Someone do the maths on this.

[STARP_Social_Officer]

It's 10pm, I've had a long and busy day, and I suck at maths. These are three very good reasons why I couldn't be bothered working out the maths on this question. Can someone help me, please? Remember that I'm timing you, and no slide rules or MP3 players are permitted. If you need help, raise your hand and do not leave your seat for any reason.

The campaign world I'm developing has two moons, Luna Alpha and Luna Beta. Luna Alpha, the larger of the two, orbits the planet in the traditional time of one month. Luna Beta, however, is smaller, slower and further away, and it orbits the planet in three months. Assuming the sun is 50 million miles from the planet, Luna Alpha is 77,000 miles from the planet, and Luna Beta is 257,000 miles from the planet (I just made those up, but we'll consider them canon now), how often will both moons be full at the same time?

The person that gives me the correct, or at least most plausible answer, wins a trip for 2 to Barbados. Unless they already live there, in which case they don't need a free trip.

Turn over your papers. You may now begin.

 

[Percivellian]

Your question does not contain enough information for a proper response to be formulated. The phases of the moon depend on where from the planet you are viewing them from, so we'd need a specific vantage point. After that we would also need the orbital axis of each moon.

 

[STARP_Social_Officer]

What did I say about talking while the exam is in progress!?

OK, you've raised two points that I would have thought of if, you know, I knew anything about this stuff. Let's say we're viewing from the north pole, just for the sake of argument. We'll sa Luna Alpha's axis is the same as the Moon's axis around Earth, and Luna Beta's axis is about...I dunno...five degrees higher? I have no idea if this is even possible...I'm just making this up. If I've got the astronomy completely wrong, someone else can feel free to make up a more realistic scenario using correct figures.

 

[Nightchilde-2]

Whatever's dramatically appropriate.

 

[Dioltach]

Surely the easy answer is that the closer moon rotates around the planet three times as often as the farther moon, and since they must both be in the same position relevant to the planet to be full, therefore they will both be full at the same time once every three months?

 

[AE020704]

Whatever's dramatically appropriate.

QFT.

Don't get yourself too hung up on the mechanics of a given thing; that's just not fun. Also, you run the risk of coming up with a scenario in which they are both full maybe once in every million years, or something ridiculous like that.

Unless you have a player that's likely to turn round and ask, "Wait. What did you say the orbital axis of the smaller moon was?" (and quite frankly, they need killing), then it's no real issue.

Forget it. You'll be much happier that way.

 

[Castellan]

The phases of the moon depend on where from the planet you are viewing them from, so we'd need a specific vantage point.

Ummm... No.

The phases are based on where the moons are, not where a person is on the planet. Whatever the moon phase, it'll appear the same from every point on the planet that can see the moon.

Since the original poster isn't interested in the math, I'd say do some hand-waving and ignore the orbital mechanics and go with the previous answer of "once every three months."

 

[Nadaka]

I assume your orbital periods stated are synodic? (in relation to the sun)
In this case, they are both full exactly once every 3 "months" when the longer period moon is full because the shorter peroid divides the longer exactly.

Or never, for exactly the same reason.

The shorter period moon will have exactly the same phase every time the longer period moon reaches full.

 

[Pbartender]

I assume your orbital periods stated are synodic? (in relation to the sun)
In this case, they are both full exactly once every 3 "months" when the longer period moon is full because the shorter peroid divides the longer exactly.

Or never, for exactly the same reason.

The shorter period moon will have exactly the same phase every time the longer period moon reaches full.

More precisely, the two moons will share indentical phases (be in conjuction, to use the proper term) -- which may not necessarily be "full", as that depends on where the conjunctions happen -- every 1 1/3 months.

So, for example, if the two moons start out together full...

0 months = both full
1 1/3 months = both 1/3 full (waning)
2 2/3 months = both 1/3 full (waxing)
3 months = both full (again)

Do a search for "Galilean Moons" to see a real world example of what you're trying to achieve.

 

[Nadaka]

More precisely, the two moons will share indentical phases (be in conjuction, to use the proper term) -- which may not necessarily be "full", as that depends on where the conjunctions happen -- every 1 1/3 months.

So, for example, if the two moons start out together full...

0 months = both full
1 1/3 months = both 1/3 full (waning)
2 2/3 months = both 1/3 full (waxing)
3 months = both full (again)

Do a search for "Galilean Moons" to see a real world example of what you're trying to achieve.

your math is off.

the position unit vector function for a circle that completes one revolution in a time of 1 is:
cos(2*pi*t)i + sin(2*pi*t)j

the position unit vector function for a circle that completes one revolution in a time of 3 is:
cos(2/3*pi*t)i + sin(2/3*pi*t)j

The moons would be in conjunction when the value of the position unit vector functions are equal, this occures every 1.5t.

example: assume they are in conjunction at t = 0.
at t = 4/3
cos(8/3*pi) = 0.98932941888317171735655614235434
sin(8/3pi) = 0.2520993595498590952085472195367
and
cos(8/9*pi) = 0.99881250041313123913009776422591
sin(8/9*pi) = 0.048719493208249901314582764598032
not equal.

at t = 1.5:
cos(3*pi) = -1
sin(3*pi) = 0
and
cos(pi) = -1
sin(pi) = 0
equal.

Note that if the slower of the two moons shares its full phase with the faster, it would also share its new moon phase.

edit: this coincedentally, is very similar to the problems on the cal 3 test I just took.