Moons of Moons of Moons

Question

My question is simple: How many nested moons are physically possible?

If our moon had a moon, that would be a nesting of 1.

I'm assuming it's easily possible for a really big moon to be orbiting a gas giant and have its own moon. If the dimensions were right, that moon could also have a moon?

How many moons deep can we go? Let me know if I'm missing something. I would like answers with calculations not just random guesses! I'm asking what is physically possible, not what is realistically plausible due to the difficulty of such a system forming.

Definitely related (possibly duplicate): How can you make a stable.... There are no answers there yet, but I suspect any answer there could be extrapolated to an answer here. — Frostfyre, Oct 01 '15 at 14:27
Not the same question, but the answers end up giving similar results: http://worldbuilding.stackexchange.com/questions/15577/life-on-a-moon-of-a-moon?rq=1 — Dan Smolinske, Oct 01 '15 at 14:28
My last sentence negates both of those questions. I did however just find this: http://astronomy.stackexchange.com/questions/8694/theoretical-limits-for-natural-satellites-having-natural-satellites?lq=1 which is indeed the same question but has one answer which again ignores the last sentence and doesn't answer my question. — Varrick, Oct 01 '15 at 14:34
We can't mark cross-site duplicates, but I would say that Do moons have moons? on [astronomy.se] would be a pretty strong candidate if it was possible (particularly considering the answers to it). — user, Oct 01 '15 at 14:34
This is essentially a maths/mechanics/physics question I guess. — Varrick, Oct 01 '15 at 14:37
Sadly the last sentence is the one that stops there being any meaningful answer to this question except 'as many as you'd like'. If we ignore what is realistically plausible (i.e.: likely) for a given system then it's possible for every particle in my body to spontaneously gain enough energy from those particles nearby that I achieve escape velocity. Sadly that's so horrendously unlikely that I will not be going to space today (and I brought my favourite suit too...) If we're ignoring what's likely and going with what's possible then you can pretty much say what you like! — Joe Bloggs, Oct 01 '15 at 15:40
@JoeBloggs yes I realize that in hindsight. I kind of came up with this question after seeing the challenge in the meta without giving it much thought and now I'm unable to delete it. — Varrick, Oct 01 '15 at 15:53
We already have something like this today. The Moon orbits Earth. Earth orbits the Sun. The Sun orbits the Galaxy (which has a supermassive black hole at its center). The Galaxies of the Local Group orbit and interact with each other. That last part might not be stable, but I just wanted to bring some giant perspective into this. — DrZ214, Oct 01 '15 at 20:55

score 52 · Accepted Answer · edited Jun 16 '20 at 11:03

52

Let's make a bunch of assumptions:

The largest primary is about 3 times bigger than Jupiter.
To really be a parent, the barycenter of a parent-satellite system must be within the parent.
Everything has approximately the same density
Orbital stability will magically work itself out (this will give us an upper bound)

Let's call the twice the distance between the barycenter of a parent satellite system and the farthest extent of that system $D_p$ then the corresponding diameter for the subsystem $D_s$

Now if the mass of the parent is $M_p$ and the mass of all the sub satellites together sum to $M_s$ then the requirement that barycenter be inside the parent yields:

$$\frac{M_s}{M_P}(D_p-\frac12D_s)<\left(\frac{3M_p}{4\pi\rho}\right)^\frac13$$

Now we know that for each parent none of the satellites can pass within the roche limit of the parent (the limit would actually be farther out due to the fact that the satellite system isn't solid but this will get us an upper bound) Lets call the diameter of the satellite system $D_s$ and the diameter of the parent system $D_p$. The Roche Limit gives:

$$\frac12 D_p>2.4\left(\frac{3M_p}{4\pi\rho}\right)^\frac13+D_s$$

If we claim that each subsystem is proportionate to the parent system then we have:

$$\left(\frac{D_s}{D_p}\right)^3=\frac{M_s}{M_P+M_s}$$

Now if we're trying to maximize the ratio of satellite mass to parent mass both of these inequalities should be equalities.

Solving the system yields:

$$D_p \approx 2.6 D_s$$

Which means each successive moon would weigh $17$ times as much as the previous one.

Now to get from a single atom moon to something 3 times the size of Jupiter would take: $$\frac{\ln\left(3\frac{1.89813 × 10^{27} kg}{1.6726219 × 10^{-27} kg}\right)}{\ln(17)}=42$$

So a system could have a maximum of 42 layers if we stopped at planets as the primary body. Note however, this doesn't consider orbital stability and I have no doubt that even a system with 10 layers would be unstable on the time scale of a century.

Bigger

If we went up larger and larger, we could eventually incorporate black holes and then relativity plays havoc with the equations. However, I think that at the extremely large end, the expansion of the universe would distort and pull apart any orbits with radii on the order of billions of light years. So if we said that was the limit, then you could nest about $85$ layers, which is a lot, but I would hardly call that infinite.

edited Jun 16 '20 at 11:03

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answered Oct 01 '15 at 20:08

Rick

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I'm not sure you can even get to 10 before it becomes untenably chaotic. Pendulums become chaotic if you chain two of them together (which is analogous to a planet/moon/sub moon)!! – Joe Bloggs Oct 01 '15 at 20:38
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This is the kind of answer I was looking for, well done sir! – Varrick Oct 01 '15 at 21:19
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@JoeBloggs: Sheesh, you're not kidding. – Kevin Oct 02 '15 at 00:07
"So a system could have a maximum of 35 layers" - sorry to ruin your day but atoms cannot be gravitationally ripped apart and are not subject to the Roche limit. Also you can imaginably have a collection of 33 atoms held together in a single molecule (or a single atom with atomic mass of 33). Speaking of which - non-inert gasses always stay in molecules. You can't have single atoms roaming around. But that it's roaming around in space that already has ${H_2}$ floating around probably is a problem (and not just for the lowest tier)... – John Dvorak Oct 02 '15 at 06:04
@JanDvorak The fact that an atom could orbit inside the Roche limit does not alter the answer much. - Because it can not have a moon if it's inside the Roche limit. – Taemyr Oct 02 '15 at 08:36
@JanDvorak Well if you switch the Roche constraint to a constraint saying the "moons" can't hit each other, then the ratio goes from 2.6 (sorry I found another error) to 2.3 which raises the planet limit to 50. So like Taemyr said, it doesn't change the answer much. The limit is still bound to be less than 100, and as Joe points out, is probably not even as big as 10 do to stability constraints. – Rick Oct 02 '15 at 11:39
Oh, you can't... Oh well! – Varrick Oct 02 '15 at 11:58
http://curious.astro.cornell.edu/our-solar-system/44-our-solar-system/the-moon/general-questions/104-can-moons-have-moons-intermediate points out an interesting concept - the Hill sphere - which might have more of an effect on your calculation than the barycenter location assumption. – childofsoong Oct 02 '15 at 19:10
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I like this answer but really I think you need a more realistic mininimum size moon than a single atom :p – Tim B Oct 02 '15 at 21:21
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I love the fact that the answer here (at least to one question) is 42. – Alfe Oct 02 '15 at 23:40
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@Alfe: Does that mean that next Thursday the Vogons will arrive to demolish our planet? – celtschk Oct 03 '15 at 16:58

score 27 · Answer 2 · answered Oct 01 '15 at 14:52

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Theoretically infinite, though the classification would get interesting.

Consider the Moon. It orbits us, the Earth. That's a nesting of 0, right? Now consider the Earth. What's to say that the Earth isn't just a moon of the Sun, apart from arbitrary human classification systems?

On this logic, you could have theoretically infinite moons. If you re-classify any orbiting body as a moon, and you start with a massive enough body, then you can have entire stellar systems orbiting it - giving you big numbers for the nesting. Think: the Sol system, orbiting another larger body, which in turn orbits an even larger body. That gives a nesting of 4 (I think.)

If you're not up for reclassifying, then the potential is small for nesting moons. According to Wikipedia/Natural satellite, the definition of a moon is:

a celestial body that orbits another body (a planet, dwarf planet, or small Solar System body), which is called its primary, and that is not artificial.

That's limiting, because the biggest object you can have a moon orbiting is a planet, which are (comparatively) small. With each orbiting moon, you have a smaller object, and eventually you're left with nothing.

answered Oct 01 '15 at 14:52

ArtOfCode

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Sorry but you've encouraged me to change my question a little bit and put some bounds to stop the answer "infinite" :P – Varrick Oct 01 '15 at 15:01
Ok, unedited... This is a bit of a botched question really, hopefully it prompts someone to ask a better one. – Varrick Oct 01 '15 at 15:10
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Good idea with going up instead of down. If you follow the 'smaller and smaller' route you actually hit the planck length surprisingly quickly. – Joe Bloggs Oct 01 '15 at 15:44
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Could you be able to show mathematical proofs of how it's possible to have infinite nesting? – John Odom Oct 01 '15 at 16:52
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Disagree - stars have definitions, as do planets (stars have fusion, planets have cleared their neighborhood, etc). – corsiKa Oct 01 '15 at 16:54
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@corsiKa: And definitions can change (see Pluto). – celtschk Oct 01 '15 at 18:08
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I believe at very large distances the expansion of the universe would prevent one body from orbiting another due to the distances between them growing faster than the speed of light. Thus there is an upper limit on size. – Rick Oct 01 '15 at 20:11
@Rick - Is there some material somewhere that shows the Universe expands faster than light? I find that quite hard to believe. – ArtOfCode Oct 01 '15 at 21:20
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@JohnOdom - the proof lies in the assumption that you can have an infinitely large object. Even if you halve the size each time, an infinitely large object provides infinite half-sizes. – ArtOfCode Oct 01 '15 at 21:21
Just because you can imagine orbits doesn't mean they would work. Rick poits out that a reasonable answer is closer to 60 than infinite, and that doesn't consider stability. – JDługosz Oct 02 '15 at 02:28
@ArtOfCode expansion isn't measured in velocity it's expanding at a rate of 74.3 plus or minus 2.1 kilometers per second per megaparsec (a megaparsec is roughly 3 million light-years) so taking the lower limit, two points would have to 13 billion light years apart to be moving away from each other at the speed of light (A kind of depressing consequence of this, is that we can never reach any point that's currently further away from this, and the amount of matter in this sphere is shrinking, so the theoretical amount of stuff to explore is forever shrinking.) – Rick Oct 02 '15 at 11:49
@Varrick Feel free to ask follow on questions yourself if this one has shaped your thinking. – Tim B Oct 02 '15 at 21:22
the movement of the galaxies towards each other has been graphed and is not an orbital shape. it is a convergence and divergence. – bandybabboon Oct 04 '15 at 07:15
@ufomorace Well aware. But that's our current universe. There's no reason they couldn't form orbits, in this other universe. – ArtOfCode Oct 04 '15 at 07:33

score 10 · Answer 3 · answered Oct 01 '15 at 16:30

There are some mathematical stability issues which might create sensible bounds, but I'm not qualified to quantify them. In general terms there is no guarantee of long-term stable orbits in a three-body system (such as Sun, Jupiter, Earth, even if Saturn and the rest weren't there). In fact, special cases aside (notably two Lagrange points) there's a proof that there are no infinite-term stable orbits, only chaos. Be reassured that the best astronomical measurements and computer modelling show Earth's present orbit won't change drastically for the next hundred million years or so, after which time we can't say anything about it because of the errors on the observations. It's therefore not impossible that five hundred million years hence, Earth will be a frozen rock wandering the galaxy in interstellar space. (Statistically, it's more likely to stay orbiting the sun until the sun goes nova).

Were you to try to model Asimov's "Nightfall" system (IIRC 4 stars and two planets in a complex dance) you'd find it was unstable on a timescale much shorter than that for the evolution of Terran life. Something would get ejected from the system into interstellar space, or would collide with another body, and so the story is highly improbable (for thermodynamical levels of improbability).

One approach to simplify the N-body problem's maths is perturbation theory: treat the Earth/Moon system as tightly coupled, treat our centre of mass as bound to the Sun and perturbed by Jupiter (the biggest other perturber) on a completely different scale. My guess is that there isn't space for more than around five "levels" before chaos becomes unavoidable on a short timescale. The upper scale is set by the scale of interstellar space and the large number of suns in the galaxy. The lower one, by the weakness of the gravitational force and the fact that other influences such as solar wind and atmospheric drag become dominant for tiny objects orbiting small ones.

Over to anyone with greater higher mathematical skill and knowledge than myself!

I will point out that our solar system is in fact a chaotic system, that cannot be accurately predicted. We get pretty close, but it is not possible to fully predict. — Dan, Oct 01 '15 at 23:52
@Dan: nothing can be predicted with perfect accuracy; it's a really fundamental thing in physics that every quantity has some uncertainty. But the solar system is in many a sense among the physical systems which we can model most precisely. — leftaroundabout, Oct 03 '15 at 00:03
@leftaroundabout I realize that nothing can be predicted with perfect accuracy. However, orbital mechanics are what's called a Chaotic system. The meaning of this is that even a tiny deviation in starting conditions leads to hugely different results, instead of mildly different results. As you pointed out, our ability to get accurate initial conditions for such calculations is limited, therefore we are not able to predict temporally distant future orbital states with great accuracy. — Dan, Oct 05 '15 at 15:25
@Dan: the Lyapunov exponent of the solar system is smaller than 10⁻⁶ yr⁻¹ [http://web.mit.edu/wisdom/www/measurements.pdf]. Yeah, it is technically chaotic, but I daresay you'll be hard-pressed to find any system with a smaller exponent. So in a sense the solar system is incredibly harmonic. But that would almost certainly not be the case for a system with more nested satellite orbits, like the one proposed in the accepted answer. — leftaroundabout, Oct 05 '15 at 16:15

score 3 · Answer 4 · answered Oct 02 '15 at 02:22

Similar discussions have been marked in comments. I recall previously discussing circumlunar orbits and why a long-lived natural sattelite won't have such an orbit.

Having the primary be uniform rather than lumpy improves the situation.

Having multiple bodies in resonance can stabilize the whole thing. So a pterbation that bumps one sattelite will be corrected by the sisters, on a scale shorter than the long-time averaging out of random outside influences.

Also, don't limit yourself to big-small progressions. Complex star systems, for example, don't have stable 3 or 4 star systems, but do have "hierarchical binaries". Two bodies here in mutual orbit, two bodies there in mutual orbit, and the two pairs in orbit. How does that count in your system, with no "root" object but clearly a hiarchy of what orbits what.

Moons of Moons of Moons

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