Eclipse Cycle Calculations Needed

Question

I am writing a work of fiction about a world where:

The length of a year is 335 days. The world has two Moons. Moon A is Larger and further away with a 78 day lunar orbit. Moon B is smaller and closer with a 31 day orbit.

1. How often would both moons be full at the same time, overlapping each other like a bullseye?
2. How often would lunar eclipses occur for Moon A?
3. How often would lunar eclipses occur for Moon b?
4. How often would both eclipse at the same time?

This is a work of fiction, so liberties can be taken. I just want it to be close.

Answer 1

Frankly, four bodies do not periodically line up, even if the orbits are planar, unless there exist a common multiplum. Because all your periods are integers, that holds, but be aware that your double alignments, like the double solar eclipse or double lunar eclipse is not periodic if one of the periods does not exactly follow that integer system. (so it does not hold for 31.01 days, for example. How you handle quazi-periods is talked about here).

1. How often would both moons be full at the same time, overlapping each other like a bullseye?

Generally non-periodic, but can only happen at the synodic period of the two moons, that is:

.

$\frac{2418}{47}$ for the two moons. However, that must also happen when both of them is on the same line as the Sun, in other words when The relative synodic period between one of the moons and the sun, and the moons' synodic period both divides the requested period. For the inner moon and the Sun, the synodic period is $\frac{10385}{304}$. Since 47 does not have any factors in common with 304, the first common multiplum appears at.

$$\frac{10385}{304} \cdot \frac{2418}{47} = \frac{12555465}{7144}$$, or 1757.484 days. (@Jonathan gets 2418, because his model consider the Sun to stay static in the sky, but it does in fact revolve around one time a year.) The exact same period is also valid for your question 4.

2. How often would lunar eclipses occur for Moon A?

Here, the answer is the synodic period of the Sun and the moon, already stated to be $\frac{10385}{304}$, or 34.161 days. That is a little bit longer than the Moon's orbital period, because the planet has moved a little in its orbit, slightly changing the Sun's position in the sky.

3. How often would lunar eclipses occur for Moon B?

Synodic period here too, $\frac{26130}{257}$ = 101.673 days.

Answer 2

Here is my attempt, assuming the orbits are circular, and along the same plane. Otherwise, calculations become more complex, and those variables would have to be specified. Size of the planet and size of the moons would also impact the calculation (e.g. they might never fully eclipse if the planet is fairly small / dense). Also note that Moon A may eclipse Moon B and Moon B may eclipse Moon A depending on their size. I do not consider this in my calculations (I only consider the planet eclipsing them).

1.How often would both moons be full at the same time, overlapping each other like a bullseye?

Not including the planets orbit, 31 is a prime number, and 78 breaks down to the factors of 2, 7, 7. Since there is no shared common denominator, it does not appear these will synch up more frequently than 78 * 31 = 2418 days

2.How often would lunar eclipses occur for Moon A?

About 78 days (+- about 78/335 days due to the planet orbiting the sun)

3.How often would lunar eclipses occur for Moon b?

About 31 days (+- about 31/335 days due to the planet orbiting the sun)

4.How often would both eclipse at the same time?

Same answer as #1, which I calculate at 2418 days

I welcome refinement / correction of my calculations, this is a "rough attempt".

Answer 3

NOTE: I am using a "geocentric" frame of reference, where both the moons and the sun orbit the planet, and am creating an arbitrary xy coordinate system.

We note from @Hohmannfan's answer that (answering your questions out of order for simplicity):

• Moon B will eclipse the sun every $\frac{10385}{304}$ (~ 34.16) days. In this time period, the sun completes $\frac{31}{304}$th of an orbit and Moon B completes $1\frac{31}{304}$ orbits, lapping the sun once.

• Moon A will eclipse the sun every $\frac{26130}{257}$ (~ 101.67) days. The sun will complete $\frac{78}{257}$ of an orbit, and Moon A will lap it by completing $1\frac{78}{257}$ orbits.

• Moon B will overlap Moon A once every $\frac{2418}{47}$ (~ 51.44) days, in which Moon A will complete $\frac{31}{47}$ of an orbit and Moon B will lap it by completing $1\frac{31}{47}$ orbits.

However, as @Hohmannfan notes, there's no guarantee that the moons will be full when they overlap.

There's also no guarantee that the two moons will ever both eclipse the sun at the exact same time, although they will get arbitrarily close to doing so:

In the $\frac{2418}{47}$ days between two successive lunar overlaps, the sun moves $\frac{2418}{47} \times \frac{1}{335}$ of an orbit.

As above, the moons have advanced $\frac{31}{47}$ of an orbit.

Thus, compared to the sun, the moons have advanced $\frac{31}{47} - \frac{2418}{47} \times \frac{1}{335}$ or $\frac{7967}{15745}$ of an orbit (this number is surprisingly close to $\frac{1}{2}$ but that's just a coincidence).

This happens between every pair of overlaps, so the sun's angular distance (in orbits) from the overlapping moons is $\frac{7967 n}{15745}+r$ where $r$ is the angular distance at a specific overlap and $n$ is any integer.

For the overlapping moons to eclipse the sun $\frac{7967 n}{15745}+r$ must be an integer. If $r$ is irrational, this can never happen.

However, the angular distance can get arbitrarily small, even to the point where an observer wouldn't realize the double moon eclipse isn't 100% perfect.

By a similar argument, you can show the two full moons will get arbitrarily close to overlapping.

NOW, if we make the simplifying assumption that both moons are eclipsing the sun at year 0 (perhaps your astronomer-priests have decided this unusual occurence is a good time to start numbering the years, and believe zero (not one) is a good first year), we can make some other calculations.

Since the moons line up every $\frac{2418}{47}$ days and the sun and Moon B line up every $\frac{10385}{304}$ days, all three will line up (to form a double moon eclipse of the sun) on the least common multiple of these numbers, or 810,030 days (which would be exactly 2418 of your years, and note that 2418 is the product of the two lunar orbits in days). In this time:

• Moon A will have completed exactly 10,385 orbits.

• Moon B will have completed exactly 26,130 orbits.

• As above, the sun will have completed exactly 2,418 orbits.

As it turns out, there can never be a perfect double full moon bullseye:

• Moon B will be full on day $\frac{10385}{608}$ (~ 17.08), at which point it will have completed $\frac{335}{608}$ of an orbit and the sun will have completed $\frac{31}{608}$ of an orbit, so Moon B will have gained half an orbit on the sun, which is required for a full moon. After that, the moon will be full every $\frac{10385}{304}$ days, the period of time it takes the sun to complete $\frac{31}{304}$ orbits, and Moon B to complete $1\frac{31}{304}$ orbits.

• By similar calculation, Moon A will be full on day $\frac{13065}{257}$ (~ 50.84) and every $\frac{26130}{257}$ days thereafter.

• To find when they're both full at the same time, we solve this linear Diophantine equation:

$\frac{10385 n}{304}+\frac{10385}{608}=\frac{26130 m}{257}+\frac{13065}{257}$

where n and m are integers. This reduces to:

$n\to \frac{47424 m+15745}{15934}$

Unfortunately, $47424 m$ is always even, so $47424 m+15745$ is always odd. Since the denominator ($15934$) is even, you are dividing an odd number by an even number, and the result can never be an integer.

However, this doesn't tell the full story. For example, if we compute the positions on day $\frac{34987576465283}{92766720}$ (~ 377156.55), we find:

• Moon B is at 122.5656 degrees.

• Moon A is at 122.5581 degrees, only ~ 27 arcseconds away.

• The sun is at 302.5658 degrees, 179.9998 degrees from moon B, and 179.9924 degrees from moon A (~ 28 arcseconds from opposition).

In other words, this is pretty close to a double full moon, even though it's not exact.

In a similar vein, even though double solar eclipses only occur once every 810,030 days, there are several close calls:

$ \begin{array}{cc} \text{Day} & \text{Sep (')} \\ -810030.00000 & 0.00 \\ -754313.10860 & 0.91 \\ -698596.21710 & 1.82 \\ -642879.32570 & 2.73 \\ -587162.43420 & 3.64 \\ -531445.54280 & 4.55 \\ -475728.65130 & 5.47 \\ -445735.13160 & 7.29 \\ -420011.75990 & 6.38 \\ -390018.24010 & 6.38 \\ -364294.86840 & 7.29 \\ -334301.34870 & 5.47 \\ -278584.45720 & 4.55 \\ -222867.56580 & 3.64 \\ -167150.67430 & 2.73 \\ -111433.78290 & 1.82 \\ -55716.89145 & 0.91 \\ 0.00000 & 0.00 \\ 55716.89145 & 0.91 \\ 111433.78290 & 1.82 \\ 167150.67430 & 2.73 \\ 222867.56580 & 3.64 \\ 278584.45720 & 4.55 \\ 334301.34870 & 5.47 \\ 364294.86840 & 7.29 \\ 390018.24010 & 6.38 \\ 420011.75990 & 6.38 \\ 445735.13160 & 7.29 \\ 475728.65130 & 5.47 \\ 531445.54280 & 4.55 \\ 587162.43420 & 3.64 \\ 642879.32570 & 2.73 \\ 698596.21710 & 1.82 \\ 754313.10860 & 0.91 \\ 810030.00000 & 0.00 \\ \end{array} $

The table above lists all near-eclipses within 7.5 minutes of arc, where day is the number of days from year 0 (including days before year 0), and sep is the maximum separation (in minutes of arc) of any two of Moon A, Moon B, and the sun. Note that days $0$ and $\pm 810030$ are perfect eclipses, as expected.

Similarly, the closest we get to double full moons is below. In this case, sep is (in minutes of arc) the maximum of:

• the angular distance of Moon A from opposition

• the angular distance of Moon B from opposition

• the angular distance between Moon A and Moon B

$ \begin{array}{cc} \text{Day} & \text{Sep (')} \\ -797168.29790 & 10.29 \\ -767174.80850 & 8.92 \\ -711457.91490 & 7.55 \\ -655741.02130 & 6.17 \\ -600024.12770 & 4.80 \\ -544307.23400 & 3.43 \\ -488590.34040 & 2.06 \\ -432873.44680 & 0.69 \\ -377156.55320 & 0.69 \\ -321439.65960 & 2.06 \\ -265722.76600 & 3.43 \\ -210005.87230 & 4.80 \\ -154288.97870 & 6.17 \\ -98572.08511 & 7.55 \\ -42855.19149 & 8.92 \\ -12861.70213 & 10.29 \\ 12861.70213 & 10.29 \\ 42855.19149 & 8.92 \\ 98572.08511 & 7.55 \\ 154288.97870 & 6.17 \\ 210005.87230 & 4.80 \\ 265722.76600 & 3.43 \\ 321439.65960 & 2.06 \\ 377156.55320 & 0.69 \\ 432873.44680 & 0.69 \\ 488590.34040 & 2.06 \\ 544307.23400 & 3.43 \\ 600024.12770 & 4.80 \\ 655741.02130 & 6.17 \\ 711457.91490 & 7.55 \\ 767174.80850 & 8.92 \\ 797168.29790 & 10.29 \\ \end{array} $

Other notes:

• Even though you said this was fiction, note that it's highly unlikely that the moons' orbital period will be an exact multiple of the planets day. The only exception to this is if the moon(s) are tidally locked, in which case the orbital period will equal exactly one day.

• Similarly, it's unlikely the planet's orbital period would be an exact multiple of its rotation period (ours certainly isn't).

This is an interesting problem in general, and I am writing https://github.com/barrycarter/bcapps/blob/master/MATHEMATICA/bc-orrery.m to solve a similar problem: https://physics.stackexchange.com/questions/197481/