[meteorite-list] Neutron and Protonproductioninhyper-velocityimpacts

From: Sterling K. Webb <sterling_k_webb_at_meteoritecentral.com>
Date: Sat, 29 Dec 2007 14:00:08 -0600
Message-ID: <0ae801c84a55$6a3e4300$b64fe146_at_ATARIENGINE>

Hi, Doug and Crowd,

> ...100% conversion [?]

    As far as the efficiency of the reaction, since the
atmosphere is roughly 80% nitrogen, well, the odds of
finding a nitrogen atom are pretty good if you're a
lonesome thermal neutron. But it isn't content that
determines the conversion efficiency; it's the cross
section to the reaction. Nitrogen's is very high at ~1.9
barns (the best unit name of all). Nitrogen is just
a sucker for a thermal neutron come-on.

    Here's the periodic table arranged by the thermal
neutron cross section of each element. Note please that
the other elemental constituents of the atmosphere have
much, much smaller thermal neutron cross sections than nitrogen:
http://environmentalchemistry.com/yogi/periodic/crosssection.html

    Only hydrogen could weakly complete for those thermal
neutrons, but there's not much hydrogen available except
in water vapor. Nitrogen's the thermal neutron hog.

    Geesh! Make us go back and check everything?

    I made one mistake! I said there was 5000 tons of C14
in the atmosphere, but 5000 tons is the C14 content of
the entire planet; the C14 content of the atmosphere is
70 tons. (Everybody makes one mistake, right? It's a
gimme.) Not much for all this fuss, is it? However, it
strengthens my point about how little C14 it takes to
affect the resulting isotopic record.

    Can we talk about something else now? I'm tired of
homework.


Sterling K. Webb
--------------------------------------------------------------------
----- Original Message -----
From: "mexicodoug" <mexicodoug at aol.com>
To: <meteorite-list at meteoritecentral.com>
Sent: Saturday, December 29, 2007 9:15 AM
Subject: Re: [meteorite-list] Neutron and
Protonproductioninhyper-velocityimpacts


Hi Darren and friends,

I think your geometry is mostly fine though there is an easier way to check
the nuclear reaction asserted, left in the wake of the meteoroid. It's more
accurate and you don't even need to truncate the atmosphere:

Try this:

Km^2 "nuclear reactor" meteoroid=Pi*.5^2
=0.785

Km^2 Earth shell=4*Pi*6383^2
=512,000,000

ratio= 1 to 652,000,000

So we agree (you got 660,000,000) whether you make long cylinders and
subtract spheres and cut the atmosphere, or just look at the area as above.
Your number has over 1% less 14-C because you assumed the atmosphere mass
was equally dense from sea level to the top, but in reality the atmosphere
is creamier near the center of the cake but only frosting at the edges.
Considering half of the atmosphere is under about 6 km height or even lower
if you make it colder, 5 km is much more averaged estimate and a whole lot
less thinking and math. Maybe I should have used 4 or 6 km, but between 3-8
km gives the same answer, so that's totally moot.

Oh. But there is a minor error in the balancing these nuclear reactions
which you're not going to like, I'm surprised Sterling didn't have his
thinking cap on when he answered you. 12-Carbon isn't being converted to
14-C according to this black-box scheme - it is the 14-Nitrogen that is. I
don't memorize the atmospheric ratio of nitrogen to carbon off-hand, but I
suspect it is greater than 7000 to 1. So, at your 100% conversion, you're
off by a factor that should have, say, 7000 times more 14-C in your reality
check.

There are so many things wrong with this model at this point, I'm just gonna
sigh rather than say what I really think of hypotheses made in a one
dimensional world to solve three dimensional problems. And this post has
nothing to do with the big Mmm-word.

Best wishes,
Doug


----- 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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Received on Sat 29 Dec 2007 03:00:08 PM PST