Worksheet 12.2, Problem 2: Goldilocks Zone

The Problem: 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, Tp, depends on the properties of the central star and the orbital properties of the planet.

This image gives a good overall preview to the Goldilocks Zone and the factors that we will ultimately calculate that influence it.

Source

Part A: Draw the Sun on the left, and a planet on the right, separated by a distance a.

Part 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?

We used dimensional analysis to remind ourselves that the units for luminosity is  $\frac{erg}{s}$ so: $$L_{\star} = \frac{E_{\star} }{t}$$ Based on the question, we then deduced the energy per time the planet receives would be $$\frac{E_p }{t} = \frac{L_{\star} }{4 \pi a^2} \cdot \pi R_p^2$$ Where $\pi R_p^2$ is the cross-sectional area of the planet that receives light and heat from the star (a circle). So this is   $$\frac{E_p }{t} = \frac{L_{\star} R_p^2 }{4 a^2}$$

How much energy per time does the Earth radiate as a blackbody?

Treating Earth as a blackbody would give: $$\frac{E_p }{t} = 4 \pi \sigma R_p^2 T_p^4$$ We know this from when we worked with blackbodies some weeks ago.

Part C: Set these two quantities equal to each other and solve for TP .
$$\frac{L_{\star} R_p^2 }{4 a^2} = 4 \pi \sigma R_p^2 T_p^4$$ When we isolate T we get $$T_p^4 = \frac{L_{\star}}{16 \sigma a^2}$$ Or $$T_p= \left( \frac{L_{\star}}{16 \sigma a^2} \right)^{\frac{1}{4}}$$

Part D: How does the temperature change if the planet were much larger or much smaller?

The temperature would not change. We know this 1) because the radius of the planet is in our equation and 2) logically, any change in size of a planet would be negligible compared to the distance of the planet from the star and should not make a difference.

Part 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?

The temperature lost due to reflected energy would be $$T_p^4 = \frac{L_{\star}A}{16 \sigma a^2}$$ So the total temperature would be $$T_p^4 = \frac{L_{\star}}{16 \sigma a^2} - \frac{L_{\star}A}{16 \sigma a^2}$$ Or $$\frac{L_{\star}(1-A)}{16 \sigma a^2}$$ And T would scale to $$T_p \sim (1-A)^{\frac{1}{4}}$$

I worked with Sean, April, and Barra on this problem.