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Solar spectral irradiance at earth's TOA

  1. Jul 11, 2010 #1
    I'm trying to reproduce a plot of Sun's black-body behavior like this one:
    http://en.wikipedia.org/wiki/File:Solar_Spectrum.png
    Problem is, after I convert the black-body radiance to irradiance, its curve is way too high as compared with measurement. The measurement data is taken from:
    http://rredc.nrel.gov/solar/spectra/am1.5/ASTMG173/ASTMG173.html

    The top of atmosphere (TOA) irradiance at Earth's distance is obtained in the following way:
    radiance (W/m^2/nm/Sr) L=2*h*c^2/(lamda^5*exp(h*c/(kB*lamda*T)-1))
    where:
    c=3e8 m/s (speed of light)
    h=6.625e-34 Joul Second (Planck's)
    kB=1.38e-23 Joul/Kelvin (Boltzman's)
    omega=pi*r_sun^2/D_sun_earth^2 (Sun disk solid angle as seen from Earth)
    r_sun=6.96e8 m (Sun's radius)
    D_sun_earth=1.496e11 m (1AU)
    Finally irradiance is E=L*omega (W/m^2/nm) (and one needs to multiply 1e9 to be in nm)

    My curve is roughly twice above the measurement, so if I do:
    E=L*omege*cos(67-deg)
    I can get something close to the picture in the wiki link. This 67-deg is roughly Earth's spin inclination. However I really doubt multiplying cos(67-deg) makes sense, as we are talking about TOA irradiance, not anywhere on Earth surface.

    What I'm missing here?

    Thanks!
     
  2. jcsd
  3. Jul 11, 2010 #2

    Chronos

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    Science Advisor
    Gold Member

    try square root.
     
  4. Jul 11, 2010 #3
    That doesn't work, making the spectrum broader, let alone w/o any physical meanings...
     
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