Worksheet 12.2, Problem 2: The Habitable Zone

The Earth resides in a "Goldilocks Zone" or habitable zone (HZ) around the Sun. At our semimajor axis we receive just enough sunlight to prevent the planet from freezing over and not too much to boil off our oceans. Not too cold, not too hot. Just right. In this problem, we'll calculate how the temperature of a planet, \(T_p\), depends on the properties of the central star and the orbital properties of the planet.

The habitable zone is the range of distance away from a star in which a planet might be able to sustain life. This distance depends on the size and luminosity of the star, and the relationship is shown in the following image. 

Habitable Zone
Image Credit: http://upload.wikimedia.org/wikipedia/commons/4/46/Gliese_581_-_2010.jpg

a) Draw the Sun on the left, and a planet on the right, separated by a distance a. The planet has a radius \(R_p\) and temperature \(T_p\). The star has a radius \(R_\star\), a luminosity \(L_\star\), and a temperature \(T_{eff}\). 

b) Due to energy conservation, the amount of energy received per unit time by the planet is equal to the energy emitted isotropically under the assumption that it is a blackbody. 

i) How much energy per time does the planet receive from the star?

To find the energy per time that the planet receives from the star, we must find the fraction of area the planet takes up in a spherical shell of radius a surrounding the star. The planet will receive the same fraction of energy from the star as this fraction of area. This fraction is equal to the cross-sectional area of the planet divided by the total surface area of the shell:

\(\dfrac{\pi R_p^2}{4\pi a^2}=\left(\dfrac{R_p}{2a}\right)^2\)

We also know that the luminosity of the star gives the energy radiated per time. This means that the energy received by the planet is equal to:

\(L_\oplus = L_\star\left(\dfrac{R_p}{2a}\right)^2\)

ii) How much energy per time does the planet radiate as a blackbody?

Since we are treating the planet as a blackbody, we can use the Stefan-Boltzmann equation to find the energy the planet is radiating per time.

\(L_\oplus = 4\pi R_p^2\sigma T_p^4\)

c) Set these two quantities equal to each other and solve for \(T_p\). 

Setting the two equations from part (b) equal to each other, we get:

\(L_\star\left(\dfrac{R_p}{2a}\right)^2=4\pi R_p^2\sigma T_p^4\)

Now solving for the temperature, we get:

\(T_p^4=\dfrac{L_\star}{16a^2\pi R_p^2\sigma}\)

\(T_p=\left(\dfrac{L_\star}{16a^2\pi \sigma}\right)^{1/4}\)

d) How does the temperature change if the planet were much larger or much smaller?

The temperature is independent of time, so it will not change no matter what size the planet is.

e) Not all of the energy incident on the planet will be absorbed. Some fraction, A, will be reflected back out into space. How does this affect the amount of energy received per unit time, and thus how does this affect \(T_p\)?

If some fraction, A, of the energy incident on the planet is reflected back, then \(1-A\) will be the fraction of the energy which is absorbed by the planet. So multiplying the energy incident on the planet per time by \(1-A\) will give us the amount of energy received per time:

\(L_\oplus = L_\star\left(\dfrac{R_p}{2a}\right)^2 (1-A)\)

Setting this equal to the energy radiated by the planet, and solving for temperature again, we get:

\(L_\star\left(\dfrac{R_p}{2a}\right)^2(1-A)=4\pi R_p^2\sigma T_p^4\)

\(T_p=\left(\dfrac{L_\star(1-A)}{16a^2\pi \sigma}\right)^{1/4}\)

If we were able to know the luminosity and approximate temperature of a star as well as the fraction of energy, A, reflected off the planet, then we can determine the region in which we might be able to find planets with potential for life.

I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem.