Estimations, Part III (or, Every Equation We've Ever Used, Combined!)

Consider the amount of energy produced by the Sun per unit time, also known as the bolometric luminosity, $L_s$. That same amount of energy per time is present at the surface of all spheres centered on the Sun at distances $r > L_s$. 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, $r$?

The intro to this question is a little bit difficult to parse, so let's break it down before we jump on in. What it's telling us is that if we were to expand a spherical shell outward from the sun (or any blackbody), the entire bolometric luminosity of the sun would pass through that spherical shell over a given unit of time, regardless of how far away we got. If you built a giant ball of radius $R_s + 1$ cm and stuck the sun inside of it, as much energy would hit it as if its radius were on the scale of light-years.

The idea of capturing an entire star's worth of energy is the theory behind the Dyson Sphere, a theoretical energy collector that would encompass a star to do just that. (Source)

However, because of our old familiar friend the inverse-square law, the amount of energy received by a 10 square centimeter patch of that spherical shell will vary dramatically depending on how far out from the sun it is (by a factor of $1/r^2$, as a matter of fact). Thinking about this in terms of Dyson Spheres actually makes a lot of sense--it would take $4\pi r^2$ times as much material to build one $r$ units of distance away from the star as it would to build it right on its surface, but both would collect the same amount of energy. What applies to entire hypotethetical energy collectors also applies to any area receiving energy from a star (or other source). Thus, the equation for flux is $F = \frac{L}{4 \pi r^2}$.

The propagation of light by the inverse-square law--the same amount of energy passes through a much larger area at greater distances from the source. (Source)

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 $F_s = 1.4 \times 10^6$ ergs s$^{-1}$ cm$^{-2}$. Given that the Sun's angular diameter is $t = 0.57$ degrees, what is the effective temperature of the sun?

This one is going to look like a greatest hits of basic astronomy formulae. Our first step is going to use the inverse-square law to give us the luminosity of the sun. We can rearrange the inverse-square law, solving for luminosity to get $L = 4 \pi r^2 F$. We'll plug this into the Stefan-Boltzmann law, $L=\sigma T^4 4 \pi  R_s^2$ in order to solve for temperature (after all that work looking at the math behind it, it's only fair that we get to actually use it!), but first we'll need something to plug in for the radius of the sun. We could, of course, look it up, but it's not difficult just to figure it out based on what we have here. The distance to the sun, $r = 1$ AU, and the angular size of the sun, $t = .57$ degrees, are given, so we can plug $r$ and $t/2$ into the small angle approximation and solve for $R_{s}$.

Worth remembering how to set up.

We do need to do a bit of math first, since the small angle approximation requires us to work in radians. $.57/2=.285$ degrees. If we set up a relation between degrees and radians, we get $\frac {.285}{360} = \frac{a}{2\pi}$, which solves to $a=5 \times 10^{-3}$ radians. The small angle approximation tells us that $tan(a) \approx a \approx \frac{R_s}{r}$, or $R_{s} = a \times r$.

And those are all the ingredients we need to solve the Stefan-Boltzmann law for $T$! Starting with \[L=\sigma T^4 4 \pi  R_s^2\] We can plug in $4 \pi r^2 F$ for $L$ and $a \times r$ for $R_{s}$, giving us \[4 \pi r^2 F = \sigma T^4 4 \pi  (ar)^2\] We can divide both sides by $4 \pi r^2$ even before we plug anything in, and then solve for $T$, which cuts things down to \[T = [\frac{F}{\sigma a^2}]^{1/4}\] This can be unit checked pretty easily--on the left we have just temperature, while on the right we have $[\frac{erg\:s^{-1}\:cm^{-2}}{erg\:s^{-1}\:cm^{-2}\:T^{-4}}]^{1/4} = T$. Perfect! Now, having simplified the algebra as much as possible, we can plug in values for $a$, $\sigma$, and $F$, giving us the temperature of the sun as \[T = [\frac{1.4 \times 10^6}{(5 \times 10^{-3})(5.7 \times 10^{-5})}]^{1/4} = 5600 \: K\]

For a difference of 178 Kelvin (the actual surface temperature of the Sun is 5778 K)--not bad! Credit goes out to Anne Madoff, Jennifer Shi, Louise Decoppet, and Missy McIntosh for collaborating with me on this fun problem.