4 ms·

Estimating the temperature of a flat plate in low Earth orbit (LEO)

19 points by coderdude 16y ago

jedsmith 16y ago
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 16y ago
  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^-8))+**+(1/4)&aq=f&aqi=&aql=&oq=&gs_rfai= http://www.google.com/search?hl=en&q=(1,360+/+(5.67+*+10... 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+*+10^-8))+**+(1/4)&aq=f&aqi=&aql=&oq=&gs_rfai== http://www.google.com/search?hl=en&q=(1360+/+2+/+(5.67+*... 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 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+*+10^-8))+**+(1/4)&aq=f&aqi=&aql=&oq=&gs_rfai= http://www.google.com/search?hl=en&q=(1360+/+4+/+(5.67+*... 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 http://www.tak2000.com/data/Satellite_TC.pdf

19 points by coderdude 16y ago

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 16y ago
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^-8))+**+(1/4)&aq=f&aqi=&aql=&oq=&gs_rfai= http://www.google.com/search?hl=en&q=(1,360+/+(5.67+*+10... 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+*+10^-8))+**+(1/4)&aq=f&aqi=&aql=&oq=&gs_rfai== http://www.google.com/search?hl=en&q=(1360+/+2+/+(5.67+*... 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 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+*+10^-8))+**+(1/4)&aq=f&aqi=&aql=&oq=&gs_rfai= http://www.google.com/search?hl=en&q=(1360+/+4+/+(5.67+*... 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 http://www.tak2000.com/data/Satellite_TC.pdf