OLD Simple Earth Climate Model

Which planetary characteristics contribute to producing a mean surface temperature that makes Earth habitable?


How much can we understand about the Earth’s climate by using basic physics? Do the properties of the Sun and of Earth’s orbit guarantee us a particular temperature? How does the Earth’s atmosphere affect this temperature?

We will develop a simple model that explores the physical properties that are important in determining the Earth’s mean surface temperature, and its level of stability.

Useful pedagogical simulations:

The mean surface temperature of Earth (14.5°C)[note]Çengel, Yunus A. Steady Heat Conduction. In: Heat Transfer a Practical Approach (2). New York: McGraw Hill Professional, 2003, p. 173. [/note] is ideal for supporting life. Where does this particular temperature come from and why is it constant over long periods of time? Defined, temperature is a measure of the internal energy that an object possesses. The natural question to follow then is, where does the energy that heats Earth up come from? It does not come from within Earth. Although Earth has a hot core and mantle, both are well insulated by Earth’s crust. We can also neglect the relatively small amounts of nuclear radiation emitted by deposits within Earth’s surface. Therefore, this energy must come from outside of Earth. As Earth is surrounded by a very good vacuum (space) the only way for Earth to gain or lose energy is through radiation, one of the three heat transfer mechanisms. Earth does receive radiation emitted from stars and planets throughout the galaxy (or else we would not be able to see them in the night sky) but personal experience can tell us that Earth receives almost all of her energy from the Sun. And because this energy from the Sun is absorbed by the Earth at a specific rate, we will refer to it as power rather than energy.

We know that the Sun has been shining and heating Earth for billions of years, yet Earth’s mean surface temperature has remained within a range of 10°C over this time frame[note]Aubrecht GJ. Solar Energy: Wind, Photovoltaics, and Large-Scale Installatons. In: Energy – Physical, Environmental, and Social Impact (3), edited by Erik Fahlgren. Upper Saddle River, NJ: Pearson Education Inc., 2006, chapt. 16, pp. 334,336.[/note]. Therefore Earth must have a mechanism to lose energy. And, because Earth is surrounded by a vacuum, that energy is lost through radiation. A constant temperature then requires that

Radiation in = Radiation out.

This expression is just the conservation of energy applied to Earth. The law of conservation of energy states that energy cannot be created or destroyed, but can only change forms. In other words, energy can only be transferred. Note that heat is defined as the transfer of energy, which includes radiation. But Earth doesn’t shine like the Sun does? It turns out, however, that the same physics can be used to describe both the Sun’s radiation and the Earth’s radiation.

How much power does the Sun radiate onto Earth?

 

Figure 1: The Electromagnetic Spectrum

The physical principle behind thermal radiation is that all objects at some temperature T emit electromagnetic radiation. The wavelength of this electromagnetic radiation depends on surface temperature. As the Sun’s surface temperature is 5780 K, it emits visible light (see Figure 1). Earth’s surface temperature is considerably lower than the Sun’s, and so it emits infrared radiation, electromagnetic waves that have a larger wavelength than visible light. The amount of energy that a blackbody emits is described by the Stefan-Boltzmann Law.

We will be building a numerical model to examine which factors may be responsible for Earth’s mean surface temperature of 14.5°C[note]Çengel, Yunus A. Steady Heat Conduction. In: Heat Transfer a Practical Approach (2). New York: McGraw Hill Professional, 2003, p. 173. [/note].  We need to use the Stefan-Boltzmann Law for the calculation of the incoming and the outgoing radiation and solve for the temperature of the Earth. There are two difficulties that we need to address. First, the emissivities of the Sun and the Earth are unknown. Second, Earth only receives a fraction of the emitted solar power that hits the Earth’s surface as sunlight. In regards to the first, the emissivities of the Sun and the Earth we can assume to be $ε = 1$, which would mean that both absorb and emit radiation perfectly. It turns out that this is a good assumption for the Sun although Earth has an emissivity closer to $ε = 0.8$ due to its atmosphere.  The latter is a geometric problem that we can solve using the inverse-square law that describes how the intensity decreases with increasing distance from the source. The result of this calculation is the well-known solar constant of S = 1367 W/m2 that describes how much power per square metre is available near Earth.  How much power is absorbed by the Earth? We just multiply the solar constant by the effective area that is exposed to sunlight at any given time, which corresponds to a disk having the Earth’s radius. Hence

$$\begin{eqnarray}
P_{in} & =& \pi r_e^2 S (1-A) \nonumber \tag{1}\\
& \approx & 1.23 \times 10^{17} \text{ W} \nonumber \tag{2}
\end{eqnarray}$$

The $(1 – A)$ factor in the equation is due to the fact that Earth does not absorb all of the sunlight, but rather reflects a percentage of this incident solar power back into space through a property called albedo. Earth’s albedo is ~ 0.3. Note that this incident power is the power Earth receives from the Sun averaged over our entire year.

Intensity is a more useful value to us than power as it simplifies later calculations. The mean incident solar intensity, $I_\text{in}$, is simply the power divided over the entire surface of Earth. In other words, it is the power per area that Earth’s surface would receive if the Sun was shining equally on every surface of the Earth. Thus,

$$\begin{eqnarray}
I_{in} & =& \dfrac{P_{in}}{4 \pi r_e^2} \nonumber \tag{3}\\
& =& \dfrac{S}{4}(1-A) \nonumber \tag{4}
\end{eqnarray}$$

How much power does the Earth radiate?

Now that we have considered the power coming into our system, the Earth, we must also consider the power going out. Earth also emits EM radiation (as it is an object with a temperature). Due to the difference in temperature between the Earth and the Sun, the wavelength of this radiation differs significantly between the two celestial bodies. While Earth’s incident solar power is only intersected by the portion of Earth’s sphere that faces towards the Sun, that is its projected 2D area, the Earth emits radiation from its surface in all directions, day and night. Therefore, the power emitted by Earth is simply

$$P_{out} = 4 \pi r_e^2 \sigma T_e^4 \tag{5}$$

Again we will convert power to intensity. The intensity emitted from Earth’s surface is

$$\begin{eqnarray}
I_{out} & =& \dfrac{P_{out}}{4 \pi r_e^2} \nonumber \tag{6} \\
& =& \sigma T_e^4 \nonumber \tag{7}
\end{eqnarray}$$

Now that we have determined an expression for the solar intensity on Earth and the intensity emitted by Earth we can apply the law of conservation of energy in order to find the unknown: Earth’s temperature $T_e$.  The intensity entering Earth’s system must equal the intensity exiting Earth’s system. The balanced equation for the conservation of energy on Earth’s surface is

$$\begin{eqnarray}
I_{in} &=& I_{out} \nonumber  \tag{8}\\
\dfrac{S}{4}(1-A) &=& \sigma T_e^4 \nonumber \tag{9}
\end{eqnarray}$$

In this simple Earth climate model where Earth is basically modelled as a big sphere of rock (with an albedo of 0.3) its surface temperature $T_e$ is found to be 255 K or -18°C. Although this model of Earth is not ideal as it is missing crucial elements that make Earth habitable, it is a starting model that can be refined to more closely model our planet.

In a more refined model it must be taken into account that not all of Earth’s emitted power escapes into space. Some is trapped within our atmosphere by what is referred to as the Greenhouse Effect (see Greenhouse Effect). If our atmosphere was modeled as a single-layer perfect greenhouse, Earth would be in danger of heating up so much that it would be uninhabitable at a surface temperature of 30°C. Fortunately, it is not a perfect greenhouse due to numerous reasons and rather acts as an imperfect greenhouse, producing a surface temperature of 23°C (if the model still uses a single-layer atmosphere). While the refined model up to this point illustrates how different variables effect Earth’s surface temperature, it does not develop further than deriving a surface temperature of 23°C. A model of a multi-layered atmosphere is needed.

The process we followed is the process of consecutive approximation in modelling. Very often we start with a very simple model and progress to more complicated ones getting (hopefully) closer and closer to reality. While this model begins to show why Earth’s surface temperature is 14.5°C[note]Çengel, Yunus A. Steady Heat Conduction. In: Heat Transfer a Practical Approach (2). New York: McGraw Hill Professional, 2003, p. 173. [/note], it also shows that very small changes in the variables effecting Earth’s temperature can cause drastic changes to the temperature of the Earth. This will be explored more fully in the problem sets. It is, however, suffice to say that Earth’s climate is not stable, and that this stability can be seen on the decade scale.

Share this page:

Big Ideas

  • The Earth is a closed thermodynamic system, freely exchanging energy with the rest of the universe. To keep the Earth’s surface temperature constant, the mean power coming in (Pin) must equal the mean power going out (Pout).

Themes