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To: Webster Groves who wrote (73842)	4/2/2001 9:05:37 AM
From: Zardoz	Read Replies (1) \| Respond to of 99985

F = F(up) + F(down) Why rockets go up? F(up) = v dm/dt (stuff shoots out the back) Why rockets do down? F(down) = Mg (g is gravity and says so). Try d/dt{mgh+ 1/2xmv^2)+d/ds(mgh+1/2xmv)C + F1 + d/dt{Mgh+ 1/2xMV^2)+ d/ds(Mgh+1/2xMV)C = 0 Actually your wrong, and Ahhaha was more right. You assume incorrectly that Gravity is a constant all the way to the end of outer space. Yet since it's variant due to range from the source, then your equation becomes useless. Anyways, Rocket science is based on a 4th order rieman field, and thus is not linear as ahhaha states. F:= massxGravitational flux+ mass flow rateXvelocity and when the velocity of the mass is of the order of ion being propelled at 95% the speed of light, it becomes VERY relativist. And then you must take into account the Herman Weyl conditions that ALL energy give off a gravitation field {and since EM fields are energy, as well as light} then you have a 4th order+ equations. Which is why Light bends around gravitational masses and why if you are to shoot only electrons from a rocket, you actually go nowhere.... Therefore Newtonian Mechanics is useless.... Now onto Ahhaha: You doubt gravitation is indistinguishable from acceleration? Consult the Equivalence Principle. That only a principle and is thus useless. And YES Gravity is VERY different to acceleration. Even in Einstein's mind experiment he stated that one could distinguish the difference by the Neutrino field passing through the base of the Elevator. If they are parallel, you're accelerating... and if they converge you're in a gravitational field. Hutch PS: I could go on for hours more and several more variables... like when a rocket becomes in orbit.

To: Webster Groves who wrote (73842)	4/2/2001 11:22:28 AM
From: ahhaha	Read Replies (1) \| Respond to of 99985

You are some kind of engineer. Don't swim in a physicist's soup. F=ma is currently a fact (a theory "proven" by experiment). How does one measure "F"? It isn't possible since F = ma is true by definition. "a" isn't measurable either. We get a handle on it by inference from a = dp/dt. We can measure p. Don't let ahahaha's technobabble influence you. He has the jargon but not the construction to say it correctly. Is the below your "construction"? I wonder what you're trying to show? Special and General relativity simply redefine the coordinate systems for certain conditions. Einstein would be very disappointed by this statement. It misses the crucial point of Relativity. The above is what we call a scalar, vector or tensor theory of gravity. They all fail under measurement. Einstein said in 1913 that he had to do exceedingly difficult calculations to go beyond the limitations of mere coordinate invariance. It forced him to the wild conclusion that space was curved. F = F(up) + F(down) Why rockets go up ? F(up) = v dm/dt (stuff shoots out the back) Why rockets do down ? F(down) = Mg (g is gravity and says so). This isn't germane to any comment that went before. Your "m" above is merely m = m1 + m2, where m1 is rocket mass before and m2 is rocket mass after. At an instant not one atom has left the rocket. dm/dt = 0. When the rocket has achieved orbit, F = 0. If not, the rocket would break orbit. So either v = 0 or dm/dt = 0. v isn't zero. F = 0, F(up) = 0, but F(down) isn't zero, because the rocket still has mass and the gravitational constant is still non zero. The rocket's velocity is compensating for earth's gravity by changing direction. None of this is germane. The previous comments dealt with whether the mass of the rocket in orbit, in compensated g, or zero g, changes as a function of velocity. You seem to have missed Galileo in your hasty physics. What we'd like to know is whether mg = ma. Is the Equivalence Principle true? Eotvos seemed to think so, but there are doubts.

To: Webster Groves who wrote (73842)	4/2/2001 4:16:45 PM
From: Math Junkie	Respond to of 99985

OT Re:"F=ma is currently a fact (a theory "proven" by experiment)." You can't prove a theory by experiment. At most you can confirm its predictions for the range of variables tested. There is no way to know with certainty whether the theory's predictions will continue to be true outside that range. As others have pointed out, F=ma is not even a fact in Newtonian mechanics unless the mass is constant. But is F=d(mv)/dt a fact? According to Richard Feynman's discussion of conservation of momentum in his Lectures on Physics, we can't even say that momentum is always equal to mv: "One of the propositions of Newton was that interactions at a distance are instantaneous. It turns out that such is not the case; in situations involving electrical forces, for instance, if an electrical charge is suddenly moved, the effects on another charge, at another place, do not appear instantaneously - there is a little delay. In those circumstances, even if the forces are equal the momentum will not check out; there will be a short time during which there will be trouble, because for a while the first charge will feel a certain reaction force, say, and will pick up some momentum, but the second charge has felt nothing and has not yet changed its momentum. It takes time for the influence to cross the intervening distance, which it does at 186,000 miles a second. In that tiny time the momentum of the particles is not conserved. Of course after the second charge has felt the effect of the first one and all is quieted down, the momentum question will check out all right, but during that small interval momentum is not conserved. We represent this by saying that during this interval there is another kind of momentum besides that of the particle, mv, and that is momentum in the electromagnetic field." Feynman goes on to say, "Now in quantum mechanics it turns out that momentum is a different thing - it is no longer mv. It is hard to define exactly what is meant by the velocity of a particle, but momentum sill exists. In quantum mechanics the difference is that when the particles are represented as particles, the momentum is still mv, but when the particles are represented as waves, the momentum is measured by the number of waves per centimeter: the greater the number of waves, the greater the momentum. In spite of the differences, the law of conservation of momentum holds also in quantum mechanics. Even though the law f=ma is false, and all the derivations of Newton were wrong for the conservation of momentum, in quantum mechanics, nevertheless, in the end, that particular law maintains itself!" [emphasis added] So it sounds like F=dp/dt, where p is momentum, may still be current theory, but there is no guarantee that even that will always be found to be true, or a useful model, in future experiments.