Effective Temperature

1. Problem

Given the Sun acts like a black body radiating at a constant luminosity, determine the effective temperature of the Earth. Assume the Earth has no atmosphere.

2. Overview

2.1. Computing the Approximate Temperature of the Earth

Identify. We use the Stefan–Boltzmann law for the Sun's luminosity, then considering the Irradiance of the Earth receiving this radiation, we assume the Earth is in a steady state…so its luminosity (from the Stefan–Boltzmann law) of radiation emitted equals its irradiance received from the Sun.

Setup and Execute. We use the Stefan–Boltzmann law for black-body temperature, the Luminosity of the sun is then stipulated to be:

\begin{equation} L_{\odot} = 4\pi R_{\odot}^{2}\sigma T_{\odot}^{4}. \end{equation}

The Earth is a distance \(a_{0}\) from the Sun, the irradiance received by a single point at this distance would be

\begin{equation} E_{\oplus} = \frac{L_{\odot}}{4\pi a_{0}^{2}}. \end{equation}

The radiance flux absorbed by the Earth would be given by treating the Earth as a disc ("broadside of a barn") which is hit by the Sun's rays. That is to say, integrating the irradiance by the cross-sectional area of the Earth, or

\begin{equation} \Phi_{abs} = \pi R_{\oplus}^{2}E_{\oplus}. \end{equation}

In a steady state, the Luminosity emitted by the Earth is equal to the irradiance received, giving us

\begin{equation} 4\pi R_{\oplus}^{2}\sigma T_{\oplus}^{4} = \pi R_{\oplus}^{2}E_{\oplus}. \end{equation}

We substitute in the value of \(E_{\oplus}\) to the right-hand side, giving us

\begin{equation} 4\pi R_{\oplus}^{2}\sigma T_{\oplus}^{4} = \pi R_{\oplus}^{2}\frac{L_{\odot}}{4\pi a_{0}^{2}}. \end{equation}

The luminosity of the Sun transforms the right-hand side into

\begin{equation} 4\pi R_{\oplus}^{2}\sigma T_{\oplus}^{4} = \pi R_{\oplus}^{2}\frac{4\pi R_{\odot}^{2}\sigma T_{\odot}^{4}}{4\pi a_{0}^{2}}. \end{equation}

Now we solve for the temperature of the Earth \(T_{\oplus}\)

\begin{equation} T_{\oplus}^{4} = \frac{R_{\odot}^{2}T_{\odot}^{4}}{4a_{0}^{2}} \end{equation}

which simplifies to

\begin{equation} T_{\oplus} = T_{\odot}\sqrt{\frac{R_{\odot}}{2a_{0}}}. \end{equation}

We can plug in empirical values for these quantities to get an estimate for the temperature of the Earth.

(let ((sun-temperature 5772.0) ;; K
      (sun-radius 695700.0) ;; km
      (earth-distance 149596000.0)) ;; km
  (* sun-temperature (sqrt (/ sun-radius (* 2.0 earth-distance)))))

278.3314843147163

Thus the temperature of the Earth \(T_{\odot}\approx 278\mbox{ K}\) or about 5 degrees Celsius on the surface of the Earth, assuming the Earth absorbs all of the Sun's rays and the Earth has no atmosphere.

Evaluate. Does this make sense? Well, as the distance of the Earth increases \(a_{0}\to\infty\) we see the surface temperature for the Earth decreases \(T_{\oplus}\to0\), as we would expect. Similarly, for decreasing the radius of the Sun \(R_{\odot}\to0\), we see the Earth would receive decreasing radiance hence decreasing temperature \(T_{\oplus}\to0\), as we would expect.

3. Remark: Explicit Formula

We should note that the explicit formula for the effective temperature of a planet at the top of the atmosphere may be written as:

\begin{equation} T_{\oplus} = T_{\odot} \sqrt{\frac{R_{\odot}\sqrt{(1-a)}}{2D}} \end{equation}

where:

\(a\) is the planet's albedo (a constant between 0 and 1, for Earth \(a\approx 0.3\))
\(D\) is the distance from the sun to the planet
\(R_{\odot}\) is the radius of the sun
\(T_{\odot}\) is the temperature of the sun
\(T_{\oplus}\) is the effective temperature at the top of the atmosphere of the planet.

A handy relationship that falls out of this: if we have two planets and we know \(D_{1}\), \(D_{2}\), the albedos \(a_{j}\), and the effective temperature of the first \(T_{1}\), we can obtain the effective temperature of the second by multiplying by \(T_{2} = T_{1}\sqrt{D_{1}/D_{2}}\sqrt[4]{(1-a_{2})/(1-a_{1})}\).

If we do not know the albedos, and assume there is a uniform probability of taking any value between 0 and 1, then the expected value of the quartic root of the ratios turns out to be

\begin{equation} \int^{1}_{0}\int^{1}_{0}\sqrt[4]{\frac{1-a_{2}}{1-a_{1}}}\,\D a_{1}\D a_{2} = \frac{16}{15}. \end{equation}

If we have absolutely no knowledge of the albedos, this is a fine estimate.

If we know \(a_{2}\) but not \(a_{1}\), we could use the approximation

\begin{equation} \int^{1}_{0}\sqrt[4]{\frac{1-a_{2}}{1-a_{1}}}\,\D a_{1} = \frac{4}{3}\sqrt[4]{1-a_{2}}. \end{equation}

If we know \(a_{1}\) but not \(a_{2}\), we could use

\begin{equation} \int^{1}_{0}\sqrt[4]{\frac{1-a_{2}}{1-a_{1}}}\,\D a_{2} = \frac{4}{5}\frac{1}{\sqrt[4]{1-a_{1}}}. \end{equation}

All assuming a uniform ("uninformed") distribution.

So, for example, a generic planet in the same solar system as Earth. If we do not know the other planet's albedo, we could use the estimate

\begin{equation} \sqrt[4]{\frac{1-a_{2}}{1-a_{1}}} \approx \frac{4}{5}\frac{1}{\sqrt[4]{0.7}}\approx0.874612. \end{equation}

This is just an estimate, to be used when we need some sense of what the answer should be near.