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Below a certain M1M2, gravity would get so weak that the masses wouldn't be able to orbit each other with stability, as other factors would overwhelm them, even in the quietest, stillest places in the universe. Passing gravitational waves, imperfections in symmetry, quantum effects, background particles and cosmic radiation come to mind.

So, what is the smallest M1M2 we can reasonably expect to remain in a stable orbit around each other once they get going?

Are there two boulders, or two pebbles, or two grains, or two molecules somewhere in the universe orbiting each-other gravitationally since the early universe?

I understand multiple variables exist in orbital equations and picking them for ideal conditions might be tricky! I just think this is a very interesting question and wonder if it's even possible to answer it.

Thanks for considering it. It might be good to define first how quiet the quietest/stillest and most peaceful place in the universe actually is first and if that should be a separate question, let me know. Not sure I'd know exactly how to formulate it though!

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    $\begingroup$ Partial answer, hence only a comment: At least 'large rocks' beyond the orbit of Neptune (lowest masses $10^{18}$g) can orbit each other, some are very delicately bound with binary semimajor axis/Hill-radius $a/r_H \approx 0.15$. See ui.adsabs.harvard.edu/abs/2020tnss.book..201N/abstract $\endgroup$ Feb 2, 2023 at 21:38
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    $\begingroup$ Related: space.stackexchange.com/questions/55856/… $\endgroup$
    – Connor Garcia
    Feb 2, 2023 at 22:47
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    $\begingroup$ The quietest place would probably be voids, large voids mostly devoid of any observable matter, not sure of stillest though because the universe is expanding. $\endgroup$
    – user47732
    Feb 3, 2023 at 18:17

1 Answer 1

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Here's one perspective. Newtonian gravity is symmetrical under two kinds of rescaling:

  1. If you scale lengths and velocities by $\lambda$ and masses by $\lambda^3$, the system will evolve in exactly the same way with the same time scale.

  2. If you scale times by $\tau$, velocities by $\tau^{-1}$, and masses by $\tau^{-2}$, the system will evolve in exactly the same way with the same length scale.

Generally, systems are more susceptible to external perturbations if they have longer dynamical time scales. So decreasing a system's mass by transformation (2) above is not the approach we want. However, by transformation (1), we could rescale a system's mass to be arbitrarily small while preserving its stability with respect to external effects.


This idea holds under classical Newtonian gravity. General relativity breaks these scaling symmetries, but not in a way that is problematic for us. The gravitational potential scales as $\lambda^2$ under transformation (1), so downscaling the system brings it further from the regime where general relativity is important (which is where the potential is of order $c^2$).

However, quantum mechanics would impose a lower limit to such scaling. The product of position and momentum uncertainties of particles in our system would scale as $\lambda^5$ under transformation (1), but that product is bounded from below by the uncertainty principle. But once we reach scales where quantum mechanics are important, the correct approach would be to start thinking about gravitational atoms rather than classically orbiting systems.

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    $\begingroup$ Nice answer, +1. One could add a note that given the average density of matter even in the intergalactic matter, atoms or binary bodies larger than the mean free path of bodies with similar size will be of limited lifetime due to interactions with other matter. $\endgroup$ Feb 11, 2023 at 13:27
  • $\begingroup$ @planetmaker Hmm, I wonder if that's the appropriate comparison. Generally, a system's capacity to survive tidal forces or impulsive encounters is set by its density (or equivalently dynamical time scale). Size doesn't enter there. You might think that making a system larger increases the frequency of encounters, but it doesn't really, because the distance required for an encounter to be disruptive depends on your system's density again and not its size. Admittedly my experience is more with many-body systems, not two-body. $\endgroup$
    – Sten
    Feb 11, 2023 at 16:17

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