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Comparing three different scales of $n$-body systems, atomic systems, solar systems and galactic systems, we find that all (observed) systems that exhibit long term stability have the largest mass at the center of the system. Even in the atomic case where the dominant force is not based upon mass but upon electronic charge, the nucleus is still the more massive object.

What property of these systems makes this configuration the preferred one?

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  • $\begingroup$ What are the forces for each, as function of position? $\endgroup$ – Kyle Kanos Nov 10 '19 at 3:04
  • $\begingroup$ @KyleKanos I have central forces in mind, both gravitational and electronic. So yes, the forces are a function of position. $\endgroup$ – OneMug Nov 10 '19 at 3:06
  • $\begingroup$ If you apply Boltzmann statistics, more particles will be in lower energy states, which is presumably near the common COG. If so many particles in small space tend to combine, you end up wit a larger body there $\endgroup$ – Hagen von Eitzen Nov 10 '19 at 3:25
  • $\begingroup$ That's not really what I asked... $\endgroup$ – Kyle Kanos Nov 10 '19 at 4:29
  • $\begingroup$ @KyleKanos Then what was it really that you asked? Please clarify. $\endgroup$ – OneMug Nov 10 '19 at 4:33
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Such systems revolve around their barycenter, the center of mass of the particles. If one particle is much more massive than the others, the barycenter will be very close to the massive particle, or even in it. Thus, the question becomes "why do stables systems have one particle that is so much more massive than the others."

If we look at N-body systems, we find that there are stable solutions for N=1 (lone particle) and N=2 (binary orbiting patterns), but at N=3 it starts to get tricky. The equations get chaotic and have a strong tendency to eventually eject one of the objects.

An exception to this rule is when one object's mass is so much higher than the others that you can basically ignore the effects of the other objects on it. For instance, the Earth's effect on the sun is pretty darn negligable, all things considered. This minimizes the chaotic effects which could cause ejection of objects.

So as such, you find systems with one particle (lone wolf), two particles (binary stars), or three or more particles with one substantially more massive particle (solar systems).

For a more formal handling of the topic, I recommend starting with the Wikipedia page on the Three Body Problem. It explains much of the difficulties and also points out some fascinating special cases that people have found over the years.

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  • $\begingroup$ I agree completely, especially with your third paragraph addressing the 'exception'. Is there a way to analytically show how ''This minimizes the chaotic effects which could cause ejection of objects.'' I don't have a good grasp of chaos theory, so any help from a physics point of view (vs a mathematician's POV) would be of great help. $\endgroup$ – OneMug Nov 11 '19 at 23:35
  • $\begingroup$ @OneMug The typical "physics PoV" approach is to declare M1 >> M2 and M1 >> M3, and then use that to identify "negligible" terms and approximate them as zeros. For a mote of dust orbiting the sun, this approximation is so good you'll probably never notice any error. For a planetoid, its good for a few million years. $\endgroup$ – Cort Ammon Nov 12 '19 at 5:57
  • $\begingroup$ Another interesting point of view: An extreme version of this simplification is patched conics, where not only do you assume your vehicle's effect on celestial objects like the Earth is small, but you also assume that only the most influential object pulls on you. From what I can tell, the Apollo missions were planned using patched conics. So that shows a very meaningful mission that was accomplishable with such simplifications =D $\endgroup$ – Cort Ammon Nov 12 '19 at 6:02
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A system of bodies will orbit around the system's center of mass ($\vec{x}_{COM} = \frac{\sum m_i\vec{x}_i}{\sum m_i}$). For systems where one body is much more massive than the sum of all the other bodies isotropically distributed, the center of mass will be inside that massive object. Stars are much heavier than planets and nuclei much heavier than electrons.

Binary stars are example of a system where there isn't a clear "this object is rotating around this one", because there isn't one "dominant" mass.

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  • $\begingroup$ If nature had made the protons the same mass as the electrons, would we still have stable matter? $\endgroup$ – OneMug Nov 10 '19 at 3:59
  • $\begingroup$ Why do we not expect to find planetary systems with a combined planetary mass greater than that of the central star? Granted it would take many planets, but again, why not? $\endgroup$ – OneMug Nov 10 '19 at 4:16

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