Worksheet 12, Problem 2: Habitable Zones

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.

(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.

(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?

Flux is given by the following formula: \[ F = \frac{L_{star}}{4 \pi a^2} \] To uncover the amount of energy that gets absorbed by the planet, we can multiply the flux by the area of the planet, as shown below: \[ E = \frac{L_{star}}{4 \pi a^2} \times 4 \pi R_{planet}^2 \] The energy that is radiated isotropically outward by the planet is then: \[E_{planet} = 4 \pi {R_{p}}^2 \sigma {T_{p}}^4 \]
(c) Set these two quantities equal to each other and solve for TP .

Setting these equal, we get the following equation: \[\frac{L_{star}}{4 \pi a^2} \times 4 \pi R_{planet}^2 = 4 \pi {R_{p}}^2 \sigma {T_{p}}^4\] Then, solving for \(T_p\), we get: \[ T_p = {\frac{L_{star}}{4 \pi a^2 \sigma}}^{\frac{1}{4}}\]
(d) How does the temperature change if the planet were much larger or much smaller?

If the planet were much larger or much smaller, the temperature would not change, since the amount of energy absorbed and radiated are both directly proportional to the size of the planet.

(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 amount of energy received per unit time will be decreased by the fraction A, since it is simply being lost to space. The temperature, \(T_p\) will also be reduced, but not by as much as the flux. This is because the temperature is proportional to the \( \frac{1}{4} \) power of flux.

To derive this exact expression, we can use our equations from part (b) and (c). For both the energy absorbed and the resulting \( T_p \), we can multiply the expression by (1 - A), which gives us the correct fraction for what is absorbed. These expression now become: \[ E = (1-A) \frac{L_{star}}{4 \pi a^2} \times 4 \pi R_{planet}^2 \] \[ T_p = (1 - A){\frac{L_{star}}{4 \pi a^2 \sigma}}^{\frac{1}{4}}\]

Acknowledgements: Team EE