[meteorite-list] Witnessed fall lunars?

From: Sterling K. Webb <sterling_k_webb_at_meteoritecentral.com>
Date: Tue, 7 Sep 2010 22:32:42 -0500
Message-ID: <6254D5FC37F74680ABE48BE0A4DDE5A2_at_ATARIENGINE2>

Hi, Lunar Gang, and List,

We have a situation here that needs straightening
out.

Escaping from the Moon is one thing. Getting
to the Earth is another. Here's how it starts.

An object is propelled off the lunar surface
(doesn't matter how). As soon as it's no longer
in contact with the force that impelled it, its
speed can't increase.

It can decrease, though, and it does. Lunar
gravity will pull down on it, reducing its speed
at the same rate it would gain if it fell. It goes
slower and slower. Eventually, its speed will fall
to zero and it will reverse course and start to
fall back.

UNLESS its starting velocity is above or at the
Moon's escape velocity. It takes 2380 meters/sec
to escape to the point 38,000 miles from the Moon's
center to where the gravitation pull of the Earth
and the Moon are equal. If the rock started with
2381 m/sec, it will get there moving at 1 m/sec,
a crawl. After that, the important thing is: which
way was it headed?

Surrounding the Moon is a distorted spherical
(parabolic) envelope with its "pocket" pointing
directly at Earth that outlines that balancing
point between the Earth's and the Moon's "pull."
It's called the Hill Sphere (for any body). The Hill
Sphere, or equipotential point for the Moon, is
at a radius of about 38,000 miles, still over 200,000
miles from earth.

If a Lunar escapee has enough speed to reach the
Moon's Hill Sphere and cross over, it will be under
the control of the Earth's gravitational field. The
Moon has only 1/81.3 of the mass of the Earth, so
the balance point between them is much closer
to the Moon than the Earth.

Oh, if it was going very fast, it could escape the
Earth too, but the odds against that are great. No,
that rock is dam lucky to have made it to the
Translunar Gravitational Equipotential Point for
its flight.

In general, since Lunar escape velocity is low
compared to the Earth's, if a rock just barely escapes,
by the time it crosses the Border, it would be moving
very slowly, almost standing still. From the viewpoint
of the Earth, it's like someone carried a rock 'way out
there and while "standing still" far from Earth, dropped it.

Like so many borders, once you cross it, you're in
another jurisdiction. The Moon no longer has any
say in what happens to the rock that crosses the
Hill Sphere Border.

Slowly at first, it begins to fall toward Earth, but it moves
faster and faster, eventually acquiring (up to) 11,233
meters/sec, plus any starting speed, blah, blah...
Will it curve and swerve and head straight for the
Earth's central spot?

No, not often. There are a variety of outcomes and
few of them will get a rock to land on Earth. Many will
end up co-orbiting the Sun along with the Earth and
will eventually tangle with the Big Mother Planet again.

Some, that are headed more or less toward the Earth
to begin with will scream past in an asymptotic pass,
whipping around the Earth, changing direction and
picking up speed, in a home grown version of the
"gravity well" maneuver. They will tossed far and gone,
in a gentler version of what Jupiter does to anything
gets near it.

But only if they miss...

Some of those headed our way, a small percentage,
will actually "strike" the Earth, or come in at a steep
angle. They might survive to the ground... or they
might not.

A few, we lucky few, will graze the top of the Earth's
atmosphere tangentially, in a flat trajectory roughly
parallel to the surface of the planet, at about zero
degrees of altitude (relative to us). They will be moving
between 11,186 meters/sec and 13,466 meters/sec
and their chances of landing are As Good As It Gets.

That's the simple view from Physics 101. It turns out
to be more complicated, however.

NOW, we have to turn the question around and look
at it from the Moon's and the Rock's perspective. If you're
a rock looking to get the Earth, what's the best way to
leave home? That will determine what happens to you
in the long run.

So, imagine you're an indecisive rock staring at the
black Lunar sky... If you aim for where the Earth is
NOW, it won't be there when you arrive. so which way
do I go?! There are no signposts and no obvious solution...

Now, it's time to introduce you to Barbara E. Shute. Her
work can be found at the NASA Technical Reports Server:
http://ntrs.nasa.gov/search.jsp?No=10&Ne=35&N=4294963886&Ns=ArchiveName|0&as=false

I suggest "Dynamical behavior of ejecta from the moon.
Part I - Initial conditions," a PDF of which can be found at:
http://hdl.handle.net/2060/19660021054

It's just what that rock is looking for --- a road map to
Earth! However, this is pretty heavy lifting if your orbital
mechanics are rusty, like mine, although no doubt Rob
Matson will eat it up and ask for please, another bowl, sir?

First, the Moon, OUR Moon, is odd. It's a long way from
the Earth and its orbital velocity (1022 m/sec) is much
slower than its escape velocity (2380 m/sec), so when
a rock does escape the Moon's gravity, it's in for a wild
ride, as it's often going too fast or too slow for where it is.

First, to actually escape the Moon, the rock's speed has to
be greater than mere escape velocity. Escape velocity will
only get you to the Hill Sphere Border. It seems that velocities
of 2600 to 2700 meters/sec are needed to actually escape the
gravitational environment beyond the Moon's Hill Sphere..

Second, given that you're going fast enough, the one
critical factor is the angle at which you leave the Moon's
surface. There is one critical angle for each spot on the
Moon's surface that guarantees you'll get to Earth if
your speed is right. That ideal angle for the minimum
possible velocity varies depending on where on the
Moon you are, but other angles will do the job if you
are going faster.

An intriguing conclusion that it is just as easy to get
to the Earth from the "back" side as it is from the "front"
(or facing) side. That means that all our breathless
speculation about whether a Lunar meteorite COULD
have come from the Backside is wasted. It makes
NO DIFFERENCE. Each side is an equally likely
source.

However, the Eastern Hemisphere is heavily favored, and
it seems likely that everything that makes it to the Earth
came from the Moon's "East Coast." When the rock leaves
the East Hemisphere, its velocity is added to the Moon's
orbital velocity. If it's pointed right, it's on a "fast return
trajectory" toward the Earth.

But if it pops out of the Moon's gravitational control from
the West Hemisphere, it's suddenly running too slowly
in a retrograde orbit that can't be sustained. It makes a
sharp right turn and crashes back into the Moon's surface
and makes a new (smaller) crater!

If Shute's math is too much for you (show of hands?), skip
to the charts and diagrams at the end. They make things
much clearer. Shute did numerical integrations to sample
impacts, ejecta-producing events, and concludes that as
much as 3.3% of the ejecta could get to Earth.

Surviving the landing is another matter. (Isn't it always?)
After reading this, it's my impression that the Moon likely
produces much more material that arrives at the Earth
than we usually think it does, and that the short supply
of Lunaites is a "collection selection" effect, as has been
suggested.

Another impression is that it may only be the more
powerful impacts that produce Lunaites. In that case,
deliveries to the Earth may only occur at intervals and
there may be a multitude of Lunaites delivered from
each impact (although they may be scattered), in contrast
to the steady rain of meteoroids from far beyond the Earth.

I'm too Googled out to check, but is there "clustering"
of the terrestrial ages of Lunaites at irregular intervals?


Sterling K. Webb
---------------------------------------------------------------------------------
----- Original Message -----
From: "Randy Korotev" <korotev at wustl.edu>
To: <meteorite-list at meteoritecentral.com>
Sent: Tuesday, September 07, 2010 4:06 PM
Subject: Re: [meteorite-list] Witnessed fall lunars?


>
>>MikeG asks:
>
>>"Is there a theory for why there have been no witnessed falls of lunar
>>meteorites? It seems odd to me that we have 4 Martian witnessed falls
>>(Shergotty, Chassigny, Zagami, Nakhla, and almost Lafayette) and no
>>lunars."
>
> One issue is that these 5 meteorites are 5 kg, 4 kg, 18 kg, 10 kg, and
> 0.8 kg in mass. Only 3 lunars are >4 kg in mass.
>
> Another issue (probably more important) is that lunar escape velocity
> is only 2.4 km/s and very little material ejected from the Moon is
> going much faster than that. This velocity compares with 20-40 km/s
> for asteroidal meteorites. Is a rock entering the atmosphere at 2.4
> km/s going to noticeably incandesce? I don't know. I believe that
> the space shuttle hits the atmosphere at ~7.7 km/s.
>
> Melanie asks:
>
> "I asked this a while ago on Greg Catterton's forum, and I was told
> that rocks
> from the moon aren't as solid (tough) as Mars rocks, and therefore are
> less
> likely to survive entry... yet what about all these Howardites?"
>
> Although breccias, most of the lunar meteorites are very tough rocks.
> Any rock that survives being blasted off the Moon isn't going to
> disintegrate in Earth's atmosphere any more than an asteroidal or
> martian meteorite.
>
> Steve says:
> "The moon is close to the earth and material knocked off the moon has
> a relatively short time to reach the earth."
>
> Compared to what? Some lunar meteorites took a million years or more
> to reach Earth.
>
> "Mars is farther away and not protected by a companion and its closer
> to the asteroid belt so it receives many more impacts than the moon."
>
> Not "many more." Only a factor of two greater for Mars, but the
> average velocity of the impactors is only 60% as great.
>
>
>
> Randy Korotev
> Washington University in St. Louis
>
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Received on Tue 07 Sep 2010 11:32:42 PM PDT