| 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?
-
Brad Guth
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 11:07:54 AM |
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Not that yourself and the likes of "tadchem" would ever agree upon
constructively contributing anything that might taint your pagan gods:
I've found more of the same information that's stipulating the
supersonic drag coefficient of a sphere can supposedly reach 1.0 at
roughly M2. At higher velocity the trend is towards being less than
0.85, however that's somewhat skewed because it's related to an
artificial environment of what a science lab using a freon test medium
at essentially sealevel was obtaining.
Terminal settling velocity obviously has to equal the given mass.
However, at the desired target impact velocity of 1000 m/s and if we're
using a 10 meter drag-balloon that's only having to moderate on behalf
of a 10 kg/6 = 1.66 kg mass of the incoming javelin probe seems a bit
more than likely capable of easily accomplishing the task.
Unfortunately, since we're still not the least bit smart enough to have
established squat worth of anything at LL-1, there's no good and
surefire method of folks easily and extremely affordably obtaining
actual hard-science as related to impacting our moon. In fact, there's
a great deal we don't seem to know about our extremely nearby moon, or
even for that matter about raw ice coexisting in space is yet another
somewhat testy topic/author situation of being taboo/nondisclosure to
death and/or entirely banished like the lunar gamma and hard-X-ray
dosage that'll absolutely nail the likes of frail human DNA within
minutes.
Of course, I'm also wide open for considering that a well engineered
javelin probe as incorporating a wedge or cone shape portion could even
manage to survive 2.5 km/s if using the tens of meters deep moon-dust
as a viable form of lithobreaking prior to entering the primary basalt
and sodium composite of our once upon a time icy proto-moon. Actually,
if given a low angle of attack via orbital impact should in many places
accommodate hundreds of meters if not several kilometers worth of soft
and fluffy moon-dust as lithobreaking.
The option or plan-B of a km worth of trailing wire or fiber optic fed
transponder which might have survived on or near the surface could then
transfer it's science data to/from the mother ship, or to the LL-1
station-keeping platform. Of course, we'd then start to have those
actual forbidden hard-science numbers as to the local environment of
our physically dark and nasty moon as of yesterday, if not decades ago,
and for not even one cent on the dollar nor a mere fraction of the
local pollution impact of what our cloak and dagger NASA wizards are
planning upon doing sometime within their next spendy decade, if ever.
-
Brad Guth
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| User: "The Ghost In The Machine" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 01:00:53 AM |
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On Mon, 22 May 2006 13:39:42 -0700, Randy Poe wrote:
Brad Guth wrote:
Please define this term. It is not one I can find defined anywhere
else. "32" of what to 1 of what else?
Good grief, what's to define? other than you obviously have no
imagination whatsoever. OK, lets say it's of a 3.2 meter by 0.1 meter
Do you mean 3.2 meter length and 0.1 meter diameter?
Why not say so?
spear like item, as in meaning a 32:1 ratio of javelin that's usually
pointed on at least the front end.
In answer to your original question, the moon does not have
a terminal velocity because there is no atmospheric drag.
You seem to have subscribed to tjfrazer's physics. tj thinks
he could use a balloon on the moon.
One can easily use a balloon on the moon -- just not as a lifting device.
:-)
_Enterprise Stardust_, in particular, hypothesizes a "pneumo-tent"; the
idea doesn't look all that difficult to put into practice, at least for
short duration stays (balloons on Earth leak helium; presumably
pneumo-tents on the Moon would leak oxygen). The main problem is how to
implement a reasonable airlock (it's also worth noting that it takes
energy to create a vacuum while saving the air), and there might be issues
with micrometeorites and the variance in temperature as the Moon swings
from hot to cold, and proper handling of oxygen as evidenced by the Apollo
1 disaster.
Zero pressure on the outside, but there's also pressure imparted by the
elasticity of the rubber or mylar.
What exactly makes you think there is an atmosphere on the moon?
Actually, there is. It's about 3 * 10^-10 Pascal or about 3 * 10^-15 as
strong as Earth's, and comprised roughly of equal parts helium, neon,
hydrogen, and argon.
http://en.wikipedia.org/wiki/Moon
Not exactly enough to give much buoyancy, of course. Even on Earth, 1
cubic meter of air only weighs about 1.19 kg (depending on temperature and
barometric pressure); 1 cubic meter of helium weighs about 0.164 kg.
1 cubic km of Moon atmosphere might have a few hundred micrograms' mass.
- Randy
--
#191,
It's still legal to go .sigless.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 09:06:19 AM |
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One can easily use a balloon on the moon -- just not as a lifting device.
A fairly large diameter aerobreaking drag balloon for getting the given
javelin with it's internal science payload down to a modus final impact
velocity is certainly doable.
Obviously you're still into thinking of rad-hard human DNA being
involved. Forget it.
Zero pressure on the outside, but there's also pressure imparted by the
elasticity of the rubber or mylar.
Certain mylars would take but as little as 0.1 PSI for a passive
application, although as used for a drag coefficient balloon is likely
going to demand 1+ psi. So we're still talking of mostly volume and
not of much involving internal density or mass, in which case the
larger and more easily contained element of CO2 or Argon or perhaps
even Xenon could be utilized.
Since one of the applications involves the balloon deployment as for
the final landing phase of arriving straight in from the direction of
LL-1, as such the amount of exposure time that a micro meteorite could
hit is minimized to perhaps as little as 5 minutes prior to impact.
Even a orbital landing method shouldn't have to involve more than using
10 minutes worth of balloon drag deployment prior to impact.
Actually, there is. It's about 3 * 10^-10 Pascal or about 3 * 10^-15 as
strong as Earth's, and comprised roughly of equal parts helium, neon,
hydrogen, and argon.
Actually, it's quite a bit greater, but since human DNA isn't involved
and the payload of a robust little javelin probe and of it's micro
circuitry of science instruments within are perfectly capable of being
less than 10 kg, whereas the size or volume of the aerobreaking balloon
in such a 1/6th gravity situation isn't going to be all that demanding
if we're going for a final impact velocity of 1000 m/s (even 1500 m/s
should be probe survivable).
Even if a km3 of Moon atmosphere should have but a few hundred
micrograms of mass (actually I'm thinking it's getting more likely of
several hundred milligrams/km3 by day, if not grams/km3 nearest the
surface where the likes of the near-vacuum boiled off Sodium plus local
Argon and that hefty element of Radon are available), whereas the
available drag coefficient will become more than sufficient and easily
regulated via inflation pressure. Perhaps as little as a 10 meter
balloon of fiber reinforced mylar as having a balloon net is all that's
necessary for accommodating the 10 kg javelin probe.
Of course this is making the javelin probe deployments cheaper than
NASA's clumping-moon-dirt of portland cement and cornmeal, and
certainly a whole lot more reliable without ever involving one strand
of frail human DNA.
-
Brad Guth
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| User: "Randy Poe" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 04:30:53 PM |
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Brad Guth wrote:
One can easily use a balloon on the moon -- just not as a lifting device.
A fairly large diameter aerobreaking drag balloon for getting the given
javelin with it's internal science payload down to a modus final impact
velocity is certainly doable.
"Aero" braking requires "aero" to brake in.
Actually, there is. It's about 3 * 10^-10 Pascal or about 3 * 10^-15 as
strong as Earth's, and comprised roughly of equal parts helium, neon,
hydrogen, and argon.
Actually, it's quite a bit greater,
Actually, it isn't. Now what makes you think it is?
- Randy
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| User: "The Ghost In The Machine" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 09:01:02 PM |
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On Sun, 28 May 2006 14:30:53 -0700, Randy Poe wrote:
Brad Guth wrote:
One can easily use a balloon on the moon -- just not as a lifting device.
A fairly large diameter aerobreaking drag balloon for getting the given
javelin with it's internal science payload down to a modus final impact
velocity is certainly doable.
"Aero" braking requires "aero" to brake in.
And there is plenty of "aero", although it should probably be called
"rego". The Moon is bombarded with micrometerorites, leaving behind a
regolith coating of about 20cm in depth minimum (it can be more than
100 cm in the highlands) that was easily compressed by, among other
things, astronaut's boots. The website describes it as similar to
Earthside talcum powder.
http://physics.fortlewis.edu/Astronomy/astronomy%20today/CHAISSON/AT308/HTML/AT30805.HTM
A 1 kg probe could stir up this dust, transferring momentum thereto ...
or, if one prefers, plowing through it.
The regolith has an approximate density of 2900 kg/m^3 to 3300 kg/m^3.
The main issue I see is to keep the craft from tumbling as it hits this
mixture; it would then skid, giving up its energy to the particles by
means of something that might look vaguely like a train's cowcatcher.
Once stopped, the probe can open its canister and start looking around.
This is of course quite different from Brad's balloons.
No doubt this stuff will make a mess of things without a good cleaning
program somewhere inside of a hypothetical Moonbase airlock. For
starters, it is probably easily charged...
Actually, there is. It's about 3 * 10^-10 Pascal or about 3 * 10^-15
as strong as Earth's, and comprised roughly of equal parts helium,
neon, hydrogen, and argon.
Actually, it's quite a bit greater,
Actually, it isn't. Now what makes you think it is?
Good question. I'll admit to wondering how easy it is to detect gaseous
radon on our Moon by using a spectroscopic telescope. However,
occultations should be easy to verify, and in any event the Apollo
astronauts hopefully were able to catch a sample of atmosphere that we
could analyze, along with their rocks.
- Randy
--
#191,
It's still legal to go .sigless.
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| User: "The Ghost In The Machine" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 04:00:05 PM |
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On Sun, 28 May 2006 07:06:19 -0700, Brad Guth wrote:
One can easily use a balloon on the moon -- just not as a lifting device.
A fairly large diameter aerobreaking drag balloon for getting the given
javelin with it's internal science payload down to a modus final impact
velocity is certainly doable.
Aerobraking? On the Moon? Maybe if the balloon were used as a thruster.
Obviously you're still into thinking of rad-hard human DNA being
involved. Forget it.
No, the general idea would be to deploy a payload. On Mars, at least,
there is some atmosphere. On the Moon there's a very tiny atmosphere
whose braking is insignificant.
Zero pressure on the outside, but there's also pressure imparted by the
elasticity of the rubber or mylar.
Certain mylars would take but as little as 0.1 PSI for a passive
application, although as used for a drag coefficient balloon is likely
going to demand 1+ psi. So we're still talking of mostly volume and
not of much involving internal density or mass, in which case the
larger and more easily contained element of CO2 or Argon or perhaps
even Xenon could be utilized.
Since one of the applications involves the balloon deployment as for
the final landing phase of arriving straight in from the direction of
LL-1, as such the amount of exposure time that a micro meteorite could
hit is minimized to perhaps as little as 5 minutes prior to impact.
Even a orbital landing method shouldn't have to involve more than using
10 minutes worth of balloon drag deployment prior to impact.
Actually, there is. It's about 3 * 10^-10 Pascal or about 3 * 10^-15 as
strong as Earth's, and comprised roughly of equal parts helium, neon,
hydrogen, and argon.
Actually, it's quite a bit greater,
It is? State pressure and composition, then. Unless you're talking about
something different from the Earth's Moon, in which case I'd have to
reestablish context.
Mars in particular has a pressure of 900 Pascal or 9 * 10^-3 bar (1 bar =
100000 Pascal = just shy of Earth sea level), mostly carbon dioxide.
Suitable for aerobraking but not much else.
Venus has 9.3 megaPascal or 93 bar, also mostly CO2.
but since human DNA isn't involved
and the payload of a robust little javelin probe and of it's micro
circuitry of science instruments within are perfectly capable of being
less than 10 kg, whereas the size or volume of the aerobreaking balloon
in such a 1/6th gravity situation isn't going to be all that demanding
if we're going for a final impact velocity of 1000 m/s (even 1500 m/s
should be probe survivable).
I'm not all that certain. In any event, the main issue is "g-force"
(acceleration). Many people are killed in crashes at the relatively
slow speed of 30 m/s (67.1 mph). Many people survive such crashes. The
main difference may very well be seat belts and air bags, plus automobile
design (which includes "crumple zones", among other things).
In a similar vein, probes might get "killed" in a low-speed crash if they
don't have enough cushioning and slam into something at a relatively
gentle 100 m/s, but without cushioning they may experience very high
forces, if only for an instant.
But that instant can be crippling.
Even if a km3 of Moon atmosphere should have but a few hundred
micrograms of mass (actually I'm thinking it's getting more likely of
several hundred milligrams/km3 by day, if not grams/km3 nearest the
surface where the likes of the near-vacuum boiled off Sodium plus local
Argon and that hefty element of Radon are available), whereas the
available drag coefficient will become more than sufficient and easily
regulated via inflation pressure. Perhaps as little as a 10 meter
balloon of fiber reinforced mylar as having a balloon net is all that's
necessary for accommodating the 10 kg javelin probe.
Aero-braking using the Moon is ridiculous, although bouncing on the moon
with large balloons may not be, if only because some of the momentum can
be "bled off" by transferring parts of it to the lunar dust.
It would certainly be interesting to try. One huge difference: on Mars
parachutes were part of the landing procedure; the Moon does not have that
option.
Of course this is making the javelin probe deployments cheaper than
NASA's clumping-moon-dirt of portland cement and cornmeal, and certainly
a whole lot more reliable without ever involving one strand of frail
human DNA.
-
Brad Guth
--
#191,
It's still legal to go .sigless.
.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 06:42:02 PM |
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No, the general idea would be to deploy a payload. On Mars, at least,
there is some atmosphere. On the Moon there's a very tiny atmosphere
whose braking is insignificant.
The Ghost In The Machine,
That's odd, as you can start off by at least counting upon all of those
sodium atoms, starting in at 9r being worth a wussy 5 atoms/cm3 and
obviously getting more populated as you're about to impact while
dragging along that 10 meter balloon, especially if you're on the sunny
side. Of course, wherever there's such tonnage of sodium in the
atmosphere, as such there's got to be a few other atmospheric elements
that simply haven't been accounted for.
Also, the last time I'd checked, Mars or Venus had absolutely nothing
to do with our physically DARK and GAMMA/HARD-X-RAY nasty moon, and
besides all of that, wherever there's such local and influx caused
radiation there's got to be a few atoms of those extremely hefty atoms
of radon plus a few other elements as having also been given birth by
the solar and cosmic influx, which includes those nearly countless
micro-meteorites vaporising lunar basalt that's offering nearly 50% O2.
It is? State pressure and composition, then. Unless you're talking about
something different from the Earth's Moon, in which case I'd have to
reestablish context.
Oddly, we still have no such direct hard-science about our moon. Why
exactly is that?
It's my best deductive interpretation of the best available remote and
thus soft-science about our moon, is that we'll likely have millions if
not hundreds of such millions of those atoms/cm3 to work with near the
surface, and at 1/6th gravity making those few atoms very impressive on
behalf of aerobreaking in the limited way that I'm suggesting of a
lunar mass of merely 1.67 kg arriving at 1000 m/s.
Obviously, we can see that you're going to keep this one out of context
and thus every bit as skewed as you can, because that's your job. Why
am I not surprised?
In a similar vein, probes might get "killed" in a low-speed crash if they
don't have enough cushioning and slam into something at a relatively
gentle 100 m/s, but without cushioning they may experience very high
forces, if only for an instant.
But that instant can be crippling.
And other than double duh, no kidding folks; your all-knowing
constructive topic contribution of the day is???????
Aero-braking using the Moon is ridiculous, although bouncing on the moon
with large balloons may not be, if only because some of the momentum can
be "bled off" by transferring parts of it to the lunar dust.
WMD in Iraq was "ridiculous". Wide spread collateral damage and the
carnage of mostly innocent Muslims was and still is "ridiculous". A
national debt of 10+ trillions and big-time counting of such debt
taking even more of our hard earned dollars is "ridiculous". A 104
acre spendy embassy within Iraq is being absolutely extremely
"ridiculous". The first half if not more of this javelin probe will
accomplish a far better job via lithobreaking and of the obvious
javelin probe deformation than via any wussy balloon bouncing. Or, do
you think the moon actually has but only that thin clumping layer of
portland cement and cornmeal as it's oddly nonreactive and zilch
electrostatic moon-dust that's on such an eroded guano island like moon
of ours, that's also being oddly xenon lamp spectrum illuminated
instead of raw unfiltered solar and cosmic illuminated?
It would certainly be interesting to try. One huge difference: on Mars
parachutes were part of the landing procedure; the Moon does not have that
option.
Perhaps if you could get yourself any more naysay/negative about this,
and/or further off-topic, as such we could use the antimatter core of
your personal intellectual black hole in order to accomplish this
otherwise SIMPLE TASK.
-
Brad Guth
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| User: "The Ghost In The Machine" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 09:00:19 PM |
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On Sun, 28 May 2006 16:42:02 -0700, Brad Guth wrote:
No, the general idea would be to deploy a payload. On Mars, at least,
there is some atmosphere. On the Moon there's a very tiny atmosphere
whose braking is insignificant.
The Ghost In The Machine,
That's odd, as you can start off by at least counting upon all of those
sodium atoms, starting in at 9r being worth a wussy 5 atoms/cm3 and
obviously getting more populated as you're about to impact while
dragging along that 10 meter balloon, especially if you're on the sunny
side. Of course, wherever there's such tonnage of sodium in the
atmosphere, as such there's got to be a few other atmospheric elements
that simply haven't been accounted for.
Also, the last time I'd checked, Mars or Venus had absolutely nothing
to do with our physically DARK and GAMMA/HARD-X-RAY nasty moon, and
besides all of that, wherever there's such local and influx caused
radiation there's got to be a few atoms of those extremely hefty atoms
of radon plus a few other elements as having also been given birth by
the solar and cosmic influx, which includes those nearly countless
micro-meteorites vaporising lunar basalt that's offering nearly 50% O2.
Ah, OK. So...this atmosphere, it was not detected by Luna-9 because they
weren't looking for it?
Would it be detectable by other means, such as absorption spectra? In
short, the Moon reflects light, but that light may be modified by the
stuff in the Moon's atmosphere; this could easily be picked up by any
telescope with a spectrometer, as the spectrum of radon is reasonably
well-known.
One can also use occultation -- the Moon passes in front of stars, and the
star's brightness can be monitored until it winks out.
It is? State pressure and composition, then. Unless you're talking about
something different from the Earth's Moon, in which case I'd have to
reestablish context.
Oddly, we still have no such direct hard-science about our moon. Why
exactly is that?
Because we haven't been up there! A permanent base, or at least a
specialized lunar probe designed to look for radon atoms and such, would
answer a lot of questions.
It's my best deductive interpretation of the best available remote and
thus soft-science about our moon, is that we'll likely have millions if
not hundreds of such millions of those atoms/cm3 to work with near the
surface,
Compared to about 2.469 * 10^19 atoms or molecules per cubic centimeter at
sea level on Earth -- most of them diatomic nitrogen.
and at 1/6th gravity making those few atoms very impressive on
behalf of aerobreaking in the limited way that I'm suggesting of a
lunar mass of merely 1.67 kg arriving at 1000 m/s.
Obviously, we can see that you're going to keep this one out of context
and thus every bit as skewed as you can, because that's your job. Why
am I not surprised?
Your ignorance is showing. :-)
In a similar vein, probes might get "killed" in a low-speed crash if they
don't have enough cushioning and slam into something at a relatively
gentle 100 m/s, but without cushioning they may experience very high
forces, if only for an instant.
But that instant can be crippling.
And other than double duh, no kidding folks; your all-knowing
constructive topic contribution of the day is???????
Aero-braking using the Moon is ridiculous, although bouncing on the moon
with large balloons may not be, if only because some of the momentum can
be "bled off" by transferring parts of it to the lunar dust.
WMD in Iraq was "ridiculous". Wide spread collateral damage and the
carnage of mostly innocent Muslims was and still is "ridiculous". A
national debt of 10+ trillions and big-time counting of such debt
taking even more of our hard earned dollars is "ridiculous". A 104
acre spendy embassy within Iraq is being absolutely extremely
"ridiculous". The first half if not more of this javelin probe will
accomplish a far better job via lithobreaking and of the obvious
javelin probe deformation than via any wussy balloon bouncing. Or, do
you think the moon actually has but only that thin clumping layer of
portland cement and cornmeal as it's oddly nonreactive and zilch
electrostatic moon-dust that's on such an eroded guano island like moon
of ours, that's also being oddly xenon lamp spectrum illuminated
instead of raw unfiltered solar and cosmic illuminated?
It would certainly be interesting to try. One huge difference: on Mars
parachutes were part of the landing procedure; the Moon does not have that
option.
Perhaps if you could get yourself any more naysay/negative about this,
and/or further off-topic, as such we could use the antimatter core of
your personal intellectual black hole in order to accomplish this
otherwise SIMPLE TASK.
So OK then. We use aerobraking on the Moon, presumably by ploughing
through the dust on the Moon's surface using several javelins protruding
through the probe's surface (the probe will tend to rotate so there's some
issues here).
Then what?
-
Brad Guth
--
#191,
It's still legal to go .sigless.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
29 May 2006 12:15:20 AM |
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Ah, OK. So...this atmosphere, it was not detected by Luna-9 because
they weren't looking for it?
The Ghost In The Machine,
What part of a perpetrated and otherwise orchestrated cold-war do you
NOT understand?
Would it be detectable by other means, such as absorption spectra? In
short, the Moon reflects light, but that light may be modified by the
stuff in the Moon's atmosphere; this could easily be picked up by any
telescope with a spectrometer, as the spectrum of radon is reasonably
well-known.
Sodium is picked up with optical instruments (far better yet from
space), including upon somewhat ordinary film, yet there wasn't so much
as a peep out of our NASA/Apollo rusemasters that said anything about
the 9r worth sodium, nor of the 900,000 km sodium cloud of a comet like
tail, much less about any salty moon. So, why don't you tell us
village idiots what the hell went so terribly wrong with just that
wussy element.
Because we haven't been up there! A permanent base, or at least a
specialized lunar probe designed to look for radon atoms and such,
would answer a lot of questions.
That's the ticket. All we'll need are a few of those moon-dirt-cheap
javelin or similar probes stuck into that physically dark and nasty
sucker, reporting their data back to their mothership or to the science
platform that's performing as their data transponder that's efficiently
station-keeping itself within the sufficiently nearby LL-1 zone.
Your ignorance is showing. :-)
Sorry about that (I've got more than my fair share if it should involve
returning the favor).
So OK then. We use aerobraking on the Moon, presumably by ploughing
through the dust on the Moon's surface using several javelins protruding
through the probe's surface (the probe will tend to rotate so there's some
issues here).
Inital spin deployments (say 100,000 rpm if need be) as somwhat 12
guage shot away (always two at a time in opposite directions) from
their mothership seems fairly doable, and especially if such were
dragging that big mylar balloon that could be interactively inflated so
as to suit whatever drag coefficient, along with a good tail end swivel
bearing for the tether line should keep that robust little javelin
probe pointed at the moon.
Then what?
Then hope for the best and suck up as much hard-science as per whatever
instruments had managed to survive their dusty and basalt penetrating
impact. A 10% survival ratio is obviously 10 out of 100, and that's
ten science probes upon and/or stuck into that nasty moon that are
still ticking, representing ten more viable sources of extracting
direct hard-science than we've ever had to work with. That really
can't be all that bad.
In somewhat mass production, we're talking of pretty damn cheap probes
with robust micro science instruments within (all of which having their
R&D phase as easily terrestrial test impacted in order to fully
simulate/proof-test their ability to survive lunar basalt impact), so
we can afford to toss out a good 100 of these little yet robust
javelins for a fraction the cost of yet another remote science
satellite, by which satellites seem to have their orbital limitations
as obviously due in part to the local atmospheric drag and all of those
pesky mascons, so that such satellite based instruments can't safely
cruise without momentum reaction wheels and main velocity sustaining
thrusters all that nearby without entirely losing it. I believe most
all of the really good stuff is below the 10 km mark, and whatever
radon shouldn't ever get much higher than 1 km, if ever that much
unless being solar wind blown. (by earthshine Rn becomes LRn and should
logically sink into that fluffy moon-dust of basalt and meteorite
debris, plus titanium, iron, sodium and all of it carbon/soot coated to
being nearly coal black and most likely extremely electrostatic to
boot)
-
Brad Guth
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| User: "tadchem" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
27 May 2006 03:42:30 AM |
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From a Google
http://www.google.com/search?num=100&hl=en&lr=&safe=off&as_qdr=all&q=terminal+velocity+calculator&btnG=Search
we learn that terminal velocity is a function of the weight, gas
density, frontal area, and the drag coefficient:
http://exploration.grc.nasa.gov/education/rocket/termvr.html
From another Google
http://www.google.com/search?num=100&hl=en&lr=&safe=off&as_qdr=all&q=gravitational+potential+energy&btnG=Search
we learn that the energy that can be given to an object by a simple
gravitational field is finite:
http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html
If *all* the potential energy got converted to kinetic energy by the
instant of impact, then the maximum velocity at impact could be
calculated from the kinetic energy and the mass.
(Of course, of one wishes to *add* to the kinetic energy by *adding
thrust* to the falling mass, that's another calculation - work = force
x distance. The difference is whether you want to consider that you
have 'dropped' something onto the moon or 'thrown' it to the moon. In
the latter case the only limit to the impact velocity is how hard you
can throw it - how much kinetic energy you can add with your
thrusters.)
From Wikipedia
http://en.wikipedia.org/wiki/Moon
we learn that the atmospheric pressure on the moon is about 3x10^(-13)
kPa versus earth's 101 kPa. Not only is this pressure less than
1/330,000,000,000,000th of earths atmospheric pressure, it is *so* low
that the mean free path of a gas molecule is larger than your object,
so the moon's atmosphere no longer acts like a 'viscous medium' but
more like a statistical swarm of nano-gnats to be swatted out of the
way. Rather than calculate "drag" we should calculate energy loss as
transfer of momentum to individual molecules in elastic collisions.
Have I given you enough here that you can do your own math?
Tom Davidson
Richmond, VA
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 11:00:11 AM |
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Not that yourself and the likes of Randy Poe would ever agree upon
constructively contributing anything that might taint your pagan gods:
I've found that the supersonic drag coefficient of a sphere can
supposedly reach 1.0 at roughly M2. At higher velocity the trend is
towards being less than 0.85, however that's somewhat skewed because
it's related to an artificial environment of what a science lab using a
freon test medium at essentially sealevel was obtaining.
Terminal settling velocity obviously has to equal the given mass.
However, at the desired target impact velocity of 1000 m/s and if we're
using a 10 meter drag-balloon that's only having to moderate on behalf
of a 10 kg/6 = 1.66 kg mass of the incoming javelin probe seems a bit
more than likely capable of easily accomplishing the task.
Unfortunately, since we still haven't established squat worth of
anything at LL-1, there's no good and surefire method of folks easily
and extremely affordably obtaining actual hard-science as related to
impacting our moon. In fact, there's a great deal we don't seem to
know about our extremely nearby moon, or even for that matter about raw
ice coexisting in space is yet another somewhat testy topic/author
situation of being taboo/nondisclosure to death and/or entirely
banished like the lunar gamma and hard-X-ray dosage that'll absolutely
nail the likes of frail human DNA within minutes.
Of course, I'm also open for considering that a well engineered javelin
probe as incorporating a wedge or cone shape portion could even manage
to survive 2.5 km/s if using the tens of meters deep moon-dust as a
viable form of lithobreaking prior to entering the primary basalt and
sodium composite of our once upon a time icy proto-moon. Actually if
given a low angle of attack via orbital impact should accommodate
hundreds of meters if not several kilometers worth of soft and fluffy
moon-dust lithobreaking.
The option or plan-B of a km worth of trailing wire or fiber optic fed
transponder which might have survived on or near the surface could then
transfer it's science data to/from the mother ship, or to the LL-1
station-keeping platform. Of course, we'd then start to have those
actual forbidden hard-science numbers as to the local environment of
our physically dark and nasty moon as of yesterday, if not decades ago,
and for not even one cent on the dollar nor a mere fraction of the
local pollution impact of what our cloak and dagger NASA wizards are
planning upon doing sometime within their next spendy decade if ever.
-
Brad Guth
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
28 May 2006 09:16:34 AM |
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tadchem,
Thanks much for sharing in what even a village idiot already knows.
Have I given you enough here that you can do your own math?
In other words, other than using the one and only infomercial-science
of our pagan NASA, you haven't a freaking honest clue, much less
willing to give this one your best swag. Why am I not surprised?
-
Brad Guth
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| User: "tadchem" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 04:04:34 AM |
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Brad Guth wrote:
tadchem,
Thanks much for sharing in what even a village idiot already knows.
Have I given you enough here that you can do your own math?
In other words, other than using the one and only infomercial-science
of our pagan NASA, you haven't a freaking honest clue, much less
willing to give this one your best swag. Why am I not surprised?
Clues I have aplenty regarding physics.
What I need are clues regarding exactly what you are asking for, and
why.
From what I have read, you seem to be a bit confused about whether you
are talking about free fall or forced motion, whether the fall is
through the interstellar medium (ISM) or withing the solar system
(which has a variable but much higher pressure), and whether you want
an honest answer or just an opportunity to abuse those who might know
something you don't know.
*plonk*
Tom Davidson
Richmond, VA
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
30 May 2006 01:57:32 PM |
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What I need are clues regarding exactly what you are asking for,
and why.
tadchem,
And why exactly at this point do you need to know why?
From what I have read, you seem to be a bit confused about whether
you are talking about free fall or forced motion, whether the fall
is through the interstellar medium (ISM) or withing the solar system
(which has a variable but much higher pressure), and whether you want
an honest answer or just an opportunity to abuse those who might know
something you don't know.
Confusion along with a lose cannon has been my middle name. Sorry
about that.
My mistake for talking about both free fall and forced motion within
the same context.
Let's just focus upon the "free fall" from LL-1 to the lunar deck
that's directly below.
Leaving LL-1 that's roughly 58,000 km above the lunar deck, say we
start this 10 kg deployment off at 1 m/s.
-
Brad Guth
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| User: "tadchem" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
02 Jun 2006 04:05:43 PM |
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Brad Guth wrote:
What I need are clues regarding exactly what you are asking for,
and why.
tadchem,
And why exactly at this point do you need to know why?
....because the reason *why* you think you need some information is
often essential in correcting mistakes in the questions asked in the
search for information.
It is quite common for people to ask questions and, upon receiving the
answers, discover that what they actually asked was not what they
really needed to ask, because the answers could not actually satisfy
their needs.
There is an old joke with the punch line "I know you believe you
understand what you think I said, but I am not sure you realize that
what I said is not what I meant."
If you can see the humor in that sentence, then you should be able to
understand why I asked for *why* you asked what you did.
<snip>
Let's just focus upon the "free fall" from LL-1 to the lunar deck
that's directly below.
Leaving LL-1 that's roughly 58,000 km above the lunar deck, say we
start this 10 kg deployment off at 1 m/s.
The answer is in the energy equation:
Energy Before = Energy After
let m(o) be the mass of the object, M(e) be the mass of the Earth, and
M(m) the mass of the moon. The Lagrange point L1
http://en.wikipedia.org/wiki/Lagrangian_point
is approximately f(i) = 61,500 km from the center of the moon (radius
1738.1 km)
http://en.wikipedia.org/wiki/Moon
[or about 59,762 km from the moon's surface]
and about R(i) = 384,400 - 61,500 km = 322,900 km from the center of
the earth.
Because bith gravitational fields affect the potential energy of the
object, the potential energy of the object m(0) at this initial point
is
U(i) = -G*m(o)*M(m)/r(i) - G*m(o)*M(e)/R(i)
http://en.wikipedia.org/wiki/Gravitation
assuming the object lands on the side of the moon nearest the earth,
the potential energy U(f) at this final point where R(f) = 384,400 -
1738.1 km and r(f) = 1738.1 km
U(f) = -G*m(o)*M(m)/r(f) - G*m(o)*M(e)/R(f)
The *change* in kinetic energy U(f) - U(i) appears as an change in the
kinetic energy such that the total energy is constant.
If the initial kinetic energy is the kinetic energy of m(o) at an
initial velocity v(i) = 1 m/sec
K(i) = (1/2)*m(o)*v(i)^2
conservation of energy requires
K(i) + U(i) = K(f) + U(f)
Since the kinetic energy K(f) at the impact point where the velocity is
v(f) is
K(f) = (1/2)*m(o)*v(f)^2
You now have enough information to calculate v(f) for any m(o) that is
small compared to M(m) and M(e).
If the object impacts the moon elsewhere than the point closest to the
earth, it well lose kinetic energy in climbing that much further out of
earth's gravity well. The formula you should get algebraically from
what I have told you will give you the maximum lunar impact velocity of
a body moving through L1 at the speed (1 m/sec) you have specified,
assuming no additional forces come into play.
HTH
Tom Davidson
Richmond, VA
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
03 Jun 2006 11:00:36 AM |
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Thanks for all of your mostly constructive feedback.
...because the reason *why* you think you need some information is
often essential in correcting mistakes in the questions asked in the
search for information.
OK, I'll buy into that one.
From a deployment at LL-1; I want to impact our moon with those little
10 kg javelin probes, some going in for the kill as deeply as possible,
while otherwise moderating upon the final velocity(Vf) of impact so as
to merely stick such javelin probes into the surface with their rear
end still exposed and having survived with it's internal science
instruments transmitting various data.
: Leaving LL-1 that's roughly 58,000 km (59,762 km as by your method of
: accounting) as above the lunar deck, say we start this 10 kg
deployment
: off at 1 m/s.
The answer is in the energy equation:
Energy Before = Energy After
There's the obvious problem, as for the meter by meter free-fall
aspects there's only the factors of gravity and of no other applied
energy. Not that applied energy couldn't become involved, but why are
you implying an application of such thrust energy, other than the
initial deployment effort of our having given that 10 kg mass a 1 m/s
start in the right direction (as directed essentially directly below,
as in headed towards the moon) from LL-1 that's roughly 58,000 km away
(59,762 km by way of your method of infomercial-science accounting).
assuming the object lands on the side of the moon nearest the earth,
What's not to assume? Unless otherwise directed, how the heck can it
possibly go any place other?
You now have enough information to calculate v(f) for any m(o) that is
small compared to M(m) and M(e).
In other words, you obviously don't have an honest clue as to this
complex V(f) of a free falling item that's arriving from LL-1, thus
having no possible idea as to the maximum potential of the applied
impact energy?
The formula you should get algebraically from what I have told you will
give you the maximum lunar impact velocity of a body moving through L1
at the speed (1 m/sec) you have specified, assuming no additional forces
come into play.
I believe that yourself as do others know damn good and well that I
don't have a CRAY supercomputer, as needed in order to timely calculate
the meter by meter extent of the net lunar gravity influence upon such
an application. If it's so gosh darn simple, then do you have the
answer to this V(f), or can this one be placed into the vast internet
parallel network of PCs melting down their CPUs.?
-
Brad Guth
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| User: "tadchem" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
07 Jun 2006 04:57:43 AM |
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<multiple snips>
Brad Guth wrote:
From a deployment at LL-1; I want to impact our moon with those little
10 kg javelin probes, some going in for the kill as deeply as possible,
Then thrust them!
while otherwise moderating upon the final velocity(Vf) of impact so as
to merely stick such javelin probes into the surface with their rear
end still exposed and having survived with it's internal science
instruments transmitting various data.
That will depends sensitively upon the actual material properties of
both the probe and the impact site.
The answer is in the energy equation:
Energy Before = Energy After
There's the obvious problem, as for the meter by meter free-fall
aspects there's only the factors of gravity and of no other applied
energy. Not that applied energy couldn't become involved, but why are
you implying an application of such thrust energy, other than the
initial deployment effort of our having given that 10 kg mass a 1 m/s
start in the right direction (as directed essentially directly below,
as in headed towards the moon) from LL-1 that's roughly 58,000 km away
(59,762 km by way of your method of infomercial-science accounting).
Given the absence of non-conservative forces such as friction, the
impact velocity will be exactly the same as the launch velocity *from*
the moon's surface would have to be to get your 'javelin' to L1 with a
residual velocity of 1 m/sec - that is the escape velocity of 2400
m/sec.
assuming the object lands on the side of the moon nearest the earth,
What's not to assume? Unless otherwise directed, how the heck can it
possibly go any place other?
That would be like aiming a rifle bullet at a moving dime at a fairly
great distance. The moon is only about 0.50 degrees wide as a target.
If you are off by even a little bit, the javelin could (A) strike
off-center and at an angle, (B) curve around the moon and hit the far
side, (C) orbit the moon once and come back at you, (D) stray off into
deep space completely, (E) go into an erratic orbit around the Lagrange
point, (F) head towards earth and repeat either (A), (B), or (C) with
the earth.
Even NASA allowed for mid-course navigational corrections in all the
lunar shots.
I believe that yourself as do others know damn good and well that I
don't have a CRAY supercomputer, as needed in order to timely calculate
the meter by meter extent of the net lunar gravity influence upon such
an application. If it's so gosh darn simple, then do you have the
answer to this V(f), or can this one be placed into the vast internet
parallel network of PCs melting down their CPUs.?
One of the problems inherent in advanced technology is that people tend
to rely upon their machines to do work by brute force that they
themselves could do much more easily with a little ingenuity. An
application of mathematical knowledge and celestial mechanics, peppered
with real-world engineering experience, tells us your javelins will
need mid-course corrections and will otherwise impact at about 2400
m/sec. It isn't a problem of mapping the bilobal gravitational field
of the earth-moon system and iteratively solving the resultant
differential equation.
It reminds me of the way the Inca used to cut boulders in ytable-sized
chunks with mirr-smooth surfaces and without leaving tool marks. They
used wood, leather, stone and *wetware.*
Of course, if you *insist* on doing it that way, no one will try to
stop you. I *can* tell you in advance that your impact velocity will be
2400 m/sec.
Tom Davidson
Richmond, VA
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
07 Jun 2006 09:35:48 AM |
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Given the absence of non-conservative forces such as friction, the
impact velocity will be exactly the same as the launch velocity *from*
the moon's surface would have to be to get your 'javelin' to L1 with a
residual velocity of 1 m/sec - that is the escape velocity of 2400
m/sec.
Tom Davidson,
Obviously you're now into pretending as to not having a clue, as
otherwise only a truly snookered fool that's otherwise dumbfounded
would ever take such pathetic words as being the truth. Obviously a
free-fall drop that's deployed away from LL-1, especially if initiated
a just 1 m/s, is going to nearly dead-center encounter the moon that's
directly below at one hell of a lot greater velocity than 2.4 km/s.
The rest of your contribution is simply MOS incest cultivated
infomercial-science, and otherwise of those NASA/Apollo conditional
laws of physics as having taken over your boxed brain. In other words,
another nasty butt-load of disinformation and lies. Such as
represented by the following crapolla of your's.
Of course, if you *insist* on doing it that way, no one will try to
stop you. I *can* tell you in advance that your impact velocity will
be 2400 m/sec.
The 4.6264 m/s of surface rotation velocity as representing the final
orbital velocity of probe impact would otherwise be essentially
nonorbiting, therefore arriving as the previously directed impact as
having been deployed away from LL-1, of which is never going to be as
slight or otherwise moderated to merely 2.4 km/s, which only proves
that you're just as phony and as baloney worthy as all the rest of the
gang.
If I were actually wrong and somehow you were correct, there'd be an
internet calculator as to the free-fall droping of whatever onto a
given orb without atmosphere. Obviously there's no such calculator,
and therefore easily proving the "Tom Davidson" lasest infomercial was
quite intentional.
Sorry to learn that you're one of them.
-
Brad Guth
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| User: "tadchem" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
07 Jun 2006 04:52:29 PM |
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<snipping gratuitous ad hominem attacks>
Brad Guth wrote:
Obviously a
free-fall drop that's deployed away from LL-1, especially if initiated
a just 1 m/s, is going to nearly dead-center encounter the moon that's
directly below at one hell of a lot greater velocity than 2.4 km/s.
Obviously? It was "obvious" to Aristotle that heavy objects should fall
faster than lighter ones.
Can you show your math on this? Exactly *HOW* fast would it be going?
What value are you using for the moon's gravitational acceleration?
How are you accounting for variation with distance?
The energy method works accurately and simply, as stated before. The
potential energy equations are found here:
http://en.wikipedia.org/wiki/Gravitational_potential_energy
and they are to be used as indicated in my earlier post.
If you want to do it the hard way, remember that the force obeys an
inverse-square law:
http://en.wikipedia.org/wiki/Gravitational_field
so that at L1 the moon's gravity will be considerably weaker than at
the surface. You can get the change in energy by integration of the
force with respect to distance:
delta K =3D Integral[ G*m1*m2/r^2 (dot) dr] from L1 to the surface of the
moon
The 4.6264 m/s of surface rotation velocity as representing the final
orbital velocity of probe impact would otherwise be essentially
nonorbiting, therefore arriving as the previously directed impact as
having been deployed away from LL-1, of which is never going to be as
slight or otherwise moderated to merely 2.4 km/s, ...
As it is all gibberish, your statement proves nothing. You really
should try to attach your adjectives to nouns so your stupid readers
can tell WTF a "surface rotation velocity" is all about (and why it is
exactly 4.6264 m/sec), why your impacting probe has a "final orbital
velocity", and why the impact is "directed."
There is a lovely graph of the gravitational potential of the
earth-moon system, and the location of L1 in this field to be found
here:
http://en.wikipedia.org/wiki/Lagrangian_point
The potential around L1 is a saddle-point, going uphill as you move
away from L1 in the plane perpendicular to the line between the centers
of the earth and the moon, and going downhill as you move parallel to
said line, in either direction - much like a mountain pass.
If you read a little of this article you will be introduced to the
fascinating concepts of "Halo orbits" and "Lissajous orbits."
If I were actually wrong and somehow you were correct, there'd be an
internet calculator as to the free-fall droping of whatever onto a
given orb without atmosphere.
You are committing a logical fallacy here - your conclusion does not
follow from the premises.
http://www.fallacyfiles.org/afthecon.html
Maybe nobody else with enough understanding of the problem to require
an answer finds that a calculator is necessary. It *IS* a very simple
equation, as such things go.
Obviously there's no such calculator,
and therefore easily proving the "Tom Davidson" lasest (?) infomercial was
quite intentional.
Use this one. (I have already done the integration since I doubt your
abilities to perform calculus.)
(delta E)/m1 =3D G*m2*[1/a - 1/b],
where
G =3D 6.6742=D710^-11 N-m2/kg2
m2 =3D 7.347673=D710^22 kg
a =3D 3476.2 km / 2 =3D 1738.1 km =3D 1,738,100 m
b =3D 384,400 km - 323,110 km =3D 61,290 km =3D 61,290,000 m (roughly - sin=
ce
the earth-moon distance varies, so the position of L1 does as well)
Sources: (I realize that there may be some dispute with the accuracy of
these numbers, but they are good enough to permit a ball-park
calculation to settle the dispute)
http://en.wikipedia.org/wiki/Moon
http://www.ucolick.org/~mountain/AAA/answers/moon/mo14.html
(delta E)/m1 =3D 6.6742=D710^-11 N-m2/kg2 * 7.347673=D710^22 kg *
[1/1,738,100 m - 2/ 61,290,000 m]
The 2 in the second ratio within the brackets accounts for the earth's
contribution to the gravitational potential at L1. I have ignored
earth's contribution to the potential at the moon's surface, as it
doesn't affect the results much, it is simpler, and I am in a bit of a
hurry.
(delta E)/m1 also =3D 2*v^2 so even an American high school student
should be able to get v from this.
I get 1631.4 m/sec. The 2.38 km/sec figure given at
http://en.wikipedia.org/wiki/Moon for the moon's escape velocity is the
upper limit I thought you were originally asking for (before you
mentioned "LL-1"). That would be the velocity of a javelin 'dropped'
from interplanetary space somewhere near the earth's orbit onto the
moon.
Tom Davidson
Richmond, VA
P=2ES. If you can control your temper and avoid using personal attacks, I
may continue to discuss this with you. If you continue to treat me
like my petulant teen-aged daughter does when she has PMS, I will not.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
07 Jun 2006 07:41:56 PM |
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tadchem (aka Tom Davidson),
I thought that had reasonably started off this phase of topic with the
notion or premise of simply releasing a 10 kg javelin probe away from
LL-1, at roughly or exactly 1 m/s, as being intentionally directed not
into accomplishing a lunar orbit but as thrown directly at the nearest
available center of the moon below. And you're saying that's way too
complicated to have understood.
The actual location of LL-1 is certainly worth knowing, but not hardly
worth arguing about unless you're trying to avoid the primary issue of
final impact velocity as based upon whatever's given and/or selected as
the LL-1 distance from the moon that you'd like to utilize. You pick
whatever's the distance and go from there.
As far as I can tell, there's absolutely nowhere except for that probe
heading directly for the moon, that's always aligned to the very same
exact spot or perhaps that of the most significant mascon below, as for
that free-falling javelin probe to go anywhere except directly towards
that very same if not at least the approximate center of the side of
the moon that's continually facing Earth, and thereby always
unavoidably going to be the same as facing LL-1, +/- perhaps within 0.1
degree. Thus there is no such 2.4 km/s involved unless you've intend
to force a given probe into a fully orbital dive. As for the otherwise
wussy 4.6264 m/s worth of surface rotation isn't hardly worth taking
into orbital account, especially since it remains essentially
synchronised with the orbital velocity of LL-1 that's offering 163.15
m/s of it's orbital velocity.
Please do feel free to explain if the above orbitals of LL-1 at 163 m/s
and of the lunar deck at 4.63 m/s are otherwise, as I believe it's a
relatively important and a very basic factor in dealing with our
physically dark and nasty moon, at least from the rather nifty
perspective of LL-1.
I appreciate your V(f), whereas "I get 1631.4 m/sec", which seems a wee
bit if not a whole lot too good to be true, meaning way too slight of a
final velocity to be true. So, for the moment I'll ponder that number
and do a little dyslexic rethinking on my own. Just don't expect my
next move to make any better sense than what I'd started off with.
"As it is all gibberish, your statement proves nothing." Not that I
was trying to prove anything, other than for coming up with the
free-fall V(f) of impacting our moon from the starting point of LL-1
and having that starting velocity of 1 m/s. Obviously you still
haven't a freaking clue, not even a SWAG worth of being even remotely
close to the target impact velocity.
Perhaps you'd merely forgoten that this distance from LL-1 to the lunar
deck of 161,290 -1,738 = 159,552 is in km, not meters, and certainly
I'd have to perceive as for much less being the case if "That would be
the velocity of a javelin 'dropped' from interplanetary space somewhere
near the earth's orbit onto the moon", which makes absolutely no final
velocity sense within the context of this topic/argument whatsoever.
Perhaps "you are committing a logical fallacy here - your conclusion
does not follow from the premises" of my original topic configuration,
that which apparently is simply too Klingon encrypted to understand.
Sorry about that.
-
Brad Guth
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
09 Jun 2006 11:55:41 AM |
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tadchem wrote:
Obviously? It was "obvious" to Aristotle that heavy objects should fall
faster than lighter ones.
I see where you're coming from and where you're going with this. Good
luck.
Unlike your born-again naysay self, I'm not the village idiot that's
looking for absolutes, though this one should be of nothing but chuck
full of easily obtained hard-science absolutes considering this all
pertains to our extremely nearby salty moon. And, I do believe the
you'll actually have to know about such things if to be landing
yourself safely upon that nasty orb of mascons, double IR plus a TBI
bath of gamma and hard-X-ray environment, along with surviving tens of
meters deep moon-dust that's rather a nasty dry quicksand soup that's
offering hardly any surface tension capacity, as well as having become
much like a solid dust bowl cash of a nasty Van Allen belt lethal
environment. Other than that, I guess we don't really need to know
about such things as dropping a javelin probe into our moon, much less
from LL-1.
Can you show your math on this? Exactly *HOW* fast would it be going?
What value are you using for the moon's gravitational acceleration?
How are you accounting for variation with distance?
Because the math is actually extremely complex or at least extremely
repetitious and most certainly not at all simplified as you'd
previously suggested, is why I'm still the village idiot that's asking
of help from the all-knowing wizards such as yourself, as to share and
share alike. Obviously that's becoming too much to ask for.
This computational task is a bit much for my dyslexic expertise and of
my wussy PC to work out the km by km rate of velocity as based upon the
available gravity(s) of primary influence, much less for taking on the
more accurate meter by meter worth of coming in for such an intensional
impact landing from LL-1, a drop distance of 59,562 km, that which I do
believe is going to be offering just about anything a whole lot greater
than of your absolutely wussy 1.6314 km/s.
Of course if your V(f) of 1.63 km/s as a direct drop from LL-1 were
true, as then what the heck do we need our NASA for, and especially why
bother with a spendy and energy consuming controlled deorbit and down
range effort that's having to deal with mascons and of the much nastier
2.4 km/s?
How's that for controlling my lose cannon temper?
-
Brad Guth
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
31 May 2006 01:09:39 AM |
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Apparently not one Usenet soul has any clue as to the maximum
obtainable velocity of a forced impact as related to hitting our moon.
In fact, there's not one Usenet soul that can manage tell us what even
a free fall from LL-1 might obtain in final impact velocity.
No wonder we haven't walked on that physically dark and nasty moon, as
for not a soul knows of anything that's mission critical.
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Brad Guth
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
09 Jun 2006 09:34:44 AM |
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As to terraforming the holy puck of whatever out of our salty moon
that's surrounded by such a thick sodium atmosphere, plus having loads
of much heavier elements near the surface, that obviously impose drag
or terminal velocity factors upon anything orbiting that physically
dark and nasty sucker:
However, with the usual gauntlet of topic-drift or topic evasion
tactics in full Usenet damage-control and/or wag-thy-dog mode, as per
Usenet disinformation usual it seems the hard-science of merely
dropping something of a science probe or impactor from a given distance
above our extremely nearby and still salty moon isn't available, at
least not any more so than our science lack of understanding the
survival of raw ice coexisting in nearby space is available. Instead
we have lots of the usual NASA and of their media hyped hocus-pocus
worth of conjectures as having been extracted from their spendy remote
soft-science efforts, as having since been moderated to death in order
to carefully suit the given follow-the-money agenda of their
faith-based infomercial or bust day.
http://news.bbc.co.uk/1/hi/sci/tech/4871934.stm
http://www.spacedaily.com/reports/X_Rays_Reveal_250000_Tonnes_Of_Water_Released_By_Deep_Impact.html
As reported by way of their obviously christian revised
infomercial-science, as contributed from UK/US scientists using our
spendy Swift telescope, whereas the Tempel-1 impactor of merely 370 km
is now via their religiously corrected NASA being reported as having
eventually released 250,000 tonnes worth of water from that otherwise
mostly dusty/pumice comet, therefore whatever other tonnage of physical
impact debris from the less than 10.2 km/s encounter is somewhat
missing in action. For some reason the actual hard-science pertaining
to whatever's of other than the release of water/ice isn't the least
bit important to their follow-the-money trail of whatever their
infomercial or bust God wants us heathens to hear about.
By way of many such instruments and of individuals as supposedly having
good expertise, At first impact it supposedly vaporised 4500~5000
tonnes of water, but surprisingly had also released even more tonnage
worth of dust. Yet other perfectly viable science had placed the
initial impact at taking out a firm crater mass worth of debris as
having ejected only 1000 tonnes, which seems a little wussy by
comparison of what had somehow leaked and/or emerged as amounting to
the 250,000 tonnes worth of just the raw element of water/ice ever
since that little impactor event, which might otherwise have to suggest
upon another good amount of primary and secondary tonnage that'll
pertain to whatever else got vaporised or having continued to melt and
subsequently vaporise/leak away.
That's actually suggesting quite an impressive impactor ratio of just
the water/ice element becoming worth 250,000t per 0.37t at 10.2 km/s =
675,675:1, suggesting that perhaps the grand total of everything
involved might be greater than 1e6:1, which could come in real handy
science for the task of terraforming our moon into having a bit more of
an atmosphere.
Supposedly Tempel-1 continually leaks at a rate of 16,000 tonnes of
water/ice per day, and to think that's merely 5.84e6t/year or 5.84e6 m3
worth per year and still oddly never manages to get itself any smaller
or by such an amount of having less physical mass per year. I guess
those conditional laws of physics, as they apply to such comets and
asteroids that'll manage to keep their faith-based infomercial-science
going strong will never fail to amaze us village idiots.
Now we have ESA's Don Quijote mission, with their spendy Hildago mother
ship and of it's Sancho companion impactor, which is obviously also
avoiding the usage of our nearby salty moon that was once upon a time
an icy proto-moon of an asteroid as initially coated with perhaps 262
km worth of salty ice.
I've previously suggested a very cost effective and relatively low-tech
notion of impacting our moon with large blocks or spheres of dry-ice
containing the likes of Ra and LRn as their core, or as for merely
containing salty ice, or perhaps as offering an impactor core of spent
nuclear fuel representing as good as anything since the moon is already
a gamma and hard-X-ray environment (a solid form of Van Allen zone) to
start off with. Unfortunately, at least thus far we've established
absolutely no such hard-science as to raw ice coexisting upon or even
within our moon, or even as having coexisted in nearby space, much less
after having impacted with such ice upon our physically dark and TBI
nasty moon, that which so badly needs to get terraformed, into at least
offering something of a local atmosphere made of heavy enough elements,
that which only the most robust of robotics would tend to appreciate.
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Brad Guth
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
21 May 2006 02:50:22 PM |
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Basically I'm suggesting that the local SM drag coefficient isn't all
that insignificant, nor is even the ISM drag coefficient or slug value
of far outer space travel per say isn't such a non-issue, especially if
we're trying to sustain or exceed 10%'c', somewhat especially testy if
there are nearby mascon issues to contend with.
The task of penetrating our moon with a robust javelin probe isn't
going to require even the escape velocity of 2.3657 km/s. Since our
local SM density isn't worth much unless we're talking massive solar
sail like area, however the lunar sodium atmosphere that's starting to
be of slim use at 14,000 km and otherwise as taken into account from as
far away as 900,000 km as being down-solar-wind is certainly offering
an alternative method of moderating the final approach velocity of such
a JAVELIN probe that's intended to survive it's impact.
If the JAVELIN probe/impactor were configured as a rear-ender form of
intended arrival, plus having utilized the available mass/m3 worth of
the vast trail of the solar wind driven sodium cloud to boot, whereas
such the final impact velocity isn't going to be all that horrific,
especially if involving a very large (spin deployed) drag chute.
However, this still doesn't get us to appreciating the original
question, of what is the actual maximum possible impactor velocity if
we were intentionally going in for the kill, especially after having
taken the long way around plus having been LRn-->Rn-->ion thrusted in
order to nail that dark and nasty sucker at the utmost retrograde (head
on) impact velocity.
Is 10%'c' doable?
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Brad Guth
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| User: "Sam Wormley" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
21 May 2006 01:41:51 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?
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Brad Guth
Terminal Velocity
http://scienceworld.wolfram.com/physics/TerminalVelocity.html
Escape Velocity
http://scienceworld.wolfram.com/physics/EscapeVelocity.html
Escape Velocity (Moon) = 2.3657 km/s
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
21 May 2006 02:11:26 PM |
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Sam Wormley,
Thanks for the status quo of your mainstream infomercial-science, that
obviously hasn't a freaking clue.
With yet another straight butt-crack, you're actually going to suggest
there's no such terminal velocity to space travel?
Why not give us Village idiots the vast wisdom and benefit of your very
best swag?
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?
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Brad Guth
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| User: "Sam Wormley" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
21 May 2006 05:59:43 PM |
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Brad Guth wrote:
Sam Wormley,
Thanks for the status quo of your mainstream infomercial-science, that
obviously hasn't a freaking clue.
With yet another straight butt-crack, you're actually going to suggest
there's no such terminal velocity to space travel?
Why not give us Village idiots the vast wisdom and benefit of your very
best swag?
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?
-
Brad Guth
Thanks Guth--another badge of honor! Too bad you got nothing from
my reply to you.
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
22 May 2006 02:44:26 AM |
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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
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| User: "Igor" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
22 May 2006 12:27:10 PM |
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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.
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Brad Guth
And just how is that practically non-existant medium going to slow
anything down?
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| User: "Brad Guth" |
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| Title: Re: Terminal Velocity of Impacting our Moon |
22 May 2006 03:07:28 PM |
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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
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