[meteorite-list] Searching for Meteorites

From: Matson, Robert <ROBERT.D.MATSON_at_meteoritecentral.com>
Date: Thu Apr 22 09:50:22 2004
Message-ID: <AF564D2B9D91D411B9FE00508BF1C8698E5B4E_at_US-Torrance.mail.saic.com>

Hi Sterling and list,

> I'm quoting myself here (if I don't, who will?), from my prior
> post of Dec. 9, 2000, "How Many Meteorites Fall?"

> "Taking the area of the Earth to be 5.1 x 10^8 km^2 and the
> meteorite flux to be 23,930 yr^-1, this yields the assumed
> collisional cross section of the earth to be 21,360 km^2 yr^-1.
> This rate means that one meteorite per year falls on an area
> of 21,320 square kilometers. The inverse function of this value
> is how long we have to wait for a meteorite to fall on a
> standard area, or the mean time to impact: 21,360 yr km^-2.
> To put this flux into perspective, if you owned a house with a
> half-acre yard, you would have to wait 10,552,000 years for a
> meteorite to fall in your front or back yard or on your roof!
> (On average, that is; it could happen tomorrow.)"

There is one important factor that isn't included in this analysis
and that is the "footprint" of each meteorite fall (assuming that
the 23,930 meteorite/year flux rate is the number of exoatmospheric
meteoroids per year that will produce at least one meteorite that
will survive reentry).

While many falls may only produce a single stone, others produce
thousands or even tens of thousands. Another way of looking at it
is that sometimes the meteorite gods are shooting bullets, other
times shotgun blasts. Over time, when you look at the accumulated
"target hits", the amount of "blank space" is quite a bit lower
than if the shooter was always using .22 slugs.

The probability of a meteorite *find* in a particular location
then is not so much an issue of:

       T * Flux * A
   N = ------------
          Ae

N = expected # of meteorites in area A
T = accumulation time (years)
Flux = exoatmospheric earth annual flux rate (meteorites/year)
A = a given area searched (km^2)
Ae = earth-surface-area (km^2)
 
Rather, it is an issue of the mean expected distance to the
nearest element of any fall. One could best estimate this with
Monte Carlo simulations covering a wide range of initial
meteoroid sizes, amount of fragmentation, angles of reentry,
etc. But the bottom line is that due to fragmentation, the
probability of finding a meteorite on an old surface of, say,
1 square kilometer, is better than what the above equation
would tell you.

Example: T = 5000 years, Flux = 23930 met/year, A = 1 km^2,
Ae = 5.1 x 10^8 km^2. N works out to 0.235 meteorites. My
field experience covering a great deal of "old" surface area
is that N is at least twice this value.

Best,
Rob
Received on Wed 03 Apr 2002 03:38:25 PM PST