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To: MtnClimber

m=f/a=E/c²

I think....


16 posted on 04/06/2016 8:34:45 PM PDT by onedoug
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To: onedoug

Your equation has gravitas.


20 posted on 04/06/2016 8:49:34 PM PDT by Rockpile (GOP legislators-----caviar eating surrender monkeys.)
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To: onedoug
m=f/a=E/c²

Very good. But only true in the Newtonian approximation. I always think that we should say E = gamma m c², which gives E = mc² + 1/2 m v², in the second order approximation of gamma ( which is 1 to first order in v )

It's all fun.

21 posted on 04/06/2016 8:56:21 PM PDT by dr_lew
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To: onedoug
m=f/a=E/c²
I think....

True, that solving for m in both Newton's F=ma (force = mass times acceleration) and Einstein's E=mc^2 (energy = mass times the speed of light squared) (f/a = E/c^2) can be set equal to each other. However, I don't believe neither F=ma or E=mc^2 is used in GPS calculations. The Einstein component applied is Relativity due to the high rate of speed of the satellite and also the stronger gravitational effects at the surface vs high altitude (see above)..

24 posted on 04/07/2016 1:36:40 AM PDT by ETL (You can lead a Trump supporter to critical facts & info, but you can't make him/her think)
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