# Surface Temperature of the Earth

The mean surface temperature of the earth is around $T = 288 \,\text{K}$.^{[1]}
Can this number be derived by means of a simple model?

## Heat Exchange with the Environment

A glance at the recent history of the earth's climate shows that the mean surface temparature has been relatively constant. During the past 40 million years the global mean temperature has been varying by only 10 degrees Kelvin and in the past 10.000 years the variation was only of about 1 degree.^{[2]}

Thus, even on large time scales the earth's temperature can be regarded as constant. In other words: There is no net transfer of heat from or to the earth's surface. It is in equilibrium with its environment and it therefore exhibits a constant temperature. This implies that the incident heat flux $P_\text{in}$ has to be equal to the outgoing $P_\text{out}$: ^{[3]}^{[4]}
\begin{align}
P_\text{in} &= P_\text{out} \label{equilibriumCondition} \\
\left[ P \right] &= \text{W}\,\text{m}^{-2}
\end{align}
Since the earth is essentially surrounded by a vacuum, the only way of exchanging heat with its environment is through electromagnetic radiation because in contrast to conduction and convection this process does not require a propagation medium.^{[5]}^{[6]} Indeed, there is a minor additional contribution through geothermal activity but this will be neglected in the following due to its relative weakness compared to the solar radiation.

## Black Body Radiation

However, we have to specify the expressions for the incoming and outgoing radiant flux. Obviously, the incident radiation originates basically from the sun with its high surface temperature whereas the outgoing radiation has to be related somehow to the earth's temperature.

### Origin of Thermal Radiation

What is the origin of thermal radiation? On a microscopic level an object's temperature is a measure of the kinetic energy of its constituents (eventually charged particles). Thus thermal radiation originates from the thermal motion of charged particles which implicates emission of electromagnetic waves. The kinetic energy of these particles is distributed statistically around a mean value and the same applies for its radiation spectrum.

### Planck's Law

In general it is rather difficult to predict the exact emission spectrum. But Max Planck was able to derive such an expression for so called black bodies (objects that are in thermodynamic equilibrium with their environment and perfect absorbers and emitters of radiation). It can be obtained by applying the laws of statistical and quantum mechanics to the radiation (which is treated as a photon gas). The resulting Planck law relates the energy density per frequency $u(\nu)$ and the frequency $\nu$ itself by \begin{align} u(\nu) \D \nu = \left( \frac{8 \pi h}{c^3} \right) \cdot \frac{\nu^3 \D\nu}{e^{\frac{h\nu}{k_\text{B}T}}-1} \end{align}

### Stefan-Boltzmann Law

The energy emitted from the surface of a black body can be obtained by integrating Planck's law wich yields
\begin{align}
J(T) &= \sigma \cdot T^4
\label{stefanBoltzmann}
\end{align}
where $J$ [W m$^{-2}$] is the radiated energy per time and per unit area (energy flux density), ${\sigma \approx 5.67 \cdot 10^{-8} \;\text{W}\,\text{m}^{-2}\,\text{K}^{-4}}$ is the so called Stefan-Boltzmann constant and $T$ is the surface temperature. This equation is known as Stefan-Boltzmann law. The total rate of energy transfer $P$ through any surface is the product of the perpendicular component of the surface and the energy flux density $J(T)$ as given in eq. \eqref{stefanBoltzmann}.^{[7]}

## Radiation Balance

### Incident Radiation

The total rate of incident energy $P_\text{in}$ [W] is the product of the energy flux density $J_\text{sun}$ [W m$^{-2}$] of the solar irradiance at the mean radius of the earth's orbit $d = 1 \,\text{au}$ (**a**stronomical **u**nit) and the projection surface of the earth ${ A_\text{e}^\perp = \pi r_\text{e}^2 }$.

#### The Solar Constant

The former quantity is in general referred to as the solar constant and its experimental value is
\begin{align}
S &:= J_\text{sun} \left( d \right) \\
&= 1370 \,\text{W}\,\text{m}^{-2}.
\end{align}
This number is a measure of how much energy we can receive from the sun and you can illustrate it as follows: If you had a perfect $1\text{m}^2$ solar cell that is capable of converting solar radiation into electricity without loss, it could serve as a power supply for up to 14 100W light bulbs, 3 washing machines or one hair dryer.

Thereby the total rate of incident energy can be expressed as
\begin{align}
P_\text{in} = S \cdot \pi r_\text{e}^2 .
\end{align}
^{[8]}

#### The Role of Albedo

At this point one important modification is necessary: Originally we were interested in the heat transferred to the earth. But until now we ignored the fact that a significant amount ($\approx 30\%$) of the incident radiation is reflected (e.g. by clouds or ice) immediately without transferring heat at all. This percentage is called albedo and amounts to $a = 0.3$ for the earth. Thus, only the proportion of ${1-a}$ of the incident radiation transfers heat to the earth's surface and one obtains:^{[9]}
\begin{align}
P_\text{in} = (1-a) \cdot S \cdot \pi r_\text{e}^2 . \label{pIn}
\end{align}

### Emitted Radiation

To determine the rate of outgoing energy $P_\text{out}$ we approximate the earth as a black body with surface temperature $T_\text{e}$. Then the total rate of energy emitted by the earth's surface $P_\text{out}$ is the product of the energy flux density radiated per unit area (according to the Stefan-Boltzmann law) and the earth's surface area $A_\text{e} = 4 \pi r_\text{e}^2$: \begin{align} P_\text{out} = \sigma T_{e}^4 \cdot 4 \pi r_\text{e}^2 \label{pOut} \end{align}

Maybe you wonder why we first used the projection area of the earth and now the total spherical area. The explanation is that in the first case we considered parallel radiation and in that case the perpendicular surface is plane. On the contrary now we considered radially emitted rays and in this case the perpendicular surface has a curved, spherical shape.

### Result

Now one can insert eqs. \eqref{pIn} and \eqref{pOut} into eq. \eqref{equilibriumCondition} and one obtains:
\begin{align}
P_\text{in} &= P_\text{out} \\
(1-a) \cdot S \cdot \pi r_\text{e}^2 &= \sigma T_{e}^4 \cdot 4 \pi r_\text{e}^2 \\[2ex]
\Leftrightarrow \quad T_\text{e} &= \sqrt[4]{\frac{(1-a) \cdot S}{4\sigma}}
\end{align}
^{[10]}
^{[11]}

When inserting the values as given in the previous sections, this calculation yields a surface temperature of about $T_\text{e} = 255\,\text{K}$. Even though this value is not quite bad for a very simple model, it still deviates significantly from the actual value of $T_\text{e} = 288\,\text{K}$.

What are the main flaws of this model? It was assumed that the earth is a closed system with a sharp surface surrounded by a vacuum. However, this is not valid for the earth's surface since there is additionally the atmosphere which influences the radiation balance considerably. In the next article a more sophisticated model will comprise the atmosphere's impact.

## References

[1] | An Introduction to Atmoshperic Physics Cambridge University Press 2000 (p. 5) |

[2] | Earth - Evolution of a habitable world Cambridge University Press 2013 (p. 238) |

[3] | Environmental Physics Wiley 2011 (ch. 1.2) |

[4] | Atmosphere, Clouds and Climate Princeton University Press 2012 (p. 28) |

[5] | Equilibrium Thermodynamics Springer 2013 (ch. 18.1.1) |

[6] | Atmosphere, Clouds and Climate Princeton University Press 2012 (p. 27) |

[7] | Environmental Physics Wiley 2011 (ch 2.1.1) |

[8] | Atmosphere, Clouds and Climate Princeton University Press 2012 (pp. 28, 31) |

[9] | Atmosphere, Clouds and Climate Princeton University Press 2012 (p. 31) |

[10] | Environmental Physics Wiley 2011 (ch. 1.2) |

[11] | An Introduction to Atmoshperic Physics Cambridge University Press 2000 (ch. 1.3.1) |