Worksheet 6, Problem 2; Temperature of the Sun

The problem: 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. (The Stefan-Boltzmann Constant: σ = 5.7 x $10^5 erg cm^2 s^{-1} K^4$)
(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 $F_\odot = 1.4 x 10^6 erg cm^2 s^{-1}$ . Given that the Sun’s angular diameter is θ = 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 right sides in terms of their respective distances, r, and fluxes at those distances.)


What we know:

• $F_\odot = 1.4 x 10^6 erg cm^2 s^{-1}$ 
• $L_\odot = 4 \pi R_\odot^2  σ T^4$
• $F = σ T^4$
• θ = 0.57 degrees, which we can use to calculate the actual diameter of the sun

$$tan(0.57) = \frac{x}{1.5 x 10^{13}} \: x = tan(0.57) \cdot 1.5 x 10^{13}$$ for small angles we can approximate tan(θ) as θ in radians so tan(.57) in radians is: $$.57\,^{\circ} \cdot \frac{2 \pi}{360\,^{\circ}} = \frac{.57 \pi}{180}$$ and the radius of the sun is: $$R_\odot \approx \frac{1.5 x 10^{13}}{2} \cdot \frac{57 \pi}{180} \approx 7.5 x 10^{10}cm$$

Solve:

Part A:

• We know the units of F is $\frac{erg}{s\:cm^2}$ and L is $\frac{erg}{s}$ and A is $cm^2$

We can therefore conclude that we can use the relationship $$F= \frac{L}{A_E}$$ where A is the area where the sun is isotropically emitting energy with the radius being the distance from the sun to Earth so $$A_r = 4 \pi d_E^2$$ which gives $$F = \frac{L}{ 4 \pi d_E^2}$$

Part B:
We know $L_\odot = 4 \pi R_\odot^2  T_\odot^4 \sigma$ and solving for luminosity gives $$L_\odot= F_\odot 4 \pi d_E^2$$ $$4 \pi R_\odot^2  T_\odot^4 \sigma =F \cdot 4 \pi d_E^2$$ and we can solve for temperature: $$T_\odot^4 = \frac{F_\odot 4 \pi d_E^2}{ 4 \pi R_\odot^2  \sigma }$$ $$T_\odot = \sqrt[4]{\frac{F_\odot d_E^2}{ R_\odot^2  \sigma }}$$ Plugging in all out givens allows us to solve: $$T_\odot = \sqrt[4]{\frac{1.4 x 10^6  \cdot  (1.5 x 10^{13})^2}{ (7.5 x 10^{10})^2 (5.7 x 10^{-5} )}}$$ $$T_\odot = 5.6 x 10^3 \:K$$

Acknowledgements: I worked with Barra and April on this problem.