Dubbelosix Posted December 23, 2019 Report Posted December 23, 2019 I'm going to need to break this up into a number of posts as the theory I presented now has evolved a bit. Also previous derivation of the drag in three dimensions and four was studied. It was unknown to me that gravitational acceleration as a drag coefficient has actually been explored. The model I found from Nasa was, F = m (dv/dt) = mg - ½ρv² f A f = 2F(drag) /ρv²A = A(b)/A(f) Be/Re² ρv² - a kinematic density related to the gravitational fluidity f - the drag coefficient mg - the acceleration force due to gravity At equilibrium the force is equal to zero - to extend this primary model into one that satisfies my approach will not differ much but I'd write it with an abuse of notation since there is no weight analog on GR, but we invent one fo thr sake of extension: Under the style of general relativity we get T_μν = ½ρ (u_μ u_ν) F = du/dt + mΓ^μν u_μ u_ν - A^μv f (T_μν - ½ g_μν T) If Γ has units of acceleration it becomes F = du/dt + g_μν mΓ^μν - A^μv f (T_μν - ½ g_μν T) The first equation uses the gravitational four force definition, F = mΓ^μν u_μ_ν Quote
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