[meteorite-list] Lunar velocities...

From: Kelly Webb <kelly_at_meteoritecentral.com>
Date: Thu Apr 22 09:44:43 2004
Message-ID: <3AC2CED1.7238EA1A_at_bhil.com>

Hi, Robert, Steve, List,
    Picture the moon in a paraboloidal pocket stuck nose-first into the
earth's gravity field. The center point of the paraboloid is where the
lunar and terrestrial gravity fields are equal and opposite, also known
as the L1 LaGrange point. The surface of the paraboloid is the null
sheet defined by all the points where the two fields balance, except
that all the other points except the central one have a small vector
directed toward the central point.
    If an object reached the null sheet with almost no residual
velocity, it would fall toward the earth, depending on the direction of
its vector, accelerating toward the local escape velocity, which is just
a hair less than the complete escape velocity of 11,200 m/sec, because
while 320,000 km away is a long way, it isn't infinity. The hair less is
maybe 20-25 m/sec (ok, I didn't stop to calculate it, but it's like less
than highway speeds).
    But if the object arrives at the null sheet with a small excess
velocity, say 50 m/sec, which is nevertheless greater than that "hair
less" velocity, then when it's accelerated toward the earth, it will
achieve more than earth's escape velocity (the square root of the sum of
the squares of the excess velocity and the local escape velocity).
    If that object missed the earth (and its atmosphere), it would leave
the earth-moon system on a "no-return trajectory." (These are the last
words you want to hear if the object is a capsule with you in it!)
    So, if objects are ejected from the moon's surface with less than
lunar escape velocity, they will fall back. If objects are ejected from
the moon's surface with more than a tiny excess velocity, they're gone.
Only a very, very small percentage of objects, with a small range of
low excess velocities will remain gravitationally bound to the
earth-moon system. All of these will eventually get to the earth (or its
atmosphere) because any object orbiting the earth with a semi-major axis
less than the moon's will be perturbed in eccentricity until its apogee
is lowered into the earth's atmosphere.
    I know, the moon is just hanging there over the earth. Common
"sense" (earthbound human inituition) says things just ought to fall
down. But the truth is it's not the easiest thing to jump off the moon
and land on the earth. I would estimate that <1% of lunar ejecta would
make it to the earth through the mechanism of a gravitationally bound
    What about the rock that got away? That lucky rock, off on its own
heliocentric orbit, is in for a nasty surprise. Its orbit is virtually
the same as the earth's orbit. One lousy little rock sharing a nearly
identical orbit with a planet; it's like sleeping in the same bed with a
12,000 pound bear. Chances are good the bear will roll over in its
sleep. Most "escapees" (more than 50%) will be swept up by the earth
sooner or later, 100,000 years or more. Some (15%) will end up at the
Venus station. The fastest escapees will be the hardest to catch.
    The survivors will get their own heliocentric orbit, however whacky,
most by achieving an orbit close enough to a very minor resonance with
Jupiter to get nudged out of the way, but Jupiter is likely to toss them
anywhere. A really tiny percentage (<0.1%) will end up on Mars (with a
transit time of up to 50,000,000 years), where they will, in the future,
be auctioned off on mBay by dealers who bought them in the bazaars on
the edge of the Vastitas Borealis.

Kelly Webb

Robert Beauford wrote:

> "These "stretch" tektites are a clear indication of terrestrial origin
> as how could such a form survive if they had been formed on the moon
> and then entered the Earth's atmosphere at 7+ miles per/sec?" Steve's
> comment above got me curious. It seems like ejecta from the moon need
> not reach a cosmic velocity or earth orbital velocity, just an escape
> velocity or orbital velocity of the moon. It seems, more
> specifically, that any fragment with a speed over the escape velocity
> of the moon, and a vector that points it inside the orbit of the moon
> around the earth (slightly less than 50% of all lunar ejecta?), and
> insufficient residual speed to send it back out of that orbit, would
> fall to earth. This should alow for some very slow objects reaching
> earth (and thus less ablation loss and entry alteration) What is this
> necessary and maximum speed? Please note, I am not advocating a lunar
> origin for tektites, just idly, and perhaps ignorantly, wondering if
> lunar meteorites might enter the atmosphere slower than some other
> meteorites.-Robert Beauford : )
Received on Thu 29 Mar 2001 12:57:38 AM PST

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