[meteorite-list] Witnessed fall lunars?

From: Randy Korotev <korotev_at_meteoritecentral.com>
Date: Wed, 08 Sep 2010 13:39:19 -0500
Message-ID: <201009081839.o88Id1X12214_at_levee.wustl.edu>

Dear Sterling:

Thanks so much for that enlightening explanation!

Randy Korotev


At 10:32 PM 2010-09-07 Tuesday, you wrote:
>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 Wed 08 Sep 2010 02:39:19 PM PDT