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I was wondering if there is a theoretical limit to how long a chain of orbiting objects can be, for instance you can orbit the moon which orbits the earth which orbits the sun which orbits the galaxy, i was wondering how long that chain could theoretically be.

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    $\begingroup$ I think this question was closed prematurely, since the linked answer is about the longest observed chain rather than the theoretical limit. This is a pity, since one can actually work it out using Hill spheres. That approximation generates a geometric progression of masses that eventually has to run out due to atoms and black holes. $\endgroup$ Apr 13, 2022 at 10:46
  • $\begingroup$ agreed, reversed closure $\endgroup$
    – Connor Garcia
    Apr 13, 2022 at 15:49
  • $\begingroup$ Assuming you mean a "stable" orbit, you need to define how stable, as nearly all n-body systems are unstable. $\endgroup$ Apr 13, 2022 at 15:58
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    $\begingroup$ Does this answer your question? Is there a non-trivial limit to a number of nested orbits? $\endgroup$ Apr 13, 2022 at 16:35
  • $\begingroup$ I think the one Mark linked to should be merged with this one, as Anders' answer here is much better than the one on the duplicate $\endgroup$
    – Rory Alsop
    Apr 15, 2022 at 14:02

3 Answers 3

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The key limiting factor on whether an object orbiting a heavier object can retain a moon is the Hill sphere: this is the region where its gravity dominates over the heavier object. For circular orbits this is approximately $$r_H \approx a \sqrt[3]{\frac{m}{3M}}.$$

Let us consider a chain of objects of mass $m_1 > m_2 > \ldots > m_n$ orbiting at distance $a_2,a_3,\ldots a_n$ from their principal object. If we demand that each object is in within the Hill sphere this gives the relation $$a_k < a_{k-1}\sqrt[3]{\frac{m_k}{3m_{k-1}}}.$$ Note that for the (unrealistic) case $m_k=m_{k-1}$ that produces the largest $r_H$ this gives $a_k < 3^{-1/3} a_{k-1}$, and hence $\sum_{k=1}^\infty a_k < \frac{a_1}{1-3^{-1/3}}\approx 3.26 a_1$ (by summing a geometric series). So we have a limit to the length of the chain, if not the number of objects.

If we assume the same density $\rho$ of the spherical objects then each object needs to have radius $r_k = (3/4\pi \rho)^{1/3}m_k^{1/3}$, and clearly $a_k > r_{k-1}+r_k = (3/4\pi \rho)^{1/3}[m_{k-1}^{1/3}+m_k^{1/3}]$. Indeed, the Roche radius gives a stricter bound that $a_k > 2^{1/3}r_{k-1} = (3/2\pi \rho)^{1/3} m_{k-1}^{1/3}$.

So it seems that we can get away with a very long chain by making the masses minuscule - which in practice is not possible because (1) there smallest stable uncharged masses available are hydrogen atoms (yes, yes, neutrinos might count too but gravitationally bound neutrino chains would be ripped apart by the expansion of the universe), (2) there are numerous disrupting forces in the universe from e.g. galactic tides, nearby stars, sunlight, etc., and (3) equal-mass chains would tend to interfere gravitationally with themselves fairly strongly due to resonances (not covered by the basic Hill estimate).

Furthermore, once the mass becomes larger than $M=\sqrt{3c^6/32\pi G^3\rho}\approx 10^{39}$ kg for molecular density objects, it will be a black hole (at which point the Newtonian approximation breaks down, plus the overlap relation becomes much stricter since $r\propto m$ for black holes).

Hence, for a somewhat plausible chain of molecular density objects that do not overlap and contain something moon-like in the middle the chain will be finite.

A plausible ratio $a_k/a_{k-1}=m_k/m_{k-1}=(3/2\pi\rho)^{1/3}\approx 0.0782$, and that would descend from $10^{39}$ kg to $10^{-27}$ kg in 59 steps. There is a bit of handwaving here since we are free to play around with the parameters a fair bit.

In nature I would be surprised to see chains much longer than about 7 (SMBH, supermassive star, orbiting stellar companion, planet, moon, asteroid, asteroid moon). The main reason is that as the mass ratio $m_k/m_{k-1}$ becomes large there are many forms of instability that come into play from the previous steps in the chain and tends to make them fall apart. Another reason is that the natural sizes of objects do not follow this geometric progression - there are mass gaps due to how objects form that makes a high mass ratio chain very unlikely. It is still a fun question.

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Here is a link to a similar question which has some (conflicting) answers which may help you get a feel for the problem.

What is the longest natural bound orbit chain observed?

There are disagreements about what constitutes an object orbiting another object. But you should be able to get a chain with eight objects in some cases.

  1. Whatever object might be in the gravitational center of a young open star cluster >
  2. a massive young star >
  3. a much less massive star orbiting it >
  4. a brown dwarf orbiting it >
  5. a giant exoplanet orbiting it >
  6. a large exomoon orbiting it >
  7. a small moon of a moon orbiting it >
  8. a dust mote orbiting the moon of a moon.
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If we allow ourselves to start with an empty universe, I don't see how there can be an upper limit. If A is a chain of orbiting objects, then you can postulate an object B, more massive than the total mass of A and far away from its center of mass. For any specified timescale, and any mass of B, you can always make B far enough from A so that the system is stable over the desired timescale. Now let A' be the union of A and B, and repeat the process. At some point, B might be a black hole, but I don't know why that would be a problem.

This answer might be more appropriate if this were a question in Physics SE. If you want to know the longest theoretically possible chain in this universe, where other galaxies will eventually get in the way, then the answer will be fuzzy and based on Hill spheres and the amount of available space to set up these delicate chains.

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  • $\begingroup$ I wrote this before I realized it was a duplicate. $\endgroup$ Apr 13, 2022 at 16:36

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