## [geocentrism] Re: Moving Earth deception

• From: Paul Deema <paul_deema@xxxxxxxxxxx>
• To: Geocentrism@xxxxxxxxxxxxx
• Date: Tue, 31 Jul 2007 18:42:26 +0000 (GMT)

```Philip M
I'm having trouble with some of your numbers. I've sprinkled comments in
colour. Can we sort these out early? It saves trouble later!
My experiment.
Figures I used.
orbit: 384,400 km (tick) from Earth
diameter: 3476 km (3474.8 -- trivial?)
2416228 kilometers travelled in 24 hours (The Geo position is one revolution in
24h 52m. Neither figure results in 2 416 228 km) or 709 hours (29.54d -- a Luna
day. One revolution however is 27.322d)
= 100,676kph (100 636 -- trivial) or 3,408kmh (3683) respectfully.
Earth moves forward 30km/sec or does not move at all. (tick)
Between Marc, and Paul, their discourse opens up some points of experiment that
need simplification, clarification, and confirmation.. I would like to
ennumerate these, hoping to enable all to get an actual mental image of what is
happening when we shoot a beam at the moon. I want to ignore the special
reflectors Paul speaks about for the moment, and Marcs ball and trains analogy,
which can cause confusion as indicated by Paul.
Now these are my presumptions, and I hope any errors will be corrected and
weaknesses resolved.
1. We will call the transit time earth to moon for simplicity 1 second each
way. This has no bearing on the relationships we are exploring. It means our
"photon" has just one second of existence, or more for those reflected back off
the surface of the moon.
2. Once a beam is transmitted, it travels in a straight line, independent of
any forces whatsoever. This means the beam as such during its travel time is
stationary in space. Its path is a fixed and stationary true straight line
radial from the earth, provided the transmitter is stationary. It is essential
this property be recognised. Note: If the transmitter is moving for 10ms then
we essentially still have a sequence of parallel lines of "photons".
3. To make the movement of the transmitter irrelevant and relatively
stationary, or independent of any alleged rotation of the earth or its orbital
speed around the sun we must make the beam have no more than a few, nominated
10 milliseconds of duration. (At 30 km/s this represents a movement of 300 m).
4. We need to know how wide the laser beam will be by the time it reaches the
moon. What size spot will it make. NASA's beam was 2km across. Only one in 30
million photons were reflected from their reflector array. se pic below .
Another said:When pointing a well focussed laser at the moon, its beam will
widen to well over 100 meters in radius by the time it reaches the surface.
5. For our purpose we need to shoot during a new moon to make the flash/spot
more visible. We can use a killer laser to make sure it is visible. Telescope
and camera. (I can't see that it will be visible from Earth as it is recognised
that even with a 3 m telescope, we still need a photomultiplier to see the
return from a very efficient and perfectly aligned reflector. I think that what
you will need is a carpet of reasonably sensitive detectors on the Moon
blanketing the expected area of irradiance and telemetry to a measuring station
on Earth).
6. As we have effectively stopped any rotation of the earth by using only a few
milliseconds, we can assume the moon to be travelling radially, as observed,
for aiming purposes. (We are not measuring reflection times, but visible
patterns on the surface)
7. The reality of the moons speed will be proven because we are making a
comparison of the light pattern drawn on the moon by a truly stationary 10ms
beam in space, with either a moon travelling daily at 168,000mph (60 352 mph?)
or travelling monthly at 5,680mph (2289 mph?) in space.
8. It should be a simple calculation to compare the distance the beam moves
across the moonscape in 10milliseconds at either speed.
9. If the firing was done on a trajectory that was at right angles to the
forward motion of the earth around the sun, then again considering the absolute
stationary property of our light beam, during the time of our 1 second journey
, both the moon and earth travelling at 30km/sec. will have moved forward 30km.
from the firing line. (no pun) by the time it marks the moon.
10. From (9) then the spot will hit the moon 30 km behind where it would hit
during an inline with the earths orbit shot, 90 degrees of moon orbit earlier.
Also the 10ms beam will have a more elongated oval shape.
I look forward to any comments.
Actually, I think this is all unnecessary complication. Let us shoot our laser
at the target on the Moon when it is at 90 deg to the Sun-Earth line ie half
phase. In the helio scenario, the Earth and the Moon are both travelling at 30
km/s in the same direction so there is no differential. In the Geo scenario,
this velocity does not exist so again there is no differential.
However, in the helio case, the Moon, at a distance of 384 400 km is moving
tangentially at 1.023 km/s. To hit a spot at this distance and 1 s in the
future, we need to lead the target by asin 1.023/384 400 -- about 152.5 u deg.
In the geo case, the Moon 1 s in the future will be 26.98 km ahead of our
aiming point and so we will need to lead the target by atan 26.98/384 400 --
about 4 m deg. Thus if the beam is about "...well over 100 m radius ...", the
precision available would produce a totally unambiguous answer. Even at 2 km
radius, there is sufficient precision available to make it worth wagering half