[meteorite-list] Neutron and Proton productioninhyper-velocityimpacts

[Sterling K. Webb]

Hi, Darren, List,

    The natural abundance of Carbon 14 on Earth is
one part per trillion or 10^-9. Your example, which
produces an atmosphere with a C14 abundance of
1.51x10^-9, is a 50% enrichment of the natural terrestrial
abundance, which would be a whopping big spike in
the curve.

    Living critters that absorbed and incorporated it
would be a red flag in the archaeological record for
thousands of years. But they'd have company, because
it would blend into the 100% enrichment spike in the
1960's from above-ground nuclear bomb testing!

    And even one tiny nuclear war would produce a
spike that would tower over everything...


Sterling K. Webb
--------------------------------------------------------------------
----- Original Message ----- 
From: "Darren Garrison" <cynapse at charter.net>
To: <meteorite-list at meteoritecentral.com>
Sent: Friday, December 28, 2007 3:35 PM
Subject: Re: [meteorite-list] Neutron and Proton 
productioninhyper-velocityimpacts


On Fri, 28 Dec 2007 14:09:36 -0600, you wrote:

>the Hulk comic character!  I am the first to respect a thought experiment:
>But what scientific experiment could you propose (or has one already been
>done?) to follow through?

Okay, I've done a bit of thought experiment on this too-- and bear with me
because though hard math has never been my strong suit, I think I have this
right.

Okay, let's just for a moment imagine that a hypervelocity bolide DOES 
generate
C14.  Let's say that a 100 meter bolide converts 100 percent of all carbon 
into
C14 in a column 1,000 meters in diameter.  For the sake of simple math, 
let's
say that the top of the atmosphere is defined as 100 km up
http://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line and the bolide came in at 
a
90 degree angle.  That hypothetical meteoroid would convert all the C to C14 
in
a volume of around 78.5 km3 by the formula for the volume of a cylinder 
being
pi*r2*h.

Now, to determine the volume of the atmosphere, I took a figure for the 
diameter
of the Earth as 12,756 km.  I then added the 100 km for the atmosphere on 
either
side to get a figure of 12,956 km for the Earth plus the atmosphere.  Then
plugged both numbers into the formula for the volume of a sphere, 4/3*pi*r3.
Subtracted the second from the first to get the volume of the Earth's
atmosphere, 51,924,386,581 km3.

So, after all that, you take the volume of the Earth's atmosphere, compare 
it to
the volume of C14 converted atmosphere in that hypothetical column of air
created by the bolide, and you get around 1.51*10e-9 of the Earth's 
atmosphere
converted, or around 1 part in 660,000,000.

Earth diameter = 12,756 km
Earth volume = 1,086,783,833,910 km3

Earth+a diameter = 12,956 km
Earth+a volume = 1,138,708,220,492 km3

Atmosphere volume = 51,924,386,581 km3

meteoroid path = 78.53975 km3

1.51*10e-9

660,000,000
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