If the moon landing was faked...

In article .com,
"Brad Guth" wrote:

You're the one that doesn't believe in using the regular laws of
physics, nor believing in the replicated hard-science that easily
proves we haven't walked on that salty and otherwise gamma plus extra
hard-X-ray by day worth of a physically dark and nasty moon.

Brad,

A word of advice:

Before you can cite the "regular laws of physics," you first must
understand them! Orbital mechanics and rocket propulsion are part of
these laws, too!

Just where in hell did the "salty moon" stuff come from?

Before you can cite the "regular laws of physics," you first must
understand them! Orbital mechanics and rocket propulsion are part of
these laws, too!
Orval Fairbairn,
I fully agree, thus how the heck did that massively inert slug of our
Saturn V plus having loads of unusable fuel tonnage and even a few
extra tonnes worth of ice manage what can't seem to get accomplished at
half the payload using the most new and improved composites of the very
best rocket-science upon Earth?

Just where in hell did the "salty moon" stuff come from?
From NASA, bless their little perpetrated cold-war hearts. However,
the thick icy proto-moon part was all me, though based upon other
accepted science that seems perfectly fine and dandy as long as such
icy moon science doesn't pertain to our moon, especially if it's salty
ice.

If the moon landing was faked...
As within that context; how exactly does one safely go about
accomplishing such an extensive deorbit and down-range past such pesky
mascons and supposedly land upon that absolutely dark and nasty moon
without having involved a good set of momentum reaction wheels, and
without a beforehand freaking clue as to programming those onboard
rad-hard CPUs for continually calculating their constantly shifting CG
as well as the free-fall potential and of their final velocity upon
whatever landing, or rather impact should one component out of tens of
thousands if not hundreds of thousands pitch a suckerpunch?

Free fall simulators or the available calculators for such are mostly
terrestrial, which is downright terribly odd since we've supposedly
been to and walked upon that physically dark and nasty moon of ours,
and for that accomplishment you'll certainly need to know beforehand on
behalf of all of those pesky mascons and that of your untested and thus
unproven fly-by-rocket landers that didn't even have benefit og
momentum reaction wheels, of such a R&D testy suckers that you'd need
to realize beforehand of exactly whatever it is that you'll be dealing
with, yet the free fall of anything upon our moon is oddly limited to
at best infomercial-science and via easily fabricated video, that for
all sorts of good reasons simply don't even remotely look as though
having been situated upon our moon as raw solar illuminated.

Here's a good little free fall and graphic animation demo for those
NASA/Apollo video clips of stuff dropping from a meter above the lunar
deck.
http://physics.bu.edu/~duffy/java/Freefall2.html
at 1.623 ms/s = 1.11 seconds, which is actually involving quite a few
video frames (33.3 to being exact) that which never once quite seemed
to record upon any slower action than whatever a 9.81 m/s/s environment
had to offer.

Solving Free-Fall Problems : this one's sufficiently good enough for
the task of dropping a javelin probe into that extremely dusty, salty
and otherwise gamma and hard-X-ray nasty moon of ours.
http://www.batesville.k12.in.us/phys...l_problems.htm
Without getting my dyslexic self too technical, in other words a little
averaging and involving zilch worth of friction, whereas we're going to
start off by using the following examples as based upon this previous
link:

1000 second free fall as based upon 1.62 m/s/s
V(f) becomes 1000 * 1.62 = 1620 m/s
Distance traveled = 3.24 km

10,000 sec free fall as based upon using 1.6 m/s/s
V(f) becomes 16 km/s
Distance becomes 32 km (that's just using up 1.84% of 1r)

100,000 sec free fall as based upon the average of 1.25 m/s/s
V(f) becomes 125 km/s
Distance = 250 km (that's just having used up 14.4% of 1r)

1e6 sec free fall as based upon the average of 0.541 m/s/s
V(f) becomes worth 541 km/s
Distance = 1082 km (that's using up 62% of 1r)

Obviously it's worth a whole lot more complex set of calculations that
should be processed as second by second and meter per meter, whereas
otherwise you may change those numbers around in order to suit and/or
moderate whatever game plan you'd like to end up with. However, and no
matters what you'd like to ignore or exclude, if to be incoming as a
free-fall from the moon L-1 that's roughly 59,562 km above, as nearly
directly aligned with the moderating gravity influence of mother Earth,
whereas it's going to take considerable time and, upon arrival is where
that javelin probe is still going to be making damn good velocity,
especially since the starting point of L-1 represents a mere 163 m/s
worth of orbital velocity, and that orbital influence gets down to a
wussy 4.6264 m/s upon impact.

Even though folks here in Usenet naysay land have been doing all they
can to snooker if not fully assimilate the likes of myself, please go
right ahead and use the very most conservative numbers you can imagine,
as in no matters what, lo and behold it's still offering an impressive
V(f) worth of final velocity that we're having to deal with.

Even the volumes upon volumes of our official NASA web pages offers us
village idiots nothing, not even so much as an external link as to
calculating a free-falling object as pertaining specifically to that of
our moon, much less as having been deployed away from LL-1. Everything
is pretty much sequestered as being terrestrial related, exactly as
though they've never been to the moon (robotically nor in person).
-
Brad Guth