## [meteorite-list] More on meteorite temperature- Mail actions: [ respond to this message ] [ mail a new topic ]
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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? Larry Received on Fri 01 Jul 2016 01:47:36 PM PDT |
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