Black body planet

A model of the planets as black bodies is surprisingly accurate, except in one interesting case¹.

In theory

A black body is an ideal object which absorbs all radiation which falls on it, and then reemits it as thermal radiation. Real objects are, of course, not black bodies, but they are often surprisingly close. One nice thing about black bodies is that there are is a nice equation which relates the amount of power they radiate to their temperature:

\[ \frac{P}{A} = \sigma T^4 \]

Where \(P\) is power, \(A\) is area and \(\sigma\) is the Stefan-Boltzmann constant which is \(\sigma \approx 5.67\times10^{−8}\,\mathrm{Wm^{-2}K^{−4}}\): this formula tells you the power per unit area that a black body radiates, at a given temperature.

Well, obviously you can derive this formula:

\[ T = \left(\frac{P}{A\sigma}\right)^{1/4} \]

In other words temperature goes as the fourth root of power.

If you consider a ball of radius \(r\) then its surface area is \(A = 4\pi r^2\), so a perfectly spherical black body of radius \(r\) at a uniform temperature \(T\) radiates a total output power \(P_O\)

\[ P_O = 4\pi r^2 \sigma T^4 \]

or equivalently

\[ T = \left(\frac{P_O}{4\pi r^2 \sigma}\right)^{1/4} \]

OK, so consider a planet, which is a perfect black-body, orbiting at a radius \(R\) from the Sun². Let the input power flux from the Sun, at the point directly facing the Sun, be \(S\). For Earth, \(S \approx 1360\,\mathrm{Wm^{-2}}\). We can calculate two things from this.

The total output power of the Sun, \(P_S\) is given by the integral of \(S\) over a sphere of radius \(R\):

\[ P_S = 4\pi R^2 S \]

The total power falling on the planet, \(P_I\), is given by the integral of \(S\) over the surface of the disk of the planet which faces the Sun, and this is

\[ P_I = \pi r^2 S \]

So the first of these equations can be used to work out \(S\) in terms of \(P_S\) and then substituted into the second one:

\[ P_I = P_S\frac{r^2}{4 R^2} \]

If the planet is at equilibrium, then \(P_O = P_I\) or, in other words, output and input power is the same³. So

\[ T = \left(\frac{P_S}{16\pi \sigma R^2}\right)^{1/4} \]

Finally, given \(S\) and \(R\) we can work out \(P_S\), and for our Sun, based on \(R \approx 1.50\times 10^{11}\,\mathrm{m}\) for Earth we get \(P_S \approx 3.85\times 10^{26}\,\mathrm{W}\).

In practice

So, we can compute the surface temperatures for the rocky planets, assuming they were black bodies.

Earth. \(R \approx 1.50\times 10^{11}\mathrm{m}\), giving \(T \approx 278\,\mathrm{K}\) or \(5^\circ\mathrm{C}\). Actual mean surface temperature is \(287\,\mathrm{K}\) or \(14^\circ\mathrm{C}\): this is reasonably accurate.

Mercury. \(R \approx 5.79\times 10^{10}\,\mathrm{m}\) giving \(T \approx 448\,\mathrm{K}\) or \(175^\circ\mathrm{C}\). Actual mean surface temperature is \(452\,\mathrm{K}\) or \(179^\circ\mathrm{C}\). This is also OK.

Mars. \(R \approx 2.28\times 10^{11}\,\mathrm{m}\), giving \(T \approx 226\,\mathrm{K}\) or \(-47^\circ\mathrm{C}\). Actual mean surface temperature is \(226\,\mathrm{K}\) or \(-47^\circ\mathrm{C}\). This is pretty spooky: it has no right to be this good, and is probably only this good by chance.

Pluto. \(R \approx 5.91\times 10^{12}\,\mathrm{m}\), giving \(T \approx 44\,\mathrm{K}\) or \(-229^\circ\mathrm{C}\). Actual mean surface temperature is \(44\,\mathrm{K}\) or \(-229^\circ\mathrm{C}\). The same is true for this: it is probably only this good by chance.

Venus. \(R \approx 1.08\times 10^{11}\,\mathrm{m}\), giving \(T \approx 328\,\mathrm{K}\) or \(55^\circ\mathrm{C}\). Actual mean surface temperature is \(730\,\mathrm{K}\) or \(460^\circ\mathrm{C}\). This is hopeless, as you would expect it to be.

These are astonishingly accurate, except for Venus, which is out by a factor of nearly two: Venus is the hottest planet in the Solar system because of a runaway greenhouse effect.

Notes

Data came from Alpha which, while I have considerable qualms about Wolfram products, is a reasonably good source of this sort of thing. I have been very casual about rounding, generally rounding to integers for both \(\mathrm{K}\) and \({}^\circ\mathrm{C}\). The temperatures don’t come with an uncertainty but I suspect that for Venus is less accurate than the others. I didn’t deal with the gas giants because they don’t have well-defined surfaces and I was just too lazy.