Re: Saturday Morning Puzzle, Part 2

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Sun, 7 Oct 2001, brian whatcott wrote:

> At 09:01 10/6/01 -0700, John Mallinckrodt wrote:
>
> > /snip/ I'd be interested in knowing how a distribution that places
> > most of the mass in one object at arbitrarily large distances (not
> > to mention in directions that cover a solid angle of nearly 2*pi)
> > from any given portion of the other object could end up giving a
> > larger net force than a distribution that keeps all of the mass
> > within a small distance *and* a smaller range of solid angles.
>
> John provided this most interesting argument for the optimality of
> two spheres each of constant mass m whose proximal surfaces are
> separated by a modest distance x, as compared with two disks each
> of mass m and separated on their proximal flat faces by a
> distance x in respect to maximizing gravitational
> attractive force.

I never claimed it as an argument for optimality.  It is nothing
more than a devastatingly clear argument (IMNSHO) for the *vast*
superiority of the spherical configuration over the laminar
alternative.

> I wonder if he would help me evaluate the following configuration:
>
> two hemispheres each of mass m separated on their flat faces
> by a distance x.

I'll think about that; maybe someone else will beat me to the
punch.  In the meantime, perhaps it would help if I simply
presented the details of the two sphere vs. two lamina comparison:

=============

First consider two spheres, both having volume V, both having mass
M, with centers as close as possible, (i.e. 2R where R is
determined from V.)

Force of attraction from Newton's law of gravity:
   F_spheres = G M^2/(4 R^2)

Volume:
   V = (4/3) Pi R^3

Eliminating R from the expression for F:
   F_spheres = (Pi/6)^2/3 [G M^2/V^(2/3)]

=============

Next consider two thin lamina, ("thin" means having a thickness
d << V^1/3), both having volume V, both having mass M, separated
by a distance similar to their thickness.

Gravitational field produced by either disk in vicinity of the
other from Gauss' Law:
    g = 2 Pi G M/area

Force of attraction
    F_disks = M g = 2 Pi G M^2/area

Volume:
   V = area * d

Eliminating area from the expression for F:
   F_disks = 2 Pi d/V^(1/3) [G M^2/V^(2/3)]

=============

Finally, find the ratio of the two forces:

   F_disks/F_spheres = (288 Pi)^(1/3) d/V^(1/3) = 9.67 d/V^(1/3)

This number becomes arbitrarily small as d -> 0.

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm