4
$\begingroup$

Does the gravitational attraction near the surface of dense celestial objects (neutron star, white dwarf itc) diverge (to infinity) from inverse square?

This question is inspired by the similarity between EM and gravity (inverse square force.). The paper by John Lekner here (doi:10.1098/rspa.2012.0133) shows that there is an electrostatic attraction between charged spheres no matter the polarities of the charges and that it diverges at close separation until electrical short for almost all charge ratios. I am wondering if there is a similar kind of gravitational inverse square divergence for anything other than a black hole.

Actually make this for a black hole as well, although I know that a black hole is not thought to have a normal surface.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Lekner's paper deals with spheres where charges redistribute themselves on the surface. The corresponding gravity thing would redistribute mass and deform the objects. That definitely complicates things compared to 1/r^2 $\endgroup$
    – Anders Sandberg
    Mar 15, 2020 at 0:12
  • $\begingroup$ I'm not expert enough to answer properly, but I think it does, due to general relativity, but to describe how it does, you need to first define your radius and time coordinates, and there is no single particularly privileged choice. $\endgroup$
    – Steve Linton
    Mar 15, 2020 at 1:22
  • $\begingroup$ Classical departure from inverse square happens when the "charges" (electrostatic or otherwise) are no longer spherically symmetric and Newton's Shell theorem no longer applies. Any time one or both gravitational bodies depart from spherical symmetric mass distribution the force between them can no longer be assumed to be inverse square. The bodies may naturally depart from spherical symmetry due to rotation, but they can also induce it. There will be a very tiny effect on the Moon's orbit due to the the motion of Earth's ocean tides it induces. $\endgroup$
    – uhoh
    Mar 15, 2020 at 1:25
  • 1
    $\begingroup$ @uhoh I clarified the question above with the two words (to infinity). $\endgroup$
    – DMac
    Mar 16, 2020 at 17:24

1 Answer 1

Reset to default
6
$\begingroup$

Leckner's paper deals with the effect of induced polarization on the spheres. Electrons are redistributed, making the force different from what one would expect. The gravitational counterpart is tidal distortion: since the gravitational field is non-radial when you have two heavy masses close to each other, matter will move to make the surface an equipotential surface. This means that the gravitational acceleration at the surfaces will not be constant at all locations.

Doing an analytic solution of how two ellipsoids attract each other and deform seems to be tractable (e.g. see this question) but algebraically very tedious and likely involves lots of special functions. See the addendum below for an approximate numeric model.

Black holes produce another complication: since spacetime nearby is curved and expanded the meaning of the distance in the inverse square law becomes problematic. The Paczyński–Wiita potential is an approximation of the potential, and it deviates from the $U=-GM/r$ as $U_{PW}=-GM/(r-R_S)$ (where $R_S$ is the Schwarzschild radius). It makes the force increase faster than the classical potential as we approach $r=R_S$. enter image description here

Addendum: I did a numeric exploration of the force between two ellipsoidal, self-gravitating masses with centres of mass separated by a given distance. To find the shape I started with spheres and adjusted the semi-major axis (while preserving volume) so the potential along the surface became more equal at the poles. After a few iterations this gives a self-consistent shape. Then I calculated the force (the derivative of the potential) due to this shape on the other mass.

Numerically estimated semi-major axis, force, and force times squared distance.

The result is indeed that the force increases faster than $1/r^2$ as the bodies approach each other, since they elongate and eventually merge (a bit before this they will deviate from my ellipsoidal assumption). If one multiplies the force by the squared distance the product should be constant for pure $1/r^2$ forces, but it starts to increase as they approach enough. Note that this is a non-rotating model: with rotation the numbers will change and the ellipsoids will become tri-axial, but I suspect the qualitative behaviour stays the same.

Improve this answer
$\endgroup$
2

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .