Supernovas and Sandra Bullock: Blog 5, Worksheet 2.1, Problem 4

A supernova goes off and you can barely detect it with your eyes. Astronomers tell you that supernovae have a luminosity of 1042 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. 

The light exploding isotropically and hitting our eye would look something like this:

So we can begin to set up this problem.

What we know:

• We know the speed of light is \(3 \cdot 10^{10} \frac{cm}{sec}\)
• E = hν
• \(P = \frac{\delta E}{\delta t}\)
• Visible light falls into a range between 390 - 700 nm, which an average could be 6.0 x \(10^{-7}\) cm.
• The radius of an eye is about 1 cm.
• The "refresh rate" of a movie is 50 Hz.
• The luminosity of the supernova is \( 10^{42} \) erg/s.

Solve:

Based on the fact that the light is emitted isotropically (shown partially in the diagram above), we can set up a ratio between the power emitted by the supernova over the area the light reaches to the power received by your eye to the area of your eye.\[\frac{P_{s}}{a_{light}} = \frac{P_{eye}}{a_{eye}}\]plugging in area formulas gives:\[\frac{P_{s}}{4 \pi d^2} = \frac{P_{eye}}{\pi r_{eye}^2}\] Solving for distance gives :\[d^2 = \frac{P_{s}\cdot r_{eye}^2}{4 \cdot  P_{eye} }\]

Now we all we need to find is the power reaching our eye, starting with finding the energy.  \[E = hv \: or \: E = h\frac{c}{\lambda} = 6.6 x 10^{-27} \cdot \frac{3 x 10^{10}}{6.0 x 10^{-7}} \approx 3.3x10^{-10} ergs\]

Now we need to find time. Something new I learned last semester, when we solved a similar problem, was  refresh rate, or the rate at which movies emit light so that the brain does not notice any flickering in the movie. If the refresh rate of a movie is 50 Hz (\( \frac{1}{s}\)) and 10 photons must hit the eye every \( \frac{1}{50}\)s, so we need 500 photons a second. So the power would be  \[ P = \frac{500 photons}{sec} \cdot 3.3x10^{-10} ergs \approx 1.65x10^{-7} \frac{erg}{s}\] Plugging this into the original equation (and keeping in mind power is just luminosity) we get: \[d^2 = \frac{10^{42} \cdot 1^2}{ 4 \cdot 1.65x10^{-7} } \] \[ d = 1.2 \times 10^{24} cm \: or \: 3.6 \times 10^5 pc\] This seems like a reasonable distance for a supernova.