Molten Blobs Create Moon Flashes - Science News

[Sam Wormley]

http://www.sciencenews.org/view/generic/id/337908/title/Molten_blobs_create_moon_flashes

> Meteorites colliding with the moon sometimes set off tiny lights dancing across its surface. Now scientists think they know what powers these lunar lightbulbs, in the absence of any atmosphere that would otherwise set incoming meteors ablaze: The flashes result from superhot material kicked up by the tiny objects striking the moon’s surface.

[Brad Guth]

On Jan 27, 9:29 pm, Sam Wormley <sworml...@gmail.com> wrote:
> http://www.sciencenews.org/view/generic/id/337908/title/Molten_blobs_...

>
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> > Meteorites colliding with the moon sometimes set off tiny lights dancing across its surface. Now scientists think they know what powers these lunar lightbulbs, in the absence of any atmosphere that would otherwise set incoming meteors ablaze: The flashes result from superhot material kicked up by the tiny objects striking the moon’s surface.

1.5 kg impactors at 72 km/sec should be fairly visible, especially on
the nighttime surface of that otherwise physically dark moon.

I wonder how much solar radiation is currently hitting our moon.
Should be quite lethal.

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[Sam Wormley]

On 1/28/12 9:11 PM, Brad Guth wrote:
> I wonder how much solar radiation is currently hitting our moon.
> Should be quite lethal.

Not good for children and other living things.

[Brad Guth]

Actually, those arriving mg meteor specks at 72 km/sec would also be
kind of moon-suit lethal as hell, not to mention any 3 g cm3. Our
best long-range S-band radar detections of items less than 10 cm
really isn't very good (1 cm3 items would be nearly invisible). So,
how do we give even an hour's notice (259 km or 161 miles away) of any
pending mg impactors, as well as their secondary impact shards?

Remember that we're talking about having to scan the whole freaking
sky plus a few degrees over the horizon in all 360 degree directions
in order to continually update those potentially exposed to anything
of a mg or larger and 100 m/s (224 mph) or faster that’ll be in need
of continuous M or W-Band radar detection.

[Chris L Peterson]

On Sun, 29 Jan 2012 06:03:02 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>Actually, those arriving mg meteor specks at 72 km/sec would also be
>kind of moon-suit lethal as hell, not to mention any 3 g cm3.

They won't be detected in advance. But the reality is that the actual
density of such particles is very low, so statistically there is only
a very small chance of an astronaut being hit. It is always going to
be viewed as an unavoidable risk- small, but anybody can have a bad
day. The Moon isn't a "safe" place.

[Davoud]

Chris L Peterson:

> ...anybody can have a bad day. The Moon isn't a "safe" place.

Good reason to tie Newt to the roof of Mitt's car and launch the pair
of them to the Moon.

--
I agree with almost everything that you have said and almost everything that
you will say in your entire life.

usenet *at* davidillig dawt cawm

[Brad Guth]

On Jan 29, 6:43 am, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Sun, 29 Jan 2012 06:03:02 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >Actually, those arriving mg meteor specks at 72 km/sec would also be
> >kind of moon-suit lethal as hell, not to mention any 3 g cm3.
>
> They won't be detected in advance. But the reality is that the actual
> density of such particles is very low, so statistically there is only
> a very small chance of an astronaut being hit. It is always going to
> be viewed as an unavoidable risk- small, but anybody can have a bad
> day. The Moon isn't a "safe" place.

The average meteorite density is not low, unless you're including
carbon buckyballs and/or crystals of ice or even dry-ice.

[Chris L Peterson]

On Sun, 29 Jan 2012 11:05:29 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>The average meteorite density is not low, unless you're including
>carbon buckyballs and/or crystals of ice or even dry-ice.

The micrometeorite flux is sufficiently low that an object with the
cross section of an astronaut on the Moon is very unlikely to be hit
during any typical excursion.

[Brad Guth]

On Jan 29, 11:34 am, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Sun, 29 Jan 2012 11:05:29 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >The average meteorite item density is not low, unless you're including

> >carbon buckyballs and/or crystals of ice or even dry-ice.
>
> The micrometeorite flux is sufficiently low that an object with the
> cross section of an astronaut on the Moon is very unlikely to be hit
> during any typical excursion.

So, as long as it's not your moon-suit butt on the line, the risk and
or expendable nature of any given astronaut is a non-issue. That's
good to know, because their body recovery would likely cost us at
least tens of billions.

Btw, I've been told the moon has gravity, at least considerably more
so than the shuttle or ISS, and perhaps that little extra gravity
might tend to attract items that would otherwise effortlessly fly past
without ever encountering the moon or even grazing near its physically
dark and naked surface at worse than lethal velocities.

Any 3 g/cm3 density item of even a one mm size is capable of being
lethal as all get out.

On any given hour, how many million of those one mm and larger items
impact the moon at 2.4 or greater km/sec?

[Chris L Peterson]

On Sun, 29 Jan 2012 12:46:36 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>> The micrometeorite flux is sufficiently low that an object with the
>> cross section of an astronaut on the Moon is very unlikely to be hit
>> during any typical excursion.
>
>So, as long as it's not your moon-suit butt on the line, the risk and
>or expendable nature of any given astronaut is a non-issue. That's
>good to know, because their body recovery would likely cost us at
>least tens of billions.

As I said, the Moon (and space in general) is dangerous. But being hit
by a micrometeoroid is probably one of the lesser dangers. There's not
much that can be done to mitigate the risk for somebody moving about
on the surface. Living quarters and expensive assets can, of course,
be shielded to a reasonable degree.

>Btw, I've been told the moon has gravity, at least considerably more
>so than the shuttle or ISS, and perhaps that little extra gravity
>might tend to attract items that would otherwise effortlessly fly past
>without ever encountering the moon or even grazing near its physically
>dark and naked surface at worse than lethal velocities.

This sort of gravitational focusing is not very significant. The
meteoroid flux on the surface isn't much different than it is in
space.

>Any 3 g/cm3 density item of even a one mm size is capable of being
>lethal as all get out.
>
>On any given hour, how many million of those one mm and larger items
>impact the moon at 2.4 or greater km/sec?

It doesn't matter how many hit the Moon, but how many are (a) large
enough to puncture a space suit, and (b) how many of those hit any
given area of concern- say a square meter. This ends up being a small
number- although not one that can simply be ignored.

[Brad Guth]

On Jan 29, 1:12 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Sun, 29 Jan 2012 12:46:36 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >> The micrometeorite flux is sufficiently low that an object with the
> >> cross section of an astronaut on the Moon is very unlikely to be hit
> >> during any typical excursion.
>
> >So, as long as it's not your moon-suit butt on the line, the risk and
> >or expendable nature of any given astronaut is a non-issue.  That's
> >good to know, because their body recovery would likely cost us at
> >least tens of billions.
>
> As I said, the Moon (and space in general) is dangerous. But being hit
> by a micrometeoroid is probably one of the lesser dangers. There's not
> much that can be done to mitigate the risk for somebody moving about
> on the surface. Living quarters and expensive assets can, of course,
> be shielded to a reasonable degree.

True, such as once those TBMs are working sufficiently underground
would become relatively safe.

>
> >Btw, I've been told the moon has gravity, at least considerably more
> >so than the shuttle or ISS, and perhaps that little extra gravity
> >might tend to attract items that would otherwise effortlessly fly past
> >without ever encountering the moon or even grazing near its physically
> >dark and naked surface at worse than lethal velocities.
>
> This sort of gravitational focusing is not very significant. The
> meteoroid flux on the surface isn't much different than it is in
> space.

So, you really don't have any objective science, such as from any
science instruments on the surface of our moon.

>
> >Any 3 g/cm3 density item of even a one mm size is capable of being
> >lethal as all get out.
>
> >On any given hour, how many million of those one mm and larger items
> >impact the moon at 2.4 or greater km/sec?
>
> It doesn't matter how many hit the Moon, but how many are (a) large
> enough to puncture a space suit, and (b) how many of those hit any
> given area of concern- say a square meter. This ends up being a small
> number- although not one that can simply be ignored.

Perhaps each and every individual impact is going to generate at least
on average a million secondary shards moving at lethal velocity for
quite some radii. A 1 kg impactor arriving at 72 km/sec is likely
going to produce at the very least a 10 km radii of death, because
there's hardly any atmosphere to slow those secondary shards down, nor
enough gravity to otherwise pull them down, and there’s even a darn
good chance of some shards temporarily going into orbit.

[Chris L Peterson]

On Sun, 29 Jan 2012 13:39:42 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:


>> This sort of gravitational focusing is not very significant. The
>> meteoroid flux on the surface isn't much different than it is in
>> space.
>So, you really don't have any objective science, such as from any
>science instruments on the surface of our moon.

In fact, the micrometeoroid flux in the near-Earth environment (which
extends past the Moon) is pretty well understood- both observationally
and theoretically.

>Perhaps each and every individual impact is going to generate at least
>on average a million secondary shards moving at lethal velocity for
>quite some radii. A 1 kg impactor arriving at 72 km/sec is likely
>going to produce at the very least a 10 km radii of death, because
>there's hardly any atmosphere to slow those secondary shards down, nor
>enough gravity to otherwise pull them down, and there’s even a darn
>good chance of some shards temporarily going into orbit.

1 kg impactors are sufficiently rare that they can largely be ignored.
The risk of an impact by one is extremely small compared with other
risks. Micrometeorites are more of a concern, but they tend to behave
more like bullets- they cause punctures, but no shrapnel or damage
outside of their immediate impact zone- a few centimeters at the most.

[Brad Guth]

On Jan 29, 2:14 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Sun, 29 Jan 2012 13:39:42 -0800 (PST), Brad Guth
>

It kinda depends on what they run into, and at what angle.

If you shot a little thorium or titanium bullet at 2.5 km/sec (say
using a wee bit of Acetone Peroxide crystals that'll yield a 5+ km/s
rate of explosion), and besides being knocked on your moon-suit butt
by the terrific recoil, there's a good chance if you didn't move aside
that you'd get nailed by that same bullet after it having gone around
the moon.

[Chris L Peterson]

On Sun, 29 Jan 2012 15:57:42 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>If you shot a little thorium or titanium bullet at 2.5 km/sec (say
>using a wee bit of Acetone Peroxide crystals that'll yield a 5+ km/s
>rate of explosion), and besides being knocked on your moon-suit butt
>by the terrific recoil, there's a good chance if you didn't move aside
>that you'd get nailed by that same bullet after it having gone around
>the moon.

You might want to go back and work on the math of that one again...

[Peter Webb]

"Chris L Peterson" <c...@alumni.caltech.edu> wrote in message
news:do8ci7lcvnrq5cap0...@4ax.com...

Although it seems wrong to me ...

The escape velocity of the moon is 2.38 kms
(http://en.wikipedia.org/wiki/Moon#Gravity_and_magnetic_fields) which gives
it an orbital velocity of 2.38/1.414 = 1.7 kms/sec.

This not much more than a sniper rifle achieves (typically 1.2 kms/sec), and
is achieved by some larger high performance projectile weapons. I could
quite believe that if you replaced the lead/tungsten/whatever bullet in a
sniper rifle with something lighter it would achieve orbital velocity.
Indeed with a fast propellant and a lightweight bullet you could probably
achieve escape velocity with a hand held weapon, and shoot bullets into the
earth's atmosphere.

Like I said, my heart says no, but my head says yes. Would be happy to find
out my gut feeling was wrong.

[Brad Guth]

On Jan 29, 8:57 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Sun, 29 Jan 2012 15:57:42 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >If you shot a little thorium or titanium bullet at 2.5 km/sec (say
> >using a wee bit of Acetone Peroxide crystals that'll yield a 5+ km/s
> >rate of explosion), and besides being knocked on your moon-suit butt
> >by the terrific recoil, there's a good chance if you didn't move aside
> >that you'd get nailed by that same bullet after it having gone around
> >the moon.
>
> You might want to go back and work on the math of that one again...

2.5 km/sec is more than escape velocity, and acetone peroxide offers a
maximum of something like 5.3 km/sec. Assuming a pair of bullets
discharged at exactly 180 degrees from one another would nullify that
recoil, and get each bullet moving up to 2.5 km/sec (retrograde to
each-other).

It would be fun to create this kind of gun or cannon, and fire it from
the top of a crater rim or whatever peak elevation on the moon.

No doubt those terrific mascons would pull each bullet off course and
cause their early orbital demise, but it would still be interesting to
see just how far they each got.

[Chris L Peterson]

On Mon, 30 Jan 2012 16:52:32 +1100, "Peter Webb"
<r.peter...@gmail.com> wrote:

>Like I said, my heart says no, but my head says yes. Would be happy to find
>out my gut feeling was wrong.

What your head is missing, perhaps, is that the Moon is not smooth.
You can't turn a pure physics problem into a practical one as Brad is
trying to do. I doubt there is any place on the Moon where you could
shoot a bullet at orbital velocity and have it come around and hit you
on the back- whether it's made of thorium or titanium. And of course,
Brad isn't suggesting that the bullet be shot at orbital velocity, but
at escape velocity. That means the bullet is going to be much higher
when it comes around again.

[Chris L Peterson]

On Mon, 30 Jan 2012 16:52:32 +1100, "Peter Webb"
<r.peter...@gmail.com> wrote:

>The escape velocity of the moon is 2.38 kms
>(http://en.wikipedia.org/wiki/Moon#Gravity_and_magnetic_fields) which gives
>it an orbital velocity of 2.38/1.414 = 1.7 kms/sec.

Of course, that's for a circular orbit. Launch the projectile faster
and it will be in a more eccentric orbit, with an orbital velocity
that changes with position in the orbit.

There are projectile launchers on Earth capable of achieving local
escape velocity... although I don't think it is feasible in practice
to actually launch something that way because of atmospheric
resistance and heating.

[Brad Guth]

On Jan 30, 7:25 am, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 16:52:32 +1100, "Peter Webb"
>

Good freaking grief. Haven't you got anything better to do than topic/
author stalk and perpetual naysay everything. Is your real name
Sheldon Cooper?

How far in orbit would you get if we stuffed your "Big Bang Theory"
butt with a maximum load of acetone peroxide?

It seems any 5.3 km/second explosive force should do a rather damn
fine job of it, though perhaps Sheldon Cooper would still be in denial
even after reaching lunar orbit.

[Brad Guth]

On Jan 30, 8:37 am, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 16:52:32 +1100, "Peter Webb"
>

And launched over the horizon while on the naked surface of our
physically dark moon, exactly how much atmospheric friction and joules
worth of heating per orbit are we talking about? (by day a sodium
bullet would likely vaporize, but would an aluminum bullet melt?)

It seems those local mascons are going to be a much greater factor in
the demise of any such close to surface orbit.

[Chris L Peterson]

On Mon, 30 Jan 2012 11:46:45 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>It seems those local mascons are going to be a much greater factor in
>the demise of any such close to surface orbit.

In practice, it is the lack of a smooth surface, combined with
mascons, that would make it essentially impossible to shoot a
projectile all around the Moon at surface level and have it return to
the launch point.

[Brad Guth]

On Jan 30, 12:13 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 11:46:45 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >It seems those local mascons are going to be a much greater factor in
> >the demise of any such close to surface orbit.
>
> In practice, it is the lack of a smooth surface, combined with
> mascons, that would make it essentially impossible to shoot a
> projectile all around the Moon at surface level and have it return to
> the launch point.

I'd had said that you'd need to be standing on a tall crater rim or
some other hill/mountain top, but of course your pure negativity
filtered that out.

The point, is that meteors hitting that naked moon could easily
generate a lethal radii of secondary shards, some of which given
sufficient escape velocity.

Are you actually suggesting there is no significant tonnage of lunar
basalt rocks on Earth?

[Chris L Peterson]

On Mon, 30 Jan 2012 12:40:50 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>The point, is that meteors hitting that naked moon could easily
>generate a lethal radii of secondary shards, some of which given
>sufficient escape velocity.

Yes, but only large meteoroids. We were discussing micrometeorites-
the only objects that really pose any significant risk. Those just act
like little bullets. One could hit a meter from an astronaut, and he'd
be in no danger. Stuff large enough to fling material about is also
very rare.

Certainly, large meteoroids can knock material off the Moon... that's
why we find lunar meteorites on Earth.

>Are you actually suggesting there is no significant tonnage of lunar
>basalt rocks on Earth?

What is "significant tonnage"? Lunar meteorites are extremely rare.
The total recovered mass of lunars isn't even 100 kg. There must be a
few tons of unrecovered lunar meteorites. Personally, I wouldn't call
that "significant"; it represents a vanishingly small percentage of
the total basalt mass on Earth's surface.

[Brad Guth]

On Jan 30, 12:48 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 12:40:50 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >The point, is that meteors hitting that naked moon could easily
> >generate a lethal radii of secondary shards, some of which given
> >sufficient escape velocity.
>
> Yes, but only large meteoroids. We were discussing micrometeorites-
> the only objects that really pose any significant risk. Those just act
> like little bullets. One could hit a meter from an astronaut, and he'd
> be in no danger. Stuff large enough to fling material about is also
> very rare.
>
> Certainly, large meteoroids can knock material off the Moon... that's
> why we find lunar meteorites on Earth.
>
> >Are you actually suggesting there is no significant tonnage of lunar
> >basalt rocks on Earth?
>
> What is "significant tonnage"?

Several trillion tonnes, mostly from that 2500 km crater.

> Lunar meteorites are extremely rare.

No, they are not rare, unless you can explain the highly paramagnetic
basalt that can be found where there's no special kind of iron, nickel
or other special kinds of paramagnetic ores are located.

>
> The total recovered mass of lunars isn't even 100 kg. There must be a
> few tons of unrecovered lunar meteorites. Personally, I wouldn't call
> that "significant"; it represents a vanishingly small percentage of
> the total basalt mass on Earth's surface.

Most of those teratonnes of lunar bedrock basalt went into the oceans.

According to our NASA/Apollo era (that which you have to believe each
and every word of), there's hardly any significant layer or
collections of secondary crater debris on the moon, and what little
there is of it is highly reflective, totally inert, monochromatic
colorless, not hardly the least bit paramagnetic, as well as not at
all anticathode worthy and yet offers terrific surface tension to
boot.

[Chris L Peterson]

On Mon, 30 Jan 2012 13:10:21 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>> >Are you actually suggesting there is no significant tonnage of lunar
>> >basalt rocks on Earth?
>>
>> What is "significant tonnage"?
>Several trillion tonnes, mostly from that 2500 km crater.

I think not.

[Brad Guth]

On Jan 30, 1:35 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 13:10:21 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >> >Are you actually suggesting there is no significant tonnage of lunar
> >> >basalt rocks on Earth?
>
> >> What is "significant tonnage"?
> >Several trillion tonnes, mostly from that 2500 km crater.
>
> I think not.

There you go again, being so perfectly negative that you might as well
stop pretending and just be the intellectual black hole (aka Sheldon
Cooper) that you are.

Obviously you can't do the volumetric displacement math on behalf of
that 2500 km crater, much less adding up all those other sufficiently
big ones that should almost double that volume. So, where the hell
did all that enormous volume of lose basalt shards go, if not
extensively towards Earth?

Remember, that within just a few cm depth, those NASA/Apollo guys
struck fused bedrock, as in they couldn't seem to force those probes
for any further penetration (nothing the least bit lose under their
magic clumping moon dirt that was so monochromatic, highly reflective
and not hardly the least bit paramagnetic or much less anticathode or
otherwise metallicity worthy).

So, do tell us what sort of inert fluff-ball smacked into our moon and
made that 2500 km crater, which apparently gave off hardly any shards?
(at least most of which didn't seem to stick with the moon)

[Mike Collins]

So now you admit your errors and have finally decided that men landed on
the Moon.

[Brad Guth]

On Jan 30, 2:55 pm, Mike Collins <acridiniumes...@gmail.com> wrote:

According to you guys that have nothing truly objective to go by, and
where so much of those laws of physics don't seem to apply or add up,
and where as much science obfuscation gets applied in order to suit
whatever NASA/Apollo had to say. Sure thing, we've been there, but to
save your pathetic souls can't seen to figure out how the hell it was
accomplished, nor much less able to help anyone else accomplish it.

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[Brad Guth]

On Jan 30, 1:35 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 13:10:21 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >> >Are you actually suggesting there is no significant tonnage of lunar
> >> >basalt rocks on Earth?
>
> >> What is "significant tonnage"?
> >Several trillion tonnes, mostly from that 2500 km crater.
>
> I think not.

There you go again, being so perfectly negative that you might as well
stop pretending and just be the intellectual black hole (aka Sheldon
Cooper) that you are.

Obviously you can't do the volumetric displacement math on behalf of
that 2500 km crater, much less adding up all those other sufficiently
big ones that should almost double that volume. So, where the hell
did all that enormous volume of lose basalt shards go, if not
extensively towards Earth?

Remember, that within just a few cm depth, those NASA/Apollo guys
struck fused bedrock, as in they couldn't seem to force those probes
for any further penetration (nothing the least bit lose under their
magic clumping moon dirt that was so monochromatic, highly reflective
and not hardly the least bit paramagnetic or much less anticathode or
otherwise metallicity worthy).

So, do tell us what sort of inert fluff-ball smacked into our moon and
made that 2500 km crater, which apparently gave off hardly any shards?
(at least most of which didn't seem to stick with the moon)

Each month the atmosphere of Earth deflects and/or vaporizes a fairly
large number of meteors and smallish asteroids (roughly 100 per day),
not to mention fending off the hundred tonnes of smaller stuff per
day, that which our naked moon doesn’t have the same built-in or
automatic defensive option. There’s well over a hundred fireballs
spotted per day, and that obviously doesn’t include the other
thousands per day that go unnoticed.

http://www.astronomycafe.net/qadir/q896.html
“In a recent article in the journal Nature, March 28, 1996 vol. 380,
page 323, Dr.s A.D. Taylor, W. J. Baggaley and D. I. Street at the
University of Adelaide in Australia discuss the results of their 1
year radar monitoring of incoming meteors. When meteorites slam into
the atmosphere, they produce ionization in the atmosphere. Radar echos
from this momentary ionization allow the velocity, altitude and
distance to be determined if you have two or more such installations
for triangulation. The AMOR radar in New Zealand was used for a year
in this fashion to detect 350,000 faint echos from very small
meteorites with sizes between 10 - 100 microns. This works out to
nearly 1000 every day, just from this site alone! Over 1508 of these
meteorites ( 0.9 percent) were found to be traveling at speeds up to
several hundred kilometers per second!”

“On any given day, the estimates are than the Earth intercepts about
19,000 meteorites weighing over 3.5 ounces, every year of which fewer
than 10 are ever recovered. About 2800 meteorites are in museums from
previous 'falls' and are chemically found to represent about 20 or so
distinct parent-bodies. The Earth acquires about 100 tons per day of
dust-sized micro-meteoroids.”

So, it's quite a wonder how all of these billions of years that our
physically naked and dark moon has been missing out on collecting its
fair share of dust, as well as having been avoiding further
significant impacts.

[Brad Guth]

On Jan 30, 1:35 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 13:10:21 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >> >Are you actually suggesting there is no significant tonnage of lunar
> >> >basalt rocks on Earth?
>
> >> What is "significant tonnage"?
> >Several trillion tonnes, mostly from that 2500 km crater.
>
> I think not.

Let us suggest that the moon only attracts or runs itself into a
highly conservative 10 tonnes per day as of the last billion years.
That’s only 3.65e12 tonnes of dust, not to mention averaging another
ten fold that much for each billion years before that, nor having
mentioned those bigger than dust meteorites of 3.5+ ounces, plus that
otherwise absolutely monstrous 2500 km crater plus having to account
for roughly that much lose crater debris again from all those other
big craters combined.

The notion that Earth doesn’t have those trillions of tonnes worth of
lunar basalt is kind of silly, that is unless you and your friends can
manage to tell us where else it went (because the vast bulk of it is
apparently not on the moon). How about you do your own math without
being artificially conservative like myself, telling us how many
10~100 micron plus larger hits per m2 does our naked moon get to deal
with per day.

Also while you’re at it. We’d like to know how that naked moon
manages to avoid being just as nasty as the Van Allen belts, and why
has it been such a good source of gamma if it supposedly doesn’t have
hardly any surface metallicity to work with.

[Chris L Peterson]

On Mon, 30 Jan 2012 14:31:16 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>Obviously you can't do the volumetric displacement math on behalf of
>that 2500 km crater, much less adding up all those other sufficiently
>big ones that should almost double that volume. So, where the hell
>did all that enormous volume of lose basalt shards go, if not
>extensively towards Earth?

The vast majority falls quite quickly back onto the surface of the
Moon. What little is actually ejected ends up in a complex orbit
within the Earth-Moon system, and is largely ejected. Only a tiny
fraction will end up on the Earth.

>So, do tell us what sort of inert fluff-ball smacked into our moon and
>made that 2500 km crater, which apparently gave off hardly any shards?
>(at least most of which didn't seem to stick with the moon)

A comet or asteroid. What else?

[Chris L Peterson]

On Mon, 30 Jan 2012 15:42:29 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>Each month the atmosphere of Earth deflects and/or vaporizes a fairly
>large number of meteors and smallish asteroids (roughly 100 per day),
>not to mention fending off the hundred tonnes of smaller stuff per
>day, that which our naked moon doesn’t have the same built-in or
>automatic defensive option. There’s well over a hundred fireballs
>spotted per day, and that obviously doesn’t include the other
>thousands per day that go unnoticed.

Your point?

The divide thousands of meteoroids per day into the surface area of
the Moon, and you'll see why the probability of any one spot getting
hit is still very small- especially for larger impactors.

[Brad Guth]

On Jan 30, 4:12 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 14:31:16 -0800 (PST), Brad Guth
>

Okay, so your conditional physics and science obfuscation are each
that good.

I still don't buy it. But then my income (what little there is of it)
isn't at risk for saying otherwise.

[Brad Guth]

On Jan 30, 4:15 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 15:42:29 -0800 (PST), Brad Guth
>

One more time: (not that you'll ever change your closed mindset)

There you go again, being so perfectly negative and unable to prove
that anyone else is even half as smart as yourself, in that you might

as well stop pretending and just be the intellectual black hole (aka
Sheldon Cooper) that you are.

Obviously you still can't do the volumetric displacement math on

behalf of that 2500 km crater, much less adding up all those other

sufficiently big ones that should almost double that volume of lose
material. So, where the hell did all that enormous volume of lose

basalt shards go, if not extensively towards Earth?

Remember, that within just a few cm depth, those NASA/Apollo guys
struck fused bedrock, as in they couldn't seem to force those probes

deeper for any further penetration (nothing the least bit lose under
their magic clumping moon dirt that was so nicely monochromatic,

highly reflective and not hardly the least bit paramagnetic or much

less anticathode or otherwise metallicity worthy). It’s almost as
though they’d mistakenly landed on a certain isolated private guano
island, that badly needed a spendy UN ticket to fly (that’s an inside
little sarcastic joke that’s actually not meant to be funny).

So, do tell us what sort of inert fluff-ball smacked into our moon and

made that 2500 km crater, which apparently gave off or ejected hardly
any shards (at least most of which didn't seem to stick with the
moon).

-

Each month the atmosphere of Earth deflects and/or vaporizes a fairly
large number of meteors and smallish asteroids (roughly 100 per day),
not to mention fending off the hundred tonnes of smaller stuff per
day, that which our naked moon doesn’t have the same built-in or
automatic defensive option. There’s well over a hundred fireballs
spotted per day, and that obviously doesn’t include the other
thousands per day that go unnoticed.

http://www.astronomycafe.net/qadir/q896.html
“In a recent article in the journal Nature, March 28, 1996 vol. 380,
page 323, Dr.s A.D. Taylor, W. J. Baggaley and D. I. Street at the
University of Adelaide in Australia discuss the results of their 1
year radar monitoring of incoming meteors. When meteorites slam into
the atmosphere, they produce ionization in the atmosphere. Radar echos
from this momentary ionization allow the velocity, altitude and
distance to be determined if you have two or more such installations
for triangulation. The AMOR radar in New Zealand was used for a year
in this fashion to detect 350,000 faint echos from very small
meteorites with sizes between 10 - 100 microns. This works out to
nearly 1000 every day, just from this site alone! Over 1508 of these
meteorites ( 0.9 percent) were found to be traveling at speeds up to
several hundred kilometers per second!”

“On any given day, the estimates are than the Earth intercepts about
19,000 meteorites weighing over 3.5 ounces, every year of which fewer
than 10 are ever recovered. About 2800 meteorites are in museums from
previous 'falls' and are chemically found to represent about 20 or so
distinct parent-bodies. The Earth acquires about 100 tons per day of
dust-sized micro-meteoroids.”

So, it's quite a wonder how all of these billions of years that our
physically naked and dark moon has been missing out on collecting its

fair share of dust, as well as having been avoiding its share of all
those further significant impacts.

Let us suggest that the moon only attracts or runs itself into a
highly conservative 10 tonnes per day as of the last billion years.
That’s only 3.65e12 tonnes of dust, not to mention averaging another
ten fold that much for each billion years before that, nor having

mentioned those bigger than dust meteorites of 0.1+ kg, plus that

otherwise absolutely monstrous 2500 km crater plus having to account
for roughly that much lose crater debris again from all those other

big craters combined, is hardly suggesting our moon isn’t pulverized
with many meters depth of dust and lose shards.

My crude conservative math has 4.055e15 tonnes of that 10~100 micron
dust accumulated within 4 billion years, and that’s not even including
those 0.1+ kg meteors as impactors, or much less worthy of whatever
generated those enormous craters. That’s a lot of lose influx when
there’s only 3.8e13 m2 of surface area to work with (roughly 106.7
tonnes/m2), most of which seems to be missing in action, because it
sure as hell didn’t get washed down into muds that compacted into
rock.

The notion that Earth doesn’t have those trillions of tonnes worth of
lunar basalt is kind of silly, that is unless you and your friends can
manage to tell us where else it went (because the vast bulk of it is

apparently not found on the moon). How about you do your own math

without being artificially conservative like myself, telling us how
many 10~100 micron plus larger hits per m2 does our naked moon get to
deal with per day.

Also while you’re at it. We’d like to know how that naked moon

manages to avoid being just as nasty as those Van Allen belts, and why

has it been such a good source of gamma if it supposedly doesn’t have
hardly any surface metallicity to work with.

The average solar wind as is, blows a 900,000 km trail of heated and
ionized sodium away from our moon. But how much nearby solar wind
density and velocity does it take in order to blow or extract those
typically heavier particles of dust off the naked surface of our moon?
(perhaps a recently nearby planet creation did the trick, unless that
moon had otherwise bounced off Earth).

[Peter Webb]

"Chris L Peterson" <c...@alumni.caltech.edu> wrote in message

news:p1ddi79fmfaaorga7...@4ax.com...

> On Mon, 30 Jan 2012 16:52:32 +1100, "Peter Webb"
> <r.peter...@gmail.com> wrote:
>
>>Like I said, my heart says no, but my head says yes. Would be happy to
>>find
>>out my gut feeling was wrong.
>
> What your head is missing, perhaps, is that the Moon is not smooth.

So, shoot it into a slightly elliptical orbit. If the point at which you
shoot and the antipodial point are relatively flat, then it won't hit
anything.

> You can't turn a pure physics problem into a practical one as Brad is
> trying to do. I doubt there is any place on the Moon where you could
> shoot a bullet at orbital velocity and have it come around and hit you
> on the back- whether it's made of thorium or titanium.

Of course there is. On any great circle there is a highest point; shoot it
horizontally from there.

>And of course,
> Brad isn't suggesting that the bullet be shot at orbital velocity, but
> at escape velocity. That means the bullet is going to be much higher
> when it comes around again.

No, he talked about orbital velocity, and no, if it is escape velocity it
won't be coming back, higher or otherwise.

[Peter Webb]

"Chris L Peterson" <c...@alumni.caltech.edu> wrote in message
news:nbudi7t2m1gi414tr...@4ax.com...

Why?

There are an infinite number of points where this can be done. Just fire
your gun from the highest point on any great circle.

I thought you were challenging the concept. Now you seem to be arguing about
practicalities. You may as well argue that nobody would be bothered trying;
that's a more obvious practicality.

And your stated objection was "check the math", which seems absolutely fine
to me.

And BTW I suspect I have made an error when I said that you could probably
fire a rifle from the moon into the earth's atmosphere. You also have to
bleed off the horizontal component of the moons velocity relative to the
earth. That's another 1 km/sec or so, meaning that you would need a rifle
with a muzzle velocity of about 4 km/sec. That would be quite difficult to
achieve with a hand held weapon. Probably about the best that you could get
from a rifle in practice would be putting the bullet into earth orbit.

Ohh, and yes Brad Guth is an idiot. But that doesn't mean *everything* he
says is wrong; you should get your facts straight before claiming he is
wrong.

[Chris L Peterson]

On Tue, 31 Jan 2012 16:07:03 +1100, "Peter Webb"
<r.peter...@gmail.com> wrote:

>I thought you were challenging the concept. Now you seem to be arguing about
>practicalities. You may as well argue that nobody would be bothered trying;
>that's a more obvious practicality.

I was never challenging the concept. That's why I repeatedly have
referred to the practical problems.

[Chris L Peterson]

On Mon, 30 Jan 2012 16:55:11 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>Obviously you still can't do the volumetric displacement math on
>behalf of that 2500 km crater, much less adding up all those other
>sufficiently big ones that should almost double that volume of lose
>material. So, where the hell did all that enormous volume of lose
>basalt shards go, if not extensively towards Earth?

Why should it go towards Earth? It is ejected hemispherically, most of
it with less than escape velocity. Only the material ejected straight
towards the Earth (which is only a tiny fraction of the total) stands
any chance of actually striking the Earth.

[Peter Webb]

"Chris L Peterson" <c...@alumni.caltech.edu> wrote in message

news:fauei7pl505tpbofa...@4ax.com...

No, here is your response in full when this thought experiment was
described:

"You might want to go back and work on the math of that one again..."

No mention of practicalities, you objected because you thought the math was
wrong.

It was only when I and some others re-did the "math" and it proved to be
correct that you changed your story. To something incredibly lame. Your new
objection is "practicalities"; given the premise is that you are on the Moon
with a high speed sniper rifle, complaining about other practicalities is
obviously very lame.

You boobed. Nothing wrong with the math. Why not just admit it?

[Peter Webb]

"Chris L Peterson" <c...@alumni.caltech.edu> wrote in message
news:6cuei7dc5l8ud1bln...@4ax.com...

Again, your display your lack of knowledge of orbital mechanics. I recommend
you go back and "check your maths". Material ejected straight towards the
earth has no chance of hitting it, as the rotation of the moon around the
earth provides a perpendicular (horizontal) velocity component of about 1
km/sec. You would need to aim well in front of the earth (at an angle) to
cause it to hit the earth.

As you appear to know nothing at all about this subject, I can only
recommend that you stop displaying your ignorance with every post you make.

[Chris L Peterson]

On Tue, 31 Jan 2012 18:55:28 +1100, "Peter Webb"
<r.peter...@gmail.com> wrote:

>Again, your display your lack of knowledge of orbital mechanics. I recommend
>you go back and "check your maths". Material ejected straight towards the
>earth has no chance of hitting it, as the rotation of the moon around the
>earth provides a perpendicular (horizontal) velocity component of about 1
>km/sec. You would need to aim well in front of the earth (at an angle) to
>cause it to hit the earth.

I was dumbing down the concept for Brad by ignoring the motion
vectors. The point is that only material ejected along one very narrow
trajectory (at lunar escape velocity) has any possibility of hitting
the Earth- a very tiny percentage of the total. That's the only
important concept here, not the details of the actual trajectory,
given that Brad has some bizarre notion that the Earth is going to
magically suck up everything excavated from a large impact crater on
the Moon.

FYI, I'm a professional orbit dynamicist... I do know something about
this subject.

[Chris L Peterson]

On Tue, 31 Jan 2012 18:50:11 +1100, "Peter Webb"
<r.peter...@gmail.com> wrote:

>No, here is your response in full when this thought experiment was

>described...

Back to your usual self again, preferring to twist words rather than
listen to clarifications and engage in intelligent conversation. No
surprise, I guess.

[oriel36]

On Jan 31, 3:05 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
.
>
> FYI, I'm a professional orbit dynamicist... I do know something about
> this subject.

I had a professional meteorologist last week who couldn't equate the
temperature rising and falling within a 24 hour period with the
rotation of the Earth and that I find remarkable so there are many
self-proclaimed professionals out there who are impressed with
themselves yet find it difficult to handle orbital dynamics and
especially the distinction between lunar orbits and planetary orbits.

The moon doesn't rotate around the Earth,the moon orbits the Earth
and doesn't have an intrinsic rotation or ,what amounts to the same
thing,variations in latitudinal speeds.A planet does turn to the
central Sun as it moves along its orbital circumference as opposed to
a lunar orbit which doesn't,you will know this as a professional
dynamicist,if you look out your window tonight as see that the moon
has rotated for as long as you have looked at it.

The sheer fall of Western astronomical tradition,and it is a
tradition,is astonishing and it has a domino effect as it weakens all
surrounding disciplines that exist as outriggers of astronomy such as
terrestrial sciences of geological and biological evolution and the
interaction with climate and each other from fossil and rock records
to the interaction of all life on the planet.

If you wish to be an orbital dynamicist then act like one.

[Chris L Peterson]

On Tue, 31 Jan 2012 07:20:56 -0800 (PST), oriel36
<kellehe...@gmail.com> wrote:

>I had a professional meteorologist last week who couldn't equate the
>temperature rising and falling within a 24 hour period with the

>rotation of the Earth...

I rather doubt that.

[Sam Wormley]

On 1/31/12 9:20 AM, oriel36 wrote:
> I had a professional meteorologist last week who couldn't equate the
> temperature rising and falling within a 24 hour period with the

> rotation of the Earth...

I suspect your "professional meteorologist" quickly "turned off"
when you started spewing your nonsense, Gerald.

[Brad Guth]

On Jan 30, 9:07 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> "Chris L Peterson" <c...@alumni.caltech.edu> wrote in messagenews:nbudi7t2m1gi414tr...@4ax.com...

Our "Sheldon Cooper" that's going by the Usenet name of Chris L
Peterson, as such can't afford to being wrong about anything. He
seems borg like programmed to only consider the utmost negatives about
everything that wasn't his idea to begin with.

A 2.5 km/sec projectile should gradually gain altitude, although the
minimal atmosphere along with the affects of mascons should prevent
that item from escaping the moon. This relates to the still lethal
secondary shards or ejected material from the creation of craters, not
to mention broken shards from the impactors that didn't vaporize.

By now, 4+ billion years since having formed its crust, most every
horizontal m2 (including +/- 22.5 degree slopes) of that physically
dark moon should be covered in meters depth of loose debris. 45+
degree slopes should be exposing nearly bare basalt, which should be
nearly as dark or perhaps even darker than coal, unless the
electrostatic and paramagnetic properties of that basalt is holding
onto the loose material.

The Kodak film obtained science from the NASA/Apollo era simply
doesn't add up to what that moon should have looked like. They
obviously doctored the living hell out of those images, and otherwise
having obfuscated most everything imaginable in order to make or force
their interpretation of everything to seem proper and logical to the
mostly uneducated Americans that still have no idea what their
artificially perpetrated cold-war was actually doing.

[oriel36]

On Jan 31, 3:38 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Tue, 31 Jan 2012 07:20:56 -0800 (PST), oriel36
>

> <kelleher.ger...@gmail.com> wrote:
> >I had a professional meteorologist last week who couldn't equate the
> >temperature rising and falling within a 24 hour period with the
> >rotation of the Earth...
>
> I rather doubt that.

Why would you doubt it ?,The dominant view doesn't keep the rotation
of the Earth in step with the daily temperature fluctuations and as a
professional dynamicist you should know that there are 1461 AM/PM
events enclosed in 4 orbital circuits reflecting the proportion of
rotations to orbital circuits.

http://groups.google.com/group/uk.sci.weather/msg/a79233d89bf9fddb

Introducing a false assumption based on clocks keeping in step with
stellar circumpolar motion would normally only amount to a short
detour into astronomical forensics instead of the decade old episode
it has become however it is shown that the AM/PM natural noon events
correspond with the AM/PM signatures of the 24 hour day take center
stage as the working principles for daily and orbital cycles.In
short,if right ascension leads to a belief in 1465 rotations in 1461
days,the onus is to track down where the error originates and it turns
out to be a misuse of the human devised calendar system.

Instead of a street fight over an analemma hoax or trying to justify
daily and orbital motions using stellar circumpolar motion,the real
issues are elsewhere and should be dealt with properly.As far as I
understand it,the nature of the empiricist approach is that they won't
find anything wrong if they are not looking for it as their interests
reside with life on Mars,black holes,goldilocks planets,dark matter
and all those other novelties whereas an astronomer would ask those
questions which are closer to home and ,as always,the answer is only
as good as the question.

So,asking how there are 1461 rotations in 1461 days is an excursion of
astronomy from the antiquity of the calendar system to the advent of
modern watches and on into planetary dynamics,asserting 1465 rotations
in 1461 days is a worthless exercise unless the readers really wants
to make the effort to enjoy astronomy.

[Brad Guth]

On Jan 30, 9:20 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Mon, 30 Jan 2012 16:55:11 -0800 (PST), Brad Guth
>

Perhaps roughly a third stayed with the moon, and two thirds got away,
and most (more than half) of that two thirds was likely attracted
towards Earth. So, only one third of ejected material from that one
2500 km crater is still going to represent teratonnes of basalt
arriving at Earth. Using 5% volume of a 2500 km sphere as
representing the total crater displacement, whereas that one third
portion of 5% is roughly 1.35e17 m3, or even a third of 1% being worth
2.7e16 m3 is hardly an insignificant volume or tonnage, especially if
the average paramagnetic basalt density were 3.5 g/cm3.

BTW; for this estimate, we should be using a sphere of at least 5000
km to start off with, or better yet is to use Earth as the impactor.

[Chris L Peterson]

On Tue, 31 Jan 2012 09:42:32 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>Perhaps roughly a third stayed with the moon, and two thirds got away,
>and most (more than half) of that two thirds was likely attracted
>towards Earth.

Please provide some rigorous support for that opinion.

[Brad Guth]

On Jan 31, 9:46 am, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Tue, 31 Jan 2012 09:42:32 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >Perhaps roughly a third stayed with the moon, and two thirds got away,
> >and most (more than half) of that two thirds was likely attracted
> >towards Earth.
>
> Please provide some rigorous support for that opinion.

I've found other research that sufficiently supports my speculation
theories, although I've certainly given this my own personal deductive
interpretation because, it seems others haven't been as willing to
suggest or swag as to what impacted our moon in order to create that
2500 km crater, nor willing to further stipulate as to what might have
created our Arctic ocean basin and perhaps having established the
seasonal tilt of Earth at the same time. I can't even find any
carvings or paintings of our big old and extremely vibrant moon until
after that last ice age abruptly terminated (via ice core science) as
of roughly 11, 712 years ago (apparently before then Earth was always
100% cloud covered, because otherwise they had no problems whatsoever
with their carvings and/or paintings of much smaller details of most
everything else that mattered to their survival, and that terrific
nighttime moonlight should have been one of the really big items on
their list of what truly mattered).

[Chris L Peterson]

On Tue, 31 Jan 2012 10:50:12 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>> Please provide some rigorous support for that opinion.
>
>I've found other research that sufficiently supports my speculation

>theories...

Please reference.

[oriel36]

Astronomy exists on a very secure footing and my works will not die
with me as they originate in my own daily experiences and those of my
astronomical ancestors,the astronomical picture and its terrestrial
effects will never be complete and in many cases the topics are just
beginning in much the same way as the complicated process of
terrestrial biology once existed in infancy but is now complex and in
most cases people enjoy things like human physiology and can handle
complex interwoven processes,planetary dynamics being no different in
itself and in its effects on terrestrial sciences.

It is not that the meteorologist 'turned off',it is a snapping out
process where the focus shifts away from 'the page and the theory' and
people notice their surroundings,often for the first time.The poet and
painter William Blake captured the condition that so afflicts this age
-

http://upload.wikimedia.org/wikipedia/commons/0/0e/Newton-WilliamBlake.jpg

There is no reason why a person cannot appreciate that daily
temperatures go up and down in a 24 hour period because of a rotating
Earth and keep it that way as the 1461 rotations and 1461 days of the
calendar system ,why the moon orbits the Earth and doesn't rotate,what
causes the seasons or any other answer people care to ask in linking
the motions of the planet with all the experiences we see and feel and
the wider community needs these things more than ever as theorists
have spun off into novelties.You hung in there for a long time with
the Fomalhaut system and the issue where the orbital distance from the
central star does not match the incremental motion of Fomalhaut b in
its orbital circumference yet all these things have yet to be tidied
up and will be when men snap out of the condition they have found
themselves in.

http://apod.nasa.gov/apod/ap081114.html

If you cannot accept the proportion of rotations to orbital circuits
of our own planet then forget analysing any other planet or moon as we
should not be stuck with this issue for over a decade without any
clear resolution towards assigning the correct facts to the Earth and
its motions.This actually requires people to think things through and
make the effort and that I have yet to see.

[Brad Guth]

On Jan 31, 10:55 am, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Tue, 31 Jan 2012 10:50:12 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >> Please provide some rigorous support for that opinion.
>
> >I've found other research that sufficiently supports my speculation
> >theories...
>
> Please reference.

I’ve lost track of most of what I’ve come across over the past decade.
Read between the lines:
Volume of Impact Crater Fallback Ejecta on the Earth:
http://adsabs.harvard.edu/full/1979LPI....10.1113S
http://www.apl.ucl.ac.uk/lectures/3c11/impacts1.pdf

Crank in some glancing blow encounters and keep in mind the less than
2.5 km/sec escape velocity that makes it relatively easy for those
secondary shards of bedrock basalt to escape away from the moon. With
further internet searching, I’m certain to find those same or better
research examples.

Of course, if I were only half as negative and/or naysay closed
mindset as yourself, I probably couldn’t find a damn thing.

If the majority of considerable crater ejected material plus whatever
impactor shards isn’t found on the moon, as supposedly documented to
death by our NASA/Apollo era that doesn’t require any sort of
independent peer review or much less independent confirming research,
then it obviously went somewhere else. Clearly the lack of atmosphere
and low gravity makes it extremely easy for crater ejected material to
travel great distances, as well as getting away from the moon, whereas
on average half of that escape velocity material is likely to become
part of Earth.

Anything encountering that moon at greater than 5 km/sec can cause
crater ejected material to exceed escape velocity. It gets much less
problematic for those items arriving at only 2.5 km/sec, unless it’s
of a purely shallow glancing angle whereas most of the impactor itself
doesn’t necessarily even stick with the moon unless the lithobraking
encounter had managed to pull sufficient velocity from the impactor
(not likely if it were the Earth causing that 2500 km crater).

[Chris L Peterson]

On Tue, 31 Jan 2012 12:13:49 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>I’ve lost track of most of what I’ve come across over the past decade.
>Read between the lines:
>Volume of Impact Crater Fallback Ejecta on the Earth:
> http://adsabs.harvard.edu/full/1979LPI....10.1113S
> http://www.apl.ucl.ac.uk/lectures/3c11/impacts1.pdf

Thank you. Unfortunately for your position, the first of these
specifically points out that very little material reaches escape
velocity (the current view, even though this is a very old abstract),
and the second has nothing to say about the matter.

In fact, if you look a little harder, you can find a number of papers
that detail modeling of impacts, and provide a good discussion on the
mechanics involved in ejecting material from various bodies following
impacts- and why very little material is actually ejected.

In addition to modeling the actual escape of material, recent papers
also discuss the specific orbital dynamics of the inner Solar System
that determine the likelihood of that material ending up on other
bodies.

>Anything encountering that moon at greater than 5 km/sec can cause
>crater ejected material to exceed escape velocity.

References?

[Peter Webb]

"Chris L Peterson" <c...@alumni.caltech.edu> wrote in message

news:bp0gi71nrhabudlvm...@4ax.com...

> On Tue, 31 Jan 2012 18:50:11 +1100, "Peter Webb"
> <r.peter...@gmail.com> wrote:
>
>>No, here is your response in full when this thought experiment was
>>described...
>
> Back to your usual self again, preferring to twist words

Funny, you snipped your own words, which comprised exactly one sentence.

Why did you want to remove your own words? Because I obviously didn't twist
them?

> rather than
> listen to clarifications and engage in intelligent conversation. No
> surprise, I guess.

ROFL. Your contribution (your "clarifications and intelligent conversation")
comprised one sentence:

"You might want to go back and work on the math of that one again...".

Mine comprised an analysis of the orbital velocity of the moon (with
calculation provided), a discussion of muzzle velocities, propellant speed
and basically a full work-down of the problem. I won't repeat it here;
unlike your single sentence claiming the maths was wrong, mine was a
clarification of the maths and intelligent conversation on the applicability
of this to other scenarios (like shooting a bullet from the Moon into the
earth's atmosphere).

Clearly, I did not twist your words. Clearly, you didn't provide
"clarification and intelligent conversation", and just as clearly by
actually doing and explaining the maths I did.

So, lets add "lying" and "stupidity" to the other quality you have
demonstrated in this thread ("lack of knowledge of orbital mechanics").

[Brad Guth]

On Jan 31, 1:41 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Tue, 31 Jan 2012 12:13:49 -0800 (PST), Brad Guth
>

Physics and logic. Apparently you've never played a game of pool, or
with glass marbles or metal bearing spheres?

Damn little atmosphere and 1/6th gravity doesn't do much for holding
onto fast moving stuff.

Again, our NASA/Apollo era found hardly any loose surface materials
(only a few cm depth and nothing that wasn't fused solid below), so
where the hell did it all go?

[Sam Wormley]

On 1/31/12 2:13 PM, Brad Guth wrote:
> Crank in some glancing blow encounters and keep in mind the less than
> 2.5 km/sec escape velocity that makes it relatively easy for those
> secondary shards of bedrock basalt to escape away from the moon. With
> further internet searching, I’m certain to find those same or better
> research examples.

The likelihood that a rock escaping the moon, hitting
the earth, is extremely low.

Surface Escape Velocity for some solar system objects:

>
> Sun (Photosphere) 617.5 km/s
> Mercury 4.3 km/s
> Venus 10.3 km/s
> Earth 11.2 km/s
> the Moon 2.4 km/s
> Mars 5.0 km/s
> Jupiter 59.5 km/s
> Saturn 35.6 km/s
> Uranus 21.2 km/s
> Neptune 23.6 km/s
> Pluto 1.229 km/s

[Chris L Peterson]

On Tue, 31 Jan 2012 17:16:53 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>> References?
>
>Physics and logic.

"Physics" is not a reference, and "logic" is frequently anything but.

>Apparently you've never played a game of pool, or
>with glass marbles or metal bearing spheres?

I have. But I lack the strength to hit two balls together such that
the force of their impact exceeds the material strength of their
constituents... so this example isn't really relevant to the
discussion. And it demonstrates how your "logic" fails when it comes
to actual physical phenomena.

[Brad Guth]

Your parrot-speak is noted. Now tell us something new and
interesting.

Go right ahead and tell us where all that magic clumping and highly
reflective dust plus ejected basalt bedrock went, because our NASA/
Apollo era doesn't seem to have a clue. As is, according to our NASA/
Apollo data, there's only at most a few percent (possibly less than
1%) of what should be there if that moon is as old and as deeply
pulverized as they've specified.

[Brad Guth]

On Jan 31, 7:30 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Tue, 31 Jan 2012 17:16:53 -0800 (PST), Brad Guth
>

Unfortunately, your mainstream logic on this is also entirely
subjective or conjecture. Do let us know when something objective
pops into your theory of nothing of any significance surviving the
impact of encountering our moon, as well as leaving hardly any dust or
loose debris behind.

[Chris L Peterson]

On Tue, 31 Jan 2012 20:40:47 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>Unfortunately, your mainstream logic on this is also entirely
>subjective or conjecture.

I'm not depending on logic, either, but on actual scientific evidence,
both theoretical and observational.

>Do let us know when something objective
>pops into your theory of nothing of any significance surviving the
>impact of encountering our moon, as well as leaving hardly any dust or
>loose debris behind.

I haven't argued that nothing of any significance survives a lunar
impact. I argued that only a tiny fraction of the material that is
excavated from the crater produced by a lunar impact stands any chance
of landing on the Earth. Those are very different things.

[Brad Guth]

On Jan 31, 9:06 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Tue, 31 Jan 2012 20:40:47 -0800 (PST), Brad Guth
>

Then where did all the rest of it go?

We're not talking about anything less than a thousand teratonnes, that
had to go somewhere.

[Chris L Peterson]

On Tue, 31 Jan 2012 21:28:06 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>Then where did all the rest of it go?

Much is vaporized, meaning there are no "rocks" left to fall anywhere.
Some of the vapor ends up in space, some falls back onto the Moon. Of
the material that survives as larger pieces, most falls back to the
surface and forms the large ejecta blanket seen around craters. Of the
large pieces that escape the Moons's gravity, most end up being
ejected from the Earth-Moon system completely, and eventually end up
in the Sun (although that may take millions of years). Of the little
bit that makes it to Earth, the vast majority burns up in the
atmosphere. The little that's left lands on the ground as Lunar
meteorites- which we know to be exceedingly rare.

[Peter Webb]

"Brad Guth" <brad...@gmail.com> wrote in message
news:1039036a-f73e-47ba...@ra5g2000pbc.googlegroups.com...

______________________________________
The overwhelming majority will just fall back onto the moon. As particles
ranging in size from microscopic to cubic kilometers.

As to where it is on the moon, the answer is probably all over. The moon
does have lots of rocks on it.

I am not even sure that meteor strikes are now creating (on balance) more
debris on the Moon. It is quite plausible to me that in addition to throwing
up debris, the heat and overpressure of the meteor strikes could fuse
existing material so some steady state condition eventuates where the amount
of debris stays pretty much constant..

[Brad Guth]

On Jan 31, 9:38 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Tue, 31 Jan 2012 21:28:06 -0800 (PST), Brad Guth
>

That's just filling in the blanks, of which 99.9% of what we
objectively know about our moon is still a blank. You just don't want
to do or say anything that's a challenge to whatever the NASA/Apollo
bible had to say.

[Brad Guth]

On Jan 31, 9:41 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> "Brad Guth" <bradg...@gmail.com> wrote in message

>
> news:1039036a-f73e-47ba...@ra5g2000pbc.googlegroups.com...
> On Jan 31, 9:06 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
>
>
>
>
>
>
>
>
>
> > On Tue, 31 Jan 2012 20:40:47 -0800 (PST), Brad Guth
>
> > <bradg...@gmail.com> wrote:
> > >Unfortunately, your mainstream logic on this is also entirely
> > >subjective or conjecture.
>
> > I'm not depending on logic, either, but on actual scientific evidence,
> > both theoretical and observational.
>
> > >Do let us know when something objective
> > >pops into your theory of nothing of any significance surviving the
> > >impact of encountering our moon, as well as leaving hardly any dust or
> > >loose debris behind.
>
> > I haven't argued that nothing of any significance survives a lunar
> > impact. I argued that only a tiny fraction of the material that is
> > excavated from the crater produced by a lunar impact stands any chance
> > of landing on the Earth. Those are very different things.
>
> Then where did all the rest of it go?
>
> We're not talking about anything less than a thousand teratonnes, that
> had to go somewhere.
>
> ______________________________________
> The overwhelming majority will just fall back onto the moon. As particles
> ranging in size from microscopic to cubic kilometers.

I'm not talking about little craters. Stick with those 50+ km
craters, and especially deal with that 2500 km crater.

>
> As to where it is on the moon, the answer is probably all over. The moon
> does have lots of rocks on it.

Not according to those Apollo missions and their rad-hard Kodak film
that offered such terrific dynamic range regardless of the extremely
high contrast of that physically dark terrain, whereas there was
hardly any depth of dust and not 10% as many crater ejected debris
rocks and/or meteorites as depicted by the images from Mars depicts.

>
> I am not even sure that meteor strikes are now creating (on balance) more
> debris on the Moon. It is quite plausible to me that in addition to throwing
> up debris, the heat and overpressure of the meteor strikes could fuse
> existing material so some steady state condition eventuates where the amount
> of debris stays pretty much constant..

That moon should have been collecting roughly 10% as much dust as
Earth receives. Of course most of the incoming dust and small
meteorites incoming to Earth vaporize before hitting the surface,
whereas that NASA/Apollo pastel or off-white surface of extremely
fluffy dust that had no problems with providing terrific surface
tension, that reflected at three times the average albedo and having
managed to erode everything in sight to that of soft rolling hills of
such a nicely clumping dust covered terrain, was in most places only a
few cm deep, under which is that fully fused basalt crust of a mostly
grayish basalt that isn't the least bit paramagnetic or otherwise
metallicity infused (apparently only the least metallicity saturated
meteors and those of mostly pastel gray and off-white got collected by
the moon, because all the darker and much greater density ones got
sent to Mars and Earth.

[Sam Wormley]

On 1/31/12 10:21 PM, Brad Guth wrote:

> Wormley wrote:
>> Mercury 4.3 km/s
>> Venus 10.3 km/s
>> Earth 11.2 km/s
>> the Moon 2.4 km/s
>> Mars 5.0 km/s
>> Jupiter 59.5 km/s
>> Saturn 35.6 km/s
>> Uranus 21.2 km/s
>> Neptune 23.6 km/s
>> Pluto 1.229 km/s

> Your parrot-speak is noted. Now tell us something new and
> interesting.

Awk -- These are scientifically determined escape velocities
from the surface of these bodies, Brad. Awk! Pay attention, Brad!

[Brad Guth]

You brown-nosed parrots are always such entertaining FUD-masters,
because for a treat you'll always say the same old words over and
over, and look cute at the same time.

Just because you can't explain any number of things about our moon,
you think putting on another silly parrot show is good enough.

[Dr J R Stockton]

In sci.astro.amateur message <76agi713pcvk7a1ul...@4ax.com
>, Tue, 31 Jan 2012 10:46:22, Chris L Peterson <c...@alumni.caltech.edu>
posted:

According to <http://www.merlyn.demon.co.uk/gravity2.htm#OEV>, escape
velocity is root 2 g r where r is the radius and g the local
acceleration.

In units for which Earth g & r are both 1 / root 2, escape velocity from
the surface of the Earth is root 1. The moon's radius is 1/4 of
Earth's, its local gravity is 1/6 of Earth's, so its escape velocity is
root 1/24, near enough 1/5, of Earth's.

At the moon's distance, r is 60 times bigger but g is 60*60 times
smaller, so escape velocity is root 1/60, say 1/8, of Earth's.

So anything leaving the Moon's surface at more than a little over the
Moon's escape velocity will escape from the Earth-Moon system, unless in
that small fraction of trajectories that hit Earth on the way out.
Approximately none of that, however, will escape from the Solar System,
so may return to hit Earth eventually.

Anything (on or) leaving the Moon's surface will of course be attracted
by the Earth; but, except for that which escapes the Earth-Moon system
or hits Earth immediately, or grazes the atmosphere to re-enter not very
much later, it will enter a stable elliptical orbit around the Earth -
until that orbit, reaching as it must as far out as the Moon's orbit, is
perturbed by a close passage with the Moon.

In fact, the orbit of anything which leaves the Moon but does not
immediately escape or hit the Earth must, long-term, be more or less
chaotic - so the object has a fair chance of hitting the Earth -
eventually, to quote (out of context) James White.

--
(c) John Stockton, nr London, UK. ?@merlyn.demon.co.uk Turnpike v6.05 MIME.
Web <http://www.merlyn.demon.co.uk/> - FAQqish topics, acronyms and links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.

[Chris L Peterson]

On Wed, 1 Feb 2012 23:38:50 +0000, Dr J R Stockton
<repl...@merlyn.demon.co.uk> wrote:

>In fact, the orbit of anything which leaves the Moon but does not
>immediately escape or hit the Earth must, long-term, be more or less
>chaotic - so the object has a fair chance of hitting the Earth -
>eventually, to quote (out of context) James White.

I don't think so. Of course some material will hit the Earth- we do
have lunar meteorites. But most will be end up being caught by the
Sun. The chances of that are far greater than ever hitting the Earth.
Any ejected material is likely to be small enough that drag effects
will result in a lifetime within the Solar System of no more than a
few million years, which isn't enough to make a collision with Earth
likely.

[Brad Guth]

On Feb 1, 3:38 pm, Dr J R Stockton <reply1...@merlyn.demon.co.uk>
wrote:
> In sci.astro.amateur message <76agi713pcvk7a1ulcrln452spogn89...@4ax.com>, Tue, 31 Jan 2012 10:46:22, Chris L Peterson <c...@alumni.caltech.edu>

>  (c) John Stockton, nr London, UK. ?...@merlyn.demon.co.uk  Turnpike v6.05  MIME.

>   Web  <http://www.merlyn.demon.co.uk/> - FAQqish topics, acronyms and links;
>   Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
>  No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.

Roughly 50% of those lunar orbital trajectory items with their initial
escape velocity of 2.4+ km/sec are going to be headed toward the orbit
of Earth more likely than anyplace else.

Chris seems to think that hardly any significant items of secondary
debris moving away at 2.4+ km will ever leave the moon. Of course I'm
only referencing those of 50+ km crater ejected shards, because
craters of 5 km or less diameter might not generate a high percentage
of secondary items, even though a 1 km crater should give off a few
percent of something moving away faster than 2.4 km/sec, especially if
the impact was 10+ km/sec to start with.

[Peter Webb]

"Chris L Peterson" <c...@alumni.caltech.edu> wrote in message
news:hlkji7decgs675d7o...@4ax.com...

______________________________________
An opinion devoid of any analysis which might sustain it.

By my calculations, the outcomes are as follows:

1. Material ejected at less than the orbital velocity of the moon - 1.7
kms/sec - will always fall back.

2. Material ejected at velocities of between 1.7 kms/sec and 2.4 kms/sec
(the escape velocity of the moon in isolation) will either fall back to the
moon or go into orbit around the moon, depending on the angle at which it is
ejected.

3. The delta V to achieve earth escape velocity for an object in the same
orbit as the moon is about 0.4 km/s. So material ejected with a velocity of
between 2.4 km/s and 2.8 km/s will enter a chaotic orbit bound to the earth,
and very likely eventually hit the earth (and less likely eventually hit the
moon).

4. The delta V to achieve solar escape orbit from earth orbit is (if I
recall correctly) about 12 kms/sec if we ignore the fact that the other
planets will likely cause the orbit to become chaotic. So velocities ranging
from about 2.8 km/s and 14 km/s will most likely enter a chaotic solar
orbit. (of course, there is also a slight chance that it will hit the earth
before completing a full orbit).

5. Higher ejection velocities allow the possibility of the material leaving
the solar system entirely.

So much for orbital mechanics 101. That much is straightforward.

What makes it (I suspect) impossible to calculate what percentage hits the
moon or earth is that nobody knows the distribution of speed or angles for
material ejected from the Moon, so it is impossible to get to square one on
this analysis, which requires a knowledge of the ejection speeds.

[Brad Guth]

On Feb 1, 6:41 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> "Chris L Peterson" <c...@alumni.caltech.edu> wrote in messagenews:hlkji7decgs675d7o...@4ax.com...

Thank you so very much. However, by far the greatest gravity well for
those secondary basalt shards that get away from our moon, is still
Earth. It's not unreasonable to think, up to 50% of whatever gets
away from that moon will eventually find its way towards Earth,
although even 1% of it will be a considerable volume.

[Peter Webb]

"Peter Webb" <r.peter...@gmail.com> wrote in message
news:jgct6u$bk1$1...@news.albasani.net...

>
> "Chris L Peterson" <c...@alumni.caltech.edu> wrote in message
> news:hlkji7decgs675d7o...@4ax.com...
>> On Wed, 1 Feb 2012 23:38:50 +0000, Dr J R Stockton
>> <repl...@merlyn.demon.co.uk> wrote:
>>
>>>In fact, the orbit of anything which leaves the Moon but does not
>>>immediately escape or hit the Earth must, long-term, be more or less
>>>chaotic - so the object has a fair chance of hitting the Earth -
>>>eventually, to quote (out of context) James White.
>>
>> I don't think so. Of course some material will hit the Earth- we do
>> have lunar meteorites. But most will be end up being caught by the
>> Sun. The chances of that are far greater than ever hitting the Earth.
>> Any ejected material is likely to be small enough that drag effects
>> will result in a lifetime within the Solar System of no more than a
>> few million years, which isn't enough to make a collision with Earth
>> likely.
>
> ______________________________________
> An opinion devoid of any analysis which might sustain it.
>
> By my calculations, the outcomes are as follows:
>
> 1. Material ejected at less than the orbital velocity of the moon - 1.7
> kms/sec - will always fall back.
>
> 2. Material ejected at velocities of between 1.7 kms/sec and 2.4 kms/sec
> (the escape velocity of the moon in isolation) will either fall back to
> the moon or go into orbit around the moon, depending on the angle at which
> it is ejected.
>

Correction:

Whilst this material has sufficient speed to go into orbit, its elliptical
orbit will intersect the surface of the moon, and hence virtually all
material ejected at less than 2.4 kms/s will fall back onto the moon.

[Chris L Peterson]

On Thu, 2 Feb 2012 13:41:28 +1100, "Peter Webb"
<r.peter...@gmail.com> wrote:

>By my calculations, the outcomes are as follows:
>
>1. Material ejected at less than the orbital velocity of the moon - 1.7
>kms/sec - will always fall back.

I agree.

>2. Material ejected at velocities of between 1.7 kms/sec and 2.4 kms/sec
>(the escape velocity of the moon in isolation) will either fall back to the
>moon or go into orbit around the moon, depending on the angle at which it is
>ejected.

Yes... but all such material will rather rapidly fall back to the Moon
in any case, because lunar orbits are unstable. That's why no
spacecraft has ever been put into a long term lunar orbit.

>3. The delta V to achieve earth escape velocity for an object in the same
>orbit as the moon is about 0.4 km/s. So material ejected with a velocity of
>between 2.4 km/s and 2.8 km/s will enter a chaotic orbit bound to the earth,
>and very likely eventually hit the earth (and less likely eventually hit the
>moon).

I'm curious about your calculations on this one. Chaotic three-body
orbits don't typically result in the low mass object hitting one of
the larger masses, but are much more likely to end up with the low
mass object being ejected from the system entirely. In that case, the
debris will end up in a solar orbit, and various drag effects will
result in it ultimately hitting the Sun.

>4. The delta V to achieve solar escape orbit from earth orbit is (if I
>recall correctly) about 12 kms/sec if we ignore the fact that the other
>planets will likely cause the orbit to become chaotic. So velocities ranging
>from about 2.8 km/s and 14 km/s will most likely enter a chaotic solar
>orbit. (of course, there is also a slight chance that it will hit the earth
>before completing a full orbit).

Why would you expect a solar orbit to be chaotic? Of course, in the
strictest terms every body in the Solar System is in a chaotic orbit,
but their regions of stability are very wide, so these orbits tend to
stay about the same for billions of years. I'd expect a meteoroid
sized body in solar orbit at around 1 AU to be substantially stable,
except for the gradual decay from drag effects.

>5. Higher ejection velocities allow the possibility of the material leaving
>the solar system entirely.

Models and experiments suggest that the maximum ejection speed for
bulk material disrupted by an impact isn't much more than about 5
km/s. Above that, the forces involved will vaporize the material, so
you'll just end up with a lot of gas and dust- no meteoroids to speak
of (certainly, nothing that will survive intact to the surface of the
Earth, other than fine dust).

There is an exception to this, which involves the ejection at high
speed of surface material because of high pressures just beneath the
surface during an impact. This can only account for a tiny fraction of
the total volume of the final crater, but does explain why lunar
meteorites are less shocked than more simple theory would suggest, and
why it would seem possible that meteorites from Venus might exist
(although none has ever been found).

[Chris L Peterson]

On Wed, 1 Feb 2012 16:28:48 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>Roughly 50% of those lunar orbital trajectory items with their initial
>escape velocity of 2.4+ km/sec are going to be headed toward the orbit
>of Earth more likely than anyplace else.

Why?

[Brad Guth]

On Feb 1, 9:52 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> "Peter Webb" <r.peter.webb...@gmail.com> wrote in message

Yes, under 2.4 km/sec stays with the moon.

[Brad Guth]

On Feb 1, 10:02 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Wed, 1 Feb 2012 16:28:48 -0800 (PST), Brad Guth
>

> <bradg...@gmail.com> wrote:
> >Roughly 50% of those lunar orbital trajectory items with their initial
> >escape velocity of 2.4+ km/sec are going to be headed toward the orbit
> >of Earth more likely than anyplace else.
>
> Why?

Because most orbital stuff keeps returning to it's source, and Earth
is a much better gravity influence than the moon where that ejected
material got away from. Half of it would likely head away from the
Earth+moon influence. A public owned supercomputer run simulation
would help get this better understood.

[Peter Webb]

"Chris L Peterson" <c...@alumni.caltech.edu> wrote in message

news:r78ki7loggo50443i...@4ax.com...

> On Thu, 2 Feb 2012 13:41:28 +1100, "Peter Webb"
> <r.peter...@gmail.com> wrote:
>
>>By my calculations, the outcomes are as follows:
>>
>>1. Material ejected at less than the orbital velocity of the moon - 1.7
>>kms/sec - will always fall back.
>
> I agree.
>
>>2. Material ejected at velocities of between 1.7 kms/sec and 2.4 kms/sec
>>(the escape velocity of the moon in isolation) will either fall back to
>>the
>>moon or go into orbit around the moon, depending on the angle at which it
>>is
>>ejected.
>
> Yes... but all such material will rather rapidly fall back to the Moon
> in any case, because lunar orbits are unstable. That's why no
> spacecraft has ever been put into a long term lunar orbit.
>

Worse, the "orbit" will intersect the moon's surface, as I stated in my
correction.

>>3. The delta V to achieve earth escape velocity for an object in the same
>>orbit as the moon is about 0.4 km/s. So material ejected with a velocity
>>of
>>between 2.4 km/s and 2.8 km/s will enter a chaotic orbit bound to the
>>earth,
>>and very likely eventually hit the earth (and less likely eventually hit
>>the
>>moon).
>
> I'm curious about your calculations on this one. Chaotic three-body
> orbits don't typically result in the low mass object hitting one of
> the larger masses, but are much more likely to end up with the low
> mass object being ejected from the system entirely.

Typically. And you might be right. But the moon has such a low orbital speed
(is so far away) from the earth and its gravity is so weak that there will
be limitted gravitational slingshot effects. But as I say, you may well be
right.

> In that case, the
> debris will end up in a solar orbit, and various drag effects will
> result in it ultimately hitting the Sun.

Dunno about that. It will continually cross the earth's orbit, plenty of
scope for other things to happen. If it is in the plane of the ecliptic,
there are also other planets to consider.

>
>>4. The delta V to achieve solar escape orbit from earth orbit is (if I
>>recall correctly) about 12 kms/sec if we ignore the fact that the other
>>planets will likely cause the orbit to become chaotic. So velocities
>>ranging
>>from about 2.8 km/s and 14 km/s will most likely enter a chaotic solar
>>orbit. (of course, there is also a slight chance that it will hit the
>>earth
>>before completing a full orbit).
>
> Why would you expect a solar orbit to be chaotic? Of course, in the
> strictest terms every body in the Solar System is in a chaotic orbit,
> but their regions of stability are very wide, so these orbits tend to
> stay about the same for billions of years. I'd expect a meteoroid
> sized body in solar orbit at around 1 AU to be substantially stable,
> except for the gradual decay from drag effects.

I wouldn't, at least not if stays in the plane of ecliptic. The only stable
orbits a 1 AU are at the two Lagangian points at +- pi/3. I would expect
that the objects would at some point intersect the earth's orbit near the
earth and be kicked into a more eccentric orbit, and eventually get evicted
entirely (as you point out, this is the usual outcome of three body problems
where one has negligible weight). But I don't really know. This stuff is
very difficult to do numerically on a computer with any confidence, you have
to balance step sizes vs rounding errors, very unstable calculation.

>
>>5. Higher ejection velocities allow the possibility of the material
>>leaving
>>the solar system entirely.
>
> Models and experiments suggest that the maximum ejection speed for
> bulk material disrupted by an impact isn't much more than about 5
> km/s. Above that, the forces involved will vaporize the material, so
> you'll just end up with a lot of gas and dust- no meteoroids to speak
> of (certainly, nothing that will survive intact to the surface of the
> Earth, other than fine dust).

OK. Sound plausible.

>
> There is an exception to this, which involves the ejection at high
> speed of surface material because of high pressures just beneath the
> surface during an impact. This can only account for a tiny fraction of
> the total volume of the final crater, but does explain why lunar
> meteorites are less shocked than more simple theory would suggest, and
> why it would seem possible that meteorites from Venus might exist
> (although none has ever been found).
>

Without a moon, Venus doesn't have as many opportunities to kick material
into different orbits. And we are in wrong direction. It would have to have
enough speed to not only achieve Venus escape velocity but do also to get up
to 1 AU. Venus's escape velocity is about 11 kms/s, and you can add the
delta V to get to 1 AU on top of that. If your 5 km/s number is correct, it
ain't going to happen.

[Brad Guth]

On Feb 2, 1:44 am, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> "Chris L Peterson" <c...@alumni.caltech.edu> wrote in messagenews:r78ki7loggo50443i...@4ax.com...

There's also way too much atmospheric drag for anything to get away
from Venus, regardless of the impactor velocity.

[oriel36]

On Feb 1, 11:38 pm, Dr J R Stockton <reply1...@merlyn.demon.co.uk>
wrote:

> So anything leaving the Moon's surface at more than a little over the
> Moon's escape velocity will escape from the Earth-Moon system, unless in
> that small fraction of trajectories that hit Earth on the way out.
> Approximately none of that, however, will escape from the Solar System,
> so may return to hit Earth eventually.
>
> Anything (on or) leaving the Moon's surface will of course be attracted
> by the Earth; but, except for that which escapes the Earth-Moon system
> or hits Earth immediately, or grazes the atmosphere to re-enter not very
> much later, it will enter a stable elliptical orbit around the Earth -
> until that orbit, reaching as it must as far out as the Moon's orbit, is
> perturbed by a close passage with the Moon.
>

You still insist in following the ideology of a guy who actually
believes the moon rotates !!! -

http://books.google.ie/books?id=gB2-Hqdx_LUC&pg=PA580&dq=newton+moon+rotates&hl=en&ei=SQJ5TJP1FYTKswadoL2yDQ&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDkQ6AEwAw#v=onepage&q&f=false

Setting aside the orbital inputs which actually caused the observed
phenomena of libration,it is unthinkable that anyone who remotely
values their intelligence imagines the moon has variations in
latitudinal speeds indicative of intrinsic rotation.

You can see the Earth rotating from any point in space and in the
following time lapse footage you can actually see the moon orbiting
the Earth -

http://www.youtube.com/watch?v=YXCnxoixb-s

I have to shake my head,men who would normally just take account of
point masses without mixing it up with orbital traits of celestial
satellites and planets,and their respective orbital motions are
entirely distinct,insist in something as mindnumbing as intrinsic
lunar rotation on account of some guy in the late 17th century.

Again,I still refuse to believe that people can state as a fact that
the moon rotates when all they have to do is look out their window and
see it doesn't.

[Brad Guth]

On Feb 2, 2:37 am, oriel36 <kelleher.ger...@gmail.com> wrote:
> On Feb 1, 11:38 pm, Dr J R Stockton <reply1...@merlyn.demon.co.uk>
> wrote:
>
> > So anything leaving the Moon's surface at more than a little over the
> > Moon's escape velocity will escape from the Earth-Moon system, unless in
> > that small fraction of trajectories that hit Earth on the way out.
> > Approximately none of that, however, will escape from the Solar System,
> > so may return to hit Earth eventually.
>
> > Anything (on or) leaving the Moon's surface will of course be attracted
> > by the Earth; but, except for that which escapes the Earth-Moon system
> > or hits Earth immediately, or grazes the atmosphere to re-enter not very
> > much later, it will enter a stable elliptical orbit around the Earth -
> > until that orbit, reaching as it must as far out as the Moon's orbit, is
> > perturbed by a close passage with the Moon.
>
> You still insist in following the ideology of a guy who actually
> believes the moon rotates !!! -
>

> http://books.google.ie/books?id=gB2-Hqdx_LUC&pg=PA580&dq=newton+moon+...

>
> Setting aside the orbital inputs which actually caused the observed
> phenomena of libration,it is unthinkable that anyone who remotely
> values their intelligence imagines the moon has variations in
> latitudinal speeds indicative of intrinsic rotation.
>
> You can see the Earth rotating from any point in space  and in the
> following time lapse footage you can actually see the moon orbiting
> the Earth -
>
> http://www.youtube.com/watch?v=YXCnxoixb-s
>
> I have to shake my head,men who would normally just take account of
> point masses without mixing it up with orbital traits of celestial
> satellites and planets,and their respective orbital motions are
> entirely distinct,insist in something as mindnumbing as intrinsic
> lunar rotation on account of some guy in the late 17th century.
>
> Again,I still refuse to believe that people can state as a fact that
> the moon rotates when all they have to do is look out their window and
> see it doesn't.

Actually it does rotate, though quite slow. From the outside looking
in towards Earth with its moon is where it's clearly rotating in
addition to orbiting Earth.

[Chris L Peterson]

On Thu, 2 Feb 2012 20:44:04 +1100, "Peter Webb"
<r.peter...@gmail.com> wrote:

>> Why would you expect a solar orbit to be chaotic? Of course, in the
>> strictest terms every body in the Solar System is in a chaotic orbit,
>> but their regions of stability are very wide, so these orbits tend to
>> stay about the same for billions of years. I'd expect a meteoroid
>> sized body in solar orbit at around 1 AU to be substantially stable,
>> except for the gradual decay from drag effects.
>
>I wouldn't, at least not if stays in the plane of ecliptic. The only stable
>orbits a 1 AU are at the two Lagangian points at +- pi/3. I would expect
>that the objects would at some point intersect the earth's orbit near the
>earth and be kicked into a more eccentric orbit, and eventually get evicted
>entirely (as you point out, this is the usual outcome of three body problems
>where one has negligible weight). But I don't really know. This stuff is
>very difficult to do numerically on a computer with any confidence, you have
>to balance step sizes vs rounding errors, very unstable calculation.

Models that deal with these sorts of interactions are now very good,
and very accurate.

The reason I said "about" 1 AU was to make it clear that I'm talking
about a meteoroid near, but not at, that distance. Unless it is meters
across (unlikely), typical meteoroid debris will move inwards towards
the Sun fairly quickly, so the significance of later interactions with
the Earth/Moon decreases over time.

>Without a moon, Venus doesn't have as many opportunities to kick material
>into different orbits. And we are in wrong direction. It would have to have
>enough speed to not only achieve Venus escape velocity but do also to get up
>to 1 AU. Venus's escape velocity is about 11 kms/s, and you can add the
>delta V to get to 1 AU on top of that. If your 5 km/s number is correct, it
>ain't going to happen.

I think you missed the point of the exception. While the majority of
the material from an impact crater experiences forces too high to
survive in bulk form, material right at the surface can be ejected at
a speed comparable to the incoming speed of the impactor, due to the
way the high pressure shock from beneath is coupled to it. This is
probably the mechanism by which debris is ejected from a planet or
large moon after an impact. It also allows for material to come from
planets with escape velocities well above 5 km/s (like Venus), or from
planets deep in the Sun's gravity well (like Mercury). Recent
publications have placed quantitative values on the number of Mercury
and Venus meteorites that should fall on the Earth.

[Chris L Peterson]

On Wed, 1 Feb 2012 22:26:28 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>Because most orbital stuff keeps returning to it's source, and Earth
>is a much better gravity influence than the moon where that ejected
>material got away from. Half of it would likely head away from the
>Earth+moon influence. A public owned supercomputer run simulation
>would help get this better understood.

The source is the Moon, and that's where most stuff will end up. At
the Moon, the Moon exerts more gravitational influence on bodies than
the Earth does.

You might want to review the literature. Lunar impacts have been
modeled, and the results published.

[Brad Guth]

On Feb 2, 6:53 am, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Wed, 1 Feb 2012 22:26:28 -0800 (PST), Brad Guth
>

About a third of the crater ejected material should have stuck with
the moon. Unfortunately, going by our NASA/Apollo data, not even 3%
managed to stick with the moon.

Remember those volumes of ejected and/or displaced material from that
2500 km crater, plus everything else that's cratered isn't
insignificant. There's also something like 10 tonnes of those
extremely small meteors (under 0.1 kg) and dust that's still arriving
every day.

[Dr J R Stockton]

In sci.astro.amateur message <jg7svj$qai$1...@news.albasani.net>, Tue, 31
Jan 2012 16:07:03, Peter Webb <r.peter...@gmail.com> posted:

>There are an infinite number of points where this can be done. Just
>fire your gun from the highest point on any great circle.

Perhaps you should be told that the Moon rotates, with an equatorial
velocity of about 2 pi 1000 miles per month - about 9 mph - ale that low
lunar orbit takes a couple of hours. So, for a transpolar hit, the
GuthButt would need to stretch for about 18 miles to the side of the
Great Divide.


>And BTW I suspect I have made an error when I said that you could
>probably fire a rifle from the moon into the earth's atmosphere. You
>also have to bleed off the horizontal component of the moons velocity
>relative to the earth. That's another 1 km/sec or so, meaning that you
>would need a rifle with a muzzle velocity of about 4 km/sec. That would
>be quite difficult to achieve with a hand held weapon. Probably about
>the best that you could get from a rifle in practice would be putting
>the bullet into earth orbit.


So you think that to escape from the Moon with a relative velocity of 1
kph, it is necessary to fire at 1 kph over escape velocity? Before
writing about elementary physics, you should learn some elementary
physics.

--
(c) John Stockton, near London. *@merlyn.demon.co.uk/?.?.Stockton@physics.org
Web <http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, and links.
Correct <= 4-line sig. separator as above, a line precisely "-- " (RFC5536/7)
Do not Mail News to me. Before a reply, quote with ">" or "> " (RFC5536/7)

[Chris L Peterson]

On Thu, 2 Feb 2012 09:27:06 -0800 (PST), Brad Guth
<brad...@gmail.com> wrote:

>About a third of the crater ejected material should have stuck with
>the moon.

So you say, but you provide no support for the assertion, either your
own work or something published. A number pulled from your ass this
way is not convincing.

In fact, the vast majority of material ejected from an impact crater
has been vaporized, and such material will not land on the Earth as
anything other than microscopic dust (what little actually is ejected
into an orbit which intersects Earth to begin with).

[Sam Wormley]

On 2/2/12 11:27 AM, Brad Guth wrote:
> About a third of the crater ejected material should have stuck with
> the moon. Unfortunately, going by our NASA/Apollo data, not even 3%
> managed to stick with the moon.

Please show calculation or cite publications supporting this
assertion of your, Brad.

[Brad Guth]

I seem to recall those Apollo missions supposedly couldn't probe or
hand-drill much over a few cm depth into that magical clumping and
terrific surface tension capable dust, before getting stuck by hitting
fused bedrock.

In many locations that fluffy dust should have been nearly
bottomless. Did any of the Apollo missions fail to land upon and
otherwise hit bedrock with their hand operated probes? (sure as hell
couldn't have been any compacted hard soil)

[Brad Guth]

On Feb 2, 6:43 pm, Chris L Peterson <c...@alumni.caltech.edu> wrote:
> On Thu, 2 Feb 2012 09:27:06 -0800 (PST), Brad Guth
>

That 2500 km crater wasn't mostly vaporized.

[Peter Webb]

"Dr J R Stockton" <repl...@merlyn.demon.co.uk> wrote in message
news:muYGy4Iw...@invalid.uk.co.demon.merlyn.invalid...

> In sci.astro.amateur message <jg7svj$qai$1...@news.albasani.net>, Tue, 31
> Jan 2012 16:07:03, Peter Webb <r.peter...@gmail.com> posted:
>
>>There are an infinite number of points where this can be done. Just
>>fire your gun from the highest point on any great circle.
>
> Perhaps you should be told that the Moon rotates, with an equatorial
> velocity of about 2 pi 1000 miles per month - about 9 mph - ale that low
> lunar orbit takes a couple of hours. So, for a transpolar hit, the
> GuthButt would need to stretch for about 18 miles to the side of the
> Great Divide.
>
>
>>And BTW I suspect I have made an error when I said that you could
>>probably fire a rifle from the moon into the earth's atmosphere. You
>>also have to bleed off the horizontal component of the moons velocity
>>relative to the earth. That's another 1 km/sec or so, meaning that you
>>would need a rifle with a muzzle velocity of about 4 km/sec. That would
>>be quite difficult to achieve with a hand held weapon. Probably about
>>the best that you could get from a rifle in practice would be putting
>>the bullet into earth orbit.
>
>
> So you think that to escape from the Moon with a relative velocity of 1
> kph, it is necessary to fire at 1 kph over escape velocity? Before
> writing about elementary physics, you should learn some elementary
> physics.
>

Touche and well spotted. Only an additional 0.12 k/s by my calculation - is
that what you get?

[oriel36]

On Feb 2, 7:16 pm, Dr J R Stockton <reply1...@merlyn.demon.co.uk>
wrote:
> In sci.astro.amateur message <jg7svj$qa...@news.albasani.net>, Tue, 31
> Jan 2012 16:07:03, Peter Webb <r.peter.webb...@gmail.com> posted:

>
> >There are an infinite number of points where this can be done. Just
> >fire your gun from the highest point on any great circle.
>
> Perhaps you should be told that the Moon rotates, with an equatorial
> velocity of about 2 pi 1000 miles per month - about 9 mph - ale that low
> lunar orbit takes a couple of hours.

People who imagine that the moon rotates are not in possession of
their senses,it doesn't require an explanation because not a single
person who has ever lived has seen a rotating moon,they is no mention
of it in any texts both ancient or recent up to the time Isaac decided

the moon has variations in latitudinal speeds indicative of intrinsic
rotation.

The Earth rotates,the moon orbiting the Earth does not rotate and
orbital motion is separate to intrinsic rotation,people know this and
why ?,because they can actually see it directly -

http://www.youtube.com/watch?v=JxrIMHKobk0

The insistence in sticking with something so astronomically perverse
as a 'lunar moon' arrives from a certain personality who is impelled
to follow Newton at all costs.Neither Poe nor his contemporaries,those
before him or after him ever understood the ideology of Newton but I
certainly have and it is far worse than anyone can think.

"To explain: — The Newtonian Gravity — a law of Nature — a law whose
existence as such no one out of Bedlam questions — a law whose
admission as such enables us to account for nine-tenths of the
Universal phænomena — a law which, merely because it does so enable us
to account for these phænomena, we are perfectly willing, without
reference to any other considerations, to admit, and cannot help
admitting, as a law — a law, nevertheless, of which neither the
principle nor the modus operandi of the principle, has ever yet been
traced by the human analysis — a law, in short, which, neither in its
detail nor in its generality, has been found susceptible of
explanation at all — is at length seen to be at every point thoroughly
explicable, provided we only yield our assent to —— what? To an
hypothesis? Why if an hypothesis — if the merest hypothesis — if an
hypothesis for whose assumption — as in the case of that pure
hypothesis the Newtonian law itself — no shadow of à priori reason
could be assigned — if an hypothesis, even so absolute as all this
implies, would enable us to perceive a principle for the Newtonian law
— would enable us to understand as satisfied, conditions so
miraculously — so ineffably complex and seemingly irreconcileable as
those involved in the relations of which Gravity tells us, — what
rational being could so expose his fatuity as to call even this
absolute hypothesis an hypothesis any longer — unless, indeed, he were
to persist in so calling it, with the understanding that he did so,
simply for the sake of consistency in words?" Allan Poe

This was not immediately apparent in the mid 19th century what that
was written but it sure is now as every single known fact,including
the one where we landed on the moon and look out from a non rotating
moon,is still under the tyranny of the late 17th century strain of
empiricism.

Is this what astronomy comes down to ?,an entire generation prepared
to remain silent on something so easy to understand as what a rotating
object looks like from any place,distance or angle such as the
rotating Earth and what is orbiting and not rotating like the moon.

[Brad Guth]

Except viewed from outside the Earth-moon orbital trek, the moon is
spinning right along with Earth that's also spinning, as is our sun
and every object discovered thus far has some kind of spin or rotation
to deal with. If you didn't want spin or rotations, it would be a
very depressing universe.

[oriel36]

I look at the old texts whether Kepler or Galileo, Plutarch or
Poe,Blake or Pascal and they are brimming with that intuitive
intelligence that makes the heart sing when they are encountered and
far from this contemporary pestilence where there are individuals who
imagine a rotating moon.One of the nicest ancient commentaries that is
well worth reading is Plutarch as with all those who have an abiding
love of putting observations into perspective and even with only a
limited amount of information compared to this era,they do not distort
things to suit a conclusion but work with whatever information they
have to assemble a picture.

The magnification guys are far to involved with the night sky,a glance
at the Sky and Telescope website is enough to affirm this,to be drawn
into the type of reasoning where the idea of objects above the Earth
are set aside for the perceptions of a celestial arena and motions in
that arena and insofar as magnification astronomy becomes little more
than a celestial pantomime with objects turning into view and out
again,only lip service is paid to individual details where the
intuitive mind feels most comfortable.So where the magnification guys
look up,the people who make sense of details look out and that is the
way it has always been -

"And we do not go wrong, I think, when we assign to those bodies above
denominated such immense depth and distance, and leave to that which
is below a certain circular course and broadway as much as lies
between earth and the moon: for neither the man who pretends the
summit of heaven to be the sole ‘above,’ and denominates all the rest
as ‘below,’ is reasonable in his definition; nor yet is he who
circumscribes ‘below’ by the limits of Earth, or rather by the Center,
to be listened to: but even moveable. . . . inasmuch as the universe
allows of the interval required by reason of its own extensiveness.
But in reply to such as demand that all which is separate from earth
shall be consequently ‘above’ and ‘on high,’ another directly responds
with the contrary axiom, that all which is reckoned from the fixed
circumference is to be considered as ‘below.’

XI. “And, finally, in what sense, and in reference to what thing is
Earth said to be ‘intermediate?’ For the universe is infinite; now
that which is infinite hath neither beginning nor limit, so it does
not belong to it to possess a middle: for infinity is the deprivation
of limits. But he who makes out Earth to be the middle not of the
universe, but of the world, is ridiculous for his simplicity if he
does not reflect that the ‘world’ itself is liable to the very same
objections: for the universe hath not left a middle place for it also,
but it is borne along without house or home in the boundless vacuum,
towards nothing cognate to itself; perhaps it has found out for itself
some other cause for remaining fixed, and so has stood still, but
certainly not owing to the nature of its position." Plutarch

http://thriceholy.net/Texts/Moon.html

Living in an era where people try to impose initial states to time and
space is one of the saddest intellectual conditions ever faced by any
generation,the people who propose limits will not consider these
things themselves for they know they would quickly lose their minds if
they had to consider a time without time or a space without space yet
they do not mind dictating these horrors to the wider population as
achievements and especially to the vulnerable student.

That treatise by Plutarch rightly considered irradiance as the reason
why the face of the moon appears as it does as opposed to any
intrinsic light such as the Sun and it considers many,many other
aspects of the moon but never have any of the astronomical authorities
ever once considered intrinsic rotation,not Galileo,not
Copernicus,Plato or any the ancient people for the simple reason that
it does not have any as you look out at it orbiting the Earth.

[Brad Guth]

Venus meteorites falling on Earth would be astronomically unlikely.
However, from our extremely nearby and massive moon that's naked, it
should be relatively commonplace for lunar basalt and other displaced
crust to land on Earth.

Those simulators are not very good because they do not involve
lithobraking encounters, so we can argue about the numbers, but not
about the likelihood of such moon rock easily landing on Earth.

BTW; the impactor that tilted Earth was perhaps the same diameter and
mass as our moon.

[Dr J R Stockton]

In sci.astro.amateur message <jggd37$unc$1...@news.albasani.net>, Fri, 3
Feb 2012 21:30:51, Peter Webb <r.peter...@gmail.com> posted:

No, not exactly; but you do not say what inputs you are using. If you
are taking lunar escape velocity as 3 kps and lunar orbital speed as 1
kps, then I think that an extra Root(3.0^2 + 1.0^2) - 3.0 is needed
: just over 162 m/s more. And a little bit less since the Earth is not
a point target.


>> --
Delenda est.

[Peter Webb]

"Dr J R Stockton" <repl...@merlyn.demon.co.uk> wrote in message

news:g9x3mCUS...@invalid.uk.co.demon.merlyn.invalid...

I used the same formula but with 4 k/s as a base. Our answers differ by 400
meters/sec for this reason.