Re: [meteorite-list] The ASU - WMAT Meteorite Expedition and Recovery

[Dolores Hill via Meteorite-list]

Hearty congratulations to all involved!

Best regards,
Dolores Hill

On 6/29/2016 5:16 PM, Ruben Garcia via Meteorite-list wrote:

Hi all,

I have a few of the most recent articles regarding our hunt here.

http://www.mrmeteorite.com







--
Dolores H. Hill
Sr. Research Specialist
Lunar & Planetary Laboratory
Kuiper Space Sciences Bldg. #92
The University of Arizona
1629 E. University Blvd.
Tucson, AZ 85721
http://www.lpl.arizona.edu/

OSIRIS-REx Asteroid Sample Return Mission Communication & Public Engagement Team
Lead OSIRIS-REx Ambassadors program
Co-lead OSIRIS-REx Target Asteroids! citizen science program
Co-coordinator Target NEOs! observing program of the Astronomical League

http://osiris-rex.lpl.arizona.edu/
http://osiris-rex.lpl.arizona.edu/?q=target_asteroids
http://www.astroleague.org/files/u3/NEO_HomePage.pdf

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[meteorite-list] AD: Glorieta, Seymchan, Camel Donga, Agoudal, 869 and More

[Meteorite World via Meteorite-list]

Still have some pieces and lots left from my meteorite collection. All 
reasonable offers considered
http://meteoriteworld.com/meteorites-for-sale/ 


Take all price $2500 shipped (USA only)

Regards,
Eric__

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Re: [meteorite-list] More on meteorite temperature

[Matson, Rob D. via Meteorite-list]

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: lebof...@lpl.arizona.edu [mailto:lebof...@lpl.arizona.edu] 
Sent: Friday, July 01, 2016 7:47 AM
To: Matson, Rob D.
Cc: meteorite-list@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?

Larry
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Re: [meteorite-list] More on meteorite temperature

[Larry Lebofsky via Meteorite-list]

Hi Rob:

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

Larry

> Hi All,
>
> Playing Devil's Advocate, I decided to try coming up with a scenario that
> attempts to maximize the
> thermal equilibrium temperature of a chondritic meteoroid just prior to
> encountering the earth's
> atmosphere. The typical formula for computing the thermal equilibrium
> temperature for an
> object without an atmosphere is:
>
> Te = [S0 * (1-A) / (4*epsilon*sigma)] ^ (1/4)
>
> where the body is assumed to be spherical (the source of the 4 in the
> denominator), S0 is the
> solar constant (mean value 1361 W/m^2), A is the bolometric Bond albedo,
> epsilon is the
> meteoroid's emissivity, and sigma is the Stefan-Boltzmann constant (5.670
> x 10^-8 W/m^2-K^-4).
> A, in turn, can be estimated from the following equation:
>
> A ~= q * pv
>
> where q is the phase integral and pv is the visible albedo. Using Bowell's
> H, G magnitude system,
> we can compute q from:
>
> q = 0.290 + .684*G
>
> The commonly used value for the slope parameter, G, is 0.15, in which
> case:
>
> q = 0.393
> A = 0.393 * pv
>
> For very dark asteroids (e.g. Trojan asteroids, Hildas, Cybeles), the
> albedo can be 5% or lower.
> However, most NEOs have semi-major axes less than 3 a.u. and albedos
> averaging closer
> to 20%.
>
> The final missing value is the emissivity. For regolith, a range of
> 0.9-0.95 is often mentioned.
> However, emissivity and albedo work hand-in-hand (epsilon + pv ~= 1). So
> if we're going
> to choose an emissivity of 0.9, we should set the albedo, pv, to 10%.
>
> So what is a typical equilibrium temperature for a spherical NEO with 10%
> albedo, 0.9
> emissivity, 1 a.u. from the sun?
>
> A = .393*10% = .0393
>
> Te = [1361 * (1-.0393) / (4*0.9*5.67 x 10^-8)]^0.25 = 282.9 K or about
> 49.6 F
>
> So, cool, but certainly not freezing. How can we get a warmer answer?  One
> way is to pick the
> time of year when the earth is closest to the sun (early January) and the
> solar constant is
> higher:  about 1414 W/m^2.  This raises the temperature in the above
> example to 285.6 K,
> or 54.4 F. Still not warm, but warmer. Lowering the emissivity will help,
> too. Let the albedo
> increase to 20%, and set the emissivity to 0.8. With the perihelion solar
> constant, the
> equilibrium temperature is now up to 291.1 K (64.3 F). Lowering the
> emissivity further
> is probably not realistic for most earth-crossing asteroids, so we're at
> the limit of what
> we can achieve via S0 and emissivity.
>
> However, there *is* a way to get a big increase in the equilibrium
> temperature which
> I'll cover in the next installment.  --Rob
> __
>
> Visit our Facebook page https://www.facebook.com/meteoritecentral and the
> Archives at http://www.meteorite-list-archives.com
> Meteorite-list mailing list
> Meteorite-list@meteoritecentral.com
> https://pairlist3.pair.net/mailman/listinfo/meteorite-list
>


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[meteorite-list] More on meteorite temperature

[Matson, Rob D. via Meteorite-list]

Hi All,

Playing Devil's Advocate, I decided to try coming up with a scenario that 
attempts to maximize the
thermal equilibrium temperature of a chondritic meteoroid just prior to 
encountering the earth's
atmosphere. The typical formula for computing the thermal equilibrium 
temperature for an
object without an atmosphere is:

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

where the body is assumed to be spherical (the source of the 4 in the 
denominator), S0 is the
solar constant (mean value 1361 W/m^2), A is the bolometric Bond albedo, 
epsilon is the
meteoroid's emissivity, and sigma is the Stefan-Boltzmann constant (5.670 x 
10^-8 W/m^2-K^-4).
A, in turn, can be estimated from the following equation:

A ~= q * pv

where q is the phase integral and pv is the visible albedo. Using Bowell's H, G 
magnitude system,
we can compute q from:

q = 0.290 + .684*G

The commonly used value for the slope parameter, G, is 0.15, in which case:

q = 0.393
A = 0.393 * pv

For very dark asteroids (e.g. Trojan asteroids, Hildas, Cybeles), the albedo 
can be 5% or lower.
However, most NEOs have semi-major axes less than 3 a.u. and albedos averaging 
closer
to 20%.

The final missing value is the emissivity. For regolith, a range of 0.9-0.95 is 
often mentioned.
However, emissivity and albedo work hand-in-hand (epsilon + pv ~= 1). So if 
we're going
to choose an emissivity of 0.9, we should set the albedo, pv, to 10%.

So what is a typical equilibrium temperature for a spherical NEO with 10% 
albedo, 0.9
emissivity, 1 a.u. from the sun?

A = .393*10% = .0393

Te = [1361 * (1-.0393) / (4*0.9*5.67 x 10^-8)]^0.25 = 282.9 K or about 49.6 F

So, cool, but certainly not freezing. How can we get a warmer answer?  One way 
is to pick the
time of year when the earth is closest to the sun (early January) and the solar 
constant is
higher:  about 1414 W/m^2.  This raises the temperature in the above example to 
285.6 K,
or 54.4 F. Still not warm, but warmer. Lowering the emissivity will help, too. 
Let the albedo
increase to 20%, and set the emissivity to 0.8. With the perihelion solar 
constant, the
equilibrium temperature is now up to 291.1 K (64.3 F). Lowering the emissivity 
further
is probably not realistic for most earth-crossing asteroids, so we're at the 
limit of what
we can achieve via S0 and emissivity.

However, there *is* a way to get a big increase in the equilibrium temperature 
which
I'll cover in the next installment.  --Rob
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[meteorite-list] AD: Magura, Buzzard Coulee, Nyirábrány, Knyahnya, Kaba, cheap Thin Sections etc in EBay

[cbo via Meteorite-list]

Dear Meteorite Collectors!


Ending soon in Weekend my few auctions in Ebay. Lots of pieces are in very
reduced price!   

See them here: http://stores.ebay.com/eurodome

or here http://www.ebay.com/usr/cbo891



Magura IAB-MG iron etched full-slice w. cohenite, 1840, Slovakia,33.4gr
999USD 
Small Magura 0.35 gr 60USD

Buzzard Coulee H4, Canada, nice Fusion Crust individual 39.76gr, 499USD

NWA 6069 BIG ureilite slice for 249USD   

Kölked H5, newest Hungary official - first on the Market  big slice also!  
Pieces available from 70 - 734USD 

Nyirábrány L/L4-5, 1914, Hungary, super rare 80USD   

Kaba CV3, 1857, Hungary, micros from 150USD

Knyahinya L/LL5, 1866, Ukraine 0.1gr 85USD   

Regmaglypted H5 Gaos   

Highly brecciated Chelyabinsk slice  - very nice   

Sidi Ali Ou Azza L4, witnessed fall from 2015, 98 gr piece, its very big  
  
Fresh NWA xxx Eucrite, black FC 30.95 gr - cheap  

Oridinary chondrites from the NWA , oriented also  

Hand selected Agoudals in Collector Box
  

Rizalite, Moldavites


Thin Sections - perfect mirror polished each of them 35-199 USD:

Neugrund, Estonia impactite - super rare 50USD off EBay
NWA 6069 Ureilite - super colorful 69USD
NWA 7309 CM2 - rare type 40USD   
NWA 10669 LL3.2-LL3.4 - very nics chonrule tonned  49USD  
Gao-Guenie H5 - nice  and cheap 35USD off Ebay   
Sidi Ali Ou Azza L4 - very nice chondrules 60USD off Ebay 

 
HunPol2000 portable polarizing/reflected microscope 2in1 model 270 USD   


and many more
  
  
Best Regards!  
Zsolt Kereszty
Hungary
IMCA#6251, MetSoc

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[meteorite-list] More on meteorite temperature

[Rob Matson via Meteorite-list]

Hi All,

Posted this from work over 7 hours ago but seems not to have worked, so
resending from
home...--Rob

- - - -

Hi All,

Playing Devil's Advocate, I decided to try coming up with a scenario that
attempts to maximize the
thermal equilibrium temperature of a chondritic meteoroid just prior to
encountering the earth's
atmosphere. The typical formula for computing the thermal equilibrium
temperature for an
object without an atmosphere is:
 
Te = [S0 * (1-A) / (4*epsilon*sigma)] ^ (1/4)
 
where the body is assumed to be spherical (the source of the 4 in the
denominator), S0 is the
solar constant (mean value 1361 W/m^2), A is the bolometric Bond albedo, epsilon
is the
meteoroid's emissivity, and sigma is the Stefan-Boltzmann constant (5.670 x
10^-8 W/m^2-K^-4).
A, in turn, can be estimated from the following equation:
 
A ~= q * pv
 
where q is the phase integral and pv is the visible albedo. Using Bowell's H, G
magnitude system,
we can compute q from:
 
q = 0.290 + .684*G
 
The commonly used value for the slope parameter, G, is 0.15, in which case:
 
q = 0.393
A = 0.393 * pv
 
For very dark asteroids (e.g. Trojan asteroids, Hildas, Cybeles), the albedo can
be 5% or lower.
However, most NEOs have semi-major axes less than 3 a.u. and albedos averaging
closer
to 20%.
 
The final missing value is the emissivity. For regolith, a range of 0.9-0.95 is
often mentioned.
However, emissivity and albedo work hand-in-hand (epsilon + pv ~= 1). So if
we're going
to choose an emissivity of 0.9, we should set the albedo, pv, to 10%.
 
So what is a typical equilibrium temperature for a spherical NEO with 10%
albedo, 0.9
emissivity, 1 a.u. from the sun?
 
A = .393*10% = .0393
 
Te = [1361 * (1-.0393) / (4*0.9*5.67 x 10^-8)]^0.25 = 282.9 K or about 49.6 F
 
So, cool, but certainly not freezing. How can we get a warmer answer?  One way
is to pick the
time of year when the earth is closest to the sun (early January) and the solar
constant is
higher:  about 1414 W/m^2.  This raises the temperature in the above example to
285.6 K,
or 54.4 F. Still not warm, but warmer. Lowering the emissivity will help, too.
Let the albedo
increase to 20%, and set the emissivity to 0.8. With the perihelion solar
constant, the
equilibrium temperature is now up to 291.1 K (64.3 F). Lowering the emissivity
further
is probably not realistic for most earth-crossing asteroids, so we're at the
limit of what
we can achieve via S0 and emissivity.
 
However, there *is* a way to get a big increase in the equilibrium temperature
which
I'll cover in the next installment.  --Rob


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[meteorite-list] Meteorite Picture of the Day

[Paul Swartz via Meteorite-list]

Today's Meteorite Picture of the Day: JAH 395 TS

Contributed by: Anne Black

http://www.tucsonmeteorites.com/mpodmain.asp?DD=07/01/2016
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