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(meteorobs) Re: average meteor (was: Snakes and meteors)

> First you have to determine what magnitude of meteor is considered "typical".
> There are more smaller meteors than the next size larger. For me I plot more
> +2 magnitude meteors than anything else it seems. So, for +2 magnitude meteors
> when they were meteoroids, I guess their size is something near that of a
> grain of sand. 

I wouldn't agree with this. If you take for avg. +2 magnitude, mass of
the meteoroid should be of the order (1/100g<) m~1/10g (<1g) (see below),
which makes it about the size of pea, peppercorn or kernel of wheat, say.
There is general agreement about order of magnitude of meteoroid mass. Or,
could it be that sandy beaches in California are not that 'sandy' as they
appear in 'Bay Watch', so George's definition of sand grain is a bit
different from mine ;>>>?

> Velocity has something to do with brightness however. The
> faster a meteor is, the more energy it has when it encounters the atmosphere.
> Thus a particle will seem a little brighter. But I think you will be safe to
> think of +2 meteors as being about sand grain size.

Brightness (and intensity of radiation) depends on velocity as ~ v^5, and
on mass as ~m^(2/3) (I use ~ for proportional). Here is simplified reason
how we come about that (you may skip to the next paragraph if it's too
techical). Mass lost from the meteoroid (dM) is proportional  to the
amount of heat transfered to meteoroid from atmosphere. Heat  released in
unit time (by friction) is proportional to the kinetic energy of the air
relative to the meteoroid. This is ~v^2 * dm, but dm~v (dm = mass of air
cylinder encountered by meteoroid in unit time ~v). So, it's dM~v^3, by
now. Now, energy spent on radiation is proportional to the  kinetic energu
of the evaporated mass. Thus, E~v^2 * dM, or in total E~v^5! Finally,
M^(2/3) dependence comes because all the heat is transfered through the
surface of the meteoroid, so amount transfered is proportional to the
surface area, which is S~M^(2/3). Once we have I ~ v^5 * M^(2/3), its easy
to see that the magnitude goes as mag ~ 2.5 * (5 log v + 2/3 log M). 

Now, here is the point: magnitude is crucialy dependent on velocity, and
weakly dependent on mass. Capricornid (v=23km/s) and Leonid (v=71km/s) of
the same mass may differ by more than five magnitudes! Or these two of the
same magnitude differ in mass by factor 1000! (Ouch, this is really
awfully lot...am I making some mistake? I'm improvising this out of my
head, and I don't have time to check it now. Correct me if I got smth.
wrong. Anyway, the difference should be huge! I think it's OK, but note
that you can apply this model only for small masses, and Capricornids
probably need some corrections.) And therefore comes the ambiguity about
the mass of average meteor, whatever your definition of average is. It may
have a sense if you specify the shower; however even then you'll get only
approximate answer, because coeficients in the theory are not well known. 

If you want to play with this, pick up the source code of the program that
will give you mass and magnit. of the meteoroid as funct. of time for
given set of initial conditions (velocity, altitude, mass, and zenith.
dist.), at http://alas.matf.bg.ac.yu/~mn92187/marina.txt. If you read
Russian, excellent reference for this is Bronsten's 'Physics of Meteor
Phenomena' ('Fizika meteornyh yavlenij'), as I already mentioned here
couple of times. If you know any good refernce for physics of meteors in
English, please let me know.

Hmmm...well, that's it. When I think it over, could be that in some
extreme case (Leonids) size could be about the sand grain..dot californian
one, perhaps. ;) 

Cheers, everybody! :)

Vladimir
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