Taking the Sun’s Temperature

We are bounded in a nutshell of Infinite Space: Worksheet # 6, Problem #2: Taking the Sun’s Temperature

2. Consider the amount of energy produced by the Sun per unit time, also known as the bolometric luminosity, $L_{\odot}$ . The same amount of energy per time is present at the surface of all spheres centered on the Sun at distance $r > R_{\odot}$ . However, the flux at a given patch on the surface of these spheres depends on r. (the Stefan-Boltzmann Constant: $\sigma = 5.7 \times 10^{-5} erg~ cm^{-2}~ s^{-1} ~ K^{-4}$ )

               

            a. How does flux, F, depend on luminosity, L, and distance, r?

               

            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 $F_{\oplus} = 1.4 \times 10^{6} ergs~s^{-1}~cm^{-2}$ . Given that the Sun’s angular diameter is $\theta = 0.57$ degrees, what is the effective temperature of the Sun? (Hint: start with the mathematical version of the first sentence of this problem, namely: $$L_\odot (r = R_\odot ) = L_\odot (r = 1 AU)$$ and then expand the left and tight sides of their respective distances and fluxes at those distances. )

a. From the several problems we have already worked on, as well as a simple analysis of the units of flux and luminosity, we know they are related by a factor of the area covered. We also know celestial sources emit light isotopically (in all directions) and their light seems dimmer the farther they are away. Therefore, it is just an understanding that flux (the brightness we see) is lessened by the distance (and area) covered by the photons is best described as:   $$F = \frac{L}{4\pi r^2}$$

b. Knowing the base values described in the problem, we can start probing into the actual uses of the equations which describe blackbodies, and objects that are very close to them (i.e. stars). One of the first things we will need before using the relationships of Flux, Luminosities and Radii, we need to solve a bit of trigonometry.

As you can see in the image, and in the problem’s descriptions, the angular size of the Sun in our sky is 0.57 degrees. It’s being so small means we can use a small angle approximation to find its diameter, the opposite side to a right triangle, where the adjacent side is the distance from the Earth to the Sun, So, we have: $$\theta = \frac{D_{\odot}}{d_{\odot - \oplus}}$$ and the angle needs to be used in radians: $$\theta = 0.57 degrees \times \frac{\pi ~rad}{ 180 degrees} = 0.01 ~rad$$ Now we can solve for the diameter of the Sun: $$\theta = \frac{D_{\odot } }{1 AU}$$ $$0.01 =\frac{D_{\odot}} {1 AU}$$ $$0.01 AU = D_\odot ,$$ and with a simple division by 2: $$0.005 AU = R_\odot$$

Now we get into the more interesting part. Using equations which describe blackbodies, we know Luminosity is intrinsically related to Temperature. But first, we use the equations relating luminosity and flux. From the problem, we have: $$L_\odot (r = R_\odot ) = L_\odot (r = 1 AU),$$  and $$F = \frac{L}{4\pi r^2} ,$$ and these two can be combined as: $$F_{R_\odot} (4\pi R_\odot ^2 ) = F_{ d_{\odot - \oplus}} (4 \pi d_{\odot - \oplus} ^2 ),$$ and after cancelling out some constants:   $$F_{R_\odot} (R_\odot ^2 ) = F_{ d_{\odot - \oplus}} (d_{\odot - \oplus} ^2 ),$$ we can start plugging in $$F_{R_\odot} (0.005~AU)^2  = F_{ d_{\odot - \oplus}} (1~AU)^2 ,$$ and with some facts retrieved from the problem statement: $$F_{R_\odot} (0.000025) =  1.4 \times 10^{6} ergs~s^{-1}~cm^{-2} ,$$ and the Flux at the surface of the Sun is: $$F_{R_\odot} = 5.6 \times 10^{10}  ergs~s^{-1}~cm^{-2}.$$

Now, we employ the equation derived in a previous blog post (), which gives a clear relationship between Luminosity and Temperature, but in this case we will be replacing it with the Flux at the surface of the Sun: $$L = 4\pi R^2  ~\sigma T^4 ,$$ and changing it into Flux: $$F_{R_\odot} \cdot 4\pi R_{\odot}^2 = 4\pi R_{\odot}^2 ~ \sigma T^4 ,$$ now we cancel some constants and begin to plug in: $$F_{R_\odot} = \sigma T^4$$ $$T = \left( \frac{ F_{R_\odot} }{\sigma }\right)^{\frac{1}{4}}$$ $$T =\left( \frac{ 5.6 \times 10^{10} ergs~s^{-1}~cm^{-2} }{ 5.7 \times 10^{-5} erg~ cm^{-2}~ s^{-1} ~ K^{-4}}\right)^{\frac{1}{4}} ,$$ and the final solution, with a bit of approximation, is: $$T_\odot = 5600 K$$