Appendix E

Some Thoughts on the

THE MYSTERY OF GRAVITATION

From Chapter III of the Papers on Neutral Currents

It is a well known fact---especially among divers---that the
gravitational pressure under the water increases at the rate of about 15 pounds per square
inch (i.e. one atmosphere) for every 34 feet we descend into a body of water (tank,
reservoir, lake, river, ocean, etc.). A pound is a measure of weight, mg, where g is the
local acceleration due to gravity. Pressure, therefore, is force or weight (mg per area r^{2})
that is to say: p = mg/r^{2}. But in the case of our gravity equations,

F = mg = mGM/r^{2}

And it's the pressure potential (M/r^{2} or mass per area) that
constitutes the variable in each separate application, for the acceleration due to gravity
must be supplied from the constant G. This constant G is measured as 6.6732 x 10^{-8}
centimeters cubed per gram per second squared: cm^{3}/gm-sec^{2}. But if
we analyze this structure, we find that it separates readily into two parts:

(1) The units (cm^{2}/gm) which are to be divided into the
variable M/r^{2 }so as to measure them in the c.g.s. System;

(2) The constant acceleration due to the gravity of the universe as a
whole---6.6732 x 10^{-8} cm/sec^{2} ---a measure that should be constant
during any particular century throughout the universe's surface as a whole, provided that
the hypersphere is nowhere oblate and hence that all points along its surface are at an
equal distance R_{u} from its origin-center.

For example, By factoring out the small mass m, the equation for the local acceleration due to gravity here on the earth becomes:

g = GM_{earth / }r^{2}_{earth}

g = __6.6732 x 10 ^{-8} x 5.977 x 10^{27}__

(6.37747 x 10

Whence, g = 980.6646 cm/sec^{2}, which agrees very well with
the measure given in my dictionary of 980.665 cm/sec^{2}. But just what did we do
to derive this answer? We had to supply the two local measures, M_{earth} = 5.977
x 10^{27} grams and r_{earth} = 6.37747 x 10^{8} centimeters to
obtain our local terrestrial pressure potential M/r^{2}. But the constant
universal acceleration, distributed through this pressure potential, 6.6732 x 10^{-8}
cm/sec, had to be supplied from the universal constant of gravity G.

Now this is a very important point. In the universal case of any given
point at the surface of the universe in relation to no other masses than the mass of the
universe M_{u} itself regarded as being centered at the Origin-Center exactly one
radius of the universe R_{u} away, we should not have to find the measures of
these universal factors, for at the surface of the universe with its constant pressure
these also should always be constant and therefore already figured into G.

Consequently, as long as we use the water-density
centimeter-gram-second measuring system in which one cubic centimeter of water always
weights one gram, we can affirm that the universal constant of gravity G itself should
give us the acceleration due to the gravity of the universe as a whole as 6.6732 x 10^{-8}
centimeters per second squared (the cm^{2}/gm now being reduced to unity, remember
that we are calculating this from the view of the Hypersphere where the density of water
is in terms of cm^{2}/gm and not cm^{3}/gm).

But we also know that the velocity of a satellite revolving around another body can always be found by equating the outward thrust of its centrifugal force with the inward pull of gravity towards the body:

F_{c} = mv^{2}/r = F_{g} = mg

Hence, v^{2} = gr.

But suppose that we now wish to find the velocity of a photon traveling
at the velocity of light c around the surface of the universe at the distance of one
radius of the universe R_{u} from its origin center.

v -----------limit-----------> c

g -----------limit-----------> G

r ------------limit----------> R_{u}

Therefore, c^{2} = GR_{u}.

But we have just discovered the formula for the radius of _the universe
R_{u }!!!

R_{u} = c^{2}/G

But if G = 6.6732 x 10^{-8} cm/sec^{2, }and if c =
2.997,925 x 10^{10} cm/sec, then R_{u} = 1.346,81 x 10^{28}
centimeters (or since 1 light-year = 9.46053 x 10^{17} cm.), then

R_{u} = 14.2361 billion light-years !!!

So the radius of the universe is some 14 billion light-years, which
means, of course, that it must have been created some 14 billion years ago, for R_{u}
= cT_{u}.

But just how do you know for sure that this is really true?

We know it to be true heuristically because, as the sequel will show, it leads to one
discovery right after another; and we know it pragmatically because all these discoveries
in turn prove to be true.

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