Week 3: WS6 #2

2. Consider the amount of energy produced by the Sun per unit time, also known as the bolometric luminosity, $L_{\odot}$. That same amount of energy per time is present at the surface of all spheres centered on the sun at distances r > $R_{\odot}$. However, the flux at a given patch on the surface of these spheres depends on r. 
a) How does flux, F, depend on luminosity, L, and distance, r? 

Since the luminosity remains constant, the flux must decrease as the light spreads out from the Sun in a sphere (essentially the photons are spreading out as they move away from the Sun). The relation, then, is just the luminosity divided by the surface area of the expanding sphere: $F=\frac{L}{4\pi r^2}$

b) The Solar flux at the Earth-Sun distance has been measured to high precision, and for the purposes of this exercise is given by $F_{\oplus}=1.4\times10^6\frac{ergs}{cm^2s}$. Given that the Sun's angular diameter is $\theta=0.57^{\circ}$, what is the effective temperature of the Sun? 

Since the luminosity remains constant, we can set the luminosities at the two distances equal. This means:
$4\pi R_{\odot}^2F_{\odot}=4\pi a^2F_{\oplus}$, where $R_{\odot}$ is the radius of the Sun and a is 1AU, or the distance from the Earth to the Sun ($1.5\times10^{13}cm$).
$R_{\odot}^2F_{\odot}=a^2F_{\oplus}$
$F_{\odot}=\frac{a^2F_{\oplus}}{R_{\odot}^2}$
Now that we've figured out what the flux at the surface of the Sun should be, we can use that to find the temperature using $F=\sigma T^4$, where $\sigma=5.7\times 10^{-5}\frac{erg}{cm^2sK^4}$, the Stefan-Boltzmann constant.
$T=(\frac{F_{\odot}}{\sigma})^{1/4}$
$T=(\frac{a^2F_\oplus}{R_\odot^2\sigma})^{1/4}$
We have a and $F_\oplus$, but we need to find $R_\odot$. We can do this based on the angular diameter of the Sun, using trig:

$R_\odot=7.46\times10^{10}cm$

Finally, we can plug all the values in and solve to get a final answer $T\approx5600K$.