The Earth's Gravitational Field
quantified from theory

I   
would like to put a number on the strength of the earth's gravitational field designating its force ... and ... have that number generated from basic principles as developed in my own theory. Because the total number of baryons in the visible universe is at present only reasonably guessed at, the answer can only be approximate. I will settle for an order of magnitude, ballpark figure. If I can obtain an answer like 106 times too large or small, I will consider the venture a failure. If 10 to 100 times too small or large, I shall be satisfied that my theory has not been discredited (as yet).

We must then calculate the distance that the center of a baryon will drop vertically when traveling parallel to the Earth's surface over the distance of its Compton wavelength at sea level ... in a time period of 10-60 seconds. 10-60 th seconds represents the total number of new baryons coming into existence relative to the test baryon at the Hubble radius per second, i.e. 1052 in a one meter shell times about 108 meters per second (light speed = the speed at which the Hubble radius advances encompassing new baryons).

Extrapolating from measurement, we know that the answer must be 10-120 meters (approximate) because of the distance equation and known gravitational force at the earth's surface.

S = 1/2 at2

which will yield,

~ 5 x 10-120 meters = 1/2 x 9.8 meters/sec2 x [10-60]2

10-120 is a huge number to deal with ...

From ntx15.htm,


The bend influencing the lateral component of the reaction velocity is

B = { (r^2 + D^2)^1/2 } - r

Where B is the bend, r is the distance to the concentration and D is the particle confinement (Compton wavelength).


The radius of the earth is 6 x 106 meters = r
The Compton wavelength of the proton is 1.3 x 10-15 meters = D

then,
[36 x 1012 + 1.7 x 10-30] 1/2 - 6 x 106 = approx. 1.4 x 10-31 meters

We must reduce this due to the fact that the earth must compete with all the rest of the mass of the universe. From the earth's mass of 6 x 1024 kg ... and the mass of the proton at 1.67 x 10-27 kg ... we get

3.6 x 1051 protons in the earth

We shall abitrarily take the baryon number at 1079, meaning that there are about 1028 earth masses to compete with in establishing a reference frame. All other masses combined form a general flat Euclidean reference frame. Therefore, the amount by which the baryon path is curved in the above manner is reduced by another 10-28.

1.4 x 10-31 x 10-28 = ~10-59 meters (drop in 10-60 seconds)

This distance is far too large but appears to be the square root of the number initially required. So, either I'm way, way off ... or ... I am missing a "squaring" parameter.

The "Other Half"

There are two parties in any gravitational attraction. M1 and M2 which we must multiply together to obtain the measure of the force. Also from ntx15.htm ...


A particle accelerating in a gravitational field can increase its rate of acceleration only to the extent allowed by its own field which is left behind owing to the fact that no signal can be propagated at infinite velocity. The field cannot follow its center instantaneously and is therefore distorted in such a way as to prevent further increase in the rate of acceleration by self-interaction. Two different test masses will fall at the same rate in a given gravitational field since it is the rate of acceleration which causes the distortion and not the mass itself.


We see here that the accelerated body has its field swept back proportional to that acceleration ... and ... that distortion lessens the effect of the field we wish to calculate by just the amount we are seeking ... namely, an almost equal and opposite force which nearly cancels it out leaving only the barest observed force operating.

the effect of M1  (times)   the effect of M2 = 10-59 x 10-59 = 10-118

which is just what the doctor ordered.




Next Page


Ebtx Home Page