### Re: Challenge to Jim Scotti

```Article: <6kl07u\$ga5@dfw-ixnews7.ix.netcom.com>
Subject: Re: Challenge to Jim Scotti
Date: 29 May 1998 00:41:34 GMT

In article <6kgd61\$rh3@bgtnsc02.worldnet.att.net> Eric George writes:
> Oh boy, where to start...

Thanks for rescuing me, Eric :-)  I'll use your figures and recap for
the readership, and then move on to the point the Zetas have been dying
to make, via or through math challenged me.  The recap:

Earth diameter of 12756.27 km.
Moon diameter of  3444.193 km.
using volume of sphere as 4/3*Pi*R^3
volume of Moon is 0.019683 of Earth
Earth is 1,237,857,886,976 km^3 in volume
Moon is 21,392,457,765.53 km^3 in volume

a metric ton is 1,000,000 grams
granite density as 2.64-2.76 g/cm^3, close enough to 2.5
assuming (Nancy's) 2.5 density,
Earth: 2,707,894,750,000,000,000,000 Metric Tons
Moon: 54,997,342,400,000,000,000 Metric Tons

Eric's reference Earth density: 5.5170 g/cm^3
Eric's reference Moon density: 3.3411 g/cm^3
(1.34 times granite density used by Nancy)
so
Earth: 5.9763e+24 kg ( 5.9763e+21 Metric Tons)
or  2.70789475e+21 Metric Tons
Moon: 7.3508e+22 kg ( 7.3508e+19 Metric Tons)
or  5.49973424e+19 Metric Tons
or 73,696,438,000,000,000,000 Metric Tons <==========

gravitational equation F = G*m1*m2/r^2
acceleration of the moon due to gravity toward the Earth
F = 2.20228796E+16 metric ton force
or =  20,228,796,000,000,000 metric tons-force <=====

C  = Pi*2*R = Pi*D
C =  2415768.45 km
gives a circular orbital speed of: 1023.183 m/s <=========

At this point the Zetas want to comment.  First, my comments on what
the Britannica states to be the supersonic speed that the Concord
flies, when breaking the sound barrier.  1,200 mph, or 1,920 km/hr,
which is 1,920,000 m/hr, which is 32,000 m/minute or 533.33 m/second,
is it not?  Second, if the diameter of the Earth is 12,756.27 km, then
stationary satellites must travel 4,635.7155 m/second.

(Begin ZetaTalk[TM])
Here we have the Moon, weighing the EQUIVALENT of some 73 tillion
trillion metric tons if on the surface of the Earth.  Bring that Moon
up due to friction and with a minimal amount of fuel required to keep
it moving.  To eliminate absurd arguments mixing Newton's theories in
with reality, let us state that the Moon does NOT weight what it would
close to the Earth's surface in point of fact, but only what you have
calculated it to weigh while out in space at the distance it rides, the
gravitational force mass of a mere 20 million trillion metric tons
based on your inverse square law of gravity.  This is the ADJUSTED
weight, the equivalent weight, the actual weight at that distance, and
thus is the REAL weight if one is to place it close to Earth.

Eric has kindly calculated the pace of your Moon at some 1023 meters
per second..  Your Concord flies at less than HALF the speed that the
Moon is traveling, and it is scarsely about to fly off into space due
to centrifugal force!  In fact, it requires a full tank of fuel and a
shape to eliminate air friction as much as possible in order to stay
above ground!  Would your Concord fly off into space if the speed were
doubled?  Do your satellites fly off into space, when going at faster
speeds?  Satellites that are stationary above the Earth are circling
the Earth each day, and this speed is far in excess of what the Moon
experiences.  Do they fly off into space?  The answer to both questions
is NO, and so much for Newton's short sighted theories.

Your Moon, adjusted for the distance it is, and going at the rate it
does while carrying this adjusted weight, is going at only 1/4 THE
SPEED OF YOUR SATATIONARY SATELLITES!  Do they, the lighter body, fly
off into space?  Your astronomers, unable to bring all their equations
and physics together on one page, are telling you that your Moon, at an
adjusted weight of 20,228,796,000,000,000 metric tons, could cruise
along at 1/4 the speed of your stationary satellites, or only twice the
speed of your Concord jets, and maintain it's place DUE TO CENTRIFUGAL
FORCE!  The theatre of the absurd is about to open.
(End ZetaTalk[TM])

Newton's laws as applied to circular motion: Ar = V^2/R
where Ar is the acceleration in the radial direction
and towards the center of the circle.

Using the orbital velocity on radius from above:
Ar = (1023.183 m/s)^2 / 384,403,000 m
Ar = 0.00272345 m/s^2

F = mA = mAr
F = 7.35085E+22 kg * 0.00272345 m/s^2
F = 2.00197E+20 kg*m/s^2(Newtons)
This is the centripetal force due to the orbital velocity

The force due to gravity calculated previously was
F = 1.98296e+20 N
The error between these calculation is 0.91%

Does F=mA=mAr mean that you've pre-balanced your equation by saying
mass equals mass, and thus the aceleration OUTwards equals the
aceleration DOWNward - ah, Newton works?  At this point, the Zetas
which to interject again.

(Begin ZetaTalk[TM])
You've cheated again, following your master cheater, Newton.  He has
described what he saw, and fiddled with the math long enough to get
this or that factor, when inserted on this or that side of the
equation, to balance.  Not difficult when the BASE figure, mass, is
placed on both sides of the equation, in the same influential position.
For Newton to have credence here, he must address why the Moon can
float up where the satellites do, at the ADJUSTED (not true) weight
when reduced for distance from the Earth that it normally rides, and go
only twice as fast as the Concord, or only 1/4 as fast as your
stationary satellites, and not fall to Earth!

The reason, as we have explained, is NOT due to Newton's centrifugal
force, but due to the outbursts of gravity particles, as we have
explained.  We will ask our emissary, Nancy, to repost our Repulsion
Force explanation, along with Gravity Particles and Gravity Flow, so
that the readership can see how this better meets the situation they
observe every night when they gaze upward at their monster Moon.
(End ZetaTalk[TM])```