a) How does flux, F, depend on luminosity, L, and distance, r?
Since the luminosity remains constant, the flux must decrease as the light spreads out from the Sun in a sphere (essentially the photons are spreading out as they move away from the Sun). The relation, then, is just the luminosity divided by the surface area of the expanding sphere: \(F=\frac{L}{4\pi r^2}\)
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 \(F_{\oplus}=1.4\times10^6\frac{ergs}{cm^2s}\). Given that the Sun's angular diameter is \(\theta=0.57^{\circ}\), what is the effective temperature of the Sun?
Since the luminosity remains constant, we can set the luminosities at the two distances equal. This means:
\(4\pi R_{\odot}^2F_{\odot}=4\pi a^2F_{\oplus}\), where \(R_{\odot}\) is the radius of the Sun and a is 1AU, or the distance from the Earth to the Sun (\(1.5\times10^{13}cm\)).
\(R_{\odot}^2F_{\odot}=a^2F_{\oplus}\)
\(F_{\odot}=\frac{a^2F_{\oplus}}{R_{\odot}^2}\)
Now that we've figured out what the flux at the surface of the Sun should be, we can use that to find the temperature using \(F=\sigma T^4\), where \(\sigma=5.7\times 10^{-5}\frac{erg}{cm^2sK^4}\), the Stefan-Boltzmann constant.
\(T=(\frac{F_{\odot}}{\sigma})^{1/4}\)
\(T=(\frac{a^2F_\oplus}{R_\odot^2\sigma})^{1/4}\)
We have a and \(F_\oplus\), but we need to find \(R_\odot\). We can do this based on the angular diameter of the Sun, using trig:
\(R_\odot=7.46\times10^{10}cm\)
Finally, we can plug all the values in and solve to get a final answer \(T\approx5600K\).
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