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The fact that the apparatus is using outer space as a heat sink or just using the atmosphere should not have any significant impact on radiative cooling properties. Is this reasoning correct? No, it is not. The cooling properties of the apparatus depend on the heat sink to which the heat is aimed, as the final result will be an equilibrium of ...

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How opaque is that -- would we be able to see a couple of meters, some kilometers, or nothing at all? The photosphere of our sun is somewhere on the order of 500 km thick. For a quick ballpark, you can imagine an exponential decrease in the transmission of light which about this characteristic thickness. It might be a little less, but it's still ...

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Well, I can give you a definitive answer to Q1, but my answer to Q2 would only be educated speculation. Perhaps some of the astrophysicists on here can be more help with that one. However, before I tackle Q1, a very important disclaimer: Temperature is a measure of the average kinetic energy of the particles of an object, and cannot be used all by itself ...

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Actually I think heuristics are very important and understanding them correctly is critical. I would point point out a Feynman statement that we really don't understand the Pauli Exclusion Principle because we don't have a good heuristic for it. Hell go ask several physicists to describe the proof starting with the Wightman Axioms and they can't do it! In ...

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Is there a minimum wavelength in a black body's radiation? No... Or rather: Yes, Zero. In this image from Wikipedia, it does seem that the curve is touching the x-axis rather than being an asymptote It's not. The curve approaches zero rather quickly. So, it's hard to illustrate.

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If you mean 'accelerate' rather than 'optimize' and if you are in an environment where there is no radiation coming back (as in empty space for example), then increasing the surface area will increase the radiated energy since the process depends on the temperature of the surface as well as the surface area. The orientation of the surface should be such that ...

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If your surface would be a perfect black body, there is pretty much nothing to do. As you said, more radiation is initially generated from the surface, but also more is self-absorbed. These two effects exactly compensate each other. Indeed, if you would put this object into to a thermal radiation bath, the II law of thermodynamics tells us that the object ...

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Simple black body radiation theory (adapted from http://www.atmos.washington.edu/2002Q4/211/notes_greenhouse.html): The earth receives a certain amount of heat per unit area from the sun - this amount is about 1370 W/m$^2$ for parts of the earth facing the sun when there is no atmosphere. But the earth presents a "disk" with area $\pi R^2$ to the sun, when ...

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It is actually very easy to consider this. We don't need any models. we are blessed with a sizable rock at exactly 1 AU from the Sun, a rock with no trace of any atmosphere at all: Moon (Source: Wikimedia Commons) Temperatures on the moon vary from 70K to 390K. Average temperatures, depending on location, vary from 130K at the poles to 220K at the ...

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Black body radiation is given by Planck's formula (see link for variables) Here is the measured irradiance of the sun and the attempt to fit it with the black body formula: The effective temperature, or black body temperature, of the Sun (5,777 K) is the temperature a black body of the same size must have to yield the same total emissive power. ...

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Maybe the simplest way to think about this is that the Sun is in approximate thermal equilibrium and would absorb any photon, of any frequency, that is incident upon it. This is essentially the definition of a BB. There are many radiative processes that can absorb (and hence emit) radiation at all frequencies, not just those corresponding to atomic ...

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