Worksheet 12.2: Problem 3

3. In this problem, we'll figure out how the habitable zone distance, \(a_{HZ}\), depends on stellar mass. Recall the average mass-luminosity relation we derived earlier, as well as the mass-radius relation for stars on the main sequence. If you don't recall, this is a good time to practice something that will very likely show up on the final!

a) Express \(a_{HZ}\) in terms of stellar properties as a scaling relationship, using squiggles in stead of equal signs and ditching constants.
The habitable zone is determined by the temperature of the planet, which has to be within a certain range to support life as we know it. We know from our previous calculations that the temperature of the planet, \(T_{P}\), is equal to \((\frac{L_{*}}{\pi \sigma a_{HZ}^{2}})(1-A)\), where A is the the fraction of energy reflected by the planet. Removing constants we see that:
\[T_{P}^{4}\:\sim\:\frac{L_{*}}{a_{HZ}^{2}}\]

b) Replace the stellar parameters with their dependence on stellar mass, such that \(a_{HZ}\:\sim\:M^{\alpha}\). Find \(\alpha\).
Since \(T_{P}\) has to be in a specific range for the planet to be habitable, we can consider it to be constant. Thus, \(a_{HZ}^{2}\:\sim\:L_{*}\). We know from our average mass-luminosity relation that \(L_{*}\:\sim\:M_{*}^{4}\). Substituting we see that:
\[a_{HZ}^{2}\:\sim\:M_{*}^{4}\]
\[a_{HZ}\:\sim\:M_{*}^{2}\]
Thus, \(\alpha\) is equal to 2.

c) If the Sun were half as massive and the Earth had the same equilibrium temperature, how many days would our year contain?
In order to solve this problem, we need to know how a change in the Sun's mass would affect the Earth's orbital period. We can solve that using Kepler's Third Law. However, before we can solve for the new period, we need to know the new habitable zone distance. Luckily, part b set up a nice relationship between the habitable zone distance and the mass of the star. Since \(a_{HZ}\:\sim\:M_{*}^{2}\), halving the Sun's mass would cause the habitable zone distance to be one-fourth as large. Using this information and Kepler's Third Law we get:
\[P^{2}\:\sim\:\frac{a^{3}}{M}\]
\[xP^{2}\:\sim\:\frac{\frac{1}{64} a^{3}}{\frac{1}{2}M}\]
\[xP^{2}\:\sim\:\frac{a^{3}}{32M}\:\rightarrow\:xP^{2}\:\sim\:\frac{1}{32}P^{2}\]
\[x\:\sim\:\frac{1}{32}\]
Thus, the square of the orbital period will be reduced by a factor of \(\frac{1}{32}\). Thus the new orbital period will be equal to the square root of \(\frac{P^{2}}{32}\). Since the square of Earth's orbital period, in square days, is \(365\:\times\:365\:=\: 133225\), the square of the period divided by 32 is equal to approximately 4,163 square days. The square root of this is equal to approximately 64.5 days.

Thus, if the Sun were half as massive, there would be 64.5 days in an Earth year.