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Re: Saturday Morning Puzzle, Part 2



At 13:09 10/7/01 -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



I found this straightforward development helpful.
It leads to the second configurational question (the force on
two half spheres each of mass m being the first):

The force ratio of two laminae versus two spheres of equal mass becomes
progressively small as the distance between their adjacent surfaces
approaches 0.
In the other direction, what is the distance between proximal surfaces
at which the force between thin laminae exceeds that between two spheres
(if ever?)

...This just gets more and more interesting.




brian whatcott <inet@intellisys.net> Altus OK
Eureka!