Week 2, Post 1

Worksheet 1, Problem 1

We are given the following clues to help us determine the radius of the Earth ($R_{\oplus}$).

Clue:  My frequent flier statement says that the distance between Los Angeles (LAX) and Boston (BOS) is 3000 miles.
Clue:  At the start of class today, it was 12:00pm in Los Angeles.
Clue:  When I was driving my car the other day,  I looked at my speedometer and noticed I was going 100!  Oh, wait, that was in km/hr.  After setting my digital speedometer back to mph, I was only going 60.
Clue:  The width of your thumb held at arm’s length subtends $\approx 1 \circ$ on the sky.  The moon takes up about half this width.
Clue:  Johannes Kepler found that the period of a planet is proportional to its distance from the central mass such that $P^2 \propto a^3$.  Newton modified this to be $P^2 = 4\pi^2 a^3 /(GM)$ where $M$ is the mass of the central body, $G = 6.7 \times 10^{-8} cm^3 g^{-1} s^{-2}$ and $\pi^3$
Tool:  A rock
Tool:  A scale
Tool:  A measuring cup half–filled with water

After reading the question (step 1 of the Expert Methodology), we make a drawing of the problem situation:

From the drawing, we can see that we might want to make use of two equal ratios to find the circumference of the earth. We know the time zone difference between LAX and BOS is $T_1 = 3 h$, and the total length of a day on Earth is $T_{tot} = 24 h$. We know the distance from LAX to BOS is $D_1 = 3000 mi$, so we can determine the circumference ($C_{\oplus}$) around the Earth. Given the circumference, we can determine the radius ($R_{\oplus}$).

Next, we write out the variables that we know and the ones we don't know:
$T_1 = 3 \ hours$
$T_{tot} = 24 \ hours$
$D_1 = 3000 \ miles$
$D_{tot} = C_{\oplus} = ? \ miles$
$R_{\oplus} = ? \ cm$

Judging from the variables we have, we can set up a ratio, since we know the time zone difference between LAX and BOS, and the duration of a day, as well as the distance between LAX and BOS, so we can solve for the distance around the entire globe. Using this circumference, we can solve for the radius using the circle equation $C = 2 \pi r$.

Ratio:
$\frac{T_1}{T_{tot}} = \frac{D_1}{D_{tot}}$

Solve for $D_{tot}$:
$D_{tot} = \frac{D_1 T_{tot}}{T_1}$

Circle equation:
$D_{tot} = C_{\oplus} = 2 \pi R_{\oplus}$

We can now solve for $R_{\oplus}$ (taking $\pi \approx 3$):
$R_{\oplus} = \frac{D_{tot}}{2 \pi} = \frac{D_1 T_{tot}}{T_1 2\pi} \ miles = \frac{3000 / miles \times 24 \ hours}{3 / hours \times 2 \pi} = 4000 \ miles $

Now we need to convert our answer into cm. We know that 60 miles is 100 km, so:
$1 / mile = 100 km/60 \approx 1.6 km$

$R_{\oplus} (cm) = R_{\oplus} \times 1.6 km/mile = 4000 \ miles \times 1.6 km/mile = 6400 \ km$

We need to convert from MKS to CGS:
$6400 \ km = 6400 \times 10^5 cm = 6.4 \times 10^8 \ cm$

$$R_{\oplus} = 6.4 \times 10^8 \ cm$$

Acknowledgment: worked on this problem in class with Elton and Simon.

Worksheet 2, Problem 2

We need some information from Problem 1 to solve this. Here's what we know:

The eye must receive ~ 10 photons in order to send a signal to the brain that says, "Yep, I see that."

Note: The energy of a photon is $E = h\nu$, for a frequency $\nu$ (Greek letter 'nu'), and Planck's constant $h = 6.6 \times 10^{-27} erg \ s$. The speed of light is $3 \times 10^{10} cm s^{-1}$. The visible part of the electromagnetic spectrum is at approximately 0.5 microns.

Another note is that the eye has approximately 1% efficiency; that is, for every 100 photons that reach the eye, only 1 is registered.

After reading the question (step 1 of the Expert Methodology), we make a drawing of the problem situation:

It looks like we can make use of a ratio in this problem as well. Light energy is emitted from the star in a sphere - the emitted energy is spread out over the surface area of a (very) large sphere, the radius of which is the distance $D$ between the star and the observer. Meanwhile, the human eye has a surface area on which the photons will land and be perceived, which is dependent on the radius of the eye's pupil $R_e$. We know how much energy it will take for a human eye to perceive light, which is 10 photons (we need to remember that the eye is 1% efficient), so we can determine the total energy emitted from the star. From that energy, we can find power, which is energy per unit time.

We know:

$c = 3 \times 10^{10} cm/s$

$h = 6.6 \times 10^{-27} \ erg \ s$

The surface area of the large sphere is:

$A_s = 4 \pi D^2$

$D = 100 \ ly$

A light year (ly) is:

$ly = c \times 1 \ year = (3 \times 10^{10} cm/s)(60 \frac{s}{min} \cdot 60 \frac{min}{hour} \cdot 24 \frac{hour}{day} \cdot 60 \frac{day}{year} \cdot 1 \ year) $

$= (3 \times 10^{10} cm/s)(3.2 \times 10^7) \approx 10 \times 10^{17} cm = 1 \times 10^{18} cm$

The area of the eye is:

$A_e = \pi R_e^2$, where $R_e$ is the radius of the pupil.

$R_e \approx 0.5 cm$

Energy received by the eye, given 1% efficiency:

$E_{rec} = 10 \ photons \times 100 = 1000 h\nu = \frac{1000hc}{\lambda}$

$\lambda = 0.5 \ micron = 0.5 \times 10^{-6} m = 5 \times 10^{-5} cm$

Energy emitted by the star:

$E_{em} = P \times t_{eye}$

$t_{eye}$ is given by the latency of the eye's perception. Let's say that the eye can perceive 100 Hz (frames per second). The latency is therefore:

$t_{eye} = 1/100 Hz = 0.01 s = 1 \times 10^{-2} s$

Let's set up our ratio:

$$\frac{E_{em}}{E_{rec}} = \frac{A_s}{A_e}$$

Rearrange to isolate $E_{em}$:

$$E_{em} = \frac{A_s \cdot E_{rec}}{A_e}$$

Divide energy by time to get power:

$P = \frac{A_s \cdot E_{rec}}{t_{eye} \cdot A_e}$

$ = \frac{4 \pi D^2 \cdot 1000 hc}{t_{eye} \cdot \lambda \cdot \pi R_e^2}$

$ = \frac{4 \pi (1 \times 10^{18} \ cm)^2 \cdot 1000 (6.6 \times 10^{-27} \ erg \ s)(3 \times 10^{10} cm/s)}{1 \times 10^{-2} \cdot 5 \times 10^{-5} cm \cdot \pi (0.5 cm)^2}$

$ = \frac{4 (1 \times 10^{36} cm^2)(1 \times 10^3)(20 \times 10^{-17} \ erg \ cm)}{1.25 \times 10^{-7} \ cm^3 \ s}$

$ = \frac{8 \times 10^{23} erg}{1.25 \times 10^{-7} s}$

$ = 6.4 \times 10^{30} \ erg/s = 6.4 \times 10^{23} W$

$$P = 6.4 \times 10^{23} W$$

Acknowledgment: worked on this problem in class with Elton and Simon.

Worksheet 2, Problem 3

The speed of sound in a gas $c_s$ is related to the pressure $P$ and density $\rho$ of the gas. Use dimensional analysis to figure out the form of this relationship.

Let's start by writing out the dimensions of the three quantities that we're working with: $c_s$, $P$, and $\rho$.

Speed is distance per unit time:

$$c_s \Rightarrow Dist/Time \Rightarrow cm/s \Rightarrow cm \cdot s^{-1}$$

Pressure is force applied per unit area. Also, let's make sure to switch from MKS to CGS:

$$P \Rightarrow Force/Area \Rightarrow N/m^2 \Rightarrow kg \cdot m^{-1} \cdot s^{-2} \Rightarrow g \cdot cm^{-1} \cdot s^{-2}$$

Density is mass per unit volume:

$$\rho \Rightarrow Mass/Vol \Rightarrow g/cm^{-3} \Rightarrow g \cdot cm^{-3}$$

We need to manipulate $P$ and $/rho$ to get the units of $c_s$. We notice that $P$ has units of time ($s^{-2}$), but not $\rho$. This is a hint that we might need to square root in order to get units of $s^{-1}$ in $c_s$. 

We can see that both $P$ and $/rho$ have units of mass (g), but $c_s$ does not. This is a hint that we should divide one of the quantities by the other to cancel out mass. If we divide $P$ by $/rho$, we'll cancel out mass, and also get distance in the numerator.

Let's work this out:

$$cm \cdot s^{-1} = (P \cdot \rho^{-1})^{1/2} = [(g \cdot cm^{-1} \cdot s^{-2})(g \cdot cm^{-3})]^{1/2} = [cm^2 \cdot s^{-2}]^{1/2} = cm \cdot s^{-1}$$

The units work out! Therefore, we know that the relationship is of the form:

$$c_s \propto (P \cdot \rho^{-1})^{1/2}$$