Answers to How does the Sun's gravity have so much force and pull on the solar system? How does it scale? Newton's law of gravity scales on orbits so that we can theorize very small orbits with very small masses.

Could two hydrogen molecules orbit one another or would external gravity prevent them from having a stable orbit? We know there are quite small orbital systems in our own asteroid belt, but is there a practical limit to how small an orbital system (with the little orbital system also orbiting the Sun) can be in terms of total mass of the orbital system?

  • $\begingroup$ The mass of orbital systems depends primarily on the tidal force of the space they are in so it would depend where in space you were. You can get much lower mass orbits as you move away from the Sun. It might be possible to work out a rough calculation but I'd wager it would be much more massive than a pair of atoms, where their electrical fields might be more relevant than any gravitational attraction between them. $\endgroup$
    – userLTK
    Nov 25, 2020 at 7:21
  • $\begingroup$ That should be the low end mass of the more massive of the two bodies in a 2 body orbital system. The smaller of the two body can be quite tiny. $\endgroup$
    – userLTK
    Nov 25, 2020 at 7:41

2 Answers 2


If we only consider gravity, one answer may be found using the Hill sphere. This is the distance the gravity of a body dominates over the sun: $$r_H \approx a \left(\frac{m}{3M_\odot}\right)^{1/3}$$ where $a$ is the semi-major axis, $m$ the mass and $M_\odot$ the sun's mass.

Now, an actual body has some nonzero density $\rho$ and $m=(4\pi/3)\rho r^3$. If the Hill sphere is inside the body there will not be any orbits around it (they would be dominated by the sun's gravity). So, we get the equation $$r = a\left(\frac{(4\pi/3)\rho r^3}{3M_\odot}\right)^{1/3}$$ which simplifies to $$\rho = \frac{9M_\odot}{4\pi a^3}.$$ Objects less dense than this have Hill spheres inside themselves: at 1 AU this density is $4.3\cdot 10^{-4}$ kg/m$^3$ (a thin gas), while at 0.1 AU it is 0.4255 kg/m$^3$ - about a third of sea level air density.

For hydrogen atoms, if we calculate the density for a 25 picometer atomic radius I get a density of 25,570 kg/m$^3$ (in actual hydrogen gas the atoms are spread out way more). Hence there the Hill sphere argument actually allows them to orbit each other!

In practice this does not happen. The orbital period at (say) 3 atomic radii is $\sqrt{4\pi^2r^3/Gm}\approx 3.4$ hours and the binding energy is $1.5\odot 10^{-27}$ J. This is $4\cdot10^{-5}$ of the thermal energy of the cosmic background radiation: even if there wasn't any sunlight or other radiation from inside the solar system it would jostle the atoms enough that they would split.

This suggests an apparent way of answering the question: if the binding energy $Gm/r$ is less than typical disrupting energy the orbit will not be possible. Actually calculating the forces is non-trivial (there are many kinds, from Jupiter's gravity to solar heating) and weaker forces can sum up over time. Knowing the disruptive background also just gives an upper bound for $m/r$, one could have smaller orbits.

So the true answer will be given by how small dense objects we are willing to consider, and (as the other answer points out) local forces. In the solar system the most relevant may be electromagnetic charging due to the solar wind: if the objects are metallic and close they can even attract each other if they have the same charge (!). Things like magnetic fields, infrared radiation and solar wind will play a role, making the true answer somewhat undefined.

  • $\begingroup$ So a bunch of hydrogen atoms orbiting each other is out, but objects like bowling balls might actually work? $\endgroup$
    – Mast
    Nov 25, 2020 at 21:03
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    $\begingroup$ I think the bowling balls could work. It would be a really slow orbit (I get about 40 hours), $\endgroup$ Nov 26, 2020 at 0:47

It is difficult to work out an actual numerical answer, but let me point out some things that I think would determine the lower bound.

If two objects are close enough, they are attracted by Van der Waals forces. That only operates over a very close range, but it sets a minimum distance before something besides gravity dominates. That is relevant because low-mass objects orbit each other very slowly. I don't know how close two hydrogen atoms have to be in order for their orbital period to be shorter than the lifetime of the universe, but it is a value to check.

I don't know the physics of atomic hydrogen in the presence of magnetic fields, but gravity is so weak that I would expect even weak magnetic fields to dominate for something as small as an atom.

I think other gravitational fields are less of an issue than electromagnetic fields along with the possibility of collision before an orbit can be completed.

For all these reasons, I suspect a pair of hydrogen atoms can't be gravitationally bound in such a way that you will see multiple orbits over the lifetime of the system. If that is correct, then there must be some lower threshold, but it would depend on the local conditions.

After Edit:

Another factor to consider is light pressure. Again, I'm not doing the calculation, but, if I understand correctly, individual photons can transfer momentum to an atom. Gravity is so weak that I would expect any orbit to be disrupted by a single photon impact to one of the atoms.

  • $\begingroup$ The Van der Waals force between atoms (and perhaps molecules) is something that can be looked up somewhere, but I'm not sure what your discussion of magnetic fields is about. Do you mean each molecule's (or each atom's) magnetic magnetic moment upon the other would be attractive or repulsive? Or do you mean they somehow interact with the interplanetary magnetic field? $\endgroup$
    – uhoh
    Nov 25, 2020 at 0:28
  • $\begingroup$ Also, remember that angular momentum is conserved, so that unless two molecules have zero angular momentum wrt their center of mass to begin with, they can't "fall into" each other unless they can "radiate angular momentum" through photon emission. You can look at that semi-classically or quantum mechanically. If they are in a given QM state they will stay that way until a QM transition happens. I have a hunch that this is well-studied by Astronomers in the context of stellar formation. $\endgroup$
    – uhoh
    Nov 25, 2020 at 0:30
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    $\begingroup$ If the Van der Waals force is strong enough, could two molecules or atoms orbit one another due to these forces alone, rather than gravity? I would be delighted if the answer to this question ends up being a set of non-intersecting range intervals. $\endgroup$
    – Connor Garcia
    Nov 25, 2020 at 1:06
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    $\begingroup$ @ConnorGarcia - Van der Waals forces are more complicated than gravitation, and in general they don't follow an inverse square law. You need that to have nice repeating elliptical orbits. I think the force increases more rapidly as you get closer, so they would tend snap together once they got within a certain capture range (the way many people mistakenly believe gravity works). $\endgroup$ Nov 25, 2020 at 17:49

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