I'm trying to show using Planck's radiation law:

$$B( \lambda ) = \frac{2h c^{2} }{ \lambda^{5} } \frac{1}{\exp( \frac{hc}{ \lambda k T} - 1 )} $$

whether radiation from the Earth or the Sun dominates for a given wavelength.

I'm using distance $r = 1 \text{ m}$ from the Earth, $r = 1.5 \cdot 10^{11} \text{ m}$ from the Sun, and scaling $B$ by a factor of $\dfrac{1}{r^2}$.

I've had a go at plugging values in, using $T=255 \ K$ for the Earth and $T=5800 \ K$ for the sun, however my answers don't look right (Sun dominating long and short wave). I was wondering if someone might be able to point out a problem with my approach?


One problem with the approach of using Planck's radiation law alone is that different portions of the Earth's atmosphere contain different concentrations of elements and molecules (in particular O$^2$ and O$^3$) and different portions of the Earth are different colors resulting in different absorption and reflection.

View of the Solar spectrum from the top of the atmosphere

There are several transitions of each atom or molecule that contribute to its spectral absorption and reflection which are relevant to the understanding of its role in the spectrum seen on Earth's surface, even for an apparently simple triatomic molecule like CO$^2$ its IR spectrum is quite complex.

Spectrum above the atmosphere vs. on the Earth's surface

Cloudyness is another filter affecting the spectrum at the Earth's surface.

Cloud effect on Solar spectrum

At the very least you would want to use Wein's displacement law in combination with Planck's to bin estimates for the top of the atmosphere. That won't be accurate 1M from the surface of the Earth.

Update: March 6, 2018

Planck's law states that:

$$ {\displaystyle B_{\nu }(T)={\frac {2h\nu ^{3}}{c^{2}}}{\frac {1}{e^{\frac {h\nu }{kT}}-1}},} $$


$B_{\nu }(T)$ is the spectral radiance (the power per unit solid angle and per unit of area normal to the propagation) density of frequency $ν$ radiation per unit frequency at thermal equilibrium at temperature $T$.

$h$ is the Planck constant;

$c$ is the speed of light in a vacuum;

$k$ is the Boltzmann constant;

$ν$ is the frequency of the electromagnetic radiation;

$T$ is the absolute temperature of the body.

The air mass coefficient defines the direct optical path length through the Earth's atmosphere, expressed as a ratio relative to the path length vertically upwards, i.e. at the zenith. The air mass coefficient can be used to help characterize the solar spectrum after solar radiation has traveled through the atmosphere.

The AMC is divided into ranges of values for each case:

  • AM0

The spectrum outside the atmosphere, approximated by the 5,800 K black body, is referred to as "AM0", meaning "zero atmospheres". Solar cells used for space power applications, like those on communications satellites are generally characterized using AM0.

  • AM1

The spectrum after travelling through the atmosphere to sea level with the sun directly overhead is referred to, by definition, as "AM1". This means "one atmosphere". AM1 ( z {\displaystyle z} z=0°) to AM1.1 ( z {\displaystyle z} z=25°) is a useful range for estimating performance of solar cells in equatorial and tropical regions.

  • AM1.5

Solar panels do not generally operate under exactly one atmosphere's thickness: if the sun is at an angle to the Earth's surface the effective thickness will be greater.

The AMC value continues up to 38, representing the Sun at an angle of 90°, the horizon. The "Reference Solar Spectral Irradiance: Air Mass 1.5" (ASTM G173-03) is commonly used to specify the performance of solar cells angled to 37° tilt toward the equator, facing the sun (i.e., the surface normal points to the sun, at an elevation of 41.81° above the horizon). See that link for tables in HTML and MS Excel spreadsheet format, along with links leading to source code to perform the calculations.

The plot you are trying to make is on page 58 of "The Principles of Atmospheric Science", it looks like this:

Thermal Emission Spectra of 5780K and 255K


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