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Topic: gravity begins with the proton
Replies: 11   Last Post: Apr 1, 2015 10:52 AM

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 haroldj.l.jones@gmail.com Posts: 67 Registered: 3/17/12
gravity begins with the proton
Posted: May 17, 2014 1:39 PM
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Where 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.

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