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Topic: gravity begins with the proton
Replies: 7   Last Post: Aug 16, 2014 12:29 PM

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haroldj.l.jones@gmail.com

Posts: 61
Registered: 3/17/12
Re: gravity begins with the proton
Posted: May 24, 2014 1:31 PM
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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.
>
>
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> 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
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> without us realising it.
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>
>
> By using 2/c as substitute for G we get the following:
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>
>
> (1). In the formula for the Planck mass, (ch/4G)^0.5, when substituting 2/c for G
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> we arrive at 2.72837394x10^-9, quite recognisable as the Planck mass except it is out by a factor of (GC/2)^0.5.
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>
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> (2). The same principle can be applied to the Planck radius, formula (Gh/c^3)^0.5.
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> This time we end up with 4.050436x10^-34.
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>
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> (3). The ratio of Strong force over gravity, 2.660725x10^38, based on
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> (Mpl/Mpr), where Mpl is Planck mass and Mpr is proton mass. Appying the above criterion the Gm template becomes 2.66079655x10^36.
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>
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> (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
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> 1.10194453xGm/4 x Mpl=1.503278583x10^-10 J. The total energy of the proton.
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> (1.503278583x10^-10)(1.009721668x10^7)=1.51789295x10^-3.
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> 2/1.51789295x10^-3=1.317615975x10^3. The square root of this is 3.629898x10.
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> Accepting there might be a small drift from the theoretical qa it can still be assumed that Gm/2 is around 1/100.
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>
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> (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
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> 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.






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