Re: Hi Nancy :-))

In Article <[email protected]> John Savard wrote:
> Your Zetas are mistaken. The Moon perfectly satisfies the
> equation a=(omega^2)*r=(v^2)/r for centrifugal force balancing
> the force of gravity, as diminished by the inverse square law,
> at the Moon's distance.
> I mean, it's true some scientists once did come up with an
> equation that said bumblebees couldn't fly, but that's because
> they oversimplified and made false assumptions.

I was hoping someone would bite on that one. This was the concluding
argument in the last sci.astro debates, and all the math, YOUR math,
leading to support the Zeta statement is there as a link off the
ZetaTalk web site to be revisited for a refresher, if you wish. These
are links from the 
June 5, 1998 page (http://www.zetatalk.com/usenet/use06058.htm)
but for the ease of the readerhip, here’s the posted wrap-up:

.............. (from http://www.zetatalk.com/usenet/use00679.htm)

The Zetas have stated that humans cannot get their math formulas to sit
on the same page. To demonstrate this, they forced math challenged me
to grope though the formulas, and fortunately I was rescued by Eric
George and M.C. Harrison who inserted the RIGHT figures. They show a
massive Moon, per the inverse square law, of millions of trillions of
metric tons of EQUIVALENT weight if on the surface of the Earth, moving
at a rate of only 1023 m/s which is 1/4 the rate of our stationary
satellites or only twice the speed of the Concord.

To refresh the readership on where these calculations arrived, I'll
repeat last weeks posting, article
<[email protected]>, below:

...........
In article <[email protected]> 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
down to where your supersonic jets, your Concord, fly without burning 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 is 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 explaintion, 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])