Week 4, Post 3

Worksheet 4, Problem 1

This question reviews Young's double-slit experiment, which demonstrates the wave behavior of light. When light is shined from a distance onto a pair of very thin slits, they form an interference pattern on a screen far away from the slits. If light were a particle, we would expect to see two bright lines on the screen, corresponding to the two slits. However, we actually see multiple bright lines, as illustrated in this diagram from Physics StackExchange:

Intensity of Diffraction on Double Slit

a)

To make things easier to understand, I'll refer to a drawing from my high school physics textbook (Giancoli 6th Ed.), which is much clearer than my attempt in Paint.

Figure 1: Wave behavior of light

We can label angle $\theta$ as the angle between the ray of light coming out of the slit and the horizontal. Figure 1d illustrates the geometry of the light rays. We see that there is a small triangle on the left, also with angle $\theta$ (because of similar triangles). The screen is placed at a distance L away from the slits, where L >> d, the distance between the slits. Because the screen is so far away, we can approximate the two rays $d_1$ and $d_2$ as parallel. The difference in path length between the two rays is therefore $d_2 - d_1$.

$\sin(\theta) = (d_2 - d_1)/d$

$d_2 - d_1 = d \sin(\theta)$

When the difference in path length $d_2 - d_1$ is a whole number multiple of the wavelength $\lambda$, we see constructive interference. When the difference in path length is off by half a wavelength, we see destructive interference.

Figure 1a shows that when $\theta = 0$, we get constructive interference, as there is no path length difference. The sine function is a phase shift of the cosine function, and $\cos(0^{\circ}) = 1$, so I am convinced that the brightness pattern of light is a cosine function.

b)

When we add a second set of slits, we get another cosine function brightness pattern. When we move the slits closer together, the brightness pattern becomes stretched out. As explained in Monday's crash course on Fourier transforms: when something is skinny in one domain, it Fourier transforms to something wide in the other domain.

Since we have two sets of slits now, their resulting interference patterns will be superimposed, adding up the two cosine functions. We can see this illustrated in the following graph:

Blue: $\cos(x)$ (first set of slits)
Red: $\cos(0.8x)$ (second set of slits, just inward)
Green: $\cos(x) + \cos(0.8x)$ (sum of both patterns)

The addition of the second set of slits causes the central peak to be brighter, but the peaks on the sides become attenuated as we go out further from the middle.

c)

Let's add even more cosine plots together and see what we get:

Sum of cosines

$\cos (x)+\cos(0.95x)+\cos(0.9x)+\cos(0.85x)+\cos(0.8x)+\cos(0.75x)$

This pattern resembles the sinc function, which is $\frac{\sin(x)}{x}$:

Sinc function

d)

The Fourier transform of top hat function (which is basically the ideal low pass filter) is the sinc function:

$sinc(x) = \frac{\sin(x)}{x}$

This finding agrees with the result from part c). Let's plot the two graphs together:

Red: sum of cosines
Blue: sinc function

e)

Following what we learned about Fourier transforms, a skinnier top hat would result in a wider sinc function, and vice versa. This means that the width of the top hat is inversely proportional to the distance between the two nulls.

The top hat function is basically a square wave, which is composed of a sum of cosines. To make a narrower top hat, we will need cosines of higher frequency, or lower wavelength. Likewise, higher wavelengths result in a wider top hat. This means that wavelength is directly proportional to width of the top hat.

We can combine these relations to get the following proportionality:

$d \propto \frac{\lambda}{D}$

Where $d$ is the width between the nulls.

f)

A telescope is basically a top hat in two dimensions, rather than just the 1-dimensional line that we've been working with up to this point. Now we have light coming in as a 2D plane to the telescope. From the examples we looked at, a wider top hat results in a narrower peak in the sinc function. Likewise, a wider telescope mirror would focus light into a sharper area, providing better resolving power so we can resolve smaller (farther away) objects.


Acknowledgements: Worked on this problem in class with Awnit, Simon, and August.

Worksheet 4, Problem 2

Here, we are comparing telescopes and their angular resolutions. We don't need to draw a diagram out for this one.

For each telescope, we are given its size and the wavelength it is tuned to observe:

• CCAT: $d_C = 25$m, $\lambda_C = 850$ microns
• MMT: $d_M = 6.5$m, $\lambda_M = 1.2$ microns

We are told that MMT operates in the J-band, which is a sub-band of IR, ranging from 1 to 1.4 microns. For the purposes of determining angular resolution, we'll use the middle of the J-band, which is 1.2 microns.

Using the Rayleigh's criterion, we can determine angular resolution for each telescope, given their respective $d$ and $\lambda$.

CCAT:

$\theta_C = 1.22 \frac{\lambda_C}{d_C} = 1.22 \frac{850 \times 10^{-6} \ m}{25 \ m} = 4.1 \times 10^{-5}$ rad

MMT:

$\theta_M = 1.22 \frac{\lambda_M}{d_M} = 1.22 \frac{1.2 \times 10^{-6} \ m}{6.5 \ m} = 2.3 \times 10^{-7}$ rad

We see that MMT is a smaller telescope, but its angular resolution is much smaller than that of CCAT because the wavelengths it observes are much smaller.

Acknowledgements: Worked on this problem in class with Awnit and Simon.

Largest telescope you can find 

and maximum resolution it is able to achieve

According to WorldAtlas.com, the biggest telescope on Earth is the Gran Telescopio Canarias (GTC). Their website states that it is a 10.4 m telescope with a segmented primary mirror.

The page on the GTC's optics says that its useful wavelength ranges from 365 nm to 25 µm.

The maximum resolution it can achieve comes from the smallest objects it can resolve, so let's use the 365 nm figure.

$\theta = 1.22 \frac{\lambda_{min}}{d_{GTC}} = 1.22 \frac{365 \times 10^{-9}}{10.4 \ m} = 4.28 \times 10^{-8}$ rad

We find that the GTC's maximum resolution is $4.28 \times 10^{-8}$ rad.

General Post

https://www.calcalistech.com/ctech/articles/0,7340,L-3756562,00.html

This article discusses the interesting and unique concerns over intellectual property rights in space.
IP rights are meant to incentivize innovation - patents grant the holder a limited monopoly, allowing them to profit from innovation and recoup the development costs that were spent on that new technology. Patents are especially important in sectors where research and development has high upfront costs, such as biopharmaceuticals.

Therefore, we would expect the field of space technology to have similar protections, since it has high risk and cost of development. However, it turns out that this field may not have as robust protection, because of loopholes that enable infringing parties to sidestep regulation. For example, consider a foreign company that develops a product that infringes on US patents. The device is sent to space on a Russian ship and brought on board the International Space Station (ISS). The device passes through the US modules of the space station before reaching the European modules. Whether or not this case presents an infringement of US patents is up for debate. Because patent protections are territorial, space makes things complicated. There are no international patents, so a US patent can only protect the holder from infringement in the US, but not in China, for example.

The US has been expanding its patent protections beyond its own territories. The Supreme Court found that if components of an infringing product are supplied from the US, the exporter of those parts can be liable for that infringement, even if the original infringement did not take place in the US. Specifically with regards to space, we can refer to 35 U.S. Code Section 105(a), protecting “a space object or component thereof under the jurisdiction or control of the U.S.” However, following precedent from maritime law, it stands to reason that a company can circumvent these regulations by registering with another foreign country. Private companies in the field of space technology are still new, so it remains to be seen if these unclear regulations have yet had an adverse impact on innovation.

Week 3, Post 2

Worksheet 3, Problem 1

The question asks us to find the difference between the sidereal and solar day on Mars. As the planet orbits the Sun, the right ascension that it was pointing at the day before (towards the center of its orbit, the Sun) would no longer be pointing towards the center. We can see this illustrated in the following drawing:

To solve this, we need to determine the period ($P_M$) of Mars's revolution around the Sun. This will allow us to determine how much of an angle ($\theta$) is covered during Mars's 24-hour rotation period. Then, we can determine how much time it takes to make up for that angle $\theta$ such that the right ascension is once again pointed towards the Sun.

We can calculate $P_M$ using the following relation:

$$\frac{P_\oplus^2}{P_M^2} = \frac{r_\oplus^3}{r_M^3}$$

Where:
$P_\oplus$ is the period of Earth's orbit around the Sun, or a year.
$r_\oplus$ is the distance between the Earth and the Sun.
$P_M$ is the period of Mars's orbit around the Sun, or a year.
$r_M$ is the distance between Mars and the Sun.

Let's solve for $P_M$:

$P_M = (\frac{P_\oplus^2 \cdot r_M^3}{r_\oplus^3})^{1/2}$
$= (\frac{(1 \ year)^2(1.5 \ AU)^3}{1 \ AU}^3)^(1/2)$
$=(1 \ year^2 \cdot 3.375)^{1/2}$
$=(3.375)^{1/2} \ year \cdot \frac{365 \ days}{1 \ year}$
$\approx 670.54 \ days$

Next, we can use the period $P_M$ (the Martian year) to determine the angle $\theta$ that is traveled in a 24-hour period. Note that a 24-hour rotation period is 1 day.

$$\frac{\theta}{360^{\circ}} = \frac{1 \ day}{P_M \ days}$$
$\theta = \frac{360^{\circ} \cdot 1 \ day}{P_M \ days}$

This angle $\theta$ can then be used to calculate the amount of time $t$ it takes to rotate $\theta$ degrees, given that we know Mars has a 24-hour rotation period.

$$\frac{\theta}{360^{\circ}}=\frac{t}{24 \ hours \cdot 60 \frac{mins}{hours} \cdot 60 \frac{secs}{mins}}$$

Let's plug in $\theta$ and solve it.

$t = \frac{\theta \cdot 86400 \ secs}{360^{\circ}}$
$= \frac{\frac{360^{\circ} \cdot 1 \ day}{P_M \ days} \cdot 86400 \ secs}{360^{\circ}}$
$= \frac{\frac{360^{\circ} \cdot 1 \ day}{670.54 \ days} \cdot 86400 \ secs}{360^{\circ}}$
$= 128.85 \ secs$

That's about 2 minutes, which seems reasonable.

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

Worksheet 3, Problem 2

This question asks us to determine the Local Sidereal Time (LST) at different times. We didn't use any diagrams to solve this question. We did make sure to read the question.

It's important to understand the difference between the solar day and the sidereal day. I found this website to be helpful:

• A solar day is the time it takes for Earth to rotate around its axis. 
• This is measured from noon to noon, so that the Sun appears in the same place in the sky, defined as 24 hours.
• A sidereal day is the time it takes for Earth to rotate around its axis relative to the celestial sphere.
• This is measured by it time it takes a particular star to pass directly overhead (celestial meridian) on two successive nights.

The website included this neat diagram that illustrates the concept (similar to the one we used in Problem 1):

Sidereal day, from http://astronomy.swin.edu.au/cosmos/S/Sidereal+Day

On Earth, a sidereal day takes about 23 hours and 56 minutes, which is 4 minutes less than the solar day. This means that the sidereal day is 4 minutes faster than the solar day, so a given star will rise 4 minutes earlier every night.

Also, as discussed in class, the equinoxes and solstices serve as benchmarks for LST. At noon:

• Vernal Equinox: 00:00:00
• Summer Solstice: 06:00:00
• September Equinox: 12:00:00
• Winter Solstice: 18:00:00

To determine LST at a certain date and time, we use a 3-step process:

1. Find the LST at noon on that date, using the LSTs of the equinoxes and solstices. With each month, LST moves by about 2 hours.
2. Add or subtract the number of hours off from noon of the time that you're concerned with.
3. Add or subtract the number of minutes to account for the difference of 4 minutes between solar day and sidereal day. This means a gain of 1 sidereal minute every 6 hours.

a) What is the LST at midnight on the Vernal Equinox?

At midnight after the Vernal Equinox, we are 12 hours past noon. The LST at noon is 00:00:00, so adding 12 hours brings us to 12:00:00. Add 2 minutes (1 minute/6 hours, so 2 minutes/12 hours) to account for the sidereal day being 4 minutes shorter than the solar day. This gives 12:02:00 LST.


b) What is the LST 24 hours later (after midnight in part 'a')?

Starting from 12:02:00 LST at midnight after the Vernal Equinox. 24 hours (solar day) corresponds to 4 sidereal minutes. Therefore, 4 minutes after 12:02:00 LST (from part 'a') is 12:06:00 LST.


c) What is the LST right now (to the nearest hour)?

Today is February 20, 2019, at 8:00 PM, so 20:00 EST.

Winter Solstice: 18:00:00
Vernal Equinox: 00:00:00

There are 3 months between the Winter Solstice and the Vernal Equinox, so that means there are 2 sidereal hours/month. The Vernal Equinox is on March 20, so we're 1 month away, or 2 sidereal hours. Then, we'll subtract 2 hours from 00:00:00, so we get LST of 22:00:00. 8 PM is 8 hours after noon, so add 8 to 22:00:00 to get 06:00:00. 8 hours is 1/3 of a day, so $1/3 \times 4$ min = 1 min 20 sec. Add this to get 06:01:20 LST.


d) What will the LST be tonight at midnight (to the nearest hour)?

At midnight on that day, we'd start with 22:00:00 LST at noon. Next, add 12 hours to get 10:00:00. Then, add 2 minutes to account for the 12 hours after noon, giving 22:02:00 LST.


e) What LST will it be at Sunset on your birthday?

My birthday is September 9. Let's count it as 6 PM EST (for a dinner party?).

Summer Solstice (June 21): 06:00:00
September Equinox (September 23): 12:00:00

September 9 is 2 weeks (half a month) before September 23. Since there are 2 sidereal hours/month, there is 1 sidereal hour for half a month. LST would be 11:30:00. 6 PM is 6 hours later, so add 6 hours to get 17:30:00. Then, 6 hours is a quarter of a day, which means we gain 1 sidereal minute. This leaves us with 17:31:00 LST at dinner time.


Acknowledgment: worked on this problem in class with Awnit and Elton. Special mention to Elton for edifying us on a systematic approach to this question.

Worksheet 3, Problem 3

This question is tricky, so it's important to read the question.

Recall the definition of local sidereal time (LST): the right ascension that is on your meridian right now.

By definition, we can only see AY Sixteenus at LST of 18:00:00, because its RA is 18 hours. It follows that on any day (including the first of each month) the LST of AY Sixteenus would be 18:00:00.

Therefore, the star would never be visible on the meridian at midnight LST.

General Post 1

The problem of space junk is a growing one; the number of satellites being launched is increasing very quickly - projections predict that there will be over 5,600 commercial launches under 500 kg from 2017 to 2027, up 10 times from the previous decade. The problem is that satellites are very difficult to fix while in orbit, so usually they just have to be brought back down to Earth. However, many old satellites are left in orbit. There are an estimated 750,000 pieces of space junk orbiting the Earth. They travel extremely fast, at 8 kilometers per second, which means a collision could accidentally pulverize an operational satellite. This can cause a cascading chain of collisions known as the Kessler effect, which would leave swaths of low-earth orbit unusable because of the debris. This poses a risk to the future of satellite applications.

A Japanese company called Astroscale proposes a new technology to reduce the amount of space junk in low-earth orbit. Astroscale claims to be the first in-orbit debris capture and removal, which leverages a two stage process: rendezvous and magnetic capture. For their test run, they will launch “chaser” and “target” modules that separate after reaching orbit. The chaser will chase after the target, and then capture it with a magnetic capture mechanism. When the target is docked, the chaser and target will power back toward Earth and burn up on re-entering the Earth’s atmosphere.

https://www.bloomberg.com/news/articles/2019-02-13/as-launches-boom-a-venture-bets-on-bringing-satellites-down

General Post 2

Last week, Opportunity, the NASA Mars Rover, was officially pronounced dead. The rover, a robot the size of a golf cart, was originally planned for a 90-day mission, yet it ended up roving about Mars’s surface for over 14 years. It is especially impactful because it brought about a paradigm shift in space exploration - rather than a stationary lander studying a single spot, a rover can move about and check out interesting things. They also promote a deeper level of engagement with the general public - making it seem as if one were actually “walking on the surface”.

The decision to finally end the mission was not an easy one. Opportunity had been unresponsive since June, because a giant dust storm obscured its solar panels, preventing it from generating enough energy to stay awake. In the fall, NASA was only willing to spend a month trying to reconnect with the rover, but many on the team felt that that was not enough time, leading NASA to allow some more time. However, it became clear that the rover was finally dead, possibly because the solar panels were covered by too much dust to be blown away, or some electronic component failed. We may never know exactly what killed Opportunity, until one day when we have astronauts on Mars to inspect the intrepid rover.

https://www.nytimes.com/2019/02/13/science/mars-opportunity-rover-dead.html

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:

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