I've been trying to calculate the period between the conjunction of all of the moons which my planet has. They are five minor moons with orbital periods ranging from 0.5 to 1.3 earth days, and one major moon with an orbital period of six days.

I've found this neat equation, but it only works for two bodies, and I'm not sure if I'm using it correctly.

1/Psyn = 1/P1 − 1/P2

Where P1 < P2.

An answer in this question mentioned that you can calculate the time between the conjunction of more than two moons by calculating it for two moons and then substituting it as either P1 or P2. I've tried all I could, but sooner or later the results will start making no sense. I suspect it is because of the relatively short orbital periods of the moons, or because I just might be misunderstanding the equation.

Is there a better way to calculate the period between conjunctions of six different moons when viewed from the planet they orbit? Am I missing something? Do the moons have to be in some kind of resonance for the equation to work?

  • $\begingroup$ You need to run the thing through a full orbital simulator, to be exact. For "good enough" answer, you need to specify WHAT constitutes a good enough alignment for conjunction? There will be no such thing as a "perfect" conjunction, as that requires multiple, continuously incrementing real numbers to be exactly equal at some point. A thing that is, by definition, infinitely unlikely. $\endgroup$
    – PcMan
    Commented Nov 29, 2020 at 19:41

1 Answer 1


(Warning: lots of unstated approximations in the following answer. Good enough for government work, not good at all for actually pointing a telescope.)

There is no simple formula for three or more celestial bodies; in fact, while two celestial bodies are (almost, barring unnatural situations) guaranteed to align perfectly from time to time, three or more are not guaranteed to align perfectly, ever; and they usually don't, so that you must decide how closely aligned they should be for you to consider their position a good enough conjunction.

For example, consider Io, Europa, and Ganymede, three of the Galilean moons of Jupiter: their orbits are in a 4:2:1 resonance, and they never ever align perfectly.

So in the end, it all depends on:

  1. The synodic orbital periods of the celestial bodies.
  2. Their starting positions in the sky.
  3. What you consider a good enough conjuction. Perfect conjuctions may be impossible, but you may settle for an imperfect conjuction.

First, let's establish a little terminology.

The "orbital period" of a satellite can be considered from two different points of view:

  • If the orbital period is considered from the point of view of the fixed stars, then it is called "sidereal period".

  • If the orbital period is considered from the point of view of an observer on the planet's surface, then it is called "synodic period".

    They are different, because during the time the satellite completes a sidereal orbit, the planet below it has rotated a little, so that it won't appear in the same position in the sky for the observer on the surface. As an extreme example, consider a geostationary satellite used for telecommunications: its sidereal period is 24 hours, while its synodic period is infinitely long.

    So the first thing to do is to determine the synodic periods of the five moons. If you know the sidereal periods, add or subtract the fraction of planetary rotation completed in a sidereal period, depending on whether the satellite orbits the planet in a prograde or retrograde direction.

At this point you have:

  • The synodic periods of the five moons, either given or computed.
  • Their initial positions in the sky, which must be given.

Now it is only a matter of tracking their positions in the sky, and stopping when you find them covering a small enough angle for them to be considered in conjuction. You can do it in a simple program, or in a spreadsheet.

For example, assigning the following synodic orbital periods to the satellites: 0.5, 0.7, 0.9, 1.1, 1.3 and 6 days; and assigning the following starting positions (degrees counterclockwise with North = 0): 0, 60, 120, 180, 300: they will come within a sector of 36.3 degrees after 901.58 days, and will again be within 35.3 degrees 901.04 days later; the next such close alignment will be 1638 days later, when they will come within 35.3 degrees. Those may be or may not be good enough conjuctions, that's your choice.

Cheating a little

If you know the synodic orbital periods of the satellites, you can compute their least common multiple with any desired precision.

For example, continuing the example of the six satellites with synodical periods of 0.5, 0.7, 0.9, 1.1, 1.3 and 6 days: it is obvious that their position in the sky will repeat every 18,018 days; so that if right now they are in good enough conjunction, they will also be in good enough conjunction 18,018 days later. (This does not mean that there won't be other good enough conjunctions during those 18,018 days.)

  • $\begingroup$ This is great! I noticed that the alignments you've listed differ greatly, with two being 901 days and the third being 1638 days. Does this mean that alignments aren't regular and the period between them always changes? I was hoping I could construct a calendar of sorts by using these conjunctions, but if the period changes, that might not be an option. $\endgroup$
    – D. Daniels
    Commented Nov 29, 2020 at 21:34
  • $\begingroup$ @D.Daniels: There is a long-term period, there is always a long-term period; but within the long term period you will have other times of relatively close alignment. (And the long-term period may not be all that useful; astronomical cycles tend to be rather long. That's why the adjective "astronomical" has the secondary meaning it has.) (And note PcMan's comment: with such small satellites long-term regularity is not to be expected. Even the motion of our big Moon is very complicated.) $\endgroup$
    – AlexP
    Commented Nov 29, 2020 at 21:48

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