A second galactic visitor has just been observed: see this link.

For a mass which is small enough to have no noticable effect on other masses, is there any other way to infer the mass? Its own motion is a function of other masses. Obviously a sufficiently resolved photograph giving the volume, and a wild guess as to the density would work, but that would be a WAG, and in any case we don't have a real picture of this thing really.

If we could get a probe to it, we could put something in orbit around it. Is there any other way?

  • $\begingroup$ As far as I'm aware using gravity is the only tool we have, and that would require sending a probe out to it to orbit the asteroid or at least have its trajectory measurably altered by the gravitational pull of the asteroid. $\endgroup$
    – user93237
    Sep 14, 2019 at 23:19

1 Answer 1


If it is small enough and observations of its trajectory are accurate enough then one can use the fact that the relative sizes of radiation pressure (from the Sun) and gravity depend on the density and geometry.

e.g., Suppose the thing was a black sphere (NB adding an albedo just changes the arithmetic rather than the principle) of density $\rho$ and radius $a$ and was at a distance $r$ from the Sun.

The gravitational acceleration would be $${\bf g} = -\frac{GM_\odot}{r^2}e_r\ ,$$ and is independent of the mass of the object.

The acceleration due to radiation pressure would be $$ {\bf g}_{\rm rad} = \left(\frac{L_\odot}{4\pi r^2}\right)\left(\frac{\pi a^2}{4\pi a^3\rho/3}\right)e_r = \left(\frac{3L_\odot}{16\pi a\rho r^2}\right)e_r\ .$$ Which depends on some combination of the size and density of the object. In this case, we are not interested in the density and we can assume the size can be measured in some way, so $\rho = 3m/4\pi a^3$ and the gravitational acceleration due to radiation pressure is $$ {\bf g}_{\rm rad} = \left(\frac{L_\odot a^2}{4 m r^2}\right) e_r\ .$$

Thus the trajectory will be determined by an acceleration that is slightly smaller than the gravity due to the Sun by an amount that depends on the ratio the area it presents to the Sun divided by its mass.

This could well be a tiny perturbation - though obviously gets bigger the more like a "solar sail" the object is (i.e. a large $a^2/m$) - but of course the effects are integrated over the course of its passage through the Solar System.

It was exactly this type of argument that lead Bialy & Loeb (2018) to suggest that "Oumuamua" (the first example of an identified interstellar rock) had a non-gravitational acceleration that could be explained if it had a high area to mass ratio, such that it might be less than a cm thick and weigh only of order 1000 kg.

The problem with using this technique is that you have to eliminate or separate it from other sources of non-gravitational acceleration - like outgassing from cometary material.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.