I am deleting all the other stuff because I believe the crux of the problem is right here. If we can settle this, then we can move on to the other stuff, if you still feel like it.
Quote:
On 2002-03-24 10:40, Gary Redmond wrote:
WEIGHT = the magnitude of the gravitational force exerted on a body by the nearest astronomical body.
I believe the weight of the body mentioned means the sum of the weights of all the particles of the body. Here is the rub. Because all those weights are directed/accelerated toward the center of gravity of the other body they create internal torques, but here again as in any mass (system of particles) the sum of all the internal forces equal zero. Therefore any external gravitational mass cannot cause a torque within another mass.
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First of all, a torque is not a force. It's a *torque*. R *cross* F. It has units of Newton-meters, not Newtons (or kg-(m/s)^2 rather than kg-m/s^2, if you prefer). Therefore, your second premise is inapplicable, and your syllogism fails.
If you still don't see this, please actually *try* to do the integral and see that it's non-zero. I don't see any point in arguing this when you haven't even done the calculation. To make it even simpler, forget about the integral. Consider a long rod (length = r) with a masses on each end of mass m. (The mass of the rod is negligible. There is a mass a distance R away (r << R), and the axis of the rod is tilted with an angle theta with respect to the line between the center of the rod and the distant mass.
By your argument, this system should be static. You say that "internal torques" will sum to zero. If there is no torque on the system, the rod will not move. Ever hear of the Cavendish experiment? Well, never mind the fact that the rod will move, and this movement has been measured (you can use it to measure the value of G - heck, I did that experiment myself in college), we can actually do the calculation. Since it's harder to write the notation in this format than it is to actually do the calculation, I will leave it to you, but you will find that the torque is non-zero. Simply write the two F vectors and the two r vectors, take the cross products and sum them. You *will* get a non-zero answer.
If you don't see how your heuristic argument is flawed, please do the math and see what the answer is. Once you see that there is a net non-zero torque, come back and we'll talk about the rest of it.
Don