Equilibrium temperature

The equilibrium temperature is the temperature of an open system in constant equilibrium of the input and output power - not to be confused with “in thermodynamic equilibrium ”, as the concept of temperature provides. In astronomy, the equilibrium temperatures of gas clouds and planetary surfaces are important.

Equilibrium temperature of a planet's surface

Under the assumptions that

1. the planet does not have its own energy sources, does not heat up or cool down,
2. the part of the radiation power hitting the planet is reflected with the same wavelength, i.e. only the part is absorbed ${\displaystyle \eta }$ ${\displaystyle (1-\eta )}$
3. the surface has a uniform temperature (fast rotation, efficient heat transport in polar regions),
4. the surface has an emissivity of 1 in the infrared, so it radiates like a black body,

and with the designations

${\displaystyle L=L_{\text{sun}}=3{,}83\cdot 10^{26}\,\mathrm {J/s} }$ for the luminosity of the sun,

${\displaystyle d}$for the distance from the sun ,

${\displaystyle R}$ for the planetary radius,

${\displaystyle T}$ for the temperature and

${\displaystyle \sigma =5{,}67\cdot 10^{-8}\,\mathrm {W} /(\mathrm {m} ^{2}\mathrm {K} ^{4})}$for the Stefan-Boltzmann constant ,

initially applies to the absorbed radiant power

${\displaystyle L_{\text{in}}=(1-\eta )\pi R^{2}{\frac {L}{4\pi d^{2}}}}$

and for the power of thermal radiation

${\displaystyle L_{\text{out}}=4\pi R^{2}\sigma T^{4}}$.

The equilibrium condition gives ${\displaystyle L_{\text{in}}=L_{\text{out}}}$

${\displaystyle T={\sqrt[{4}]{\frac {(1-\eta )\cdot L}{16\pi d^{2}\sigma }}}}$

For the earth this results with in value , with in , but is observed . This higher temperature is due to the greenhouse effect . This is particularly strong on Venus , where the temperatures calculated with (cloud surface, measured) and (fictitious ocean, ice-free, cloudless) are 250 and 318 K, respectively, but the ground temperature is 740 K. ${\displaystyle \eta =0}$ ${\displaystyle T=277\,\mathrm {K} }$ ${\displaystyle \eta =0{,}3}$ ${\displaystyle T=255\,\mathrm {K} }$ ${\displaystyle T=288\,\mathrm {K} }$ ${\displaystyle \eta =0{,}76}$ ${\displaystyle \eta =0{,}1}$

Individual proof

1. ^ R. Haus et al .: Radiative energy balance of Venus based on improved models of the middle and lower atmosphere. Icarus 272, 2016, doi: 10.1016 / j.icarus.2016.02.048 .