The physics says it cannot be at the shell center of mass (COM) as the closest ½ shell segment contributes a greater share of the total force on m than the shell segment farther from m.
This is exactly why, we are using an average value at center. You're just reinventing light bulb.
Learn the integral calculus and the way, in which it works..
I know the calclus. The shell integral, see below, calculates frces from mdifferential mass segments on the shell. The fina result, F = GmM/d^2 says nothng about the shell behaving as if all the mass in the shell was concentated at the shell center located a d. Th statemet says, "the force on m from the total of a thin shell located at d." The shell theorem is not an "average" of forces or masses. You are in stark error when you say "we ar using an average
value at center". Average value of what? The average vlue of the mass on the shell? Your cmment maks no sense at all.
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This is a proof that, one may
not consider that aa shell behaves as if all the mass on the shell were concentrated at the shell center", and this is, of course, is contrary to Newton's Shell Theorem attesting to the converse (see links).
http://en.wikipedia.org/wiki/Shell_theoremhttp://www.physclips.unsw.edu.au/jw/NewtonShell.pdfIf contrary arguments of those responding here do not specifically go to an error in this note, the argument will be ignored and monitors who feel this one sided statement may act accordingly. I don't mind criticism or disagreement; I just want what I publish to be criticized, what I said,only.
This proof uses the system of rings containing differential masses on the shell surface, dM where the rings are centered on the r-axis linking the test mass m with the center of the shell located a distance d from m = 1. Adjust this 'standard' system to include two mirror image rings R1, located closest to m, and R2 located farthest from m. The mass on each ring is M1 = M2 = 1 (for the purposes of this demonstration).
Project the total force from each ring, F1 and F2, onto the r-axis, as the normal shell integral provides. It should be clear that m "sees" the two forces, F1 and F2, along the r-axis concentrated at both rings' centers, each located the same distance from the COM of the two rings (and parenthetically equal distance from the COM of the shell). Without losing generality, place one ring at 9 distance units, the other at 11 from m. To repeat, because the rings are mirror images each ring is the same distance from the shell center plane.
The calculations of the two forces are, F1 = 1/81, F2 = 1/121 and using F = GmM/r^2, where G =1, m = 1, M1 = M2 = 1. The forces F1 + F2 = 1/81 + 1/ 121 = 202/9801 = .020610 = F12.
This is the total force from both rings and if some symmetry is assumed let 10 distance units be the presumed COM of the two calculated forces.
Using F = GmM/r^2 again, but solving for M = F12(r^2)/Gm. Then M = .020610(100) = 2.0610 a mass 6% larger than actually available. Again, from observation, it is clear that the closest equal mass to m contributes the greater share of the total force on m and therefore [i]the COM of the two rings is not the center of force of the combined forces F1 and F2..
Using the values of M1 and M2 used in determining the forces F1 and F2, and
calculating r from the combined forces, F12 = F1 + F2, r^2 = 2/.02081 = 97.040, or r = 9.851 a distance off set from the COM of M1 and M2 in the direction of m. The number 2 is used in place of M in the general statement GmM/r^2, or "M1 = 1" + "M2 = 1".
Any mirror image rings will always provide a center of force as discussed here being located in the shell half nearest the test mass. The expression F = GmM/d^2 states only the force of mass M on m
from a spherical shell located a distance d from m.Symmetry arguments do not distribute mass to be other than the
actual distribution with the resultant generated forces.
