Blog Post 5, WS 2.1, Problem 4: Supernova Distance

A supernova goes off and you can barely detect it with your eyes. Astronomers tell you that supernovae have a luminosity of \(10^{42}\) erg/s; what is the distance of the supernova?
Assume the supernova emits most of its energy at the peak of the eye’s sensitivity and that it explodes isotropically.

This problem is remarkably similar to Problem 3a from WS 2.1. We can use the same equation relating distance, flux, and luminosity.

\[d=\left(\frac{L}{4\pi F}\right)^{1/2}\]

The tricky part is finding the flux.  

\[F=\frac{energy}{Area\times Time}=\frac{erg}{cm^2s}\]

We are told in the problem that the supernova is just barely visible to the human eye.  This is the key to the flux problem.  

Here are some facts about the human eye. The eye needs to receive 10 photons in order to get a light signal.  The eye refreshes itself on average 60 times per second.  The pupil has a radius of about 0.35cm.

Let's change these into variables:

Signal \(s=10 photons\), refresh rate \(f_p=60hz\), and radius \(r=0.35cm\)

The eye's peak frequency sensitivity is \(f=5.4\times 10^{14} hz\)

We can calculate the energy in a photon (\(E_p\)) using Planck's constant \(h=6.63\times 10^{-27} erg\times s\)

\[E_p=hf\]

We need 10 photons per refresh cycle, so:

\[E=s E_p=s hf\]

We need energy per second, so we should introduce the eye's refresh rate.

\[\frac{E}{t}=s hf f_p\]

This is the energy per second received by the eye.

Next, we need the eye's area.  Since the pupil is a circle, this is elementary.

\[A=\pi r^2\]

So, putting this together, we get:

\[F=\frac{E}{tA}\]

\[F=\frac{shff_p}{\pi r^2}\]

We want distance, so when we contend this with our original equation, we get:

\[d=\left(\frac{L\pi r^2}{4\pi shf_p f}\right)^{1/2}\]

It's time to plug in numbers:

\[d=\left(\frac{(10^{42}erg/s) (0.35cm)^2}{4 (10)(6.63\times 10^{-27}ergs\times s)(60hz)(5.4\times 10^{14}hz)}\right)^{1/2}\]

\[\boxed{d=3.8\times 10^{24}cm \approx 1200 kpc}\]

I worked with B. Brzycki, G. Grell, and N. James on this problem.