quantified from theory
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. 10^{52} in a one meter shell times about 10^{8} 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.
which will yield,
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 10^{6} meters = r
then,
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 10^{24} kg ... and the mass of the proton at 1.67 x 10^{27} kg ... we get
We shall abitrarily take the baryon number at 10^{79}, meaning that there are about 10^{28} 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}.
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. M_{1} and M_{2} 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 selfinteraction. 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.
which is just what the doctor ordered.
