Originally Posted By: geistkiesel
..are you convinced, not by my incessant whining, but by the math and the physics?
Of course. Everyone, who knows at least a bit about integral calculus would see immediately, Newton's shell theorem is valid only for infinitely thin shell.
after all, here we can find explicitly:
Therefore, a thin shell can be treated as a point mass, provided the second object is outside the shell
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What did you mean by "after all, here we can find explicitly:"?
It is not the fact of an infinitely thin shell , or even volume as is used here in the integral. The fact that the shell has extension means that the ½ shell closest to m contribute a greater share of the total force on m than the ½ shell segment farther from m.
That the theorem permit’s the "concentration of mass at a point" is of no problem, but where do you put the center of force of the shell? 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.
Take three equal masses located on a common axis. Say that two masses, m1 and m2 are a distance of 10 units apart and m3 13 units from m1 [3 units from m2 away from m1]. The test mass m1 is looking down the common axis and can determine the direction of the force, but where is the center of the combined m2 and m3 forces located? The answer is not at the center of mass of the m2 - m3 system, that's for sure.
This location must be determined from the law of gravity, or F = GmM/r^2.
If you can indicate, please explain how the expression for force is proof that the shell behaves as if the mass of the shell was concentrated at the COM of the shell. Wouldn't you think the imbalance of force as seen by m would off set the center of the force in the direction of the strongest applied force?
