In many tutorials in atmospheric physics, I have seen a basic method used to calculate Earth's surface temperature, assuming it was a black body in equilibrium. The method follows this outline:

  • Calculate the radiation energy per unit time received by Earth from the Sun. Say this value is $P$.
  • Since Earth is assumed to be a black-body in thermal equilibrium, it will absorb all of $P$, and re-emit $P$ equally across its surface area, giving radiative power emitted per unit area equal to $\frac{P}{4\pi a^2}$, where $a$ is Earth's radius.
  • Using the Stefan-Boltzmann Law, which relates a black body's radiative power emitted per unit area to the body's temperature, we get $\sigma T^4 = \frac{P}{4\pi a^2}$, and we can solve for $T$, and thus get Earth's surface temperature, around 272K.

I get confused only on the last bullet point: the Stefan-Boltzmann Law relates the whole body's temperature to its emitted radiative power per unit area. So, isn't this calculation really saying that all of Earth has an average temperature of 272K (which seems inaccurate, given how hot Earth is below the surface)? Or does this model only consider the Earth's surface as the black body, with everything else omitted?


2 Answers 2


Blackbody radiation from the interior of the Earth does not escape into space, because it is just re-absorbed by the opaque Earth. Only radiation from the surface escapes into space. Thus, only the temperature of the surface matters when evaluating the rate or properties of the Earth's blackbody radiation into space.

  • $\begingroup$ That last sentence makes sense -- I understand why the surface temperature is important. My main question, however, is how one could use the Stefan-Boltzmann Law to calculate this surface temperature, when it only relates the temperature of the entire black body to emitted radiative power. Or are you saying that because of the huge thermal insulation between Earth's surface and interior, the Earth's surface alone is the black body? $\endgroup$
    – araj
    Commented Sep 4, 2023 at 18:54
  • $\begingroup$ @araj Since only the radiation from the surface escapes the Earth, only the surface temperature enters into the Stefan-Boltzmann law when you are calculating the power in escaping radiation. Interior parts of the Earth are also blackbodies, but their radiation does not escape and only contributes toward heating their immediate surroundings. $\endgroup$
    – Sten
    Commented Sep 4, 2023 at 22:08

The temperature you have calculated is the effective temperature of the Earth. It is precisely defined as that temperature that would yield the luminosity of the Earth if inserted into the Stefan-Boltzmann equation for the luminosity of a blackbody.

A blackbody only radiates from its surface. It is opaque to all radiation, including its own, so the interior radiation does not directly contribute to what is finally emitted from the surface.

Indeed, the Earth is neither of uniform temperature or a blackbody. Although it is certainly opaque to radiation, that opacity occurs at different heights in the atmosphere or at the surface, depending on wavelength. Therefore the temperature one actually measures at any position will depend on geography and the atmosphere, and will vary.


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