Is it possible for a molecule or atom to orbit a star (e.g. the Sun)? Or is there always too much outward force imparted by solar radiation compared to the inward force of gravitational attraction?

  • $\begingroup$ Just some napkin-style not-very-accurate calculations here, but ignoring any quantum effects, it would seem that the away-from-Sol acceleration due to radiation pressure for a helium-4 atom at 1 AU is about 10 meters per second squared, versus the toward-Sol acceleration due to gravity is about 0.01 meters per second squared (or about 1000 times less). Likewise, at one third of the distance to Proxima Centauri from Sol, the acceleration from radiation pressure seems to be about 10^-8 m/s^2 versus the acceleration due to gravity about 10^-12 m/s^2, or again about 1000 times less. $\endgroup$ Commented Apr 4, 2018 at 21:33
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    $\begingroup$ It's not necessary to repeat the estimate at different distances, because both radiation pressure and gravitational forces go like $1/r^2$. $\endgroup$
    – user4552
    Commented Apr 4, 2018 at 21:34
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    $\begingroup$ You cannot use a classical formula for macroscopic radiation pressure to treat the interaction of atoms with photons. Possibly you could use the Rayleigh scattering cross-section for atoms and Thomson scattering cross-section if ionised. $\endgroup$
    – ProfRob
    Commented Apr 4, 2018 at 23:49
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    $\begingroup$ @Nicole It would be really great if you'd edit your own thoughts on the matter (e.g. your first comment) into the question. $\endgroup$
    – David Z
    Commented Apr 5, 2018 at 5:53
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    $\begingroup$ Pedantic answer : We're full of molecules orbiting the sun. $\endgroup$ Commented Apr 5, 2018 at 8:14

2 Answers 2


Cute idea! Thanks for posting this question. I've enjoyed thinking about it.

Geometrical absorption

Suppose we start by assuming that we're talking about a particle that just absorbs all light that impinges on its cross-section. The sun's gravitational force on a particle is proportional to its mass, and therefore to the cube $a^3$ of its linear dimension $a$. Radiation pressure is proportional to the cross-sectional area, and therefore to $a^2$. Since the exponents are different, it follows that for small enough objects, the net force will be repulsive, and there can be no closed orbits.

For objects that are just a little above that size cut-off, we could have Keplerian orbits, but they would not obey Kepler's law of periods with the same constant of proportionality as for objects such as planets that are large enough to make radiation pressure negligible.

Without having to do a numerical estimate, we can tell that atoms are below the cut-off size for closed orbits, since solar sails exist, and a solar sail is considerably thicker than one monolayer of atoms.

All of this holds regardless of the distance $r$ from the sun, because both radiation pressure and gravitational forces go like $1/r^2$. This is also why the orbits are still Keplerian: the interaction with the sun acts like gravity, but just with a different gravitational constant.

A stable, electrically neutral particle such as a neutrino or a dark matter particle can orbit, because it doesn't interact with electromagnetic radiation. In fact, I think dark matter is basically known to exist only because it's gravitationally bound to bodies such as galaxies.

Wave model

But as pointed out by Rob Jeffries in a comment, this is not right at all for objects that are small compared to the wavelength of the light. In the limit $a \ll \lambda$, we have Rayleigh scattering, with a cross-section $\sigma \sim a^6/\lambda^4$. Let


be the ratio of the radiation force to the gravitational force. If we don't worry about factors or order unity, then it doesn't matter if we're talking about absorption, reflection, or scattering. Pretend it's absorption, and let $a$ be the radius of a spherical particle. We then have

$$R=\frac{3}{16\pi^2 Gc}\cdot\frac{L}{M}\cdot\frac{1}{\rho a^3}\cdot\sigma,$$

where $\rho$ is the density of the particle, $L$ is the luminosity of the sun, and $M$ is the mass of the sun.

For a particle with $a\sim 300\ \text{nm}$, the geometrical absorption approximation $\sigma\sim \pi a^2$ is pretty good, and the result is that $R$ is of order unity.

For a particle with $a\sim 50\ \text{nm}$, the Rayleigh scattering approximation is valid, and we have $\sigma\sim a^6/\lambda^4$. The result is $R\sim 10^{-4}$.

So it seems that the result is somewhat inconclusive. For a star with the $L/M$ of our sun, there is a pretty broad range of sizes for particles, with $a\sim\lambda$, such that there is fairly even competition between radiation pressure and gravity.


Leftroundabout's answer pointed out the importance of ionization, and he estimated that effect for high-energy electrons. Actually I think UV is more important. For a 25 eV photon, which is at the threshold for ionization of helium, the cross-section is about $7\times 10^{-18}\ \text{cm}^2$. Suppose that $\sim10^{-2}$ of the sun's radiation is above this energy. For an atom at a distance of 1 AU from the sun, the result is that ionization occurs at a rate of $\sim10^{-3}\ \text{s}^{-1}$.

This suggests that there is no way an atom is going to complete a full orbit around the sun without being ionized. If we assume that our atoms(/ions) are all independent of one another, then an atom will basically spiral around the sun's magnetic field lines. One thing I don't know from this analysis is whether it's really valid to assume the atoms are independent. We could also imagine that there are parcels of gas orbiting the sun, and these are electrically neutral in bulk.


This analysis seems inconclusive for particles of baryonic matter with sizes less than about 300 nm. It seems like we need more work to understand this -- or someone could find where the subject has been treated in more detail in the astrophysics literature.

For stars off the main sequence, I think we can make some definite conclusions. Giant and supergiant stars, which have $L/M$ much higher than that of the sun, will efficiently sweep out all particles from $a\sim\lambda$ up to some upper size limit. For white dwarfs and such, with very small $L/M$, radiation pressure will never be significant.

Here is a paper (Mann et al., "Dust in the interplanetary medium," Plasma Phys. Control. Fusion 52 (2010) 124012) on dust in the interplanetary medium. It describes things like the trajectories of charged dust particles.

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    $\begingroup$ I don't understand the solar sail comparison. $\endgroup$
    – BowlOfRed
    Commented Apr 4, 2018 at 21:48
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    $\begingroup$ @BowlOfRed: The transverse dimensions of the object are irrelevant, because they have equal effects on the two forces. Only the radial dimensions (i.e., in the direction away from the sun) are relevant. The radial dimension of a solar sail is its thickness. $\endgroup$
    – user4552
    Commented Apr 4, 2018 at 21:49
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    $\begingroup$ Best result I've found so far is an osmium atom, which might have only about 100 times less gravitational acceleration than outward acceleration from radiation. Osmium is more dense than buckminsterfullerene. It looks like the cutoff point for stable orbits though might be when the area per unit mass is greater than 1306 square meters per kilogram (0.8 square picometers per atomic mass unit). That should apply to solar sails too I think (if they are too large they have too much thrust to remain in orbit). The cross-section of an osmium atom is 180,000 m^2/kg. $\endgroup$ Commented Apr 4, 2018 at 22:51
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    $\begingroup$ But might be approaching the problem wrong. If we ignore nuclear radiation, we have primarily just photon-electron interactions. The photon flux density represents a certain probability that an electron will have an interaction. Presumably, the electron cloud absorbs a photon, excites to a higher-energy state, and then reemits the photon, which causes the atom to accelerate. $\endgroup$ Commented Apr 4, 2018 at 23:04
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    $\begingroup$ I think the argument in the first paragraph only applies to macroscopic particles. Individual atoms and ions can be essentially transparent to (most wavelengths of) light. $\endgroup$
    – ProfRob
    Commented Apr 4, 2018 at 23:42

No, and radiation pressure is not the only reason.

The interplanetary space is not empty; apart from the optical photons it is in particular also flooded with the charged particles of the solar wind. This includes in particular a significant popolation of electrons with energies in (amongst others) the range of $100\:\mathrm{eV}$, where they can quite efficiently ionize helium atoms (with cross-section $\sigma\approx3\cdot10^{-16}\:\mathrm{cm}^2$). That energy range corresponds to a velocity-cube of ca. $v=6000\:\mathrm{\frac{km}{s}}$, i.e. $v^3 = 2\times10^{20}\:\mathrm{\frac{m^3}{s^3}}$.

The velocity density of such electrons at 1 AU is around $10^{-27}\:\mathrm{\frac{s^3}{cm^6}} = 10^{-15}\:\mathrm{\frac{s^3}{m^6}}$, i.e. a density of $\rho \approx2\times10^{5}\:\mathrm{m^{-3}}$ electrons with relevant energy. Moving at $v$, those electrons impinge on the atom's cross-section at a rate of $$ \nu = v\cdot\rho\cdot\sigma \approx 40\cdot 10^{6+5-20}\:\mathrm{s^{-1}} = 4\cdot10^{-8}\:\mathrm{s^{-1}} \approx \frac{1}{0.8\:\mathrm a}. $$ So, the atom is likely to get ionized before surrounding the sun once, and once it's ionized, its path is dominated by electrodynamic rather than gravitational forces. It certainly wouldn't keep a stable orbit.

I actually expected the collision frequency to be significantly higher than those $4\cdot10^{-8}\:\mathrm{s^{-1}}$; quite possibly I made an error in the calculation. If the rate is correct, then UV is actually the dominant cause of ionisation as Ben Crowell points out. It's plausible enough, seeing as the cross-section for those photons is actually not much lower than the cross-section for electron.

  • $\begingroup$ I was wondering about that. Even using the term "atom or molecule" loosely to include ionized versions would mean that they should be swept away by the magnetic field too (I forgot about that). Also, I even read about one theory of quantum gravity that claims that any objects below a Planck mass are incapable of absorbing or emitting gravitons, which would mean that most atoms could be immune to gravity, even in the absence of any other forces. $\endgroup$ Commented Apr 4, 2018 at 23:31
  • $\begingroup$ Nice. But wouldn't ionization by UV be a faster process than ionization by high-energy electrons? I'm also not sure whether the particles really end up spiraling around magnetic field lines. Certainly some do, because the sun has radiation belts. But it seems possible to me that we essentially get parcels of gas that are electrically neutral. If the frequency of ionization were really as low as $10^{-7}\ \text{s}^{-1}$, then I would think that the atoms would be swept out of the solar system by radiation pressure before they could be ionized. (Nicole estimates 10 m/s2 for the acceleration.) $\endgroup$
    – user4552
    Commented Apr 4, 2018 at 23:59
  • $\begingroup$ Some really quick and haphazard calculations seem to indicate that the magnetic force on charged particles in the Solar System might be a lot stronger than any gravitational force or forces from radiation pressure. And from LeftAroundAbout's answer it does seem like it would be difficult to remain electrically neutral in the Solar wind. Neutrinos move too fast and neutrons decay too quickly for astronomical orbits. $\endgroup$ Commented Apr 5, 2018 at 0:00
  • $\begingroup$ @Ben Crowell, the estimate of 10 m/s^2 (actually rounded down to the nearest factor of 10, from 42.1 m/s^2) is by taking the cross-section of the helium van der Waals radius to approximate surface area. Of course, a helium atom is actually a three-dimensional electron cloud and is not a solid disc facing the sun. Any actual acceleration due to photon flux density would need a quantum-mechanical treatment. $\endgroup$ Commented Apr 5, 2018 at 0:11
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    $\begingroup$ @leftroundabout: I've estimated ionization by UV, and it's much faster than what you estimated for electrons. I'll edit my answer. If you like, we could cooperate on editing your answer to make a best over-all answer. $\endgroup$
    – user4552
    Commented Apr 5, 2018 at 0:44

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