[meteorite-list] Temperature of meteorites

[Chris Peterson]

Sterling- I think you underestimate the effect of convective heat transfer 
during cold flight. A fist-sized meteorite might fall for a good three to 
five minutes through -40°C air, at around 100 m/s. That is long enough for 
the entire stone to equilibrate to that temperature. In the last minute or 
so of flight it will generally be in warmer air, and will therefore start to 
warm up- but probably not to equilibrium. The critical point here is that 
the meteorite will not maintain an interior temperature similar to its 
temperature in space. The exception would be a larger stone that remains 
hypersonic to a lower height, and therefore spends less time in dark flight.

We don't really care what the temperature was for the parent's millions of 
years in space. For any given distance from the Sun, it shouldn't take more 
than a few days to reach equilibrium, and any meteorite can be assumed to 
come from a parent that was at 1 AU for that long. So the only real variable 
is emissivity.

Chris

*****************************************
Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com


----- Original Message ----- 
From: "Sterling K. Webb" <sterling_k_webb at sbcglobal.net>
To: "Meteorite List" <meteorite-list at meteoritecentral.com>
Cc: "Chris Peterson" <clp at alumni.caltech.edu>; "Bernd Pauli" 
<bernd.pauli at paulinet.de>; "Larry Lebofsky" <lebofsky at lpl.arizona.edu>; 
<meteorite-list at meteoritecentral.com>
Sent: Tuesday, November 23, 2010 1:18 PM
Subject: Re: [meteorite-list] Temperature of meteorites


> Some points for the debate:
>
> The rapid flight through the atmosphere is very brief --
> 1-2 seconds. This is not much time to change the
> temperature of the stone.
>
> The rate at which the friction-generated heat is
> transferred to the interior of the stone is determined
> by the thermal conductivity of that rock, and rock's
> thermal conductivity is very low, so low that virtually
> none of the heat will affect temperatures deeper
> than a few millimeters or a centimeter into the stone.
>
> Most of that heat generated by friction on the outer
> surface goes into melting rock which is then is removed
> from the meteorite by on-going ablation.  The molten
> material stripped from the stone takes that heat with it
> as it becomes the particles in the trail (which have their
> own thermal evolution that does not affect the stone).
> Only a small fraction is "wasted" by warming the stone
> itself.
>
> That said, thermal equilibrium of the stone is likely
> achieved (or nearly) within a very short time once it
> lands. Its temperature will be more-or-less whatever
> it was before it encountered this obstructive planet.
> Apart from some rough treatment of the surface, the
> stone's temperature is the same as it always was.
>
> So, what temperature WAS the meteoroid in the many
> thousands or millions of years that it orbited the sun?
>
> That depends on what its orbit was, or more precisely,
> WHERE its orbit was and its emissivity and reflectivity
> and so on. Take a look at the following chart of Meteoroid
> Temperature vs. Solar Distance, supplied by MexicoDoug:
> http://www.diogenite.com/met-temp.html
>
> It is a model derived from fairly complete and reasonable
> assumptions, which were discussed on this List long ago:
> http://six.pairlist.net/pipermail/meteorite-list/2005-January/007521.html
> This is the first of three parts; follow the links for #2 and
> #3.  Those with more factors to include are welcome to
> refine the model, I'm sure.