# The shapes of gravigationally bound solid bodies

On this page I examine the shapes of bodies held together by Nrwtonian gravitation, along with the gravitational fields they produce, under various hypotheses about the structure and state of those bodies. I'll assume the body, in each case, is fluid enough to redistribute itself into a form that minimises its total gravitational + kinetic energy; you can think of this as either a non-rigid solid with just enough flexibility to redistribute itself, or as a hyper-viscous liquid, that will move in response to pressure differences, without setting up established flows. I'll also be assuming stasis: the body won't be dynamically rearranging itself. The study of real celestial bodies, with atmospheres, oceans and even the solid part flowing slowly over time, all subject to tidal influences of other celestial bodies, variable heating from at least one of those and any (other) electromagnetic influence can wait for another day and another page, if I ever get to it.

First, however, let us consider an imaginary body, whose effects we'll be able to exploit when considering the non-spinning solid bodies.

## Inside a spherically symmetrical shell

Suppose I have a spherical shell of uniform density ρ, inner radius Ri, outer radius Ro and thus mass (4.π.ρ/3).(power(3, Ro) −power(3, Ri)) = (4.π.ρ/3).(Ro −Ri).(Ro.Ro +Ro.Ri +Ri.Ri). The gravitational field inside the shell can be determined be considering the accumulated forces, on a test particle at any position inside the shell, from the various parts of the spherical shell.

Let the test particle be at distance h from the centre of the spherical shell; take the line through the test particle and the centre of the sphere as z-axis. Since the situation is symmetrical about this axis, I'll use polar co-ordinates in the planes perpendicular to this axis, with radius r; we have no need to name the angular parameter, due to the symmetry, which also implies that the force on the test particle must be, in one direction or the other, along the line between it and the centre of the sphere. The shell comprises the region Ro.Ro ≥ r.r +z.z ≥ Ri.Ri.

The distance from any [r, z] point on the plane to the [0, h] test particle is hypot([r, z −h]) where hypot is the pythagorean function reverse(power(2))∘(: sum(: power(2, u(i)) ←i :) ←u :{lists}), so power(2, hypot(r, z −h)) = r.r +(z −h).(z −h) = r.r +z.z +h.h −2.z.h. The z-component of a force directed along the line from our test particle to such a point is (z −h)/hypot(r, z −h) times the magnitude of the force. Our element of volume, in terms of z and r, is then 2.π.r.dr^dz, describing a ring around the z-axis, exerting a force on the test particle whose z-component is 2.π.G.ρ.dr^dz.r.(z −h) / power(3, hypot(r, z −h)), so the net force on our test particle is

2.π.G.ρ.integral(: dr^dz.r.(z −h) / power(3, hypot(r, z −h)) ←[r, z]; Ro ≥ hppot(r, z) ≥ Ri :{[r, z]: r, z real, r ≥ 0})
= 2.π.G.ρ.integral(: dv^du.u.u.Sin(v).(u.Cos(v) −h) / power(3, hypot(u.Sin(v), u.Cos(v) −h) ←[u, v]; Ro ≥ u ≥ Ri :{[u, v]: real u ≥ 0, angle v})/radian

by a change of co-ordinates r = u.Sin(v), z = u.Cos(v), so u = hypot(r, z), dr = du.Sin(v) +u.Cos(v).dv/radian, dz = du.Cos(v) −u.Sin(v).dv/radian, so dr^dz = u.dv^du/radian; we can now expand out hypot:

= 2.π.G.ρ.integral(: dv^du.u.u.Sin(v).(u.Cos(v) −h) / power(3/2, u.u −2.u.h.Cos(v) +h.h) ←[u, v]; Ro ≥ u ≥ Ri :{[u, v]: real u ≥ 0, angle v})/radian

in which we can now replace v by the substitution c = Cos(v), dc = −Sin(v).dv/radian

= 2.π.G.ρ.integral(: du^dc.u.u.(u.c −h) / power(3/2, u.u +h.h −2.u.c.h) ←[u, v]; Ro ≥ u ≥ Ri :{[u, v]: real u, c, u ≥ 0, −1 ≤ c ≤ 1})

directed along the x-acis. The integration with respect to c now has form

integral(: (u.c −h).dc / power(3/2, p −q.c) ←c; −1 ≤ c ≤ 1 :)

There must be a neat geometrical argument to obtain that result ! A relevant diagram, using the intersecting chords theorem, has a diameter through the test point, split as r +h on one side and r −h on the other, crossed by its perpendicular through the test point, whose two parts each have length √(r.r −h.h), establishing the equality the final ln-term's collaps results from.

## Uniform density, no spin

The simplest case is a homogenous body that isn't spinning. It strikes me as obvious that this will be spherically symmetric: any mountain peak can releas energy by tumbling down into the valleys around it. So the shape of the body is easy: it'll be a sphere. Its uniform density and its volume thus determine its total mass; hence the gravitational field outside it.

It remains to consider how the gravitational field varies within the sphere of the object's body. Let its density be ρ and surface radius R; its volume is then 4.π.power(3, R)/3 and its mass is M = 4.π.ρ.power(3, R)/3. The mass within radius r of the centre, for r < R, is likewise m(r) = 4.π.ρ.power(3, r)/3.

Written by Eddy.