
Re: gravity begins with the proton
Posted:
May 24, 2014 1:31 PM


On Saturday, 17 May 2014 18:39:09 UTC+1, haroldj...@gmail.com wrote: > We're under a cloud when it comes to gravity, the Gravitational Constant and consequently an exact given value for the Planck units. We're forever being told > > how the next series of measurements will confirm these phenomena once and for all. > > > > We do have templates to give us some idea but only templates. The most useful are the Gc templates. Many of the values in physics are factorised by Gc or (Gc/2)^0.5 > > without us realising it. > > > > By using 2/c as substitute for G we get the following: > > > > (1). In the formula for the Planck mass, (ch/4G)^0.5, when substituting 2/c for G > > we arrive at 2.72837394x10^9, quite recognisable as the Planck mass except it is out by a factor of (GC/2)^0.5. > > > > (2). The same principle can be applied to the Planck radius, formula (Gh/c^3)^0.5. > > This time we end up with 4.050436x10^34. > > > > (3). The ratio of Strong force over gravity, 2.660725x10^38, based on > > (Mpl/Mpr), where Mpl is Planck mass and Mpr is proton mass. Appying the above criterion the Gm template becomes 2.66079655x10^36. > > > > (4). Mpr/h is equal to (1.009721668x10^7)/4. The quantum adjustor, qa, is equal to 4/{4h(c/2)^4}^0.333r=3.62994678, the quantum adjustor. 4/qa is equal to 1.10194453. 1.10194453xGc. If the radius of the proton was equal to the Planck Radius then the total energy given to a falling Planck sphere would be > > 1.10194453xGm/4 x Mpl=1.503278583x10^10 J. The total energy of the proton. > > (1.503278583x10^10)(1.009721668x10^7)=1.51789295x10^3. > > 2/1.51789295x10^3=1.317615975x10^3. The square root of this is 3.629898x10. > > Accepting there might be a small drift from the theoretical qa it can still be assumed that Gm/2 is around 1/100. > > > > (5). The analogue to 1.009721668x10^7 is 1.11585396x10^35. This value multiplied by qa is the Planck radius. It is found by the formula 2Mpr/c > > and is removed from GMpr by the familiar differential Gc.
FINDING METRES FROM PLANCK TEMPLATES:
(6). By using the Gc template for Planck mass, 2.72837394x10^9, see item (1) and dividing by the proton mass then squaring the result, we get 2.66079655x10^36, see item (3).
(7). If we multiply the proton mass by 1.00972168x10^7, see item (4), we get 1.688883686x10^20. The reciprocal squared equals (5.92107087x10^19)^2 = 3.50590803x10^39, which equals (Mpl/Mpr)^2 multiplied by (qa)^2, or, 2.660725x10^38x(3.62994678)^2, see item (3).
(8). If we multiply 3.505908x10^39 by 1.11585396x10^35, see item (5), we get 3.912081282x10^4 metres.
3.912081282x10^4 metres will be the Schwarzschild radius for a body of mass equal to 2.660725x10^38 Planck masses and then multiplied by (qa),appx. 3.62994678.
(9). There is a number, around 8.179349, that when divided by c gives us the Planck mass. When divided into the Planck constant gives us the Planck length, when multiplied by c gives us the Planck energy and when multiplied by (c^2/h) gives us the surface gravity of the Planck sphere. The (Gc/2) version is 0.817945942. We know this because this quantum number formula must be:
(hc^3)/4G=(number)^2.
If we replace G with 2/c we get a formula: (hc^4)/8=0.817945942)^2.
0.817945942x(qa)=2.96910024.
3.629897995x10, see item (4), has been thoroughly investigated and is proven to be qa/Gc/2)^0.5. And;
2.96910024x3.62989799x10 is equal to 1.077753101x10^2. This is the Schwarzschild radius of the mass equivalence of (GC template) 2.66079655x10^36 Planck masses. 1.0777531x10^2 is universal. Whatever the change from the kilogram/second there is there will be a corresponding change in the actual value of the metric and a nominal change in qa. Each time the differences cancel out and the result stays at, nominally,1.0777531x10^2 m.
10. 3.912081282x10^4, see item (8), divided by 1.0777531x10^2, equals
3.629849247x10^2, which is qa divided by Gc/2. The reciprocal, 2.75493534x10^3, is equal to the escape speed from a proton surface where radius is equal to the Planck radius.

