| Topic: |
Science > Physics |
| User: |
"Brad Guth" |
| Date: |
21 May 2006 12:37:08 PM |
| Object: |
Terminal Velocity of Impacting our Moon |
We seem to realize from actual experience based hard-science and via
the regular laws of physics that the minimum velocity (w/o retrothrust
nor drag coefficient) is roughly 2.4 km/s. However, what is the
absolute maximum obtainable terminal velocity of artificially impacting
our moon?
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Brad Guth
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| User: "Igor" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
23 May 2006 11:54:59 AM |
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Brad Guth wrote:
And just how is that practically non-existant medium
going to slow anything down?
At 30,000 km/s it's certainly not helping, and for every atom there's
likely another 1e100 photons to deal with, and/or a sub-picoscopic
amount of dark matter that's likely asociated with each of those
quantum string like photons.
So, why don't you tell us something/anything specific?
BTW; It's perfectly OK if you simply don't know.
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Brad Guth
It's not that I don't know. It's that your question has no real answer
because you don't know what you're talking about. Something that
sparse is not going to supply a dissipative force even remotely close
to gravity, so your question has no meaning.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
25 May 2006 09:25:31 AM |
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It's not that I don't know. It's that your question has no real answer
because you don't know what you're talking about. Something that
sparse is not going to supply a dissipative force even remotely close
to gravity, so your question has no meaning.
Igor,
Your supposed all-knowing response has absolutely no meaning,
especially since I've given you nice folks the specifics that are
apparently too complicated for your pathetic naysay mindsets to deal
with. You're using the ruse of a minor issue in order to disallow the
remainder of what's doable. In other words, you are either another
status quo liar or merely so thoroughly snookered and summarily
dumbfounded that you can't hardly think for yourself. Which is it?
There is a lunar terminal velocity factor, just as there's a
terrestrial terminal velocity. You just refuse to admit as to how much
of an atmosphere our moon has to work with, especially effective since
there's so much less gravity per radius distances that are involved.
It's also rather interesting that whenever we village idiots should
take to discussing upon most anything associated with our moon, that
the essential science as to appreciating it's surrounding atmosphere
that's of such a sodium and most likely of many other element content,
of it's uneven gravity issues, of the LL-1 zone and of whatever's
associated with dropping items down onto that nasty sucker is either
skewed or entirely missing.
The math should actually be rather simple for that of a CRAY
supercomputer, and otherwise perfectly doable with most any good PC if
the wizards such as yourself should have known about such things, would
merely share and share alike. Obviously you're not about to share upon
anything that'll taint or otherwise infringe upon your pagan cash-cow
god.
Without so much as a basis of LL-1 science, as such I'm into thinking
there's no possible way of accomplishing any of those stealth
fly-by-rocket landings upon our physically dark and nasty moon,
especially if there's not so much as a speck of such hard-science
that's related to the simple task of having dropped a few items from
LL-1 directly onto the lunar deck, much less their having ignored
issues of mascons that'll demand the usage of those powerful momentum
reaction wheels plus having seriously computer modulated reaction
thrusters applied all the way that'll provide the only viable deorbit
and safe down-range capability that's still at best going to remain as
spendy and damn risky business, not to even mention the
gamma/hard-X-ray dosage that's potentially lethal within minutes unless
they can manage to park in a deep cave.
The direct alignment drop and final impact velocity as obtained from
the 58,000 km point of LL-1 is of absolutely critical science for any
such probe deployments onto/into our moon. Yet lo and behold there's
still nothing whatsoever stipulated about such hard-science nor even
sharing squat about the environment of the LL-1/(ME-L1) zone. It's as
though the entire worth of LL-1 doesn't matter any more so than do
mascons, nor that of whatever the gamma/hard-X-rays environment might
further suggest, and it's otherwise looking exactly as though we've
never set a moonsuit naked foot on our physically dark and nasty moon.
-
Brad Guth
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 09:19:57 PM |
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On 25 May 2006 07:25:31 -0700, "Brad Guth" <ieisbradguth@yahoo.com>
Gave us:
There is a lunar terminal velocity factor, just as there's a
terrestrial terminal velocity.
You're an idiot. Terminal velocity is SPECIFICALY related to
atmospheric drag. The moon has no atmosphere, and any stray
gatherings of gasses on its surface are only a few feet thick, and
would pose ZERO effect on a falling or thrusted body.
Get a clue, boy.
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 09:17:44 PM |
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On 25 May 2006 07:25:31 -0700, "Brad Guth" <ieisbradguth@yahoo.com>
Gave us:
Your supposed all-knowing response has absolutely no meaning,
In other words, you have no clue what the word dissipation means.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
31 May 2006 12:48:32 AM |
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Roy L. Fuchs wrote:
In other words, you have no clue what the word dissipation means.
In other words, you don't have a freaking clue as to how fast a falling
item of whatever size, volume or mass that's arriving from LL-1 that
directly impacts the lunar deck?
Why don't you just admit it's simply far too complicated for yourself
to figure out?
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Brad Guth
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| User: "" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
12 Jun 2006 12:30:49 PM |
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Brad Guth wrote:
Roy L. Fuchs wrote:
In other words, you have no clue what the word dissipation means.
In other words, you don't have a freaking clue as to how fast a falling
item of whatever size, volume or mass that's arriving from LL-1 that
directly impacts the lunar deck?
Why don't you just admit it's simply far too complicated for yourself
to figure out?
Did you even attempt to understand the links Sam gave you?
I mean, did you even click on them?
What Sam is trying to tell you is, the escape velocity of the moon
is too low for an object to achieve terminal velocity. So, any object
dropped onto the moon from whatever height will still be accelerating
when it hits the moon, and the force due to the local trace of
atmosphere will still be many orders of magnitude smaller than
the force of gravity.
In other words, in the usual meaning of the phrase "terminal velocity,"
the moon hasn't got one.
Now do try to learn something.
Socks
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
12 Jun 2006 04:11:54 PM |
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Did you even attempt to understand the links Sam gave you?
I mean, did you even click on them?
Yes I did. In fact, I'd been there and done that several times in the
last few years. As an outsider (NASA's official messenger from hell),
I'm also very much in use of the "Wikipedia" as a perfectly viable
resource of nifty graphics and of what little infomercial-science there
is about our moon, whereas sometimes the "Wikipedia" staff makes the
mistake of their actually sharing relevant hard-science than can be
replicated outside of the NASA/Apollo koran.
BTW; where's all the hard-science?
"Advanced Composition Explorer (ACE) is an Explorer satellite mission
to study matter in situ, comprising energetic particles from the solar
wind, the interplanetary medium, and other sources. It was launched in
1997 and is currently operating at the L1 Lagrange point."
http://www.srl.caltech.edu/ACE/
As per the usual infomercial need-to-know basis, and of nearly a decade
after the fact, this ongoing mission that's extremely well outfitted
yet having been kept as media nondisclosure as possible doesn't provide
squat worth of their hard-science data that should exist, and should
have been publicly accessible from the very get-go. Apparently our
cloak and dagger team ACE doesn't yet realize that our physically dark
and nasty moon is actually so freaking near by, and that all of their
L-1 data as to gravity the interactive issues is entirely missing in
action, more so than WMDs in Iraq.
ACE Level 2 (Verified) Data is performing exactly as though the moon
and of whatever's gravity related to all of the stationkeeping aspects
of coexisting within the L-1 zone doesn't exist. Why am I not the
least bit surprised, at least not the least bit any more so than as to
that supposedly stuck mirror onboard the ESA Venus EXPRESS?
In other words, in the usual meaning of the phrase "terminal velocity,"
the moon hasn't got one.
But you lie, and you're otherwise not willing to even include the
sodium element that's starting in from 9r and getting more populated as
arriving into that salty surface. Wherever there's such a thin
atmosphere of sodium, there has got to be more than a few other heavier
elements that'll obviously populate near the surface.
Obviously you're naysay fixated upon the terminal velocity aspects,
which by the way does happen to exist at greater velocities, of which
an item as having been deployed away from L-1 that is not intended to
orbit but free-fall directly onto the physically dark and nasty deck
below is going to achieve some measurable degree of interference (AKA
encountering a short period of friction) before impacting at a truly
horrific rate of velocity.
Obviously you folks have no honest intentions of your ever sharing as
to what that sort of friction might amount to, or of your ever sharing
as to what the final impact velocity might represent. Do you have
another good set of reasons for this form of taboo/nondisclosure worth
of your otherwise need-to-know infomercial basis of science that's
about our extremely nearby moon?
-
Brad Guth
puppet_sock@hotmail.com wrote:
Brad Guth wrote:
Roy L. Fuchs wrote:
In other words, you have no clue what the word dissipation means.
In other words, you don't have a freaking clue as to how fast a falling
item of whatever size, volume or mass that's arriving from LL-1 that
directly impacts the lunar deck?
Why don't you just admit it's simply far too complicated for yourself
to figure out?
Did you even attempt to understand the links Sam gave you?
I mean, did you even click on them?
What Sam is trying to tell you is, the escape velocity of the moon
is too low for an object to achieve terminal velocity. So, any object
dropped onto the moon from whatever height will still be accelerating
when it hits the moon, and the force due to the local trace of
atmosphere will still be many orders of magnitude smaller than
the force of gravity.
In other words, in the usual meaning of the phrase "terminal velocity,"
the moon hasn't got one.
Now do try to learn something.
Socks
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| User: "Sam Wormley" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
13 Jun 2006 07:04:53 PM |
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Brad Guth wrote:
Did you even attempt to understand the links Sam gave you?
I mean, did you even click on them?
Yes I did. In fact, I'd been there and done that several times in the
last few years. As an outsider (NASA's official messenger from hell),
I'm also very much in use of the "Wikipedia" as a perfectly viable
resource of nifty graphics and of what little infomercial-science there
is about our moon, whereas sometimes the "Wikipedia" staff makes the
mistake of their actually sharing relevant hard-science than can be
replicated outside of the NASA/Apollo koran.
NASA Science News for June 13, 2006
Last month, astronomers watched a meteoroid blast a hole in the lunar
Sea of Clouds. Their video of the event is a must-see.
FULL STORY at
http://science.nasa.gov/headlines/y2006/13jun_lunarsporadic.htm?list89139
Check out our RSS feed at http://science.nasa.gov/rss.xml !
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
14 Jun 2006 12:45:44 AM |
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On Wed, 14 Jun 2006 00:04:53 GMT, Sam Wormley <swormley1@mchsi.com>
Gave us:
Brad Guth wrote:
Did you even attempt to understand the links Sam gave you?
I mean, did you even click on them?
Yes I did. In fact, I'd been there and done that several times in the
last few years. As an outsider (NASA's official messenger from hell),
I'm also very much in use of the "Wikipedia" as a perfectly viable
resource of nifty graphics and of what little infomercial-science there
is about our moon, whereas sometimes the "Wikipedia" staff makes the
mistake of their actually sharing relevant hard-science than can be
replicated outside of the NASA/Apollo koran.
NASA Science News for June 13, 2006
Last month, astronomers watched a meteoroid blast a hole in the lunar
Sea of Clouds. Their video of the event is a must-see.
FULL STORY at
http://science.nasa.gov/headlines/y2006/13jun_lunarsporadic.htm?list89139
Check out our RSS feed at http://science.nasa.gov/rss.xml !
2.4 MB! Biggest animated gif I have ever seen! :-]
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
13 Jun 2006 09:28:25 PM |
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Sam Wormley,
Thanks for the science intelligence report. Very impressive show for a
wussy 25 cm rock at 38 km/s. Unfortunately, too many of our warm and
fuzzy Usenet naysayers and of their typical need-to-know and/or
nondisclosure mindset are intellectually constipated. Just like their
not having shared any of the nifty moon, L-1 station-keeping and Van
Allen belt science via ACE.
http://www.srl.caltech.edu/ACE/
http://science.nasa.gov/headlines/y2006/13jun_lunarsporadic.htm?list89139
BTW; there are a few frames from the NASA/Apollo archives of extremely
interesting images as obtained from orbit, as one of their missions
(manned or robotic) flew directly over an artificial impact, which I
believe at the time was arriving upon that physically dark and nasty
deck as somewhat closer to the 2.3 km/s mark, but otherwise likely due
to the size and considerable tonnage of the item having made an even
bigger than 14 meter crater.
-
Brad Guth
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
14 Jun 2006 12:51:28 AM |
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On 13 Jun 2006 19:28:25 -0700, "Brad Guth" <ieisbradguth@yahoo.com>
Gave us:
Thanks for the science intelligence report. Very impressive show for a
wussy 25 cm rock at 38 km/s.
As if you would know.
Unfortunately, too many of our warm and
fuzzy Usenet naysayers and of their typical need-to-know and/or
nondisclosure mindset are intellectually constipated.
You're an idiot.
Just like their
not having shared any of the nifty moon, L-1 station-keeping and Van
Allen belt science via ACE.
Hahahah!
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
14 Jun 2006 04:10:27 PM |
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Now you're into calling the regular laws of physics and of perfectly
good remote-science as having been accomplished by others, as being
"idiot" worthy?
What exactly is your intellectually constipated problem?
Roy L. Fuchs wrote:
On 13 Jun 2006 19:28:25 -0700, "Brad Guth" <ieisbradguth@yahoo.com>
Gave us:
Thanks for the science intelligence report. Very impressive show for a
wussy 25 cm rock at 38 km/s.
As if you would know.
Unfortunately, too many of our warm and
fuzzy Usenet naysayers and of their typical need-to-know and/or
nondisclosure mindset are intellectually constipated.
You're an idiot.
Just like their
not having shared any of the nifty moon, L-1 station-keeping and Van
Allen belt science via ACE.
Hahahah!
You mean to say that you actually know of nothing about mission ACE?
-
Brad Guth
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 09:16:48 PM |
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On 22 May 2006 13:07:28 -0700, "Brad Guth" <ieisbradguth@yahoo.com>
Gave us:
And just how is that practically non-existant medium
going to slow anything down?
At 30,000 km/s it's certainly not helping, and for every atom there's
likely another 1e100 photons to deal with, and/or a sub-picoscopic
amount of dark matter that's likely asociated with each of those
quantum string like photons.
So, why don't you tell us something/anything specific?
BTW; It's perfectly OK if you simply don't know.
-
Brad Guth
Idiot Photons are not going to slow the mass. Not one iota.
If they did, we would have laser weapons here on earth that push an
object on impact.
I think they have an excimer laser that can knock a dent into a
missile body from a distance.
We are talking about several orders of magnitude more photons,
however.
Your problem is that you actually think that you have some knowledge
of physics, when in fact, you have virtually none.
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 09:13:24 PM |
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On 22 May 2006 10:27:10 -0700, "Igor" <thoovler@excite.com> Gave us:
Brad Guth wrote:
If the local SM is supposedly 10~100 atoms/cm3, then obviously the SM
soup of the day gets itself a wee bit thicker as approaching the moon,
as it does approaching Earth (especially thick within the Van Allen
badlands). But then you wouldn't know anything about such matters
related to our moon.
-
Brad Guth
And just how is that practically non-existant medium going to slow
anything down?
On a scale from 1 to 100 with earth's "atmospheric drag" at 2000
feet up being 100, the moons "atmospheric drag" would still be less
than one, and only for an object dropped from a few feet. More like
0.0001
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 09:10:47 PM |
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On 22 May 2006 00:44:26 -0700, "Brad Guth" <ieisbradguth@yahoo.com>
Gave us:
Sam Wormley; Thanks Guth--another badge of honor! Too bad you
got nothing from my reply to you.
Too bad that you're so easily dumbfounded and otherwise without a clue
as to what I'd asked for.
Perhaps you missed out on basic reading-101, thus "Too bad you got
nothing" of worth to share in return. Apparently you're another one of
those all-knowing types that actually knows absolutely nothing about
our moon or of the space in between us and our moon, or of having to be
anywhere near that nasty sucker, much less with any regard to the task
of merely impacting our nearby moon.
If the local SM is supposedly 10~100 atoms/cm3, then obviously the SM
soup of the day gets itself a wee bit thicker as approaching the moon,
as it does approaching Earth (especially thick within the Van Allen
badlands). But then you wouldn't know anything about such matters
related to our moon.
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Brad Guth
You're an idiot. The only thing that is thick related to you is
your skull, and when you play with your poop. Outside observers can't
tell the difference between that and your brain tissue. Are your eyes
brown?
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 09:07:35 PM |
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On 21 May 2006 12:11:26 -0700, "Brad Guth" <ieisbradguth@yahoo.com>
Gave us:
Sam Wormley,
Thanks for the status quo of your mainstream infomercial-science, that
obviously hasn't a freaking clue.
You're an idiot. Without an atmosphere, ther is NO DRAG, and without
drag, there is no terminal velocity, you retarded fucktard.
With yet another straight butt-crack, you're actually going to suggest
there's no such terminal velocity to space travel?
You're an idiot. Your grasp of physics rests very near nil.
Why not give us Village idiots the vast wisdom and benefit of your very
best swag?
That would be Schwag, you stupid *****, and yes, you are the village
idiot.
What say if using a supper titanium plus ceramic or diamond tipped
alloy composite of a javelin, with an L/D ratio of perhaps 32:1, as
representing a 10t configured impactor that's being pumped along at
maximum retrograde velocity by a final LRn-->Rn-->ion thruster?
Is 10%'c' (30,000 km/s) physically doable?
You're an idiot.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
31 May 2006 12:53:20 AM |
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Roy L. Fuchs wrote:
That would be Schwag, you stupid *****, and yes, you are the village
idiot.
NO, it's swag, as in Scientific Wild ***** Guess. What planet are you
from?
-
Brad Guth
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| User: "Igor" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
21 May 2006 01:29:31 PM |
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Brad Guth wrote:
We seem to realize from actual experience based hard-science and via
the regular laws of physics that the minimum velocity (w/o retrothrust
nor drag coefficient) is roughly 2.4 km/s. However, what is the
absolute maximum obtainable terminal velocity of artificially impacting
our moon?
-
Brad Guth
How can a body without any atmosphere imply a terminal velocity?
I think you got your phenomena scrambled there, general.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
21 May 2006 02:00:25 PM |
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Igor,
Besides the SM population per cm3, there's also a 14,000 km gauntlet
worth of sodium, then eventually some worth of argon to encounter,
quite possibly a measurable dosage of O2 or O3, and then at least
having a thin touch of somewhat hefty radon by day. But then you still
haven't given us an answer as to the maximum possible impact velocity.
Why is that?
Say if using a supper titanium plus ceramic or diamond tipped alloy
composite of a javelin, with an L/D ratio of perhaps 32:1, as a 10t
configured impactor that's being pumped along at maximum retrograde
velocity by a final LRn-->Rn-->ion thruster?
-
Brad Guth
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| User: "Igor" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
21 May 2006 04:15:40 PM |
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Brad Guth wrote:
Igor,
Besides the SM population per cm3, there's also a 14,000 km gauntlet
worth of sodium, then eventually some worth of argon to encounter,
quite possibly a measurable dosage of O2 or O3, and then at least
having a thin touch of somewhat hefty radon by day. But then you still
haven't given us an answer as to the maximum possible impact velocity.
Why is that?
Say if using a supper titanium plus ceramic or diamond tipped alloy
composite of a javelin, with an L/D ratio of perhaps 32:1, as a 10t
configured impactor that's being pumped along at maximum retrograde
velocity by a final LRn-->Rn-->ion thruster?
-
Brad Guth
How many angels can dance on the head of the moon?
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
21 May 2006 04:41:13 PM |
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Igor; How many angels can dance on the head of the moon?
The black lunar sky is the limit. Thus you're saying that 10%'c' is
actually doable with the degree of applied rocket-science or at least
via the Rn-->ion thruster capability we have as of today?
-
Brad Guth
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| User: "Igor" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
22 May 2006 12:20:11 PM |
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Brad Guth wrote:
Igor; How many angels can dance on the head of the moon?
The black lunar sky is the limit. Thus you're saying that 10%'c' is
actually doable with the degree of applied rocket-science or at least
via the Rn-->ion thruster capability we have as of today?
-
Brad Guth
Maybe if you would learn what those terms you are using actually mean,
someone might be able to help you with a legitamate inquiry. Till
then, you're more or less just a clown.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 04:03:38 PM |
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Maybe if you would learn what those terms you are using actually mean,
someone might be able to help you with a legitamate inquiry. Till
then, you're more or less just a clown.
Igor,
Just because I'm not sufficiently smart enough and subsequently I don't
use all those right and fancy words in the exact proper context,
whereas now you're being a self certified bigot against us nice clowns?
What exactly do you have against clowns? especially against us
dyslexic clowns?
-
Brad Guth
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 09:03:45 PM |
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On 21 May 2006 14:41:13 -0700, "Brad Guth" <ieisbradguth@yahoo.com>
Gave us:
Igor; How many angels can dance on the head of the moon?
The black lunar sky is the limit. Thus you're saying that 10%'c' is
actually doable with the degree of applied rocket-science or at least
via the Rn-->ion thruster capability we have as of today?
-
Brad Guth
Ion thrusters take several hundreds of miles of "On Time" to build
speed up in the vehicle being thrusted.
It's like a person pushing a pool ball in one direction with a
fingertip at 3 grams force every two seconds. A 274 gram pool ball is
gonna take quite a while to gain speed from each of those pushes.
An ion thruster is like a one pound push constantly over several
thousand miles of space. It takes a LONG time to reach high
velocities.
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 08:59:17 PM |
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On 21 May 2006 12:00:25 -0700, "Brad Guth" <ieisbradguth@yahoo.com>
Gave us:
Igor,
Besides the SM population per cm3, there's also a 14,000 km gauntlet
worth of sodium, then eventually some worth of argon to encounter,
quite possibly a measurable dosage of O2 or O3, and then at least
having a thin touch of somewhat hefty radon by day. But then you still
haven't given us an answer as to the maximum possible impact velocity.
Why is that?
You're a goddamned idiot. Even if there was enough radon to make a
ten foot thick "radosphere" around the moon, it would STILL be too
small to affect a falling body. A spheroid requires a LARGE
atmosphere, and the body has to fall IN said atmosphere for a given
period of time before a terminal velocity can even be reached.
Say if using a supper titanium plus ceramic or diamond tipped alloy
composite of a javelin, with an L/D ratio of perhaps 32:1, as a 10t
configured impactor that's being pumped along at maximum retrograde
velocity by a final LRn-->Rn-->ion thruster?
You're an idiot.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
31 May 2006 12:40:16 AM |
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Still no numbers, no swag, nothing whatsoever. I wonder what the
problem is?
Sorry this topic is simply too complicated for your abilities to
manage.
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Brad Guth
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| User: "Roy L. Fuchs" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 08:54:50 PM |
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On 21 May 2006 11:29:31 -0700, "Igor" <thoovler@excite.com> Gave us:
Brad Guth wrote:
We seem to realize from actual experience based hard-science and via
the regular laws of physics that the minimum velocity (w/o retrothrust
nor drag coefficient) is roughly 2.4 km/s. However, what is the
absolute maximum obtainable terminal velocity of artificially impacting
our moon?
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Brad Guth
How can a body without any atmosphere imply a terminal velocity?
I think you got your phenomena scrambled there, general.
That's not all the idiot has that has been scrambled.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
11 Jun 2006 05:51:46 PM |
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Free fall simulators 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/physics/phynet/mechanics/Kinematics/solving_free-fall_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 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 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, nor 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
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| User: "" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
12 Jun 2006 08:15:49 AM |
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In article <1150066306.075241.128390@c74g2000cwc.googlegroups.com>, "Brad Guth" <ieisbradguth@yahoo.com> writes:
Free fall simulators 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.
Yes. Lunar gravity is right around 1/6 g.
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/physics/phynet/mechanics/Kinematics/solving_free-fall_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
You're off by a factor of 250, having multiplied by two where you
should have divided and having skipped over a factor of 1000 entirely.
d = 1/2 gt^2 = 1/2 * 1.62 * 1000 * 1000 = 810,000 meters = 810 km.
Sanity check: Final velocity is 1.62 km/sec. That means that during
the last two seconds of free fall about 3.24 km would have been covered.
That leaves very little distance to have been covered during the previous
998 seconds.
810 km is almost half of the moon's radius. At this distance, lunar
gravity is down to about half.
10,000 sec free fall as based upon using 1.6 m/s/s
[...]
And since gravity is down to half, and since your distance estimates
are so far off, the rest of your calculations just went up in smoke.
One can do a calculation along these lines. One way to proceed is
to compute the energy supplied going from 2 * radius to 1 * radius
(i.e. going from 1700 km up on down to the lunar surface) based on
an assumption of constant gravity of 1/6 g.
One can then do the same for 4 * radius down to 2 * radius.
And for 8 * radius down to 4 * radius.
If you make a table, you'll see that the energy contributed on each
iteration is half of the energy contributed on the previous iteration.
This is a "geometric series". The sum of an infinite geometric
series with ratio 2 is twice the value of the first term.
So no matter how high we start from, the impact energy per unit
mass is no more than 1700 km * 1.6m/sec^2 * 2
That comes to about 3.3 km/sec and is an over-estimate because we were
pretending that gravity was constant between the surface and 1700 km up.
If you want to integrate 1.6 m/sec^2 * r^2 / (r+h)^2 dh
for h [height] going from 0 to infinity, you can get a much better
estimate.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
12 Jun 2006 04:54:57 PM |
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bri,
Thanks so much for all the constructive feedback.
: 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
You're off by a factor of 250, having multiplied by two where you
should have divided and having skipped over a factor of 1000 entirely.
d = 1/2 gt^2 = 1/2 * 1.62 * 1000 * 1000 = 810,000 meters = 810 km.
In other words, the previous link (as listed below and last having been
updated as of March 8, 2006 by JL Stanbrough) that I'd used as math is
way more than 100% incorrect, yet stands as yet another educational
example of what American kids are basically learning as being physics
truth.
http://www.batesville.k12.in.us/physics/phynet/mechanics/Kinematics/solving_free-fall_problems.htm
If I understand correctly; in 1000 seconds of free-fall, from being at
zero m/s to start off with, as situated directly above the lunar deck
(aligned specifically with the gravity of mother Earth) is actually
going to become that impressive distance traveled of 810 km, and of
that 0.0 m/s deployed item having only obtained the final velocity of
1.62 km/s from such a horrific distance.
That's actually a very slow velocity of final arrival. I'm seriously
impressed with all the apparent terminal velocity that must be
involved. Even though the link of simplified math was skewed into the
nearest space-toilet, whereas actually the 1.62 km/s is only obtained
in the very final end of the very last traveled second, whereas each
and every previous second gets slower and slower.
However, from this positive and thus quite useful feedback, I'll have
to assume that you must also know exactly of what I'm driving at, and
therefore, might I further ask as to what's the final velocity of
impact from having been deployed at 1 m/s away from the realm of L-1
(59,562 km or try 60,000 km or even 36r being 60,830 km if it'll help)?
-
Brad Guth
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