA 4.0
2023-06-24T00:33:16.877Z
@DavidMcKee I'm thinking of it as a polyrhythm basically. In one "cycle" there are two orbits of the outer planet, 3 of the next, 4 of the next, etc... Down to 24. The resonance chains makes me think it should be read as x orbits per cycle. The outer planet reaches its starting position at 1/2 & 1, the inner at 1/24, 2/24... And the second at 1/15, 2/15, etc... And the third and 1/9, 2/9, etc... And so they all make it back to the start at 1.
skout
CC BY-SA 4.0
2023-06-30T08:05:07.783Z
The Q is telling you that when the inner planet has done 24 orbits, then all the other planets have executed an integer number of orbits too. The answer is 24 times the inner orbital period.
ProfRob
CC BY-SA 4.0
2022-08-26T17:40:41.360Z
2022-10-18T11:19:22.387Z
2022-10-18T11:19:22.387Z
David McKee
<p>My question is simply how long it takes for the 7 known planets in Trappist-1 to be in the same configuration in relation to their host star, and to each-other. Those seven planets are in a 2:3:4:6:9:15:24 resonance, which is just extraordinary. I thought that they would be configured in the same manner roughly every 36 days, but I am not sure I did it right. I basically thought, &quot;if all their orbital ratios are perfect, then I only need to see how long it takes for one of the planets to go through the amount of orbits listed in said ratio, and then I'll get my answer.&quot; I know that the answer I am looking for won't be exactly perfect, I am just wanting to know when they will be in practically the same place.</p>
<p>The specific calculation I did was take the orbital period of the sixth planet, say that 1/12th of that orbit was a &quot;day&quot;(instead of using earth days) and then multiplied it by the number of orbits that it has in the ratio (3) and got my answer. (36) I know this doesn't answer it in earth days, but since I am using that &quot;orbit of #6/12 as the daylength in the story I am writing in this world, I am using the measuring stick that matters to my situation. So, am I right? Does it take 36 &quot;days&quot; for Trappist-1 to reconfigure itself, or did I get the math terribly wrong?</p>

<orbit><exoplanet><planetary-transits>
How often does the configuration of the Trappist-1 system repeat
skout
# How often does the configuration of the Trappist-1 system repeat

(Asked by: username_0 on 2021-10-22)

My question is simply how long it takes for the 7 known planets in Trappist-1 to be in the same configuration in relation to their host star, and to each-other. Those seven planets are in a 2:3:4:6:9:15:24 resonance, which is just extraordinary. I thought that they would be configured in the same manner roughly every 36 days, but I am not sure I did it right. I basically thought, "if all their orbital ratios are perfect, then I only need to see how long it takes for one of the planets to go through the amount of orbits listed in said ratio, and then I'll get my answer." I know that the answer I am looking for won't be exactly perfect, I am just wanting to know when they will be in practically the same place.

The specific calculation I did was take the orbital period of the sixth planet, say that 1/12th of that orbit was a "day"(instead of using earth days) and then multiplied it by the number of orbits that it has in the ratio (3) and got my answer. (36) I know this doesn't answer it in earth days, but since I am using that "orbit of #6/12 as the daylength in the story I am writing in this world, I am using the measuring stick that matters to my situation. So, am I right? Does it take 36 "days" for Trappist-1 to reconfigure itself, or did I get the math terribly wrong?

**Tags:** orbit, exoplanet, planetary-transits

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## 2 answers

### Answer by username_1 on 2023-06-25. Score: 1 | Accepted answer

If the outer planet orbit is 18.77 d and for every 2 orbits of this planet, the inner planets executed exactly 3, 4, 6, 9, 15 and 24 orbits respectively (roughly the case for Trappist-1), then all the planets would return to their "starting positions" every 37.54 days.


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### Answer by username_2 on 2022-08-26. Score: 5

If the resonances are 2:3:4:6:9:15:24, we need to find the smallest number that is a multiple of all those numbers, so each planet has made an integer number of revolutions, and is back where it started.
The most obvious way to do this is to work out the prime factors of those numbers. They are:
2: 2
3: 3
4: 2x2
6: 2x3
9: 3x3
15: 3x5
24: 2x2x2x3

For each prime, we take the largest number of copies needed for any number: so three 2s from 24, two 3s from 9 and one 5 from 15 and multiply them together to get the least common multiple, or 360.
So, it would take 360 units of time for the planets to be in the same place relative to the stars, where the orbit of the closest planet is two units of time.
The orbital period of the closest planet is 1.5 days, so our unit is 0.75 earth days, so it takes 0.75 x 360 earth days (about 9 months) for the planets to return to their initial configuration.

> **Comments:**
> **username_0**: I don't understand this math at all. You're making some unwritten jumps.  
> **username_2**: Edited to try to make it more clear. What specifically are you struggling with?  
> **username_3**: @skout, after the edit (didn't see it before), I can follow it OK.  You might want to ask follow-up questions if you don't understand parts of it.  
> **username_4**: any thoughts on [Just how "locked" are resonant-chains of exoplanets thought to be? (e.g. K2-138 and TOI-178)](https://astronomy.stackexchange.com/q/23980/7982) (just bountied)  
> **username_2**: Nothing beyond https://en.wikipedia.org/wiki/Orbital_resonance#Mean-motion_resonances_among_extrasolar_planets  
> **username_0**: Okay, so what I am not getting is the cart nature of resonance. I would think that if this resonance is how I am thinking, every time the outer planet completes 2 full orbits, the inner planet would complete 24 full orbits, and every planet is completing their orbits in the same amount of time, so everyone would align in the time it took the inner planet to do 24 or the outer to do 2 or the middle to do 6, etc.  
> **username_2**: Whilst 24 divides by 2 nicely, it doesn't divide so nicely by 15. The 15 planet comes back to its starting position at times 15, 30, 45, 60, 75, 90, 105, 120. The 24 planet comes back to its starting position at times 24, 48, 72, 96 and 120. Only 120 divides neatly, so you have to wait until the 24 planet has been around 5 times for those two planets to match up. But this doesn't divide by 9 (so not all the planets are back to their starting positions) nor does 240, but 360 does...  
> **username_0**: @David McKee Ah, yeah, you're wrong. The orbital period of planet 1 is the unit, each planet gets back to it's starting position every orbit, which means planet one gets back to it's starting position at the end of every period of time  
> **username_0**: @DavidMcKee I'm thinking of it as a polyrhythm basically. In one "cycle" there are two orbits of the outer planet, 3 of the next, 4 of the next, etc... Down to 24. The resonance chains makes me think it should be read as x orbits per cycle. The outer planet reaches its starting position at 1/2 & 1, the inner at 1/24, 2/24... And the second at 1/15, 2/15, etc... And the third and 1/9, 2/9, etc... And so they all make it back to the start at 1.  
> **username_1**: The Q is telling you that when the inner planet has done 24 orbits, then all the other planets have executed an integer number of orbits too. The answer is 24 times the inner orbital period.  

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