Blog #5: Is that a Supernova?

4) A supernova goes off and you can barely detect it with your eyes. Astronomers tell you that supernovae have a luminosity of $10^{42} \frac{\text{ ergs}}{\text{ second}}$. 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. 

In order to solve this problem, let's consider what it means for a supernova to be barely detectable with their eyes. Since the light is barely being received by the eye, it means that even the faintest appearance of the supernova would make the eye receives at least 10 photons of light with a framerate of about 60 frames per second. Also important to note is that humans see the average wavelength of 600 nm. This information can tell us how much power the eye inputs in order to see this image.

Power is defined as units of energy over time. Energy of one photon is defined as $E = \frac{hc}{\lambda}$, where $h = 6.62 \times 10^{-27} \frac{erg}{sec}$ is planck's constant, $c = 3 \times 10^{10} \frac{cm}{sec}$ is the speed of light, and $\lambda = 6 \times 10^{-5}$ is the wavelength. Using this information, the power int of the eye is:

\begin{align}
P_{eye} &= \frac{\text{Energy of 10 Photons}}{\text{time}}\\
P_{eye} &= \frac{10 \times \frac{hc}{\lambda}}{\frac{1}{60 \text{ sec}}}\\
P_{eye} &= \frac{10 \times hc \times 60}{\lambda}\\
P_{eye} &= \frac{(10) \times (6.62 \times 10^{-27} \frac{\text{ erg}}{\text{ sec}}) \times (3 \times 10^{10} \frac{\text{ cm}}{\text{ sec}})\times (60 \text{ sec})}{6 \times 10^{-5} \text{cm}}\\
P_{eye} &\approx 2 \times 10^{-9} \frac{ \text{ ergs}}{\text{ sec}}
\end{align}

Now that we know how much power the eye is inputting, we can set up a proportion to compare the surface area of the eye, $SA_{eye}$ and the surface area of the $SA_{supernova}$ with the power input of the eye, $P_{eye}$, with the power output of the supernova, $P_{supernova}$. The proportion is as follows:

\begin{align}
\frac{P_{supernova}}{SA_{supernova}} &= \frac{P_{eye}}{SA_{eye}}\\
\frac{P_{supernova}}{4\pi D^2} &= \frac{P_{eye}}{4\pi R_{eye}^2}
\end{align}

Having this proportionality, we can rearrange the equation above to solve for the distance to the supernova. Assume that the radius of the eye is about 1 cm.

\begin{align}
\frac{P_{supernova}}{4\pi D^2} &= \frac{P_{eye}}{4\pi R_{eye}^2}\\
D &= \left(\frac{R_{eye}^2 \times P_{supernova}}{4 P_{eye}}\right)^{\frac{1}{2}}\\
D &= \left(\frac{(1 \text{ cm})^2 \times (10^{42} \frac{\text{ ergs}}{\text{ second}})}{4 \times (2 \times 10^{-9} \frac{ \text{ ergs}}{\text{ sec}})}\right)^{\frac{1}{2}}\\
D &= 1.12 \times 10^{25} \text{ cm}
\end{align}

Therefore, the supernova exploded $1.12 \times 10^{25} \text{ cm}$ away!!!