[meteorite-list] Cali chondrite fell extremely cold!
Sterling K. Webb
sterling_k_webb at sbcglobal.net
Sat Jul 28 23:06:16 EDT 2007
Dear Alex, Mike, List,
Alex said:
> several posts about this on the list in the past...
Mexico Doug has done more work on this than anyone
else I can think of. Go to the website http://www.diogenite.com/
and click on the item "Meteoroid" in the left-hand menu.
There is Doug's graph of the space equilibrium temperatures
for irons, ordinary condrites, and carbonaceous chondrites
for any distance from the sun.
From the graph, we can see that an OC orbiting at about
the same distance from the Sun as the Earth would have a
"natural" temperature of about -10 degrees C. The heating
of ablation is too brief to penetrate far into the stone, and
the cooling of dark fall is likely to completely nullify exterior
heating (as Marcin said).
Above the graph, Doug provides a link to his post to The
List where he detailed the assumptions the graph is based on,
but the List archives no longer go back that far. Mine do,
however, so here's Doug's full explanation:
Part I
There is no minimum size to absorb energy from the Sun, even a molecule can
do it - heat of everything is actually principally caused by molecular
vibrational motion in the Infrared range under normal circumstances -
molecular
vibrations. So that answers the easy question.
On the temperature of a meteoroid traveling in space, that is a complicated
question because the question is really millions of questions in one
depending on what temperature you mean - and where you measure it.
But I think I can give a shot at a satisfying at everything you wanted to
know on the first question and weren't afraid to ask, with the following
calculations you kept me awake doing. You can make a lot of interesting
observations here. I'd add a Eucrite and an Enstatite Achondrite, which I
expect the
former would not be closer to OC, and the later more on the way to the
irons.
I guess:)
Distance Energy flux <-------T (degrees C)--------> T(Planet av. Surf.)
AU W/m2 CC OC Fe-Ni "ideal" note
0.31 14,214 216 195 378 227 Mercury (p) 167
0.47 6,184 124 107 255 133 Mercury (ap) 167
0.72 2,635 48 34 154 55 Venus 464
1.00 1,366 -1 -12 89 6 Earth 15
1.52 591 -52 -62 21 -47 Mars -63
2.80 174 -110 -117 -57 -107
3.00 152 -116 -123 -64 -112
4.00 85 -137 -143 -92 -134
5.00 55 -151 -156 -111 -148
5.2 51 -154 -159 -114 -151 Jupiter -144
9.5 15 -185 -188 -155 -183 Saturn -176
19.2 3.7 -211 -214 -190 -209 Uranus -215
29.7 1.5 -223 -225 -207 -222 Pluto (p) -223
30.1 1.5 -223 -226 -207 -222 Neptune -215
49.3 0.6 -234 -236 -221 -233 Pluto (ap) -223 Kuiper,
Comets (ap)
50K 0.000001 -272 -272 -271 -272 Oort Cloud
(Table may not display well in text, but if you copy it into excel and use
text-to-columns it should look great.)
Part II
Assumptions & Discussion:
T = [(absorptivity/emissivity)*(Energy flux/sigma)*(a/A)]^(1/4)
where:
Emissivity = energy ratio emitted at a temperature = compositional
property.
Absorptivity = energy fraction absorbed at a temperature = compositional
property.
Temperature is proportional to absorptivity but proportional to the inverse
of emissivity, i.e. T^(4)=k*(absorptivity/emissivity).
The useful form of this law is called the Stephan-Boltzmann Law: Energy =
sigma*T^4 (S-B law constant, sigma = 5.67*10^-8*Wm^-2*K^-4)
That the integrated energy radiated from a black booty is proportional to
the fourth power of the temperature. This "law" is useful to convert the
energy hitting an object into the temperature and is really the scientific
key to
address your question approximately.
Kirchoff's (other) Law:
For a body in radiative equilibrium energy absorption = energy emission.
So we consider the very plausible scenario that the meteoroid's position in
orbit doesn't alter radically (it doesn't travel near light speed!)
Solar energy is mainly provided for absorption in the UV-Visible range.
Energy is emitted in the IR range (vibrational energy, the meteoroid
doesn't
emit much light energy:-).
Meteoroide is spherical in shape (OK not generality, so could vary maybe
50%
either way for example when a planar shaped meteoroid had verrry low
rotational energy w/r to the Sun)
ABSORPTIVITY = CC = 0.8, OC = 0.65, Fe-Ni = 0.80, "ideal" = 1
EMISSIVITY = CC = 0.88, OC = 0.85, weathered Fe-Ni = 0.28, "ideal" = 1
Assumed constant emissivities, but actually they vary, for example,
decreasing somewhat (and then only by a square root) as the AU's increase
for Iron,
and a lot for Nickel, though I expect the "weathered" surface might
somewhat
mitigate this.
The temperature of a meteoroid in space will of course depend on the
latitude, depth, it's cross sectional exposure vs. overall mass
distribution and
distance to the Sun, just like any other non-radioactive cooled body
whether in
orbit or passing through.
Part III
So the directional answer is best gotten to by a bunch of assumptions and
simplifications, including that the thing is rotating on a nice skewer and
isn't too big so that depth becomes an issue, which adds some calculus for
all
those onion rinds. So sticking to something say a couple of meters in
diameter... Otherwise we deal with things like what is the temperature of
Mercury
(Solar side, mass, backside, transition zone, latitude, rotation,
greenhouse
effect if any, and composition which will affect what radiation can be
absorbed
and converted into heat. A simple way to think about the latter is
thinking
about a microwave oven. If the object to be heated is made with lots of
water, it has a strong absorbance in that range, but the glass door doesn't
(even
on the inside). So a Tektite sent in orbit very well could have a lower
temperature than a cometary water containing carbonaceous type body which
in
turn creates nice effects in part due to these warmings in the estelas. An
example like Mercury, of one not rotating with respect to the Sun causes
other
complications. So best to think of an ant in a spacesuit when asking your
question. The center of the meteoroid will be cooler at its equilibrium
temperature so wherever the ant walks or burrows will depend on the
temperature.
Average temperature is a much easier proposition, but knowing that wouldn't
help
the ant's survival chances at all if he ends up in the Ant:-)arctic vs.
Sahara of the meteoroid.
If one assumes that a meteoroid has no greenhouse effect and is made of
stone or iron-nickel, and absorbs typically a full spectrum in the range of
what
the sun mainly radiates for heating i.e. UV-Visible light, lots of
simplifying assumptions can be made. You can look at Venus and see what a
Greenhouse effect does, or even Jupiter, which I suspect is somehat warmer
still than the NASA page reference I used, because it actually puts out
more
energy than it received that little stunted star...or figure out at what
distance
comets get tails (snowball's sublimation temperature). You get the idea:)
The basics would include the following, I would think, calling the
meteoroid
shape a sphere for simplification, which of course is not true but good
enough, also that the Sun is like a Black Body at 5800 degrees K from
Wein's Law.
Taken together with the idea that radiation from any source drops off as
the
square of the distance from the source. (Which is understandable by knowing
that the surface area of a sphere, ie, non-directional emmiting source,
4*Pi*r^2 increases by the square of the radius.), you can get a handle on
temperature caused by the Sun on objects floating in the Solar System. And
in the
case of the meteroid, we only get a quarter of the total area exposed to
the
Sun in the simplified case of a spherical meteoroid (area sphere = 4*pi*r^2
vs.
great circle exposed = pi*r^2, a factor of 1/4).
Using the two laws in a numer of ways, but sparing the the tedious math,
the
Sun's photosphere ("surface") clocks at near 5800 degrees K being basically
a heat source (150,000,000 km from earth minus Sun's radius, aww lets just
say the center of the Sun since we can then measure in AU and are not
dealing
generally with the inner Solar System to make a huge difference with the
Sun's
700,000 km or so radius. Using the inverse square law then you get the
energies in my table I gave you.
Then, I derived the temperature in the above table (all in an excel
spreadsheet) using the Stephan-Boltzmann Law, for you with this exact
formula:
T = [R/sigma*(a/e)*(areacrosssec/areatot)*(1/R^2)]^(1/4)
and substituded the differewnt absorptivities and emissivities...presto, a
solar meteorite thermometer. You can graph them too and it is easy to look
at, but I couldn't figure out how to graph in plain text for the list:)
[End of Doug's explanation]
Sterling K. Webb
-------------------------------------------------------------------
----- Original Message -----
From: "Alexander Seidel" <gsac at gmx.net>
To: "Michael Farmer" <meteoriteguy at yahoo.com>;
<meteorite-list at meteoritecentral.com>
Sent: Saturday, July 28, 2007 9:18 PM
Subject: Re: [meteorite-list] Cali chondrite fell extremely cold!
Mike, your comment was obviously triggered by my earlier post to the list
this day. I always thought stones of a meteorite fall would be rather "cold"
after touchdown. Then again one has to look at typical equilibrium temps for
a tumbling stone meteoroid at a typical Earth orbit cruising distance before
encounter with the Earth atmosphere. There were several posts about this on
the list in the past, may be someone can retrieve them from the archives.
The hot phase of atmospheric entry, before reaching the retardation point,
will most likely only affect the very outer zones of a meteoroid large
enough to not disintegrate, while the inner parts will remain effectively at
the former equilibriums temps due to a rather low heat transfer process for
typical stony meteorites in the few seconds or minutes of atmospheric
transit. And the final dark flight will cool down the former temporary high
temps on the outsides of the stone pretty soon.
If I ever had the chance to pick up a freshly fallen stone immediately after
the event, I would expect it to be rather "cold" to the touch.... :)
Alex
Berlin/Germany
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