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Yesterday while Green, Andy and I were recording new episodes for the podcast we stumbled upon a bit of a problem that we need someone with physics/astronomy knowledge to help out with.

We want to create an annual astronomical event for the sake of one of our species hosting a once per year festival/gathering.

The world:

  • Earth-like
  • Axial tilt is slightly more pronounced at 27%
  • The planet has two moons, their size and distance from the planet are undecided and may be tailored to fit this scenario.
  • A 12 constellation Zodiac similar to Earth

One of the constellations in the Zodiac is a cat and each year we would like some astronomical event related to the constellation to happen.

The event:

  • Must be tied to the cat constellation
  • Should take place in the fall
  • Will ideally include one or both moons
  • Can only happen once per year
  • Needs to be visible from anywhere within 20 degrees latitude of the equator
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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – James Feb 27 '18 at 17:01
  • $\begingroup$ Whose fall? Do you mean it should take place near an equinox? $\endgroup$ – Monica Cellio Feb 28 '18 at 3:50
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Two moons are in orbital resonance

You can have your two moons be in a m:n orbital resonance, the way that Io and Jupiter are. That way, at some time period $nT$ where $T$ is the orbital period of the outer moon, the two moons will line up in the sky. The orbital period of the inner moon is $nT/m$ in this case.

A further twist, which is something that will become more important later, is that the outer moon could potentially be in a retrograde m:-n orbit. There exist asteroids that are in retrograde orbital resonances (Morais and Namouni, 2013) with both Jupiter and Saturn. There are also exoplanets that orbit retrograde to the main star (Hebrard, et al, 2017). So it is plausible that your planet could have an inner main moon, like Luna, and an outer retrograde moon. Keep in mind, this situation does not have to be stable over geological time. If it just so happens that the moon is captured, modified in orbit into resonance, and stable for 10,000 years that just happen to be the same 10,000 years that your civilization is developing, no one will know the difference until the space age.

This alignment of the moons could produce some interesting optical effects, especially if the inner moon has an atmosphere of some sort, or is volcanically ejecting a dust. The outer moon could change (apparent) color or otherwise look strangely during the alignment. But in general, it would be astronomically notable if the two moons were lined up in the sky only one time per year.

Now line the orbital resonance up with the zodiac

There is no reason that $nT$, the alignment period of the moons, can't also be the orbital period of your planet around the sun. Since the orbit of the Earth drives the seasonality of the zodiac, if the period of the moon alignment lines up with the orbital period of the Earth, then the alignment of the two moons will happen at the same time, relative to the zodiac, every year.

An annual celestial event similar to what you are planning is the rising of Sirius above the horizon (as seen from Egypt) right before the Nile flooded. Sirius precessed and no longer times the flood well (and the Nile doesn't even flood anymore, thanks to dams), but that took millenia. For centuries the association between the star and the flood was stable, and highly important to the royal astronomers and what have you.

In the same way, if the orbital period of the alignment and the planet around its sun just coincidentally happen to be the same, within a few seconds at least, you will have a stable celestial event for centuries or longer, long enough to build up some cultural tradition.

Dynamics of the two moons

In this answer (also your question, by the by), I showed that there are a variety of stable orbits for a second moon, both inside and outside of Earth's orbit. In general, a moon at 1/3 of Luna's mass, that is at least 2.5 times more distant from the Earth than Luna is, will be stable. This was for prograde orbits, not retrograde; I don't have access to Rebound right now to check it but I am confident that we can find some stable orbits for a second outer moon outside of our own.

We want to put the outer moon as far away from the planet as possible to lead to a long orbital period. This will help it line up with the planetary year. The Hill sphere of a planet is that region of space where a planet's gravitational force is dominant over other forces (such as the gravity of the main star). The Hill sphere of the Earth is around 0.01 AU; but true stability won't be found in the outer parts of that region due to other forces, such as gravity from a gas giant, or radiation pressure from the main star. However, objects can be captured more readily into retrograde orbits in the outer part of the Hill Sphere (Astakhov, et al., 2003).

So, our goal is to place a smaller moon in a retrograde orbit around the Earth, such that the outer moon is in resonance with the larger moon, and its orbital period is the same as the planet's year.

The maths

The Hill sphere of Earth is about $1.5\times10^{9}$ meters. Let us say that we can have a retrograde orbit of up to $1.0\times10^{9}$ meters; this is the semi-major axis of the outer moon's orbit ($a$). Let us also say that the planet in question is the same size as Earth. Orbital period is

$$T =2\pi\sqrt{\frac{a^3}{GM}} = 2\pi\sqrt{\frac{\left(1\times10^{9}\right)^3}{6.67\times10^{-11}\cdot5.97\times10^{24}}} = 115 \text{ days}$$

If we make $n$, from our orbital resonance ratio, equal to 3, then the period of moon alignment is $3T$ = 345 days. This is just a little bit less than our Earth year. So, simply move the Earth a little bit closer to the sun, until the Earth year is 345 days. Alternately, you could increase the mass of the Earth to increase the Hill sphere until there is a stable orbit of 365/3 = 122 days.

Last, you need to set the inner moon into a stable resonance with the outer moon. Since we've settled on a 'm:3' resonance, we can make it a 7:3 resonance, like Ganymede and Callisto. This means the inner moon need an orbital period of about 49 days; noticeably more than our moon's.

There are of course pretty much an infinite number of way to change these numbers slightly to end up with a stable system. If the Earth's size goes up, its Hill sphere goes up and the moon can orbit farther away, so the orbital period goes up as well, etc.

Conclusion

Your planet is Earth-like, and orbits a star slightly smaller and less bright than our own. The day length doesn't matter in this problem, so it can be the same as on Earth. The year is 345 days long. There are two moons, one the size of our current moon, and another 1/3 the mass (which means about 70% the diameter). The inner moon orbits every 49 days; the outer moon every 115 days. The two moons line up in the sky with each other every 345 days, at the same time every year.

Given the significance of the lunar alignment, astronomers might start the zodiac with the alignment. The Month of the Cat, the first month of the lunar year, when Leo the Great Cat is high in the night sky, begins every year on the Blessed Night of the Two Moons. Truly an auspicious night for celebration!

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  • $\begingroup$ Thanks for giving such a good explanation of my idea. It's unfortunate I don't have the know-how for this kind of an answer, but I'm impressed with your explanation. $\endgroup$ – Gryphon Feb 21 '18 at 19:16
  • $\begingroup$ "There is no reason that nT, the alignment period of the moons, can't also be the orbital period of your planet around the sun." I'm not sure there would be any reason for it to be though, either. Unlike moons to moons, or planets to planets, the forces between moons and planets are considerably different magnitudes. I'm not sure it happens that moons resonate with the planet's period. Can you go further into demonstrating why that would happen? In Jupiter's case, the moons resonate with each other, but not with Jupiter's year. $\endgroup$ – Octopus Feb 21 '18 at 22:28
  • $\begingroup$ @Octopus That would just be a random coincidence. In the universe of possible planets, if this planet configuration I offer is possible, it is probably out there somewhere. I don't think there needs to be a reason that it should occur, as long as it could occur. $\endgroup$ – kingledion Feb 21 '18 at 23:36
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    $\begingroup$ I think that won't work. For a moon to have a stable orbit it's orbit should be less than one ninth as long as the planet's year. Thus the Outer moon should have a month less than 1/9th year long, and the inner moon a shorter one. But both your moons have months more than one 9th of a 345 day year and thus their orbits would both be unstable. $\endgroup$ – M. A. Golding Feb 22 '18 at 5:02
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    $\begingroup$ The inner moon is named Felis and the outer is named Mus. On the "Blessed Night of the Two Moons", the cat catches/devours the mouse. $\endgroup$ – Dennis Williamson Feb 22 '18 at 21:45
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You make me think of the Perseid Meteor Shower.

The Perseids /ˈpərsiːɪdz/ are prolific meteor showers associated with the comet Swift–Tuttle. The Perseids are so called because the point from which they appear to hail (called the radiant) lies in the constellation Perseus.

It happens every year when the earth travels through a field of debris left by a comet. It's named after the constellation Perseus because the meteors always appear to originate from that part of the sky.

You can only see them at night, but it lasts for several days so the whole planet get a chance to see it.

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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    $\begingroup$ This is not a hard science answer. $\endgroup$ – kingledion Feb 21 '18 at 17:51
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    $\begingroup$ This was my first thought too, though with all the details the OP sought to incorporate, I get the feeling they're looking for more of an alignment thing. But as for yearly celestial events, meteor showers are it! $\endgroup$ – Cort Ammon Feb 21 '18 at 17:52
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    $\begingroup$ @kingledion This is a hard science answer. It is backed up by empirical evidence. $\endgroup$ – Luke Feb 21 '18 at 18:21
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    $\begingroup$ Just as an observation guys: you are coming across as condescending know-it-alls here. Good on you for trying to keep up the standards but it might be better long term to put some effort into positive reinforcement. $\endgroup$ – Borgh Feb 22 '18 at 8:23
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    $\begingroup$ Considering that this was my exact thought when I read the question title, I'm pretty sure it fits. It's also a very well known real-world phenomena that is really easy to look up. It's in the news almost every year (well, it is here anyway) This isn't a hypothetical answer that could happen given what we know about the universe, or one that explains a scenario foreign to us (i.e. does happen, but not to us) $\endgroup$ – Baldrickk Feb 22 '18 at 11:20
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If you need something to happen in the Cat Constellation, your best best is to have the eye of the cat being a binary star system that is relatively close to the observers solar system.

If the angle of the stars is right, as the smaller star orbits the larger star it will regularly pass behind the larger star, growing fainter, and then it will become brighter as it passes in front of of the larger star. If the smaller star takes two years to orbit the larger star, in the first fall it will 'blink' and in the following year its eye will 'widen', all clearly visible to the masses. Example enter image description here

Sorry can't really help with the moons, unless you want a yearly eclipse around the same time.

Empirical evidence

we can see this process from Earth with 34 binary star systems using telescopes and the naked eye.

Thirty-four variable stars have a range of at least 0.4 magnitude and become brighter than visual magnitude 4.0, according to the authoritative General Catalogue of Variable Stars (GCVS) and its supplements the Name-Lists of Variable Stars. (This doesn't include novae or supernovae, which occasionally reach naked-eye brightness.) Among these stars are many eclipsing binaries, Cepheid variables, and semiregular red variables, as well as a few long-period stars of the Mira type and the recurrent nova T Coronae Borealis. As many as 24 do not fade below magnitude 5.1 and so remain visible to the unaided eye all the time.

For the math, please go here, it's rather detailed and involves a large number of equations and explainiations, too many for this answer unless you want to spend the next half hour reading it.

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    $\begingroup$ An elegant solution. Full marks for moving outside its planetary system. Plus one for an answer that deserves more points than I can give. $\endgroup$ – a4android Feb 24 '18 at 1:01
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Taking kingledion's suggestion of an alignment of the two moons with the special constellation, I have some suggestions.

For the alignment of moons to happen at least once a year and at the same date(s) each year, both of the moons have to orbit an exact or whole number of times during a year. Thus a moon could orbit 10.000 or 11.000 times per year, not 10.49 times per year, for example. It would be a vary rare coincidence for both the moons to orbit whole numbers of times per year, but there are so many possible rare coincidence that rare coincidences do happen.

It has been calculated that a moon needs to orbit more than 9 time's during a year of it's planet, in order to orbit close enough to the planet to have a stable orbit for astronomical and geological time frames. Thus the outermost moon must orbit at least 9.000 times per year and the inner moon must orbit at least 10.000 times per year.

If the inner moon orbits faster and more times per year than the outer moon, it should catch up with the outer moon at least once per year. In order to make the two moons lining up in the specified direction a unique yearly event, the lining up the two moons should happen as few times as possible.

One way to make that unique would be to have the orbits of the two moons highly titled relative to each other. The two orbital planes would intersect in a line through the planet with intersecting nodes 180 degrees apart. If the two moons line up and pass each when they are not at the nodes, they might pass at ten degrees, 45 degrees, possibly even 90 degrees, so their passing each other would not be very noticeable. But if the two moons pass through the nodes at the same time, they will pass very close and the inner moon might eclipse the outer moon.

So James may want to have the orbit of the outer moon be highly inclined, perhaps because it is a captured celestial object, and with one of the nodes where the orbital planes of the two moons intersect pointed at the desired constellation.

I think that I have discovered a couple of methods to make the alignment of the two moons happen only once per year.

1) make the inner moon orbit only one more time per year than the outer moon.

If the outer moon must orbit at least 9.00 times per year, make it orbit 9.00 times per year and the inner moon orbit 10.00 times per year. Then the outer moon will travel 3240 degrees in one year to come back to zero (or 360) degrees and the inner moon will travel 3600 degrees in one year to come back to zero (or 360) degrees. 10 orbits is 5 X 2 orbits. 9 orbits is 3 X 3 orbits. Since none of the factors of 9 is identical with a factor of 10, the two moons should not line up in the original direction more than once in a year.

What if the outer moon orbits 11 times a year and the inner moon 12 times a year? 11 x 12 is 132. In one year the outer moon will orbit 11 times and travel 3960 degrees to wind up at 0/360 degrees while the inner moon will orbit 12 times and travel 4320 degrees to wind up at 0/360 degrees. Thus they will only line up in their original direction once per year.

If the outer moon orbits 12 times per year and the inner moon orbits 13 times per year, the outer moon will travel 4320 degrees in 12 orbits and the inner moon will travel 4680 degrees in 13 orbits to both windup at 0/360 degrees.

2) make the inner moon orbit a prime number of times per year. A prime number cannot be evenly divided except by one and itself. All prime numbers are odd numbers, but not all odd numbers are prime numbers. The lowest prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71.

The orbits of some bodies have orbital resonance (which sometimes destabilize and sometimes stabilize orbits) if their number of orbits during a time period have a simple numerical relationship.

As near as I can tell none of the orbital relationships that allow for a once a year alignment of moons would be resonant, though some could be close to a resonant orbit.

James will have to decide if he wants the two moons to align pointed at the proper constellation during the day or the night. During the day people would not be able to see the stars of the constellation, though ancient astronomers and astrologers could calculate the position of the sun among the invisible stars in daylight over three thousand years.

This disadvantage would be offset by the possibility of the two moons eclipsing the sun, or transiting over it, if one of the nodes where their orbits intersected was pointed at the sun at the time when the two moons passed through the node.

Depending on the apparent diameters of the two moons, they might eclipse the sun or merely transit it as tiny black dots against its brightness. Thus watchers on the ground might see the sun blotted out in two directions as the two moons crossed it in different directions, or two tiny dots crossing it in different directions. The two moons could cross the son simultaneously or minutes apart.

If the two moons passed close to the sun but didn't cross it they would be invisible in the glare of the sun. If they crossed each other in the day sky far enough from the sun, they would appear as two thin crescents. The farther from the sun, the fatter the two crescents would be.

On the other hand, if the event happens in the night sky, the two moons would look like fatter crescents, or half moons, or gibbous moons, or full moons. People on the ground could see the stars of the culturally important constellation, except in so far as the light of the two moons might drown the stars out. Possibly there is a nearby open star cluster in the important constellation that has many stars that appear far brighter than Sirius does on Earth.

I think that James should prefer to have the two moons cross paths either in the day eclipsing the sun or at night opposite the sun as full moons.

Thus it seems like it should be fairly easy for James to have someone calculate the orbits to make everything work out.

.

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One of the moons is the size of Earth's moon, and has a semi-major axis of around 413,277 km. This gives it 12 orbits a year.
The moons orbit is such that it is full while it is in the constellation of the Lynx.

This moon is young in cosmic terms, and so isn't tidally locked. It has a feature where a combination of craters and exposed olivine rich rock on the surface closely resembles a cat eye.

The rotational period is slow, so that with every orbit 1/12th of a day passes, and is synced so that the cats eye face is visible during the full moon while it is in line with the constellation of the Lynx.

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    $\begingroup$ That orbit is outside of the Earth's Hill sphere. So unless your planet is much more massive than the Earth, or much more distant from its sun, you can't have a moon at that distance. $\endgroup$ – kingledion Feb 21 '18 at 20:05
  • $\begingroup$ @kingledion Thanks for pointing that out. I changed it back to my original idea of 12 orbits per year and 1/12th of a day per orbit, so that the cats eye face is only visible while it's in line with the constellation. It's still further out than Luna, meaning a smaller moon could be closer in as we talked about in chat. $\endgroup$ – AndyD273 Feb 21 '18 at 21:33
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I would give this much simpler answer.

Have a stationary asteroid belt with a specific formation of rocks in whatever shape you please. Of course it just happens to align with whatever and wherever your cat constellation is located.

Debris on the right and left side in the asteroid field can easily obscure the field of view until your super close.

You can't see through the sun so when your one the other side of the sun you still won't be able to see it even though you will be in alignment on the opposite side of the sun.

The moons are not even necessary for my scenario. Sure you could have them line up just perfectly so that each one looks like part of the cat shape (or whatever) and connected to the shapes in the asteroid field. Maybe its eyes, or an eye and a nose.

However, this is unduly complicated, and its more likely 2 moons won't stably orbit for 10's of thousands of years and either crash into each other or crash into your main planet. (in the eyes of a dog person having a moon destroy your planet of cat lovers, might be acceptable loss)

Forget about the 2nd moon, have it crash uneventfully into the first and put it out of its misery.

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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    $\begingroup$ A stationary belt of asteroids! Stationary relative to what? Kindly expand upon the orbital mechanics of your stationary asteroid belt. I keep thinking stationary asteroids must fall towards the nearest gravitational mass of sufficient size. $\endgroup$ – a4android Feb 22 '18 at 4:32
  • $\begingroup$ en.wikipedia.org/wiki/Asteroid_belt In our own system we have this belt and the Kuiper belt and neither one has fallen into the nearest gravitational mass. $\endgroup$ – cybernard Feb 23 '18 at 0:12
  • $\begingroup$ Yes, but the asteroids and Kuiper belt objects have orbital velocities. The word 'stationary" does suggest, however inadvertently, they aren't moving. I can see what you meant to write, but that isn't what you wrote. Otherwise a periodic alignment between a formation in an asteroid belt & a constellation is a good answer. $\endgroup$ – a4android Feb 24 '18 at 0:55
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The Cat Awakens.

cat with moon eyes

The moons line up at the perfect spot to play the role of the Cats eyes. True to type, the Cat does not stay awake very long.

cat: https://www.wpclipart.com/space/constellations/figures/color_2/Felis_constellation.jpg.html

moons https://news.nationalgeographic.com/news/2008/12/081202-venus-jupiter-photo.html

desert https://idahothemovie.com/2012/09/14/deserts/

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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