### Order of Magnitude Astrophysics

From the Ay20 blog, here's a solution to one of the week 1 worksheet problems.

### Estimating The Luminosity of a Sun-like Star

by: John A. Johnson, Jackie Villadsen

Abstract

We present the solution to Worksheet problem #2, from week 1, estimating the power output of a Sun-like star. Each group should submit one to two of these per week. Decide amongst your group members who will be first author, second author, etc. Acknowledge people and resources used in your solution. Cite ancillary information. State your assumptions clearly. Write your solution such that a frosh could duplicate your steps and arrive at the same solution.

Introduction

The oldest astronomical instrument is the human eye. A marvel of evolution, the eye has both high sensitivity and a large dynamic range. A classic study of the eye's response to light conducted in 1942 showed that of order 10 photons need to impinge on the eye in order for the brain to register detection (Hecht, Schlaer & Pirenne 1942). In other words, the eye has a gain of 10 photons/DN. In this contribution we use this fact as a starting point for estimating the luminosity (power output) of a Sun-like star. As additional input for our calculation we note that a Sun-like star at 100 light years is just barely visible to the naked eye if the star is viewed from a dark site. (As a side note, this corresponds to a G2V star with an apparent magnitude of V=6).

Order of Magnitude (OoM) Calculation

We start with a rough estimate of the aperture area of the eye. Fully dilated, an eye has an entrance diameter of roughly Reye = 0.5 cm, corresponding to an area of 0.25 cm^2. From here on we consider only a single eye since it is unclear how two eyes would combine for the detection of a faint star, and since we will only incur a factor-of-two error at most, which is insignificant for our OoM calculation. As an additional assumption we ignore absorption by the Earth's atmosphere and set interstellar reddening to zero.

The star is at a distance of 100 light years. Light travels at 3x10^10 cm/s, and there are (π x 10^7) seconds in a year. A light year is therefore D ~ (10 x 10^10 x 10^7) = 10^18 cm. The star emits some number Nemit photons isotropically, and the eye subtends a tiny fraction of the area of a sphere with a radius of D = 10^20 cm and receives 10 photons. This fractional area is (AD/Aeye), where Aeye is the area of the eye and Ais the area of the sphere surrounding the star. Thus

We are interested in the power output of the star, which is the energy emitted per second. We can get the energy corresponding to Nemit photons with
where \lambda is the wavelength of light. We can assume that the eye's spectral response is well-tuned to the peak of the Sun's spectral energy distribution, which corresponds to about 550 nm (we'll learn more about this after we estimate the Sun's temperature and learn about black body radiation). Thus

where I have used cgs throughout (note that 550 nm = 550 x 10^-7 cm). Now we need to figure out the time interval. The eye detects the 10 photons at a certain "readout rate." This can be estimated by noting that movies are typically shot at 24 frames per second. At a slower rate the eye would notice a distinct slowing of the movie scenes (imagine watching a movie that shows one frame every second, i.e. a slide show), and at a faster rate the movie studio would be wasting film. So the time the brain takes to "read out" the eye is about 10 milliseconds or 0.01 seconds, to an OoM. Thus, the power output of the Sun-like star is

This compares well to the actual luminosity of the Sun, which is 3.862 x 10^33 ergs/s.

Summary and Discussion

We have performed an OoM calculation of the Sun's luminosity by noting that a Sun-like star at 100 pc is barely visible to the naked eye. Our final answer is correct to within a factor of 4, demonstrating the usefulness of OoM calculations. By not worrying about the exact numbers, but instead focusing on the problem-solving process, we are free to concentrate on the physics of the problem knowing that we can perform the exact calculation using the same reasoning and a bit more time/effort.

Acknowledgements

We thank Owen and Marcus Johnson for playing nicely with each other for the 45 minutes it took Daddy to write this. We made use of WolframAlpha when our initial estimate of the photon energy was off by two orders of magnitude, and when we couldn't remember Planck's constant in cgs. WolframAlpha helped us realize that we needed the wavelength of a green photon in cm rather than meters. Duh. The equations were generated using CodeCogs online LaTeX editor.

### On the Height of J.J. Barea

Dallas Mavericks point guard J.J. Barea standing between two very tall people (from: Picassa user photoasisphoto).

Congrats to the Dallas Mavericks, who beat the Miami Heat tonight in game six to win the NBA championship.

Okay, with that out of the way, just how tall is the busy-footed Maverick point guard J.J. Barea? He's listed as 6-foot on NBA.com, but no one, not even the sports casters, believes that he can possibly be that tall. He looks like a super-fast Hobbit out there. But could that just be relative scaling, with him standing next to a bunch of extremely tall people? People on Yahoo! Answers think so---I know because I've been Google searching "J.J. Barea Height" for the past 15 minutes.

So I decided to find a photo and settle the issue once and for all.

I then used the basketball as my metric. Wikipedia states that an NBA basketball is 29.5 inches in circumfe…

### Finding Blissful Clarity by Tuning Out

It's been a minute since I've posted here. My last post was back in April, so it has actually been something like 193,000 minutes, but I like how the kids say "it's been a minute," so I'll stick with that.
As I've said before, I use this space to work out the truths in my life. Writing is a valuable way of taking the non-linear jumble of thoughts in my head and linearizing them by putting them down on the page. In short, writing helps me figure things out. However, logical thinking is not the only way of knowing the world. Another way is to recognize, listen to, and trust one's emotions. Yes, emotions are important for figuring things out.
Back in April, when I last posted here, my emotions were largely characterized by fear, sadness, anger, frustration, confusion and despair. I say largely, because this is what I was feeling on large scales; the world outside of my immediate influence. On smaller scales, where my wife, children and friends reside, I…

### The Force is strong with this one...

Last night we were reviewing multiplication tables with Owen. The family fired off doublets of numbers and Owen confidently multiplied away. In the middle of the review Owen stopped and said, "I noticed something. 2 times 2 is 4. If you subtract 1 it's 3. That's equal to taking 2 and adding 1, and then taking 2 and subtracting 1, and multiplying. So 1 times 3 is 2 times 2 minus 1."

I have to admit, that I didn't quite get it at first. I asked him to repeat with another number and he did with six: "6 times 6 is 36. 36 minus 1 is 35. That's the same as 6-1 times 6+1, which is 35."

Ummmmm....wait. Huh? Lemme see...oh. OH! WOW! Owen figured out

x^2 - 1 = (x - 1) (x +1)

So $6 \times 8 = 7 \times 7 - 1 = (7-1) (7+1) = 48$. That's actually pretty handy!

You can see it in the image above. Look at the elements perpendicular to the diagonal. There's 48 bracketing 49, 35 bracketing 36, etc... After a bit more thought we…