Proof that gravity constant G = (3/4)(H^2)/[(pi)(density)e^3] Proof that gravity constant G = (3/4)(H^2)/[(pi)(density)e^3] Proof of 1.7% accuracy: http://www.wbabin.net/physics/cook2.htm Background info: http://members.lycos.co.uk/nigelbryancook/ Summary: Nugent, Physical Review Letters (v75 p394), cites decay of nickel-63 from supernovae, obtaining H = 50 km/sec/Mps (where 1 Mps = 3.086x10^22 m). The density of visible matter at our local time has long been known to be 4x10^-28 kg/m3. However, White and Fabian in the March 1995 Monthly Notices of the Royal Astronomical Society, using the Einstein Observatory satellite data, estimate that invisible gas increases this density by 15%. Using these data, G = 3(H^2)/[4pi(e^3)ρ] = 6.783x10^-11 Nm^2kg^-2, within 1.65% of the physical measurement for G of 6.673x10^-11 Nm^2kg^-2. Background: ‘Electronic Universe’ article (Electronics World, Vol. 109, No. 1804) proves G = 3(H^2)/(4 pi ρ). [Ref: http://cdsweb.cern.ch/search.py?f=author&p=Cook,+N and http://www.wbabin.net/physics/cook1.htm. H is the Hubble constant and ρ is the density of universe responsible for causing gravity by the inward reaction of 377-ohm physical space to the outward big bang; pi is the mathematical constant. Considering the density, it is highest at early times and thus density increases in the observable space-time trajectory, as we look further into the past with increasing distance. But the increasing spread of matter with increasing distance partly offsets this increase, as proven when we put the observed Hubble equation (v = Hr) into the mass continuity equation and solve it. For spherical symmetry, dx = dy = dz = dr. Hence: dρ/dt = -div.(ρv) = -div.(ρHr) = 3d(ρHr)/dr = -3ρH. Solving dρ/dt = -3ρH by rearranging, integrating, then using exponentials to get rid of the natural logarithms (resulting from the integration) gives the increased density to be ρe^(3Ht), where e is Euler’s constant (2.718 ...). In the absence of gravitational retardation (i.e. with the cause of gravity as inward reaction of space to the outward big bang), H = 1/t when H = v/r = c/(radius of universe) = 1/t, where t is the age of the universe, so e^(3Ht) = e^3 = 20.1 and observed G = 3(H^2)/[4pi(e^3)ρ].