This is where you live

We are bounded in a nutshell of Infinite Space: Week 11: Worksheet #20: Problem #1: This is where you live 

1.

a.

b. For this problem, we are attempting to find the range of the distance around a star in which a planet may have liquid water, what we know as the habitable zone. In order to understand this, we must find the correlation of temperature and the brightness of stellar objects. For a planet, we know the Flux it receives from the star it orbits is: $$\frac{L_\star}{4\pi a^2}  = F_P ,$$ which can be re written to better represent all the energy impacting the planet in a specified time: $$F = \frac{Energy}{time \cdot area}$$ $$\frac{L_\star}{4\pi a^2} \cdot \pi R_P^2 = \frac{Energy}{time}~received~by~planet$$

Furthermore, we should understand the energy the planet then radiates back into space, which can be described with the equation for bolometric flux: $$F = \sigma T_P^4,$$ and from previous worksheets we know that the flux can be turned into luminosity by just multiplying by the surface area $$L_P = \sigma T_P^4 \cdot 4\pi R_P^2$$  

c. Next we set these equations equal to each other and find the expression for the temperature of the planet: $$\frac{L_\star}{4\pi a^2} \cdot \pi R_P^2 = \frac{Energy}{time}~received~by~planet = L_P = \sigma T_P^4 \cdot 4\pi R_P^2$$   $$\frac{L_\star}{4\pi a^2} \cdot \pi R_P^2 = \sigma T_P^4 \cdot 4\pi R_P^2,$$ we start simplifying the equation and find that $$\frac{L_\star}{4a^2} = \sigma T_P^4 \cdot 4\pi ,$$ which finally turns into: $$T_P = \left(\frac{L_\star}{16 a^2 \pi \sigma}\right)^{1/4}$$

Also, we can keep on simplifying the expression were we to consider the definition of the star’s luminosity: $$L_\star = 4\pi R_\star^2 \cdot \sigma T_{eff}^4,$$ which can be placed into the equation we just derived to further understand the factors involved: $$T_P = \left(\frac{4\pi R_\star^2 \cdot \sigma T_{eff}^4}{16 a^2 \pi \sigma}\right)^{1/4},$$ $$T_P = \left(\frac{ R_\star^2 T_{eff}^4}{4 a^2 }\right)^{1/4},$$ This results in the simplified version of the equation which expresses the relationship between the planet’s temperature, and the radius and temperature of the star: $$T_P = T_{eff} \sqrt{\frac{R_\star}{2a}}$$

d. As we can see from the equations, the radius of the planet becomes insignificant, it not being a factor in establishing the temperature of the planet, rather the radius of the star comes into play.  

e. Now, we can assume some energy gets reflected from the surface, yielding an equation similar to: $$T_P = \left(\frac{L_\star}{16 a^2 \pi \sigma}\right)^{1/4},$$ but where A is the energy per time reflected: $$T_P = \left(\frac{L_\star - A}{16 a^2 \pi \sigma}\right)^{1/4},$$ which simplifies to: $$T_P = T_{eff} \sqrt{\frac{R_\star}{2a}} - \left(\frac{A}{16 a^2 \pi \sigma}\right)^{1/4}.$$ Therefore, the temperature most definitely goes down as the reflectivity increases.