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-   -   Is the moon leaving, or are we shrinking by 38 mm/year (http://www.spacebanter.com/showthread.php?t=12787)

 Marvin November 17th 03 06:43 PM

Is the moon leaving, or are we shrinking by 38 mm/year

So again, how much kinetic energy, and/or in terms of thrust (ISP
or kg/s in terms of fuel/energy consumption), would it take for a
satellite the size and mass of our moon, to be accelerated so as to
escape Earth by a rate of 38 mm/year?

Thanks so much for all the feedback. I'll try a little more input into
my three remaining brain cells and see if the amount of recession
energy can be realized.

The amount of energy needed to raise the moon's orbit by that 3.8cm/year is
easy.. Just plug into basic orbital mechanics, out pops an answer of some
1.25 * 10^20 joules per year, easier to visualise as a constant power input
of some 4 terawatt (4 * 10^12 watt)

The problem is that this is indeed the *resultant* of all forces
applicable. Just how big are the other forces working on the moon,like
frictional drag against the not-quite-empty vacuum of local space.(keep in
mind the moon orbits earth at about 1km/sec, and the earth whizzes around

 Brad Guth November 17th 03 09:52 PM

Is the moon leaving, or are we shrinking by 38 mm/year

Marvin wrote in message ...
So again, how much kinetic energy, and/or in terms of thrust (ISP
or kg/s in terms of fuel/energy consumption), would it take for a
satellite the size and mass of our moon, to be accelerated so as to
escape Earth by a rate of 38 mm/year?

Thanks so much for all the feedback. I'll try a little more input into
my three remaining brain cells and see if the amount of recession
energy can be realized.

The amount of energy needed to raise the moon's orbit by that 3.8cm/year is
easy.. Just plug into basic orbital mechanics, out pops an answer of some
1.25 * 10^20 joules per year, easier to visualise as a constant power input
of some 4 terawatt (4 * 10^12 watt)

The problem is that this is indeed the *resultant* of all forces
applicable. Just how big are the other forces working on the moon,like
frictional drag against the not-quite-empty vacuum of local space.(keep in
mind the moon orbits earth at about 1km/sec, and the earth whizzes around

Terrific feed back, at least 4 terawatts plus something for drag,
thanks much.

BTW; here's something I picked up off the net:

http://www.europhysicsnews.com/full/.../article3.html
Hydrodynamics of planetary nebulae: 10e10 atoms per m3.
and of much lower density, typically 10e7 atoms per m3.

Perhaps the friction aspect can be roughly derived, as there's got to
be some worth in that much surface being driven through at the 1.025
km/s.

The part about keeping up with Earth I'd think is a given, as pushing
ahead (into the solar wind) should be fully counteracted by the exact
opposit of traveling down-wind while being taken along by the momentum
of Earth's traveling about the sun.

 Brad Guth November 19th 03 12:08 AM

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.

 [email protected] \(formerly\) November 19th 03 01:16 AM

Is the moon leaving, or are we shrinking by 38 mm/year

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

A real sidelight, but I don't think they would "become one". I think that
in the existing system, there is enough angular momentum that there is no
way a single "partially liquid" body wouldn't spin out one or more lobes.
You might jam the two together, but I don't think they'd stay together.

David A. Smith

 Brad Guth November 19th 03 09:57 PM

Is the moon leaving, or are we shrinking by 38 mm/year

\(formerly\)" [email protected] wrote in message news:[email protected]

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

A real sidelight, but I don't think they would "become one". I think that
in the existing system, there is enough angular momentum that there is no
way a single "partially liquid" body wouldn't spin out one or more lobes.
You might jam the two together, but I don't think they'd stay together.

David A. Smith

That's certainly a good thing to know about.

Any better notion upon my village idiot efforts at obtaining the
kinetic energy requirement for just sustaining the lunar orbit?

My guestimate of adding in another terawatt of continuous energy, as a
26 mw/m2 seems somewhat reasonable, though what do I know?

Perhaps this following context is just the tip of this tether dipole
iceberg.

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 most
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.

For the continuing entertainment sake, plus on behalf of those coming
into this topic without benefit of nearly three years worth of my
efforts, I've added another infomercial page (GV-LM-1) and edited upon
a couple of others.

If I were to be suggesting upon the sorts of wild and crazy things, as
if this is what makes your life worth living, especially if those were
to be of the sorts of horrifically spendy and somewhat lethal agendas
like the Mars or bust and ESE fiasco, in that case I've got lots other
to say about utilizing the moon as well as Venus.
http://guthvenus.tripod.com/gv-lm-1.htm
http://guthvenus.tripod.com/gv-cm-ccm-01.htm
http://guthvenus.tripod.com/earth-moon-energy.htm
http://guthvenus.tripod.com/gv-lse-energy.htm
http://guthvenus.tripod.com/vl2-iss-joke.htm
http://guthvenus.tripod.com/gv-illumination.htm
http://guthvenus.tripod.com/moon-sar.htm
http://guthvenus.tripod.com/gv-town.htm
plus a few dozen other pages, with more on their way.

 [email protected] \(formerly\) November 20th 03 12:14 AM

Is the moon leaving, or are we shrinking by 38 mm/year

om...
\(formerly\)" [email protected] wrote in message

news:[email protected]

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

A real sidelight, but I don't think they would "become one". I think

that
in the existing system, there is enough angular momentum that there is

no
way a single "partially liquid" body wouldn't spin out one or more

lobes.
You might jam the two together, but I don't think they'd stay together.

That's certainly a good thing to know about.

Any better notion upon my village idiot efforts at obtaining the
kinetic energy requirement for just sustaining the lunar orbit?

Zero additional energy should do OK. The solar wind boosts one side, and
retards the other. The earth would have to freeze solid for all eternity,
of course. All the way down to the core. Might as well tidally lock it as
well.

Otherwise, it'll just continue to gain angular momentum from the Earth.

Should pretty much stop the recession...

David A. Smith

 Brad Guth November 21st 03 05:36 AM

Is the moon leaving, or are we shrinking by 38 mm/year

\(formerly\)" [email protected] wrote in message news:[email protected]

om...
\(formerly\)" [email protected] wrote in message

news:[email protected]

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

A real sidelight, but I don't think they would "become one". I think

that
in the existing system, there is enough angular momentum that there is

no
way a single "partially liquid" body wouldn't spin out one or more

lobes.
You might jam the two together, but I don't think they'd stay together.

That's certainly a good thing to know about.

Any better notion upon my village idiot efforts at obtaining the
kinetic energy requirement for just sustaining the lunar orbit?

Zero additional energy should do OK. The solar wind boosts one side, and
retards the other. The earth would have to freeze solid for all eternity,
of course. All the way down to the core. Might as well tidally lock it as
well.

Otherwise, it'll just continue to gain angular momentum from the Earth.

Should pretty much stop the recession...

David A. Smith

I agree about the solar wind being a given "zero".

From: MLuttgens )

"To calculate the effects of the deceleration on the orbit
of planets (or satellites)"

"According to LLR data, the Moon is receding from Earth at a rate of
about 3.8 centimeters per year. Such increase of orbital distance,
attributed to tidal effects on Earth, could mask the present small
decrease of 1.87 cm/year."

Unless I've misunderstood as usual, this above page seems to be
indicating that the coefficient of lunar orbital friction is actually
quite a fair percentage upon the overall scheme of things, as 1.87
cm/year is nearly half of the 3.8 cm of reported recession, thereby
the overall energy necessary in order to impose the 3.8 cm/year
recession may in fact become a factor of 3.8 + 1.87 = 5.67 cm/year,
which in turn might further suggest 6 terawatts worth instead of my
initial guestimate of 5 terawatts that was based upon adding one
terawatt to what Marvin ) specified as to his
calculations being 4 terawatts worth of recession energy (excluding
matters of friction) that was capable of inducing the 3.8 cm/year.

A satellite of the rough area of the moon, cutting through at the very
least viscosity of 6e6 atoms/m3 at 1.025 km/s (more likely density
6e9/m3) will require some portion of said tidal energy just for
sustaining the status quo. If it were not for the tidal gravity doing
its thing, seems like an orbit degrade of 1.87 cm/year is sufficiently
reasonable (nothing is forever).

 Brad Guth December 15th 03 07:38 PM

Is the moon leaving, or are we shrinking by 38 mm/year

The moon is in fact leaving us behind in it's dust, or perhaps
clumping moon dirt. Though Earth is also shrinking by leaps and
bounds, with our precious energy leaking out in more ways than you can
shake that flaming stick at.

Unfortunately, China or perhaps even India if not ESA is going to be
actually on the moon, or of at least hovering over it, as in
establishing their first come first served as for creating their
LSE-CM/ISS, as for there being room for only one of these suckers. The
following rant is yet another infomercial on behalf of sanity, on
behalf of humanity and on behalf of accomplishing what's been doable

BTW; just in case some nice folks haven't discovered what the likes of
wizard Jay is all about, checkout some of the following and be sure to
reflect upon those two quotes offered by Lord Jay Windley.

"The moon, the Apollo ruse/sting, the snookered fools we are"

The lunar environment is obviously not moderated by any significant
atmosphere nor Van Allen belt, thus of the solar/cosmic, cosmic and
gamma ray exposures are unimpeded, and as such the radiation
environment is hardly being stabilized nor averaged over time. It's
either too damn hot or too damn cold or too freaking lethal unless
you're enjoying all of it by earthshine, though not to mention having
to avoid the somewhat pesky issue of it raining micro meteorites.

The space/solar weather of such nasty stuff includes a great deal of
the relatively passive warmth of IR, on into the somewhat lethal UV
spectrums, either of which can be fended off by relatively low
technology, although UV/c can start to be a bit penetrating unless
there's an artificial barrier of sufficient solids, such as any good
moon suit will suffice.

Higher frequency and thus high energy is not so easily stopped by any
moon suit, and of what is being slowed down and/or partially absorbed
by the suit, or by way of most any substance, is what creates those
hard x-ray class radiation issues. Actually the greater the material
density the greater the secondary impact becomes, especially at the
thickness and/or density per square centimeter of what our Apollo

Of too little shielding and you're affected by the direct radiation
impact, of thicker shield and/or of greater density obviously blocks
more of the primary influx while creating greater and even somewhat
more lethal hard x-ray class dosage. Depending upon what sort of
influx or solar flak is hitting your exterior environment, such as
cosmic and/or gamma can obviously make a rather tremendous difference
of mostly negative issues as far as protecting life as we know it.
Just like our sun can deliver relatively passive and low energy
dosages, while at times the solar output offers the capability and/or
intensity of exceeding several thousand rads per hour, which is not a
serious exposure problem if you've got a healthy Van Allen zone plus
tonnes of atmosphere per m2 as your shield, and not that thousands of
folks don't go about expiring each and every year specifically due to
their receiving too much solar and cosmic radiation.

When those several thousand rads per hour impact a substance such as
clumping moon dirt, a matrix of many things that should represent
3.4+g/cc, this is where the somewhat lethal solar flux that's just
plain old nasty becomes downright lethal within an hour's worth of
exposure. Thus the lunar surface exposed to a passive solar
environment might lull itself into creating a mere 100 rads (1 Sv) per
day (24 hours or a respectable 4.17 rads/hr), although the sun wasn't
in any passive mode nor was the solar activity sufficient as to fend
off the cosmic and gamma ray aspects, thus the combined surface impact
for whatever and/or whomever was certainly capable of creating 360
rads per day (15 rads/hr), that is if you're honestly accounting for
the secondary contributions of what the lunar surface itself was
capable of creating.

Your standard issue moon suit can cut the likes of direct solar
radiation, mostly because at least for some of the passive/thermal
solar event timeline isn't itself of lethal hard x-ray class, although
of whatever does impact the suit and mostly of what impacts the lunar
surface will be creating a fairly large TBI worthy dosage. More recent
solar events such as those of October/November 2003 were off the
scale, so strong that of our best instruments were essentially blown
away. Fortunately there were only much smaller ongoing solar events
during the Apollo mission era, which was a good thing as for fending
off some of the cosmic class radiation, though representing a truly
bad sort of thing as for any space expedition that's as close as we
were to our sun.

As for being further away from the sun, such as Mars, offers a solar
environment safety improvement, though somewhat worse off as for
allowing more cosmic radiation to impact and subsequently interact
with whatever and/or whomever is anywhere near and/or situated behind
a substance that's not sufficiently thick enough as to block and
otherwise absorb all of the influx, plus having to subdue secondary
hard x-ray class radiation before it gets to your butt. It seems we
currently have a wee bit of a problem in placing sufficient mass into
orbit, much less headed off to places like the moon or Mars, thus our
manned missions off to whatever is residing outside our Van Allen zone
of death are essentially unresolved issues as of today, though not
insurmountable.

The absolute proof that it's truly nasty beyond our Van Allen zone of
death is in the pudding, in the fact that there's been an effort to
skew and/or cloak the truthful data, as for example in providing
transparencies of these Apollo missions. At this point I'm not even
suggesting upon obtaining an actual image frame, but merely of the
leader and/or trail which couldn't possibly have betrayed and/or
impacted upon one of those infamous images, of which there are 10's of
thousands of said frames to select from, of which the public has
viewed copies and/or prints from less than 1%, leaving 99% of those
available frames (stills and movie film) nonutilized, perhaps because
those weren't all that great to look at, though of what the image
contains is rather insignificant as for otherwise determining
radiation, of which just about any portion of film, from an actual
frame or of what's between or of the leader/trailer portions would
have done just fine and dandy.

Though sadly, at this late time, there'd be no way of identifying the
film as for being actual Apollo related, unless those were of viable
lunar landscape images included. As for obtaining a trailer/leader
portion of processed film would simply be unreliable and entirely
meaningless since there'd be no certainty of it actually being what it
is.

Using an electron microscope, or even a sufficiently good digital scan
of a section of even a film leader and/or trailer could have revealed
the exact dosages of radiation exposure, down to the individual
recorded dosage could have been detected, though this would have taken
100+ millirad in order to have become observed to the human eye, of
which all such Apollo mission film should have received at least
several rads/rems if not hundreds. Human cells will for the most part
recover from such TBI dosages, though film offers a one-way recording
of the radiation accumulation, with or without ever being exposed to
taking pictures.

Of course at this point there's no simple and/or definitive method of
identifying a primary radiation impact from that of a secondary,
although the electron microscope could help to determined the various
wavelength differences affecting those film emulsion crystals. Film
crystals being mostly analog, but also somewhat digital in that every
individual crystal or photon bucket can be affected to a differing
degree, as there are far more of those emulsion crystals (photon
buckets) per square mm than our finest CCD technology of even today,
thus a great deal of information has always been available, far
exceeding the optical lens resolution, including the detection of
mostly near UV starlight upon those crystals. But oddly all access has
been avoided for the rather obvious reasons, of reasons that must
include the fact of such imaging wasn't necessarily accomplished on
the lunar surface.

This doesn't represent that our Apollo missions didn't for a time exit
the Van Allen zone of death, possibly even to orbit the moon and of
robotically deploying any number of experiments, as even a lunar orbit
would have been quite risky business and of itself somewhat TBI
worthy, although nowhere as bad off as for the actual solar and cosmic

Since there's supposedly been absolutely nothing for NASA or as for
those worshiping of Apollo folks to fear nor lose, absolutely no
possible damage to an original frame of their precious film, the only
remaining fact of the matter becomes rather too obvious. Not that
there's plenty of image contents worth arguing about, like the 50+%
reflective index that's clearly observed within so may of the images,
and for the rather odd lack of sufficient meteorites and various
impact shards strewn about the lunar morgue, of a fully exposed
surface which should have been at least as covered by such debris as
Mars is, if not a whole lot more so.

Actually, the ongoing numbers of micro meteorites impacting the lunar
surface at 5+km/s should have been at least one per m2/day, although
one per m2/hour shouldn't have been unexpected, and of any suitable
lander constructed as for fending off such an influx. We now realize
that the lander was anything but sufficiently constructed as to fend
off much more than clumping moon dirt, among many other deficiencies
which included radiation abatement that obviously wasn't worth squat,
except for avoiding a UV class sun burn.

Jay Windley wrote:
"It is simply not necessary to follow all lines of investigation to
some absolute standard of completeness in order to draw reliable
conclusions."

and

"The search for truth is not a game in which evidence is doled out
according to some strategy. It is based on full and accurate
disclosure of the facts for examination."

Jay Windley's first quote is quite true to life, although his second
quote is surely from another planet besides Earth, perhaps from
another dimension to boot.

I guess I'm still the village idiot that's thinking way outside the
box, as for our going back to the moon (if ever) may have to be for
robots, not for mankind. At least not until we have obtained a
sufficiently astronaut pilot documented and thus working lander of
sufficient shielding as for radiation as well as for fending off all
those pesky micro meteorites.

LSE-CM/ISS (Lunar Space Elevator plus Counter Mass and new ISS) or
GMDE (Guth Moon Dirt Express), plus there's lots of other related
stuff, with more on the way (incorrect math, poor grammar and my
dyslexic syntax to boot);
http://guthvenus.tripod.com/gv-cm-ccm-01.htm
http://guthvenus.tripod.com/gv-hybrid-irc.htm
http://guthvenus.tripod.com/gv-h2o2-irrce.htm
http://guthvenus.tripod.com/gv-lm-1.htm
http://guthvenus.tripod.com/gv-basalt.htm