Worksheet 6: Hydrostatic Equilibrium

A Hydrostatic Haiku:
Gravity Pulls Down
Down, Inexorably Down
P briefly impedes

Consider the Earth's atmosphere by assuming the constituent particles comprise an ideal gas,
such that P=nk$_B$T, where n is the number density of particles (with units cm $^{-3}$), $k_B=1.4\times 10^{- 16}$ erg K $^{-1}$ is the Boltzmann constant. We'll use this ideal gas law in just a bit, but first:

a) Think of a small, cylindrical parcel of gas with the axis tinning vertically into the Earth's atmosphere.  The parcel sits a distance, r, from the Earth's center, and the parcel's size is defined by a height $\Delta r<<r$ and a circular cross-sectional area A.  The parcel will feel pressure pushing up from gas below ($P_{up} = P(r)$) and down from above ($P_{down}=P(r+\Delta r)$).



Make a drawing of this, and discuss the situation and the various physical parameters with your group.  

b) What other force will the parcel feel, assuming it has a density, $\rho$, and the Earth has a mass, $M_\oplus$?

Like any other object with mass on Earth, the parcel will feel the force of gravity pulling it down.  

$F_g=-\frac{GM_{cyl}M_\oplus}{r^2}$

$M_{cyl}=\rho A\Delta r$

$F_g=-\frac{G\rho A\Delta rM_\oplus}{r^2}$  Eq. 1

c) If the parcel is not moving, give a mathematical expression relating the various forces, remembering that force is a vector and pressure is a force per unit area.

To find this expression, I added up all of the forces together.  The parcel isn't moving, so the net force on it is 0.  

$F_{tot}=F_g+(P_{up}+P_{down})A$

$0=-\frac{GM_\oplus \rho A\Delta r}{r^2}+(P(r)-P(r+\Delta r))A$

$\frac{GM_\oplus \rho \Delta r}{r^2}=-P(r+\Delta r)+P(r)$  Eq. 2

d) Give an expression for the gravitational acceleration, g, at a distance, r, above the Earth's center in terms of the physical variables of the situation.

If nothing else from introductory mechanics stuck, the fact that F=ma did.  Seriously, I might even be able to recite that in my sleep.  Eq. 1 defines the force of gravity.  If $\rho A\Delta r$ defines the parcel's mass (density times volume), then the other components of Eq. 1 must define the acceleration, or g.

                                                        $g=\frac{GM_\oplus}{r^2}$

e) Show that $\frac{dP(r)}{dr}=-gp$.  This is the equation of hydrostatic equilibrium.  What does it "say" in English?  Does it make sense?

First, we'll take Eq. 2 and substitute in g.   

$g\rho \Delta r=-P(r+\Delta r)+P(r)$

$-g\rho=\frac{P(r+\Delta r)-P(r)}{ \Delta r}$

As annoying as it was to calculate in high school calculus, 

$\frac{dP(r)}{dr}=\frac{P(r+\Delta r)-P(r)}{ \Delta r}$

Therefore, 

$\frac{dP(r)}{dr}=-gp$   Eq. 3

This says that pressure as a function of altitude decreases as density increases.  

f) Now go back to the idea gas law described above.  Derive an expression describing how the density of the Earth's atmosphere varies with height, $\rho (r)$.  Recall that $\frac{dx}{x}=d\ln x$.

If n is number density and m is the mass of an individual particle, $\rho = nm$, which means we can rewrite the ideal gas law as 

$P(r)=\rho(r)\frac{kT}{m}$

Solving Eq. 3 for $\rho$, we can rewrite an equation for $\rho$ in terms of $\rho(r)$ as

$\rho=-\frac{d\rho}{dr}\frac{kT}{gm}$    Eq. 4

If you solve Eq. 4 for dr and integrate, you get the following equation:

$r=-\frac{kT}{gm}\ln\rho+c$

This equation is really easy to solve for $\rho(r)$:

$\rho(r)= e^{-\frac{rmg}{kT}-c}=ce^{-\frac{rmg}{kT}}$

When doing problems like this in class, professors always tell us to remember to consider the initial conditions.  When r=0, the density should equal the initial density, so $c=\rho_o$.

g) Show that the height, H, over which the density falls off by a factor of 1/e is given by

$H=\frac{k_BT}{mg}$

where m is the mean mass of a gas particle. This is the "scale height." First, check the units.

Then do the math. Then make sure it makes physical sense, e.g. what do you think should

happen when you increase m? Finally, pat yourselves on the back for solving a fi rst-order

di fferential equation and finding a key physical result!

This problem is asking us to find an r such that $\frac{\rho(r)}{\rho_o}=e^{-1}$ is true.  We can rewrite this as 

$e^{-\frac{rmg}{kT}}=e^{-1}$

If we take the natural log of both sides of this equation, we find 

$\frac{rmg}{kT}=1$

When we solve this for r, we find that 

$H=r=\frac{kT}{mg}$   Eq. 5

h) What is the Earth's scale height, $H_\oplus$?  The mass of a proton is $1.7\times 10^{-24}$ g, and the Earth's atmosphere is mostly nitrogen, $N_2$, where the atomic nitrogen has 7 protons and 7 neutrons.

This can be found by substituting the values provided into Eq. 5.  

$H=\frac{1.4\times 10^{-16}}{(7)(1.7\times 10^{-24})(9.8\times 10^2)}$

$=1\times 10^4$

Midterm Discussion

When I opened the midterm for the first time, my first thought was, "Wow! I've seen all of this stuff before!"  So, good job on that, AY 16 staff!

Despite my initial enthusiasm, however, I still made a few mistakes.  I will discuss them here.

On problem 1, I lost two points-- for what I now realize was a careless mistake.  The question gave the RA of a star--13 hours-- and asked when in the year the star would be directly overhead at midnight (suposing we are at the latitude corresponding to the star's declination.  I drew the following picture 

which is my automatic first step every time I work on an RA problem.  I knew, based on my diagram, that the Southern Cross would be directly overhead at midnight when the Sun was at an RA of 1.  Because I know that each month changes the RA of the sun by 2 hours, I knew this would happen half a month after the Vernal Equinox.  Here comes the careless part.  I forgot that the Vernal Equinox is a week and a half before the end of March (and not at the end), so half a month after that would be the first week of April, not the last.  

My second mistake came in number 4, and was little more than my lack of ability to do math well without a calculator (and a little bit because I'm apparently really bad at copying what's right in front of me).  The problem asked me to calculate the semi-major axis of Io's orbit using Kepler's Third Law.  I forgot to carry the exponent over the $\pi$ from one equation to the next.  This threw off all of my subsequent calculations.  Also, when I had to take the cube root of my (incorrect) quotient, I got a little scared because I didn't get an integer.  The person who graded my exam helpfully pointed out that rough guess-and-check would have been sufficient.

The biggest chunk of my missed points came from number 6.  I knew going into this exam that Fourier Transforms were a weak spot.  Still, I thought I had a (loose) grasp on top hat functions.  As it turns out, I had a pretty strong grasp on the general concept.  The height of the top hat corresponds to amplitude while its width corresponds to the frequency of the function.  

I misinterpreted the widths, however, and thought that a wider top hat corresponded to lower frequencies.  At the time, I believe I reasoned it by saying that a wider top hat means more space to distribute the waves, and therefore a lower frequency? I know now that wider top hats correspond to higher frequencies

as if the x-axis of these graphs were in Hertz.  

On my exam, I completely switched the two graphs corresponding to the two top hats.  

My last missed point came from problem 7.  It was, again, another careless mistake that came about because I...am really careless.  I actually don't know how to explain what went wrong.  I would like to say I'll never make this mistake 

$Pv=2\pi \Rightarrow r=\frac{2\pi}{Pv}$

or anything like it again, but I know I'd be lying.  

Free Form: Creationism?

First off, I understand this post could potentially be seen as controversial.  In order to avoid that, I'm going to do my best to do nothing but present the two sides of an argument.  You, the reader, can choose for yourself what to believe.

I was scrolling through my facebook newsfeed when I saw pictures of a high school acquaintance at the Creation Museum in Petersburg, KY.  (This isn't actually as surprising or uncommon as you might think since I come from a town in Pennsylvania where confederate flags are still flown and I've actually heard people say they "don't believe in science.")  A couple weeks ago, I would have kept on scrolling, but the recent discovery of what most of the science community believes to be proof the Inflationary Theory made me stop.  I wanted to see what their argument was, so I decided to look up the creation Museum's exhibits.  And this is what I found.


The museum recently added an exhibit dedicated to the comet ISON, called Fires in the Sky.

Museum's trailer for the 23-minute long video shown in its Stargazer's Planetarium as proof of a young universe

The museum's exhibit explains that ISON is a sun-grazer, which means its orbit takes it really close to the sun.  Every time it reaches the part of its orbit that's closest to the sun*, it loses some of its mass.  Last December, ISON completed its last orbit as it passed the sun and lost too much of its mass to continue.

This is not the only way comets get destroyed, though, according to the exhibit.  They can also get shot out of the solar system if they hit an object's gravitational field in the right way, or they can get pulled in by an object's gravity and crash.

By itself, this information isn't enough to prove that the universe is only a few thousand years old, but we also have to consider that there's no source of new comets.  Astronomers claim that most of the comets we see orbiting through our solar system come from sources called the Oort Cloud and Kuiper Belt, located far beyond Pluto and just past Neptune, respectively.  We've never seen the Oort Cloud, though, we we likely never will, so we can't know that it exists.  These two facts together--that all comets will die or be shot out of our solar system and that there's no way for us to get new ones-- prove that the universe can't possibly be more than a few thousand years old.  If it were, we wouldn't still comets orbiting through our solar system.

************************************************************

Astronomers have several facts supporting their theory that the universe is, without a doubt, older than  three to four thousand years old.

1) They're pretty sure the Oort Cloud does exist.  Even if we've never seen it, a collection of comets so far away that they aren't too strongly bound by the sun's gravity and are therefore easily influenced by other objects is the most plausible explanation for long-period comets.  Long-period comets are seen relatively infrequently and have more erratic orbits (presumably due to the influence of other objects), like Halley's Comet.

Space.com

2) That little discovery (mentioned above) of gravitational waves, which were predicted to be visible 380,000 years after the Big Bang.  No matter how long ago the Big Bang happened, if something happened 380,000 years after it, the universe is at least that old.

3) This is a (relatively) simple exercise we did in class (WS 7, Problem 5) that shows a ballpark age of our solar system.  

The Virial Theorem states that half of a gravitationally-bound system's potential energy goes into kinetic energy. It turns out that for a cloud of gas, the other half of U goes into thermal radiation. We know that the Sun started from the gravitational collapse of a giant cloud of gas. Let's hypothesize that the Sun is powered solely by this gravitational contraction, as was once posited by astronomers long ago. As it shrinks, its internal thermal energy increases, increasing its temperature and thereby causing it to radiate. How long would the Sun last if it was thermally radiating its current power output, $L_\odot =4\times 10^{33}erg/s$? This is known as the Kelvin-Helmholtz timescale. How does this timescale compare to the age of the oldest Moon rocks (about 4.5 billion years, also known as Gyr)?

The first two sentences of the problem can be interpreted as 

$KE=-\frac{1}{2}U=TE$

In the previous problem on the worksheet, we figured out that potential energy, U, can be found using this equation:

$U=-\frac{3}{5}\frac{GM^2}{R}$

By combining these equations, we get:

$KE_\odot=TE_\odot=\frac{1}{10}\frac{GM_\odot^2}{R_\odot}$

If, as stated in the problem, we're assuming that the only source of the sun's energy is gravitational contraction, we can take the thermal energy of the sun (found using the equation above) and divide it by the sun's luminosity to find its age.  When we did this, we found that the sun, and therefore the solar system is about 16 million years old.

This is far from accurate because gravitational contraction contributes very little to the sun's total source of energy--most of its power is nuclear fusion.  But still, even with so little energy, the sun is at least several million years old.  

               ************************************************************

I've given you the different arguments.  Now its up to you to decide what to believe.


*I know that there's a word for this--perihelion--but that word is not used in the Creation Museum's explanation, and I wanted to keep my representation of their argument as authentic as possible.

WS 7, Problem 2-4: Application of the Virial Theorem

2. Consider a spherical distribution of particles, each with a mass m$_i$ and a total (collective) mass $\sum_{i}^{N}m_i=M$, and a total (collective) radius R.  Convince yourself that the total potential energy, U, is approximately 

$U=-\frac{GM^2}{R}$

We decided to solve this problem by finding the sum of all of the potential energies of all of the individual small masses, $\sum_{i}^{N}U_i$.  

Potential energy is usually given by 

$U=-\frac{GMm}{R}$

where M is the mass of the object being orbited and m is the mass of the orbiting object.  

To find the sum of all of the potential energies of each individual particle, we used the following equation:

$\sum_{i}^{N}U_i=-\frac{GM\sum_{i}^{N}m_i}{R}$

$=-\frac{GMM}{R}$

$=-\frac{GM^2}{R}$

There was some doubt in our group that we didn't have to sum over the different Rs.  We resolved the issue by assuming that the particles are uniformly distributed, so the particles at larger radii, because there would be more of them, would overpower the particles at smaller radii in the summation.


We were, however, wrong.  We really did have to take into account the different Rs, and this is how we did it:

At the beginning of the next class, we were told that we should assume the particles form a spherical distribution with uniform density and that the density can be written as

$\rho (r)=\frac{M(r)}{\frac{4}{3}\pi r^3}=\rho$           Eq. 1

Using this new information, we approached the problem as an integral with respect to the radius of the sphere.  We start with a radius barely more than 0.  At that radius, the potential energy, U, will depend only on the mass enclosed by that radius, r. Now, say we increase the radius by a tiny little bit.  We can find the mass of that new "shell" on the sphere, dM, by multiplying the surface area of the shell by its thickness, $\Delta r$.  

$dM=4\pi r^2 \rho \Delta r$

This approach leads us to the following integral:

$\int_{0}^{R}U=\int_{0}^{R}-\frac{GM}{r}dM$

At first, I was confused by the fact that $m_i$ wasn't included in the equation for potential energy of the shell.  But now I know it's not included because the point of the integral is to add up the potential energies of each shell, and the individual shells' energies only depend on mass they enclose.  

From there, the rest of the problem was pretty much algebra, with a little bit of calculus thrown in.  

First, rearrange Eq. 1 to solve for M and substitute that for M in the integral.

$-\int_{0}^{R}\frac{G\frac{4}{3}\rho \pi r^3}{r}4\pi r^2\rho \Delta r$

$-\int_{0}^{R}\frac{16}{3}G\rho^2\pi^2r^4\Delta r$

$=-\frac{16}{15}G\rho^2\pi^2r^5$                Eq. 2

This is the really complicated equation for the total potential energy of the particles.  But we know it can be simplified by dividing it by the equation provided in the problem (substituting the rearranged version of Eq. 1 for M).

$\frac{-\frac{16}{15}G\rho^2\pi^2r^5}{-\frac{GM^2}{r}}$

$=\frac{-\frac{16}{15}G\rho^2\pi^2r^5}{-\frac{G\frac{16}{9}\rho^2\pi^2r^6}{r}}$

$=\frac{3}{5}$  

This final fraction tells us that what we got when we found the integral is actually $\frac{3}{5}$ times the equation provided.  So the actual equation for the total potential energy of this distribution of particles is  $U=-\frac{3}{5}\frac{GM^2}{r}$

3.  If the average speed of a star in a cluster of thousands of stars is $\left \langle v \right \rangle$, give an expression for the total mass of the cluster.  

We started with the Virial Equation 

$K = -\frac{1}{2}U$

$\frac{1}{2}mv^2=\frac{1}{2}\frac{GM_* m}{R}$

To find the mass of the cluster, we rearranged the above equation to solve for M$_*$.  This is the result:

$M_*=\frac{Rv^2}{G}$

4.  The cluster M80 has an angular diameter about 10 arcminutes and resides about 10$^4$ parsecs from the Sun.  The average speed of the stars in the cluster is  $\left \langle v \right \rangle$=10 km/s.  Approximately how much mass, in solar masses, does the cluster contain?

This step, though it wasn't conceptually difficult, had many steps.  The first step was to convert the angular diameter provided into radians using the following conversion

$10'\left ( \frac{1^o}{60'} \right )\left ( \frac{\pi rad}{180^o} \right )$

The second step was to use the angular size and the distance to the cluster to find the clusters actual size.

where $\theta$ is the angular size (found using the above conversion) divided by 2.  

We can use geometry and small angle approximation to find R.

$\tan \theta = \theta =\frac{R}{d}$

$R=d\theta$

We can now use this R in the equation we found in part 3.  The only problem is that the $\left \langle v \right \rangle$ we were given is in km/s, not parsec/s, so we have to first convert the radius into kilometers using the fact that 1 parsec = 3x10$^13$km.  

Once that's done, we can find the mass of the cluster in grams by using the equation found in part 3.  Then, because we're asked to find the mass of the cluster in solar masses, we use the fact that 1 solar mass = 2x10$^33$ grams to find the mass of the cluster.  

We found that the mass of the cluster was $3\times 10^{5}M_\odot$.  

After googling the actual mass of M80, we found that we were well within an order of magnitude, and actually off by less than half the actual value.  Any error in our result was most likely due to rounding during our calculations.  

WS 7, Problem 1: Kepler's Third Law

For a planet of mass m orbiting a star of mass M$_{*}$, st a distance a, start with the Virial Theorem and derive Kepler's Third Law of motion.  Assume that m<<M$_{*}$.  Remember that since m is so small, the semimajor axis, which is formally 
a=a$_{p}$+a$_{*}$, reduces to a=a$_{p}$.


We can say that a=a$_{p}$ because the planet is so much smaller than the star that it can essentially be treated as a point mass by comparison to the star.  The distance to the planet's center is too small compared to the distance to the star's center to significantly change a.

We started by drawing this picture:

We were told to start with the Virial Theorem, which defines the relationship between the kinetic and potential energies of a system bound by potential forces.  

$K = -\frac{1}{2}U$

We substituted the equations for the kinetic and potential energies and ended up with this equation:

$\frac{1}{2}mv^2=\frac{1}{2}\frac{GM_* m}{a}$

We knew from previous worksheets that Kepler's Third Law related orbital period to the semimajor axis of the orbit.  We just couldn't remember exactly what the equation was (which was good, because that was the equation we were trying to find, and knowing it beforehand would have ruined all the fun).  Since orbital period measured in time, we decided to solve the above equation for seconds.

There was a brief moment of panic when we saw that there aren't any seconds in the equation, but that moment quickly passed when we remembered that velocity is distance per time.  We quickly realized that, in an orbit, the distance traveled in the velocity is circumference of orbit per orbital period.  That led us to this equation:

  $\left ( \frac{2\pi a}{P} \right )^2= \frac{GM_*}{a}$

where P is the orbital period and we assume a circular orbit when solving for its circumference.  Simplifying the equation gives us these equations:

$\frac{4\pi ^2a^2}{P^2}=\frac{GM_*}{a}$

$P^2=\frac{4\pi^2a^3}{GM_*}$

We compared our answer to the equation provided to us on an old worksheet and it was correct!

I've said "we" a lot, and by that, I mean Dennis Lee, Delfina Martinex-Pandiani, and myself.  

AU Lab: Putting It All Together!

In the previous three lab installments, we found the following values:

• Rotational Period of the Sun = 28.3 days
• Angular Size of the Sun = 0.56$^{o}$
• Rotational Velocity of the Sun = 0.975 km/s

The first step in finding the Astronomical Unit is to find the physical size of the sun, which we can find using the sun's rotational period and velocity.  These two quantities are a time and a velocity, which, when multiplied together, provide a distance.  In this case, that distance is the circumference of the sun, which can be used to find the sun's radius.

$P_\odot V_\odot = \boldsymbol{Circ}_\odot =2\pi R_\odot$

$R_\odot = \frac{P_\odot V_\odot}{2\pi}$

Now that we know the radius of the sun, we can use geometry to find the Astronomical Unit, d

where $\alpha = \frac{\theta}{2}$.

According to the picture above, $\tan(\alpha )=\frac{R_\odot}{d}$.  Because of small angle approximation*, we can say that $\tan(\alpha ) = \alpha $, which leads us to the equation

$\alpha = \frac{\theta}{2}= \frac{R_\odot}{d}$

$d=\frac{2R_\odot}{\theta}$

When you substitute the radius equation for $R_\odot$, you get this equation for the Astronomical Unit:

$d=\frac{\frac{P_\odot V_\odot}{2\pi}}{\frac{\theta}{2}}=\frac{P_\odot V_\odot}{\pi \theta}$

Our group found that the distance to the sun is $7.986\times 10^{12}$ cm, which is about half the actual value.  This makes sense because the rotational velocity we found was almost exactly half of the actual value.  

*Small angle approximation says that, for $\theta < < 1^{o} $, $\sin \theta = \theta$ and $\cos \theta =1$

AU Lab: Rotational Velocity

Theoretical Background

Going from the doppler shifts at different spots on a rotating object to the objects rotational velocity is fairly easy.  You use this equation:

$\frac{\mathbf{source\: velocity}}{\mathbf{velocity\: of\: light}}=\frac{\mathbf{change\: in\: wavelength}}{\mathbf{rest\: wavelength}}$

Figuring out where to measure the doppler shifts is more difficult.  In order to use the doppler shift to accurately measure an object's rotational velocity, one has to measure the shifts at opposite points along the object's diameter.  To find measurements from the most accurate (farthest apart) locations, because we don't know where the equator of the sun is, we took doppler shift measurements at four different sets of opposite points.  Whichever set gives the biggest difference in shifts is the line closest to being a diameter of the sun.  

Supplies

• Heliostat - a device that uses mirrors to angle a reflection of the sun toward a target
• Spectrometer- a device that, in this context, is used to determine the wavelengths of light present in a light source

From the Wikipedia page on Spectrometers

• Sodium Lamp

Process

The first step is to calibrate the CCD to the sodium wavelength lines by shining the sodium lamp into the receiving slit of the spectrometer.  This is a very tedious process, involving meticulously adjusting the slit width, focal length of the camera, and exposure time to get the sharpest images of the sodium lines possible.  Once that is done, and you have an accurate measurement of the wavelengths of the sodium lines, remove the sodium lamp and get measurements using the sunlight.  

DO NOT move the CCD, because it is now perfectly aligned, and if you mess things up, your group will be very angry with you.  

Using the heliostat, angle the image of the sun into the spectrometer.  Use the motor to move the image reflected by the heliostat and take measurements of the sodium lines at eight different spots (shown below) on the sun's outline. 

Measure the shifts relative to the shifts of the water (telluric) lines, which shouldn't move because their measurement doesn't depend on the motion of the sun (the telluric shift should be 0).  

Analyze the shift data to see which set of points has the greatest difference in doppler shift, as this set of points will be closest to the sun's diameter.  

Analysis

We took two sets of measurements at each point so as to get more accurate results.  Measuring the doppler shifts at each of the points indicated in the figure above provided us with the following two graphs.  

There's little differnce between the two graphs, which is good, because that means there should be little difference between the two groups' results, which will be averaged at the end. 

The particular doppler shifts we're looking at are the shifts of the sodium lines, which are located at about 325 and 665 angstroms (along the x-axis).  When we zoomed into those two spots, we found that the lines representing the shifts at the top right and bottom left of the sun had the greatest difference.  To find the exact shift between those two locations at those wavelengths, we analyzed the following graphs.  

For lower-wavelength sodium line

For the higher-wavelength sodium line

I said before that we had to measure the shift of the sodium lines relative to the shift in the telluric lines, which should be 0.  Our group actually measured a small shift in the telluric lines,

which caused some issues.  The telluric lines shifted to the left, which decreased the relative shift of our sodium lines.  

Results

Our two different groups, after using the computer software to convert from the doppler shift between the two opposite spots to the rotational velocity using an equation similar to the one in the Theoretical Background section*, found the following velocities in km/s:

1.913

1.971

2.021

1.897

Averaging these four velocities gives a result of 1.951 km/s.  This is not the final rotational velocity of the sun, however, because the change in wavelength we observed by finding the difference in doppler shifts at the two opposite locations is actually twice the real difference.  One side of the sun is moving toward us at the same speed that the other side is moving away from us, so the relative speed between the two is twice the actual value.  In order to get the rotational velocity of the sun, one must divide the above value by 2, making the rotational velocity of the sun 0.975 km/s!

Discussion

The actual rotational velocity of the sun is about 2 km/s, so our team's result is less than half what it should be.  Why is our error so large? 

It's possible that our equipment--the camera we were using--was just really noisy.  Our results were also definitely affected by the fact that we measured a shift in the telluric lines.  Also, it's not guaranteed that we found the equator by measuring the shifts at the eight different spots on the sun.  Measuring the shift differences only guaranteed that we used the diameter line closest to the equator (compared to the others we tested).




*The process used by the computer program wasn't identical to the one described by the equation, because the program had to convert from pixels to wavelength by using the CCD's plate scale, which is the physical size subtended by one pixel.  The plate scale is a quantity measured in arcsec/pixel that can be found using this formula

$\Theta =\frac{P}{F}$

where P is the length of an indivual pixel and F is the focal length of the camera. This way, a shorter focal length gives you a wider field of view (bigger plate scale).  Theta has to be converted into arcseconds because this equation gives you radians. 

AU Lab: Angular Size

Theoretical Background

As the earth rotates, the sun appears to move 360$^{o}$ around the earth in 24 hours.  By measuring the amount of time it takes the sun to move the length of its diameter, one can use the following formula

$\frac{\Delta t}{24 hrs}=\frac{\theta }{360^o}$

$\theta =\frac{360 \Delta t}{24 hrs}$

to solve for the sun's angular size.

Process

Set up a piece of paper so that the sun, when focused through a lens, falls on the page.  Use a pencil to mark the right side of the sun's image.  Start a stop watch AT THE SAME TIME because the sun's image is always moving across the page.  Stop the timer as soon as the left side of the sun's image touches the mark you made before.  The time you measured is the time it takes for the sun to move the length of its angular diameter.  

Repeat the above process multiple times.  Team Crush repeated the process thrice and got the same time measurement for each one.  Each time, it took the sun 2 minutes and 14 seconds to move the length of its diameter. 

Results

With our measurements and the above formula, we found that the angular size of the sun is 0.57$^{o}$ or $9.74\times 10^{-3}$ rad.  

AU Lab: Rotational Period

In order to determine the Astronomical Unit, the average distance from the earth to the sun, one has to know the following information

• Rotational Period of the sun
• Angular diameter of the sun
• Rotational velocity of the sun

Those values can be manipulated to find the physical size of the sun and the distance from the Earth to the sun.

The purpose of this post is to describe the process used to find the sun's rotational period.

Theoretical Background

Photo originally from NASA

The sun has sunspots (seen above), which travel along the sun's lines of latitude.  They can be used as location markers on the sun and tracked over time to determine the sun's rotational motion.  If one knows how far a sunspot moved (in degrees) over a known amount of time, one can figure out the sun's rotational period using this formula:

$\frac{P_\odot}{360^{_o}}=\frac{\Delta t}{\Delta \theta }$

Supplies

• Heliostat - a device that uses mirrors to keep sunlight reflected onto a single target
• Piece of paper and pencil
• Clock
• Latitude and longitude transparency 

Process

Use the heliostat to get a reflection of the sun onto a piece of paper by angling the mirrors to aim the light toward the target.  Trace the outline of the reflection with a pencil (the reason for this will be evident later).  Identify a sunspot and trace and label it.  Circle it so that it can be identified as the original sunspot.  Turn off the heliostat so that it no longer traces the sun's motion across the sky and the reflection moves across the paper.  Trace the sunspots' motion across the paper until a clear line of latitude emerges.  Be sure you mark the time and date.  

The next time you're in the lab, repeat the procedure described above.  You may not track the same sunspot.  

Find two sheets on which the same sunspot is tracked on different days (you might end up using other groups' sheets).  Overlay one piece of paper on the other--the most recent one on top--against a window.  Line up the outlines of the suns (this is why you traced the outline before) and rotate the papers so that the line followed by the sunspot on one page is parallel to the line it followed on the other. 

Take the transparency and line it up with the two suns' outlines.  Orient it so that the sunspot line is parallel to a line of latitude on the transparency.  Find the difference in longitude between the sunspot on the first and second day it was observed.  This difference is the $\Delta \theta$ in the formula above.  The time elapsed between the two observations is the $\Delta t$

Doing this process with multiple sunspots gave our group, Team Crush*, the following data.

By averaging the periods in the above table, Team Crush found a solar rotational period of 28.3 days.  

Why This Works

Like I said before, the sun has sunspots that move along the sun's lines of latitude (which is helpful because we don't know the angle of the rotational axis of the sun).  Once lines of latitude are established (using the sunspots), differences in longitude can be measured.

Between the first and second observations, the sunspots move a certain distance around the sun.  We don't know the actual distance, but the difference in longitude between the spots on the different observation days gives an angular distance.  If one knows the amount of time it took the spot to move this angular distance, one can use the formula above by comparing the  $\Delta \theta$ to the 360 degrees that the spot would move if it made a full rotation.  



Team Crush included Tom Leith, Dylan Munro, Sammy Mehra, Jennifer Shi, and myself.

WS 5.1, Problem 1: Lamp Luminosity

Use a lamp tted with a 100 Watt incandescent light bulb to estimate the luminosity of the Sun.
Do this with the knowledge that

• the Earth-Sun distance is 1 astronomical unit (AU), or a =1.5x10$^13$ cm
• an estimate of how far from the bulb you have to hold your hand in order for it to feel like sunny Spring day
• and the fact that incandescent light bulbs convert electrical power into radiant energy with an effi ciency of only about 3%

Our first step was to go up to the lamp and use our hands to determine the distance asked for in the second bullet point.  For me, having grown up in Pennsylvania where a sunny Spring day doesn't feel all that warm, the distance was about 10 cm.  The other people in my group were from really warm places (California and Argentina), so their distances were shorter.  We compromised and used the distance d=7 cm.

To solve this problem (and by doing the first step mentioned above), we are essentially setting the flux of the light bulb at 7 cm equal to the flux of the sun.  

$F_{bulb}=F_{\odot}$

$\frac{L_{bulb}}{4\pi d^2}=\frac{L_{\odot}}{4\pi a^2}$

$\frac{L_{bulb}}{d^2}=\frac{L_{\odot}}{a^2}$

$L_{\odot}=\frac{L_{bulb}a^2}{d^2}$

Because we're told that the light bulb is only 3% efficient, the actual luminosity of the light bulb is 3 Watts.  

Using this information, the luminosity of the sun in Watts is 

$L_{\odot}=1.2\times 10^{25}$ Watts

Also, I've started doing what I like to call "Selfies with Astro," a series of photographs in which I make ridiculous faces next to Astrophysics work.  This particular Selfie with Astro has the added bonus of containing the beautiful picture by Delfina Martinez-Pandiani (seen on the left side of the board).

WS 5.2, Problem 2: Tidal Forces

Recall the reading assignment about tides. Draw a circle representing the Earth (mass M$_{\oplus}$), with 8 equally-spaced point masses, m, placed around the circumference. Also draw the Moon with mass M$_{moon}$ to the side of the Earth. In the following, do each item pictorially, with vectors showing the relative strengths of various forces at each point. Don't worry about the exact geometry, trig and algebra. I just want you to think about and draw force vectors qualitatively, at least initially.


(a) What is the gravitational force due to the Moon, F$_{moon, cen}$, on a point at the center of the Earth? Recall that vectors have both a magnitude (arrow length) and a direction (arrow head).

(b) What is the force vector on each point mass, F$_{moon}$, due to the Moon? Draw these vectors at each point.

Both (a) and (b) are drawn in the picture below.

(c) What is the force difference, $\Delta$F, between each point and Earth's center? This is the tidal force.

(d) What will this do to the ocean located at each point?

At the points where the vectors are pointed away from Earth, the ocean will bulge.  We see this as an increase in sea level, known as high tide.  At the points where the points are pointing in towards the Earth, the ocean is slightly compressed.  We interpret this as low tide, a decrease in sea level.

(e) How many tides are experienced each day at a given location located along the Moon's orbital plane?

Along the moon's orbital plane, oceans experience exactly two full cycles of high and low tide.  The first high comes when the moon is directly overhead.  As the moon orbits Earth the tides at that original location get lower until it's time for low tide.  Low tide happens with the moon is at a position in its orbit 90$^{_o}$ from the original location.  After that, tide gets higher and higher until the moon is above the spot on the Earth exactly opposite its original location.  There begins the second tide cycle.

(f) Okay, now we will use some math. For the two points located at the nearest and farthest points
from the Moon, which are separated by a distance r compared to the Earth-Moon distance  $\Delta$r, show that the force difference is given by

$\Delta \mathbf{F}=\frac{2GmM_{moon}}{r^3}\Delta r$

This is the formula for gravitational force according to Newton's law of gravity, found in the reading on tidal waves.

$f(r)=\frac{-GmM_{moon}}{r^2}$

Below, I set up an equality between $\frac{\Delta F}{\Delta r}$ and the derivative of the gravitational force formula to find the force difference.  This equality works because the force difference, or tidal force, is actually the change in force over change in distance, r.  

$\frac{\Delta F(r)}{\Delta r}=\frac{dF f(r)}{dr}=\frac{\left ( \frac{-GmM_{moon}}{(r+\Delta r)^2} \right )-\left ( \frac{-GmM_{moon}}{r^2} \right )}{\Delta r}$

$\frac{\Delta F}{\Delta r}=\frac{2GmM_{moon}}{r^3}$

$\Delta F = \frac{2GmM_{moon}}{r^3}\Delta r$

(g) Compare the magnitude of the tidal force F$_{moon}$ caused by the Moon to F$_{\odot}$  caused by the Sun. Which is stronger and by how much? What happens when the Moon and the Sun are on the same side of the Earth?

$\frac{\Delta F_{moon}}{\Delta F_{\odot}}=\left ( \frac{2GmM_{moon}\Delta r}{r_{moon}^3} \right )\left ( \frac{r_{\odot}^3}{2GmM_{\odot}\Delta r} \right )=\frac{M_{moon}r_{\odot}^3}{M_{\odot}r_{moon}^3}$

The tidal forces caused by the moon are about 2.25 times stronger than the tidal forces caused by the sun.  This makes sense because even though the sun is a lot more massive than the moon, it's a LOT farther away.  

If the sun and moon were on the same side of the Earth, their tidal forces would add together, creating an extra high tide called Spring Tide.  When the sun and moon happen to align in such a way that they form a 90$^{_o}$ angle with the Earth, we see the lowest tides, called Neap Tides.  

(h) How does the magnitude of $\Delta$F caused by the Moon compare to the tidal force caused by Jupiter during its closest approach to the Earth (r is about 4 AU)?

$\frac{\Delta F_{moon}}{\Delta F_{jup}}= \frac{M_{moon}r_{jup}^3}{M_{jup}r_{moon}^3}= \frac{M_{moon}(4r_{\odot})^3}{M_{jup}r_{moon}^3}$

The tidal forces by the moon on the Earth are 1.5x10$^5$ times stronger than the tidal forces caused by Jupiter.  Jupiter is even lighter and farther away from the Earth than the sun is, so its effect should be really small compared to the moon.  

Freeform: G2 and Sagittarius A*

I was looking through my old texts (my entertainment method of choice when I'm bored at work) when I saw an old, unread message from my roommate.  The message contained a link to a New York Times article about a gas cloud that should enter the Schwarzschild radius of Sagittarius A*, the black hole at the center of the Milky Way.

Snack Time in the Cosmos

The article cited several scientists, including Avi Loeb, a Harvard professor, who were very excited about the opportunity to see the "spoon-feeding of a black hole."  Some scientists are predicting that the gas will be completely swallowed up by the black hole (they stuck really close to the eating analogy throughout the article) and others predict that some of the gas will escape from a "jet" out of the hole.  With all of the predictions going around, scientists guarantee that no matter what happens, they will learn something interesting.

The event, which is expected to take place any time in the next month, should happen in two roughly defined stages.  The first stage is pretty simple: the cloud, which has already spaghettified*, falls into Sagittarius A*.  When this happens, a shock wave will be created that has some pretty exciting effects.  It's most likely going to create X- and radio waves, which most scientists agree will probably be observable by ground-based telescopes.  If there's a star at the center of G2, the collision will produce light that we'll be able to see from the Earth.

Basically, this is a really exciting event for people who are interested in strong gravity phenomenon that will provide scientists with data for possibly decades to come.


*Spaghettification is the stretching effect that extreme gravity and tidal forces cause for objects near black holes.

Spaghettification GIF!!!

(It's surprisingly difficult to find good spaghettification GIFs, and even more difficult to embed them in blog posts.)

Worksheet 5.1, Problem 2: Solar Flux and Luminosity

Consider the amount of energy produced by the Sun per unit time, also known as the bolometric luminosity, L$_{\odot}$. That same amount of energy per time is present at the surface of all spheres centered on the Sun at distances r > R$_{\odot}$. However, the flux at a given patch on the surface of these spheres depends on r.

The first part of this problem asked how flux, F, is dependent on luminosity, L, and distance, r.

An object's luminosity diffuses according to the inverse square law, $L=4\pi r^2F$, which says that luminosity is equal to the flux at a given distance, r, times the surface area of the sphere with radius r.
Given the inverse square law, one can see that the relationship between flux, luminosity, and distance is:
$F\propto \frac{L}{r^2}$

The second part of the problem asked us to find the effective temperature of the sun given the following information:

• The flux felt on the Earth from the sun, which I will call the Earth Flux, F$_{\oplus}$, is $1.4\times10^6 \frac{ergs}{s\cdot cm^2} $ 
• The angular diameter of the sun is $\theta =.57^{_o}$
• $L_{\odot}(r=R_{\odot})=L_{\odot}(r=1 AU)$
• The Stefan-Boltzman constant, $\sigma$, is $5.7\times10^-5 \frac{ergs}{s\cdot cm^2\cdot K^4} $

Using the third bullet point, we set up the following equality

$4\pi R_{\odot}^2F{_\odot}=4\pi d^2F_{\oplus}$

where d is the distance between the Earth and the sun, 1.5x10$^{_13}$ cm.  As we derived on last week's worksheet (work can be seen in the post entitled Flux and Luminosity), flux is related to temperature by $F=\sigma T^4$.  This gives us the following equation:

$R_{\odot}^2\sigma T_{\odot}^4=d^2F_{\oplus}$

$T_{\odot}=\left ( \frac{d^2F_{\oplus}}{\sigma R_{\odot}^2} \right )^{1/4}$

In order to find the radius of the sun, R$_{\odot}$, we had to use the provided angular diameter.  Because the angle is so small, we were able to use small angle approximation, which says $\sin \theta =\tan \theta =\theta $.  According to this super advanced picture,

where $\alpha=\frac{\theta}{2}$, 

$\tan \alpha =\frac{R_{\odot}}{d}$

$R_{\odot}=\frac{d\alpha }{2}$

Once alpha has been converted from degrees to radians, the final equation for the temperature of the sun can be read as 

$T_{\odot}=\left ( \frac{d^2F_{\oplus}}{\sigma \left ( \frac{d\theta }{2} \right )^2} \right )^{1/4}$

After substituting all of the given values for their variables, we found that the temperature of the sun is about 10$^4$ K.  Any error in this result is due to the many times we rounded during our calculations at the end because we didn't have a calculator.  A quick google search let me know that the real temperature of the sun is closer to 1.6x10$^{7}$, so the rounding error is really significant.