[meteorite-list] More on meteorite temperature

From: Matson, Rob D. <ROBERT.D.MATSON_at_meteoritecentral.com>
Date: Fri, 1 Jul 2016 17:47:36 +0000
Message-ID: <4A4FA25E4DFE584AA580F4F069F9B440B5F53E3D_at_EMP-EXMR104.corp.leidos.com>

Hi Larry/All,

Yes -- that's where that factor of 4 comes into the denominator. For a sphere,
one hemisphere is collecting solar radiation. If the sphere's radius is R, then the
sphere presents pi*R^2 of collecting area -- same as if it were a disk of radius R
pointing normal to the sun. However, that sphere re-radiates (mostly in the
infrared) from a 4*pi*R^2 surface area. So there ends up being a geometric
factor inside the radical that is the ratio of the solar collecting area divided
by the thermal emitting area. Now, a point I did not mention yesterday is
that thermal equilibrium does *not* mean that the entire sphere is at the same
temperature. Even if the sphere was spinning fairly rapidly (barbecue mode),
the surface temperature will still have latitude dependence. Just as on earth,
points on the surface near the equator will be warmer than those near the
poles. But the overall energy balance means that the average temperature
of the entire sphere will be constant.

But this segues into how you can end up with a meteoroid that has a much
warmer equilibrium temperature: its shape. What if the meteoroid had a
shape more like a flat plate or the disk mentioned above? Then the ratio
of the collecting area divided by the total surface area is larger, and therefore
the equilibrium temperature is higher. Let's start with the case of a circular
disk that is flipping like a tiddlywink (or a flipped coin if you are too young
to know what a tiddlywink is). When the disk is normal to the sun, it
collects the maximum area (pi*r^2), but when edge-on it collects nothing.
The average collecting area over time ends up being (2/pi) * (pi*r^2), or
simply 2 r^2. But the thermally emitting surface area is 2*pi*r^2. So the
absorption-to-emission area ratio is 1/pi. So instead of:

Te = [S0 * (1-A) / (4*epsilon*sigma)] ^ (1/4)

we have

Te = [S0 * (1-A) / (pi*epsilon*sigma)] ^ (1/4)

Recall previously that if we set the albedo to 20%, emissivity to 80%, and
use the solar constant at perihelion (S0 = 1414 W/m^2), we got Te =291.1 K.
But for the case of a flipping disk-shaped meteoroid, we get:

S0 = 1414
pv = 0.2
epsilon = 0.8
A = .393*pv = .0786
Te = [1414 * (1-.0786) / (pi * 0.8 * sigma)] ^ 0.25 = 309.2 K = 96.9 F.

So, nearly body-temperature. Can we get warmer still? Sure! Change the
axis of rotation of the disk so that it is spinning like a wheel and pointed
normal to the sun. Now the collecting-area-to-emitting-area ratio
increases to 1/2, and the Te equation becomes:

Te = [1414 * (1-.0786) / (2 * 0.8 * sigma)] ^ 0.25 = 346.2 K = 163.5 F.

I would call that "hot". Of course, no meteoroid is shaped like a disk or
a flat plate, but then again no meteoroid is shaped like a sphere. All real
meteoroids lie somewhere between these extremes. But for a meteoroid
with a modestly high albedo encountering the earth in early January with
a spin axis that maximizes the amount of its surface area oriented
toward the sun, the rock could actually start out hot. --Rob

-----Original Message-----
From: lebofsky at lpl.arizona.edu [mailto:lebofsky at lpl.arizona.edu]
Sent: Friday, July 01, 2016 7:47 AM
To: Matson, Rob D.
Cc: meteorite-list at meteoritecentral.com
Subject: Re: [meteorite-list] More on meteorite temperature

Hi Rob:

Did you remember an object is only illuminated by the Sun half the time?

Received on Fri 01 Jul 2016 01:47:36 PM PDT

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