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.