Worksheet 6, Problem 2

2. Consider the amount of energy produced by the Sun per unit time, also known as the bolometric luminosity, L. That same amount of energy per time is present at the surface of all spheres centered on the Sun at distances r > R. 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 arg cm^-2 s^-1 K^-4) 


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

We touched on this in 2e in the previous worksheet. There we found that \(F\:=\:\frac{L}{4\pi R^2}\). In that equation, R was the radius of the blackbody and not the distance away from the blackbody, r, that we are trying to include in the equation in this problem. The reason that we didn't need to factor r into our first equation is that the flux of the blackbody on its surface is 100% of the flux that is emitted from the blackbody. However, an object at some distance r, will not receive all of that flux because only some of the radiation will be emitted in its direction and much will dissipate across the great distance. Thus, we know that \(L\:=\:4\pi R^{2}F_{sun}\) and \(L\:=\:4\pi r^{2} F_{received}\).

Not all flux through the Sun's surface reaches the Earth

Thus, Flux received depends on luminosity and distance because \(F_{received}\:=\:\frac{L}{4\pi r^{2}}\).

(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 = 1.4 x 10⁶ ergs s^-1 cm^-2. 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⊙(r=R⊙)=L⊙(r=1AU)

and then expand the left and right sides in terms of their respective distances, r, and fluxes at those distances.) 

 The luminosity of the Sun is a constant and is independent of the observer's distance from the sun. Since it is a constant, \(4\pi R^{2}F_{sun}\:=\:4\pi r^{2}F_{received}\). Knowing this, we can see that \(F_{sun}\:=\:\frac{F_{received} r^{2}}{R^{2}}\). Since the temperature of the sun can be determined from the flux on its surface, we can use this relationship to determine the sun's temperature. The only parameter we still need to solve for is the radius of the sun, R.

From this diagram, we can deduce that \(sin(\theta)\:=\:\frac{2R}{1\:AU}\). The law of small angles tells us that for small angles, \(sin(\theta)\:\approx\:\theta\). However we need to convert our given \(\theta\) into radians for this to work.
\[0.57\:degrees\:\times\:\frac{\pi\:rads}{180\:degrees}\:=\:9.948\:\times\:10^{-3}\:radians\]
\[9.948\:\times\:10^{-3}\:=\:\frac{2R}{1\:\times\:10^{13}\:cm}\]
 \[R\:=\:7.0\:\times\:10^{10}\:cm\]

We can then plug this value for R into the equation for \(F_{sun}\) and we find that \[F_{sun}\:=\:6.5\:\times\:10^{10}\:\frac{ergs}{s\:\times\: cm^{2}}\]

We know that \(F_{sun}\:=\:\sigma T^{4}\). Thus:
\[6.5\:\times\:10^{10}\:=\:5.7\:\times\:10^{-5}\:\times\:T^{4}\]
\[T\:=\:5800\:K\]