From the Stars to Your Eyes (2.1.02)

Question: What is the power output of a star that is 100 light years away that you can barely see from a dark site at night? Assume the star emits most of its energy at the peak of the eye's sensitivity. 

Okay, so how do we approach a problem that seems to give us so little information? Do we even have enough information to solve this? One of the goals of Astronomy is to be able to infer the various properties of the things that are involved in a problem.



To start thinking about this problem, lets figure out all the information given to us. The information that is directly given to us is:

• The distance, $r$, between the star and our eyes is about 100 light years away, which is about $10 \times 10^{20}$ centimeters
• Most of the energy emitted by the star is given at the peak of the eye's sensitivity

Okay, so not much information here. But let's unpack the second piece of information given to us, about the "peak of the eye's sensitivity". What are certain thing we know about the human eye:

• The eye can only see visible light of the electromagnetic spectrum, which is at about 500 nm, or $5 \times 10^{-5} cm$
• The eye can process about 10 frames per second (10 Hz)
• The eye can "see" something when it receives 10 photons of light
• The surface area of the eye is about $2$ $cm^2$

Now that we have a little bit more information to work with, lets think about the physical constraints we are working with, and some of the physical constants and equations that can help us come up with a solution. Oftentimes, a picture can help see how the situation looks like, so I created the following diagram:

• The star emits its photons radially outwards, meaning that at any distance, the concentration of photons can be measured by the surface area of the sphere the light reaches. The surface area that the light reaches at any distance can be calculated by $SA_{light}= 4\pi r^2$
• The speed of light is $c = 3 \times 10^{10} cm/s$ 
• The amount of energy in a photon is $E = h\nu$, for frequency $\nu$ and Planck's constant, $h = 6.6 \times 10^{-27} erg/s$
• Frequency, $\nu = \frac{velocity}{\lambda}$, where $\lambda$ is the wavelength of the photon
• Finally, since we're solving for power, the equation for power is: $Power=\frac{Energy}{Time}$ 

Whew! That was a lot of information we just gathered that is relevant to the question. So let's go about solving for the power output of the star.

In order to solve for the power output of the star, we would need to know how much energy the star is emitting, and the amount of time it is emitting at. Unfortunately, based on the information we gathered, we don't know much about the star other than the fact that it is 100 light years away. However, we have a lot of information about the eyes. We can calculate how much power the eye is absorbing when it "sees" a star. We also know how big both the eye is, as well as how big the sphere of light is that reaches the eye. Therefore, we can create a proportionality that relates the power output of the star to the power absorbed by the eye using the surface area of both the eye and the sphere of emitted light.

$
\begin{align*}
\frac{SA_{Light}}{SA_{Eye}} &= \frac{Power_{Star}}{Power_{Eye}}\\
Power_{Star} &= \frac{Power_{Eye} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{\text{Energy of 10 Photons}}{\text{Time}} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{10 \cdot h\nu}{\text{frame rate of eye}} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot h \cdot velocity}{\lambda}}{\text{frame rate of eye}} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot hc}{\lambda}}{\text{frame rate of eye}} \times SA_{Light}}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot hc}{\lambda}}{\text{frame rate of eye}} \times 4 \pi \times \text{radius of light sphere}^2}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot hc}{\lambda}}{\text{frame rate of eye}} \times 4 \pi r^2}{SA_{Eye}}\\
&= \frac{\frac{\frac{10 \cdot (6.6 \times 10^{-27} erg \cdot s)(3 \times 10^{10} cm/s)}{5 \times 10^{-5} cm}}{\frac{1}{10} s} \times 4 \pi (10 \times 10^{20} cm)^2}{2 cm^2}\\
Power_{Star} &\approx 2 \times 10^{33} erg/s
\end{align*}
$

As you can see, the power output of the star is $2 \times 10^{33}$ ergs per second.