

Estimating the temperature of a flat plate in low Earth orbit (LEO) - coderdude
http://www.grc.nasa.gov/WWW/K-12/Numbers/Math/Mathematical_Thinking/estimating_the_temperature.htm

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jedsmith
To simultaneously demonstrate my lack of training in physics and save you the
Google that I just performed:

    
    
      394 K = 249.53 deg F
      394 K = 120.85 deg C
    

To me, that temperature was a surprise.

~~~
uvdiv
It's a surprise because it's wrong. The author used an emissive power of 1,360
W/m^2 -- the intensity of direct sunlight -- as so:

[http://www.google.com/search?hl=en&q=(1,360+/+(5.67+*+10...](http://www.google.com/search?hl=en&q=\(1,360+/+\(5.67+*+10^-8\)\)+**+\(1/4\)&aq=f&aqi=&aql=&oq=&gs_rfai=)

But this isn't the right number to use. The physics of the system is: when the
plate is oriented facing the sun, 1,360 W of solar energy are absorbed per m^2
of plate area; and when it is at thermal equilibrium, the same amount are
emitted. However, the sunlight is only incident on _one_ side of the plate,
while thermal radiation is emitted in both directions! If the plate reaches a
uniform temperature (if it's not thick and insulating), then equal amounts of
radiation will be emitted from both sides -- twice as much emitting area, with
half the radiated power density (680 W/m^2). With this factor-of-2 correction,
the equation is:

[http://www.google.com/search?hl=en&q=(1360+/+2+/+(5.67+*...](http://www.google.com/search?hl=en&q=\(1360+/+2+/+\(5.67+*+10^-8\)\)+**+\(1/4\)&aq=f&aqi=&aql=&oq=&gs_rfai==)

For an equilibrium T of 331 K -- a more reasonable 58 deg C. And what's more,
this is assuming a perfect blackbody -- a perfect, black, absorber of light at
all wavelengths. A light-colored disk could be far colder. Check out the
discussion on the wikipedia article (particularly the "earth's blackbody
temperature"):

<http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law>

For a black earth (with no atmosphere), the equilibrium temperature would be
around 278 K. (Note it is different from that of a flat disk. With a sphere,
the ratio of emitting area to absorbing area (that is, the cross-sectional
area of the earth projected onto a plane -- the plane normal to the direction
of the sun) is 4 pi R^2 / pi R^2 = 4. So twice as much emitting area as a flat
disk.)

[http://www.google.com/search?hl=en&q=(1360+/+4+/+(5.67+*...](http://www.google.com/search?hl=en&q=\(1360+/+4+/+\(5.67+*+10^-8\)\)+**+\(1/4\)&aq=f&aqi=&aql=&oq=&gs_rfai=)

But with the earth's albedo accounted, (says wikipedia) the equilibrium T is
255 K, 24 degrees colder than the blackbody. Likewise, the temperature of the
satellite disk would be very sensitive to its color. (edit): This source says
(p. 26) a white _sphere_ in LEO would be 90 K colder than a black one!

<http://www.tak2000.com/data/Satellite_TC.pdf>

