[meteorite-list] Cali chondrite fell extremely cold!

From: Alexander Seidel <gsac_at_meteoritecentral.com>
Date: Sun, 29 Jul 2007 05:27:04 +0200
Message-ID: <20070729032704.77840_at_gmx.net>

Thanks a bunch, Sterling, hi "Mexico Doug", where art thou????
So it is no surprise an OC can be pretty cold at the touch upon
arrival on earth, as was experienced in Cali/Colombia - qed...

Best from early morning Berlin,
Alex



-------- Original-Nachricht --------
Datum: Sat, 28 Jul 2007 22:06:16 -0500
Von: "Sterling K. Webb" <sterling_k_webb at sbcglobal.net>
An: "Alexander Seidel" <gsac at gmx.net>, "Michael Farmer" <meteoriteguy at yahoo.com>, meteorite-list at meteoritecentral.com
Betreff: Re: [meteorite-list] Cali chondrite fell extremely cold!

> 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
>
> ______________________________________________
> Meteorite-list mailing list
> Meteorite-list at meteoritecentral.com
> http://six.pairlist.net/mailman/listinfo/meteorite-list
Received on Sat 28 Jul 2007 11:27:04 PM PDT