Why are zeros after a decimal point significant?by curiousstudent Tags: decimal, point, significant, zeros 

#37
Sep1013, 02:41 PM

Thanks
P: 5,535

As regards gravitation, there is even a model where it is constant. It works fine where it is applicable. Is it "true" or not? There are other models. Newton's model; Einstein's model (and the postNewtonian spinoff); Yukawamodified gravity, just to name a few. 



#38
Sep1013, 03:02 PM

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P: 779

Getting back. You claimed that the 2 in r^{2}, was not exactly two (i.e. not an integer 2), and when this was pointed out you claimed that was axiomatic. Newton's statement is that it is an integer 2 and not 2 point something. It just happens to fail in certain situations, and (just as you say) we can use other models. Am I interpreting you correctly? If that is so, I fail to see how your second paragraph is not also axiomatic physics. We have what the Newton's model says (the 2 is exact) and then afterwards we have whether it it satisfies the observations. You can't compare it to data unless you understand the model first. Or maybe I have this "axiomatic vs nonaxiomatic" division completely wrong? 



#39
Sep1013, 03:25 PM

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P: 5,535





#40
Sep1013, 05:17 PM

P: 754

Sure, that seems fair. 



#41
Sep1013, 05:28 PM

Mentor
P: 14,481

Discard those assumptions and you don't get Newtonian gravity. You get general relativity, which doesn't look like Newtonian gravity. In the limit of small masses, small velocities, and large distances, general relativity does simplify to Newton's law of gravitation  and the 2 is exact. 



#42
Sep1113, 02:00 AM

Sci Advisor
HW Helper
P: 4,301

Sorry to interrupt this fascinating discussion, but is this still at all relevant to the original question? Or has the original topic starter left us long ago?



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