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Old November 19th 03, 01:08 AM
Brad Guth
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Default Is the moon leaving, or are we shrinking by 38 mm/year

Obviously there's tidal gravitational forces continuously at work, as
otherwise something as massive as the Earth/Moon union would become
one.

Since I've learned from Marvin ) that lunar
recession is presumably taking roughly 4 terawatts/year, while I'm
thoroughly confused about calculating something other upon lunar drag
into similar terms of continuous energy requirement, I've decided to
do a little shock and awe reverse engineering, based upon yet another
of my village idiot guestimates of lunar drag consuming another 1e12
W, a terawatt worth of tidal forces that's having only to offset for
drag in order to sustain the lunar orbit speed at zero recession.

If we were to allot said 1e12 (terawatt) to represent the necessary
energy to overcome a lunar drag coefficient.

1e12/38e12 = 26.3 mw/m2

If we were to further suggest that the space weather environment in
which the moon travels about Earth is representing but 1/1000th that
viscosity and kinetic energy of the ISS orbit.

If we were use an ISS surface reference area of 1000 m2, as well as
the 1000 times greater atmosphere as multiplier factors:

26.3e-3 * 1e3 * 1e3 = 26.3 kw (in other words 35.8 hp)

Thus it's taking 26.3 kw worth of continuous energy applied in order
to sustain the likes of ISS in a fixed (non-recession) orbit, in other
words of overcoming friction as that based upon the consideration of
the ISS environment being of 1e3 greater viscosity than lunar space,
as well as for having to travel roughly 11 fold faster than the moon,
which may actually compute as an overall 11e3:1.

Obviously if lunar space were merely 100 times thinner, then the
reverse engineering calculations for ISS compensation drops to a mere
2.6 kw.

Perhaps 26 kw is simply too much continuous energy applied for ISS
but, I'd have to bet that 2.6 kw is not sufficient, thereby the
atmosphere and/or space weather difference may actually become 400~500
times thinner for that of the moon traveling through at 1.025 km/s.

Of course absolutely none of this is sufficiently correct but, at
least it gives myself something to go along with those 4 terawatts
worth of recession energy, making the overall tally for continuous
gravitational tidal forces per year applied onto the moon as
representing perhaps 5 terawatts.

4e12 + 1e12 = 5e12 W

If there's now 5e12 W to draw upon (disregarding solar PV conversions
and/or induced solar plasma electrons and/or EMF factors taken from
solar weather as well as from our Van Allen zone of death, all of
which should actually be worth quite a great deal, so much so that it
seems exactly the sort of tether dipole extraction potential that I'm
thinking can be safely applied into those massive counter-rotating
flywheels situated at the ME-L1 (gravity-well null).

Just for a little further shock and awe argument sake, lets say that
the additional energy input (besides the overall recession energy)
provides us with one additional terawatt resource, now all toll were
looking at 6 terawatts, whereas tapping into 50% of that energy yields
3 terawatts, enough energy to run 20 of those 100 GW laser cannons
plus another terawatt to spare. Obviously only 2 of those terawatts
are those being extracted from the recession energy.

5e12 + 1e12 = 6e12 * 0.5 = 3e12 W extracted

We certainly can't take away everything, as that could reverse things
by pulling the moon into Earth, a seriously bad sort of thing to be
doing. Although, if the bulk of energy taken is what's converted into
laser cannon energy focused upon relatively small portions of Earth
(not that Earth isn't already getting a little too hot under the
collar), say quadrant targets zones of as little as 1 km diameter
along with a 10 km safety buffer zone, at least some of this extracted
energy should return itself as a slight repulsion factor.

As we manage to run ourselves out of natural petroleum as well as
other natural resources and remain too dumbfounded to safely utilize
nuclear energy (like those smart ass French have been doing for
decades), and way too energy inefficient to rely upon wind and ocean
tidal resources (too polluted and otherwise too greenhouse for solar
PV), thereby having insufficient energy for exterminating the
remaining populations we don't happen to like; By properly using laser
cannons of either near UV or near IR, or perhaps both spectrums so
that at least humans aren't blinded while the least amount of thermal
energy is contributed upon Earth (this is actually where I'd favor the
near IR [750~800 nm] that shouldn't blind the nocturnal species), this
tactic being where those multiple 100 GW cannons can sort of light
your fire from afar, just might do the trick.

If those 20 or so 100 GW laser cannon beams are being efficiently
transmitted and converted, I'd tend to think the overall input/output
conversion that's obtained on Earth might eventually reach 25%, thus 2
terawatts of input becomes 500 GW, which isn't all that bad for the
hundreds of billions it may take to pull that one off.

Regards, Brad Guth http://guthvenus.tripod.com