Blog #27 - Habitable Zone pt.1

Habitable Zone Pt. 1

The Earth resides in a “Goldilocks Zone” or habitable zone (HZ) around the Sun. At our semi major 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, Tp, depends on the properties of the central star and the orbital properties of the planet.

(a) Draw the Sun on the left, and a planet on the right, separated by a distance a. The planet has a radius Rp and temperature Tp. The star has a radius R and a luminosity L and a temperature Teff.

We begin by putting together a picture of the setup, seen below: 

(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. How much energy per time does the planet receive from the star?How much energy per time does the Earth radiate as a blackbody?

We start by finding the amount of energy incident on the planet. This number comes from the amount of flux that hits the planet's surface. \[F_s = \frac{L_s}{4 \pi a^2}\] \[E_{in} = F_s * \pi R_p^2 = \frac{L_s R_p^2}{4 a^2} \]We then solve for energy coming out of the planet as simply the luminosity out of the surface of the Earth. \[E_{out} = L_p = \sigma T_p^4 (4 \pi R_p^2)\]

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

Setting the two energies equal to each other gives us: \[E_{in} = E_{out} \rightarrow \sigma T_p^4 (4 \pi R_p^2) = \frac{L_s R_p^2}{4 a^2}\] Rearranging for \(T_p\) we get \[T_p = (\frac{L_s}{16 \pi a^2 \sigma})^{\frac{1}{4}} \]

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

Notice that this equation does not depend on the size of the planet so the distance is inherently based on the distance from and luminosity of the star. 

(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 Tp?

By adding a term for reflectance the \(E_{in}\) becomes \[E_{in} = \frac{L_s R_p^2}{4 a^2}*(1-A)\] Meaning that our \(E_{in}\) will not reach the full flux of the star incident to the planet. This means that the energy out of the planet will also have to decrease meaning the overall temperature of the planet will decrease as well.