Orbit resonance: Is it possible to have exactly 3 months in a year?

It is possible for a planet to have orbital resonance with a sun (e.g. Mercury has a 3:2 spin resonance with the sun). It is also possible for a moon to have orbital resonance with a planet (e.g. our moon has a 1:1 resonance with earth).

Is it possible for there to be a resonance such that there is a discrete ratio between the time it takes for the moon to go around the planet (a month) and the time it takes for that planet to go around the sun (a year)? a) Is this possible in the case that there is a spin resonance with the planet as well (length of a day)? b) Is it possible if there is no spin resonance with the planet? (i.e. a day can be any length of time).

• Based on the precision of your request, I think you are far better raising the question on Astronomy SE. I'm not saying it's improper for WB. Oct 29 '21 at 9:57
• Someone here can probably hammer this out, I probably could if it wasn't nearly 2AM local time, but like Adrian says the Astronomy peeps can do it faster and better, and you may find they already have.
– Ash
Oct 29 '21 at 12:42
• He did ask the question at Astronomy SE. My answer was that it was not possible for have exactly (or approximately, for that matter) 3 sidereal months in a year. Since there are different types of months en.wikipedia.org/wiki/Lunar_month#Types it might be possible to have exactly 3 synodic months in a year but I haven'tworked that out. astronomy.stackexchange.com/questions/39401/… Oct 30 '21 at 4:33
• Ah. Question asks "is a moon with a period of 1/3 of year stable?"... Answer == no. It would be outside its parent planet's hill sphere, unless the planet itself was of stellar size. Oct 30 '21 at 7:49
• As suggested by Adrian Colomitchi, I have posted the question to Astronomy SE. Here's the link: astronomy.stackexchange.com/questions/47272/…. The question linked by M.A. Golding was asked by someone else. Oct 30 '21 at 8:01

The sidereal period $$T$$ of the orbit of two bodies with masses $$M_1$$ and $$M_2$$ is $$T = 2\pi \sqrt{\frac{a^3}{G(M_1 + M_2)}}$$ where $$a$$ is the orbital semi-major axis and $$G$$ is the gravitational constant.

The Hill radius $$r_H$$ of the smaller mass $$M_2$$ is the maximum distance at which things might reasonably be said to be gravitationally bound to it, and is $$r_H \approx a \sqrt{M_2/3M_1}$$. Note that I'm assuming the orbits are circular to keep things simple, so the semi-major axis is the orbital radius.

For an orbit to be actually stable, it should be no more than half of $$r_H$$. It might need to be even less than this (maybe as little as a third of $$r_H$$) but I'll take the optimistic approach here.

If we say that the length of the moon's orbital period (ie. the sidereal month) is $$T$$, the length of the planet's orbital period (ie. the year) will be $$kT$$ (where $$k$$ is obviously the number of months in the year), the orbital radius of the planet is $$a_p$$ and the orbital radius of the moon is $$a_m$$. If the mass of the star $$M_s$$ is significantly larger than mass of the planet $$M_p$$ and the mass of the planet is significantly larger than the moon, we can ignore the mass of the smaller body when computing the orbital period. We can now re-arrange the first equation to get

$$a_m \approx \sqrt{\frac{a_p^3 M_p}{k^2M_s}}$$

For $$a_m$$ to be stable, we need it be less than $$r_H / 2$$:

$$\frac{r_H}{2} \approx \frac{a_p}{2} \sqrt{\frac{M_p}{3M_s}}$$

with a bit of deft re-arranging, you get $$a_m^3 \approx \frac{a_p^3 M_p}{k^2M_s}$$ and $$\left( \frac{r_H}{2} \right) ^3 \approx \frac{a_p^3 M_p}{24M_s}$$. You can now eliminate the common bits on each side, just leaving the constraint $$\frac{1}{k^2} \leq \frac{1}{24}$$.

If you wanted a year three times longer than the month, $$k$$ is 3 and so $$\frac{1}{k^2}$$ is $$\frac{1}{9}$$. Because $$\frac{1}{9}$$ is obviously larger than $$\frac{1}{24}$$, you cannot have a moon in a stable orbit with a period equal to 3 times the parent planet's orbital period. If you simplify the constraint further to $$k \geq \sqrt{24}$$, then to a first approximation, the smallest integral number of months per year would be 5, independent of the masses of the bodies or the length of the year.

edit: If you take the more pessimistic restriction that stable orbits must have a radius no greater than 1/3 of the Hill radius, then the contraint becomes $$\frac{1}{k^2} \leq \frac{1}{81}$$, or more simply $$k \geq 9$$. And in writing this, I have suddenly remember the paper Transit timing effects due to an exomoon, which came to the same conclusion and citing it would have saved me a lot of algebra and typing, but it is always nice to work things out for oneself.

If you were feeling keen, you can create an example solar system in a gravity simulator like "orbit simulator". You'll see that the distantly orbiting moon, even though it is within the Hill radius of the planet, always "falls" out into a heliocentric orbit in a very short period of time. I've cobbled together a simple example here.

And to cover the rest of your sub-questions:

Is it possible for there to be a resonance such that there is a discrete ratio between the time it takes for the moon to go around the planet (a month) and the time it takes for that planet to go around the sun (a year)?

There is (see On the Occurrence of Commensurable Mean Motions in the Solar System), but this can't cause the specific effect you want, eg. 3 months in a year.

Is this possible in the case that there is a spin resonance with the planet as well (length of a day)?

I believe that all planetary moons in the solar systems are tidally locked (ie. a 1:1 spin-orbit resonance with their primaries).

Is it possible if there is no spin resonance with the planet? (i.e. a day can be any length of time).

I'm not sure that having a non-tidally locked moon would have a meaningful effect here, but the problem handily goes away as tidal locking tends to be pretty swift and strong (tidal locking timescales are of the order of millions of years).