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The sun is generally considered to be an approximate blackbody. A blackbody has to be a perfect absorber. How do we know that the sun absorbs most of the radiation that falls on it?

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How do we know that the sun absorbs most of the radiation that falls on it?

  1. The incoming radiation falling on the Sun is insignificant compared to the amount of radiation that it emits. So any incoming radiation that is reflected rather than absorbed will have an insignificant effect on the Sun's spectrum.
  2. The outer layer of the Sun (the photosphere) is a 100 km thick layer of diffuse hot gas. Theoretical models suggest that this layer is likely to absorb almost all incoming photons.
  3. Empirical observations tell us that the spectrum of the Sun (and of other stars) follows, to a first approximation, the shape of a black-body radiation curve.
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    $\begingroup$ I see. So all we have is the emission spectrum, and since that matches with a blackbody spectrum, we assume that it should be near perfect absorber as well. (Also considering the fact that the photosphere is likely to absorb almost all incoming photons). $\endgroup$
    – Sasikuttan
    Oct 4, 2021 at 10:59
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    $\begingroup$ Can you elaborate on how item #1 suggests/demonstrates that the sun absorbs that energy? $\endgroup$
    – Matt
    Oct 4, 2021 at 11:21
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    $\begingroup$ @Curiouserandcuriouser the gas doesn't radiate, the filament does $\endgroup$ Oct 4, 2021 at 11:43
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    $\begingroup$ @Curiouserandcuriouser You know how lightbulbs are transparent, except where the filament is? If you look through one, you can see what's behind it, except where the filament is? $\endgroup$ Oct 4, 2021 at 12:09
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    $\begingroup$ @Curiouserandcuriouser, The glass envelope and the fill gas play no part whatsoever in the theory of incandescent light bulbs. They only are there for practical reasons: E.G., to keep the filament from burning up. For purpose of describing how it works, the filament is the "black body." $\endgroup$ Oct 4, 2021 at 14:53
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What can you see through the surface of the Sun?

There are lots of examples of things that are eclipsed by the Sun and therefore the light they emit must have been absorbed or reflected.

In principle I suppose you can shine a laser or radar signal at the Sun to estimate a reflectivity.

However in practice it doesn't matter because the Sun isn't a perfect blackbody. When we look at the Sun we observe an anisotropic radiation field arising from material at a range of temperatures. This is a far more important departure from the assumptions behind blackbody radiation than any small amount of reflectivity.

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  • $\begingroup$ Reflectivity was the issue that bothered me. An object can be opaque even if it reflects light; it doesn't have to absorb it. But I get your point. Thanks. $\endgroup$
    – Sasikuttan
    Oct 4, 2021 at 13:10

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