I'm making a simple planetary system with a primary similar to our sun and a planet like our earth. This earth analog has a moon similar to our moon, just smaller and closer.

How do I calculate the period between eclipses, given all the orbital parameters of all bodies? There must be some fairly simple formula, right?

  • $\begingroup$ I think this would be better on maybe Physics $\endgroup$
    – bendl
    May 27, 2018 at 12:22
  • $\begingroup$ There must be some fairly simple formula, right? Usually wrong. For a start do you want lunar or solar type eclipses ? $\endgroup$ May 27, 2018 at 13:16
  • $\begingroup$ Total or partial eclipse? At a given point on a planet, or anywhere on the planet? $\endgroup$
    – cms
    May 27, 2018 at 18:20
  • $\begingroup$ Total and partial solar eclipse, anywhere on the planet $\endgroup$
    – nAUTILUS
    May 27, 2018 at 21:31

2 Answers 2


The Saros cycle

To elaborate on L.Dutch's answer, a saros is the unit of time in which a certain Sun-Earth-Moon geometry will (almost) perfectly repeat itself. For our Moon and Earth, this number is equal to 6585.3211 days (or 18 years and 11.321 days). That means if you had an eclipse (either lunar or solar) on a particular day, then you would also have an eclipse one saros (6585.3211 days) later. Each successive eclipse after a saros is called the saros series. Note that the saros only tells you when the next eclipse occurs given you have already had one. To calculate eclipses from scratch, you would need an orbital mechanics propagator, which is beyond the scope of this (already long) answer.

Since there are more than one geometry during a year that can cause an eclipse, there is more than one saros series. You can see all the saros series here. If you look at that list of series, you will see that each series has a "first eclipse" and "last eclipse". That's because the saros cycle isn't perfect. Eventually they stop and end. They tend to be active for at least ~500 years, spanning ~70 eclipses.

Note that a saros is not a whole number of days, therefore the Earth spins 6585 and a third times. This extra 1/3 of a spin means the next eclipse in the saros series won't occur at the same location on Earth; instead it occurs about 1/3 a revolution (120 degrees) in longitude to the west. For a particular eclipse then, it requires three saros to repeat at the same location on Earth.

What this means for a fictitious planet: calculate a saros

The saros cycle is the number of days that give (near) whole periods to the Moons synodic, anomalistic, and draconic periods. These months are:

  • Anomalistic Month $T_a = 27.55455$ days. The time the moon takes to come back to a particular point in it's orbit (perigee usually). This is the newtonian-esque period that most people mean when they talk of lunar periods, and can be calculated from Kepler's Law: $$(T_a / 27.55455 )^2 = (a / 384784 )^3$$ where $T_a$ is in days and $a$ is in kilometers.
  • Synodic Month $T_s = 29.53059$ days. The period the moon takes to come to the same position relative to the Sun. This is different since the Moon orbits the Sun in a prograde orbit. You can estimate it from the Anomalistic Month as: $$T_s = T_a (1 + \frac{T_a}{Y})$$ where $Y = 365.25$ days for Earth. Substitute your planet's year.
  • Draconic Month $T_d = 27.21222$ days. The time it takes the Moon to return to either an ascending or descending node. Since the eclipses only occur when the Moon is at one of these nodes, it must be included in a saros. It differs from the anomalistic month because the nodes precess (move ever so slightly). Calculating this will take us way off course, so lets just assume it has the same proportion to the anomalistic Month as Earth's does: $$ T_d = \frac{27.21222}{27.55455} T_a $$

Your goal then is to calculate $T_a$, $T_s$, and $T_d$ for your hypothetical system. Next, find the number of days that produce (near) whole numbers of periods. There's no equation to do this; you will need a spreadsheet or a computer program. Here's the algorithm:

set n = 1
set limit = 0.25
Do the following until error <= limit:
    saros_a = n * T_d
    saros_s = round( saros_a / T_s) * T_s
    saros_d = round( saros_a / T_d) * T_d
    error = largest saros - smallest saros
    increment n by 1
return saros_a

This will calculate Earth's saros number, and so should be good enough for a fictitious planet. Once you have your saros number, sprinkle a couple of starting eclipses in around your planet's equinoxes and get to calculating. Good luck!

  • $\begingroup$ Thank you for your fantastic answer, really well structured and you obviously know a great deal about this. $\endgroup$
    – nAUTILUS
    May 27, 2018 at 22:40

I am not sure if it is fairly simple, to use your own words.

I know that the periodicity of eclipses for the Earth-Moon-Sun system is described by the saros:

The saros, a period of 6585.3211 days (14 common years + 4 leap years + 11.321 days, or 13 common years + 5 leap years + 10.321 days), is useful for predicting the times at which nearly identical eclipses will occur. Three periodicities related to lunar orbit, the synodic month, the draconic month, and the anomalistic month coincide almost perfectly each saros cycle. For an eclipse to occur, either the Moon must be located between the Earth and Sun (for a solar eclipse) or the Earth must be located between the Sun and Moon (for a lunar eclipse).

After one saros, the Moon will have completed roughly an integer number of synodic, draconic, and anomalistic periods (223, 242, and 239) and the Earth-Sun-Moon geometry will be nearly identical: the Moon will have the same phase and be at the same node and the same distance from the Earth. In addition, because the saros is close to 18 years in length (about 11 days longer), the earth will be nearly the same distance from the sun, and tilted to it in nearly the same orientation (same season).

The emphasized text in the quote is the relation you are looking for.


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