Blog #13, Famous Astronomer Biography, Tycho Brahe

Astronomer Bio - Tycho Brahe 

Tycho Brahe was born December 14, 1546 in Skane, Denmark (now part of Sweden). Brahe came from an aristrocratic family and was able to get into alchemy and astronomy at an early age as an apprentice for his uncle at the University of Copenhagen. At age 20, he got into a duel with a classmate and lost part of his nose, forever to wear a metal cover on the bridge of his nose to cover the wound. Brahe is commonly known as the first person to put empirical evidence behind the claims in the world of astronomy. He was set on finding exact, precise calculations of astronomical bodies such as stars and planets, something his counterparts had only theorized about. Most, if not all, of his work was completed as a professor and researcher at the University of Copenhagen.

Brahe's largest contributions to the world of astronomy were mostly focused on proving that the universe was not stagnant, never changing, surrounded by "celestial spheres" as was the common thought of his era. He did so by observing a new star, or super novae, that formed in the constellation of Cassiopeia. During the 1.5 years that the new star was visible in the sky, Brahe also observed various comets shoot across the night sky. At the time, both were considered to be disturbances just above the atmosphere. However, Brahe saw that while the comet moved across the sky with respect to the background stars, the supernovae was actually part of the constant night sky.

 A drawing of Brahe's homemade astronomical device for measuring orbital distances

With this observation, Brahe was able to prove that the heavens were living and constantly changing with the birth and death of stars. Brahe also proposed a new model to the universe that had the sun orbiting around the Earth while the 5 other discovered planets of the time still orbited the sun. While incorrect, Brahe used his theories to measure the orbits of various celestial bodies, often accurate to within the half arcsecond, an incredible feat given the fact that he did it without the use of a telescope.

It was Tycho Brahe who took on Johannes Kepler as an assistant at the end of the 16th century. However, because Brahe was so proud of his work, he would not share many of his calculations with the young Kepler. Instead he assigned him the seemingly "impossible" task of debunking Mars' orbit. It was through these investigations that Kepler would go on to discover his 3 laws of planetary motion, a law not published until after Brahe had died.

Brahe died in 1601 after attending a dinner party. There is speculation about his death though some claim he died after drinking to excess and failing to relieve himself, causing his bladder to explode, eventually killing him. Others rumor that he was poisoned after coroners found traces of mercury in his blood. Many accused Kepler of the murder.

Sources:
http://galileo.rice.edu/sci/brahe.html
http://www.space.com/19623-tycho-brahe-biography.html
http://www.nndb.com/people/559/000024487/brahe3.JPG
http://www.atnf.csiro.au/outreach//education/senior/cosmicengine/images/cosmoimg/tychomodel.gif
https://upload.wikimedia.org/wikipedia/commons/e/e0/Tycho_instrument_sextant_16.jpg

Blog #12, Free Form Post, Super moon

Free Form Post 

Although I know that I will not be the only one from Astro 17 to post about this Sunday's supermoon, I wanted to look a bit more into the phenomenon that brought millions of people outside to stare at the moon for a few hours.

At around 10:11 (ET) on Sunday night, the full moon was shaded a dark red and looked incredibly prominent in the sky. The moon will be fantastic due to a number of terrific coincidences, the first of which is a full lunar eclipse. In the lunar eclipse, the Earth will be directly between the sun and moon meaning that there will be no direct sunlight hitting the moon, the usual cause of its shine in the night sky. Instead of its usual white shine, the moon on Sunday had a dark red tint, due to Rayleigh Scattering, the same effect that causes the sky to look blue. Basically the low amounts of light particles that do reflect off of the Earth and hit the moon will be scattered by the particles of the atmosphere only allowing certain wavelengths to continue moving an assigned path. In this case, a lot of the red light will be reflected off the moon, giving it the "Blood Moon" effect that we saw on Sunday night.

The reason the September 27 caused such a stir throughout the world was because it combined a total lunar eclipse, the 4th of the year, and the moon at its perigee, or the closest that it will ever be to the Earth. Because the Moon's orbit is elliptical, it has a furthest (apogee) and closest (perigee) approach to Earth. The moon passes through its apogee about once a month. During this pass through, the moon looks 14% larger and 30% brighter than it does at its pedigree. However because this occurrence and a full lunar eclipse do not happen too frequently, the chances of them falling on the same night is extremely rare. In fact, the phenomenon, called a Saros, only happens every 18 years, 11 days and 8 hours.

Luckily, due to the rotation of the Earth and moon as well as weather patterns, the Boston area was one of the top sights to see the super moon. It was really refreshing to walk back to my dorm in Currier and see hundreds of students and faculty standing in the quad lawn just gazing up at the sky in silence. While it may not be as evident as it is to us, the students who choose to take an astronomy class, the night sky is pretty captivating. Hopefully this inspires people to take some time to appreciate the stars a bit more often. Or maybe it will take another 18 years to get people looking up.

Sources:
http://www.space.com/11161-supermoon-full-moon-science-infographic.html
http://www.timeanddate.com/astronomy/moon/lunar-perigee-apogee.html
http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html#c2
http://www.foxnews.com/science/2015/09/27/sundays-rare-supermoon-eclipse-what-need-to-know/
http://images.latinpost.com/data/images/full/53810/mexic-moon-mission.jpg?w=600

Blog #11, Worksheet #4.1, Problem 3, Angular Quantities of Microlensing

3) When speaking about microlensing, it is often easier to refer to angular quantities in units of \(\theta_E\). Let’s define \(u = \frac{β}{θ_E}\) and \(y = \frac{\theta}{\theta_E}\)

(a) Show that the lens equation can be written as: \[u = y - y^{-1}\]We know that the lens equation takes the form \[\beta = \theta - \frac{4GM_L \cdot \pi_{rel}}{\theta c^2} \]We also showed in question 2 that \[{\theta_E}^2 = \frac{4GM_L \cdot \pi_{rel}}{c^2} \]Using these facts and rearranging the given quantities \[\beta = u \theta_E \qquad and \qquad \theta = y \theta_E \]We substitute in to find \[u \theta_E = y \theta_E - \frac {{\theta_E}^2}{y \theta_E}\]Canceling all of our \(\theta_E\) we get the desired answer of \[u = y - y^{-1}\]

(b) Solve for the roots of y(u) in terms of u. These equations prescribe the angular position of the images as a function of the (mis)alignment between the source and lens. For the situation given in Question 2(f) and a lens-source angular separation of 100 µas (micro-arcseconds), indicate the positions of the images in a drawing.

We begin by solving for y in terms of u. We start with our equation from part a) \[u = y - y^{-1}\] Rearranging for y we get \[y^2 - uy - 1 = 0 \]which solves quadratically as \[y = \frac {u \pm (u^2 + 4)^{\frac{1}{2}}}{2}\]We begin by solving for u which we know is \(u = \frac{\beta}{\theta_E}\) where \[{\theta_E}^2 = \frac{4GM}{c^2} \cdot \frac{D_s - D_l}{D_s D_l}\]Where \[D_s = 8kpc \qquad D_l = 4kpc \qquad M = .3M_\odot \]When we solved for this in question 2 we found that \(\theta_E \thickapprox 1x10^{-6} \: arcseconds\) and we know that \(\beta = 100 \mu as \) Which means that \[u = \frac{\beta}{\theta_E} = \frac{100x10^{-6}}{1x10^{6}} = 100\]We then solve for y and get \[y_+ = \frac{100 + ((100)^2 + 4)^\frac{1}{2}}{2} \thickapprox 100\]and \[y_- = \frac{100 - ((100)^2 + 4)^\frac{1}{2}}{2} \thickapprox\ -1x10^{-2}\]We know that \[y = \frac{\theta}{\theta_E} = \frac{\theta}{1x10^{-6}}\]So finally solving for theta we get \[\theta_+ = 10^{-4} \: arcseconds\] and \[\theta_- = 10^{-8} \: arcseconds\]

Blog #10, Worksheet #4.1, Problem 1, Microlensing

1) Mass bends space-time! This is a prediction of general relativity, but fortunately we can heuristically derive the effect (up to a factor of 2) using Newtonian mechanics and some simplifying assumptions. Consider a photon of “mass” mγ passing near an object of mass \(M_L\); we’ll call this object a “lens” (the ‘L’ in \(M_L\) stands for ‘lens’, which is the object doing the bending). The closest approach (b) of the photon is known as the impact parameter. We can imagine that the photon feels a gravitational acceleration from this lens which we imagine is vertical (see diagram).

(a) Give an expression for the gravitational acceleration in the vertical direction in terms of M, b and G.

To solve for the vertical acceleration we simply use our standard gravitational force equation \[F_g = \frac {GMm}{b^2} = ma\]Where b is the distance between the photon and our larger mass of the lens. We then simplify and solve for \(a\) giving us \[\frac{GM}{b^2} = a\]

(b) Consider the time of interaction, ∆t. Assume that most of the influence the photon feels occurs in a horizontal distance 2b. Express ∆t in terms of b and the speed of the photon.

We know the equation for distance with respect to velocity and time is \[\Delta d = \Delta t \cdot v\]Rearranging and solving for \(\Delta t\) we get \[\Delta t = \frac{\Delta d}{v}\] and we know that \(\Delta d = 2b\). So we end up with \[\Delta t = \frac{2b}{c}\] where \[c = v_{photon} = 3x10^8 \frac{m}{s}\]

(c) Solve for the change in velocity, ∆v, in the direction perpendicular to the original photon path, over this time of interaction.

To solve for \(\Delta v\) we simply have to combine the previous two parts using the equation \[\Delta v = a \Delta t\]Plugging in we get \[\Delta v = \frac{GM}{b^2} \cdot \frac{2b}{c}\] which we simplify as \[\Delta v = \frac{2GM}{bc}\]

(d) Now solve for the deflection angle (α) in terms of G, \(M_L\), b, and c using your answers from part (a), (b) and (c). This result is a factor of 2 smaller than the correct, relativistic result.

We just solved for the vertical velocity of the photon as \[\Delta v = \frac{2GM}{bc}\]and we know that a photon must have a horizontal velocity of simply the speed of light, c. We use these two velocities to create a triangle with the two shorter sides as the velocities and find our deflection angle. We find that \[tan (\alpha) = \frac{\frac{2GM}{bc}}{c}\]Simplifying and using the fact that for small angles \(tan(\alpha) \thickapprox \alpha\), we end up with \[\alpha = \frac{2GM_L}{bc^2}\] which is in fact a factor of 2 smaller than the general relativism answer of \[\alpha = \frac{4GM_L}{bc^2}\]

Blog #9, Free Form Post, New Pluto Images

Free Form Post: Blog #9 

New Horizons is the fastest man-made space vehicle ever created. Traveling at nearly 36,000 mph, New Horzions is changing the face of interplanetary travel. Launched on January 19, 2006, it only took the probe about a year to pass Jupiter and make its way towards the outer planets. Most people first heard about New Horizons this past July when it became the first vehicle to ever encounter the dwarf planet, Pluto, about 40 AU from the center of the solar system

This week, New Horizons was in the news again as it provided the closest, clearest pictures of Pluto to date. The pictures were originally taken on July 14th and downlinked back to Earth on September 13. The images show an incredibly advanced surface make up of the dwarf planet including icy dunes, mountain ranges, and the possibility of large ice chunks floating in dense rivers of nitrogen. Beyond the astonishing landscapes, the pictures (taken at a distance of only 1,100 miles about Pluto's surface) show complex layers of atmosphere above the planet's surface. These layers of various gasses lead researchers to believe that Pluto has daily weather patterns and possibly a "water cycle" of precipitation, evaporation, and condensation though instead of using frozen water, with frozen nitrogen. 

As New Horizons continues on passed Pluto, it has a new target in sight. By 2019, New Horizons hopes to encounter 2014 MU69, a Kuiper Belt exoplanet. If successful, New Horzions would be the first man-made object to leave our solar system. It is truly an exciting time to be studying engineering and astrophysics. 

Works used

http://www.foxnews.com/science/2015/09/17/nasa-releases-dramatic-pluto-images/

http://www.npr.org/sections/thetwo-way/2015/09/10/439232708/new-pluto-pictures-are-out-of-this-world

http://www.cnet.com/news/beautiful-new-photos-of-pluto-show-terrain-atmosphere/

Blog #8, Worksheet #3.1, Problem 1, Orbital Speeds

1. Below are the orbital distances and periods of solar system planets.

a) Calculate the orbital speed of each planet assuming that the orbits are perfectly circular. Report these speeds in AU/Year.

We calculate the orbital speed by first finding the circumference of each planet's orbit using \(d = 2 \pi a\) then dividing the circumference by the orbital period to obtain the orbital speed. We then add it to the given chart.

 
b) Recall that Kepler’s Third Law has the form:

Where P is the orbital period, a is the semimajor axis, \(M_{tot}\) is the sum of the two masses in the system. Calculate the orbital speeds of the planets predicted by Kepler’s Third Law for each planet.

We begin by solving for an equation for v, the orbital speed. We know that the orbit speed takes the form \[v = \frac{d}{p}\]We know that \[d = 2 \pi a \] and p is found from taking the square root of kepler's equation above. Thus to solve for our speed, v, we get \[v = \frac{2 \pi a}{(\frac {4 \pi^2 a^3}{G M_{tot}})^{\frac{1}{2}}}\]When simplified, this comes out to \[v = (\frac {G M_{tot}}{a})^{\frac{1}{2}}\]Solving for the the semimajor axis of each planet, we get the chart below. The orbital speeds from Kepler's equation are incredibly close to the observed speeds. 

c) Plot the observed orbital speeds against the semimajor axis of each planet. In the same graph, plot the curve predicted by Kepler’s Third Law. Describe the shape of the resultant graph. What you have plotted is a rotation curve for the solar system, and the shape you observe is characteristic of Keplerian systems where one central mass dominates (e.g. the Sun).

Blog #7, Worksheet #3.1, Problem 4, Flat Rotational Curve

4. We actually observe a flat rotation curve in our own Milky Way. (You will show this with a radio telescope in your second lab!) This means v(r) is nearly constant for a large range of distances. 

(a) Lets call this constant rotational velocity Vc. If the mass distribution of the Milky Way is spherically symmetric, what must be the M(<r) as a function of r in this case, in terms of Vc, r, and G? 

In question 3 we solved for \(v(r)\) in terms of M, r and g: \[v(r) = (\frac{GM}{r})^{\frac{1}{2}}\]All we then need to do is reorder the equation and to solve for M. \[M_{enc} = \frac{r{V_c}^2}{G}\]

(b) How does this compare with the picture of the galaxy you drew last week with most of the mass appearing to be in bulge? 

In our drawing of the galaxy, a majority of the mass of the galaxy was centered in the bulge of the galaxy, within a small portion of the entire radius of the Milky Way. However, in the flat rotation model that we just found, the mass of the galaxy is directly proportional to the distance away from the center of the galaxy, (given that Vc and G are constants). We know this is not true because a majority of the mass is concentrated at the center of the galaxy like we saw in the picture. To explain how the constant velocity is attained, we need some way to make up for this large mass that is seemingly existent as you move further and further from the center of the galaxy. This is the main proposal for why Dark Matter must exist.

(c) If the Milky Way rotation curve is observed to be flat (Vc = 240 km/s) out to 100 kpc, what is the total mass enclosed within 100 kpc? How does this compare with the mass in stars? Recall the total mass of stars in the Milky Way, a number you have been given in your first assignment and should commit to memory.

We found that \[M_{enc} = \frac{r{V_c}^2}{G}\] Plugging in the values we found as \[V_c = 240 \frac{km}{s} \qquad r = 100 kpc = 3 x 10^{18} km \qquad G = 6.67 x 10^11 \frac{m^3}{kg \cdot s^2} \]

Thus \[M_{enc} \thickapprox 10^{40} kg \]However the observed mass of the galaxy is \[10^{42} kg\]We can only see about 1% of the mass of the galaxy. The other 99% of the mass is hypothesized as being made up of "Dark Matter."

Blog #6, Worksheet #3.1, Problem 2, Shell Theory

2. For the ensuing problems and future endeavours in physics and astronomy, you will need knowledge of a very useful property of gravity called the Shell Theorem. It is a statement about how a spherical shell of uniform density gravitationally affect objects exterior and internal to the shell: 

A spherical shell of uniform density and mass Mshell gravitationally affects objects external to the shell exactly as a point mass (Mshell) would at the shell centre. The shell exerts no net gravitational effect on objects internal to the shell. 

If instead of shell, you had a solid sphere with some spherically symmetric mass distribution, what would the Shell Theorem imply about the force of a gravity on a hypothetical object inside/outside the sphere. Reason this out from Shell Theorem qualitatively with diagrams and/or squiggle math.

We assume that there is a spherical symmetry between the object and our central mass. The larger object is hypothetically broken up into infinite shells of of a very small mass. Depending on which of these shells our smaller object resides on, there will be a different ratio of gravitational pull on the object.

We learned that any shell beyond the distance of our object does not exert a gravitational force on the object. Thus only shells at a position \(<r\) will contribute to the force felt by the small object. If the object is beyond the radius, \(R\) of the larger object, then the larger object is just considered a point mass and the simple gravitational relationship still exists: \[F = ma = \frac{GmM}{r^2}\]However when our smaller mass is within the radius of the larger mass, the gravitational pull must be modeled as a ratio of the volume enclosed vs. the total volume of the body. Only a portion of the mass of the larger object contributes to the gravity and we are left with \[F = ma = \frac{Gm (\frac {r}{R})^3M}{r^2}\]

Blog #5, Worksheet #2.1, Black Bodies and Flux

*Worked with Tim Mcnamara 

2. You are floating space with Sandra Bullock. It’s dark and there is a barely discernible 1 Watt light bulb ahead of you. You know the eye must receive ~ 10 photons in order to send a signal to the brain that says, “Yep, I see that.” How far away is it? 

Note: Assume the bulb is emitting light isotropically (the same in all directions). 
Note: 1 Watt is \(10^7 \frac{erg}{s}\) 
Note: The energy of a photon is E = hν, for a frequency ν (Greek letter ‘nu’), and Planck’s constant
h = \(6.67x10^{-27} erg \cdot s\) 

We start by solving for the energy in one photon. We will approximate the wavelength of a photon as \(\lambda = 500 nm = 5x10^{-5} cm\). We then use quantum equations for light to solve for energy, E. \[E = h \cdot v\] and \[v = \frac{c}{\lambda}\] thus \[E = \frac{h \cdot c}{\lambda}\]When we solve this with our constants we find that the energy in 1 photon, \(E = 4x10^{-12}\) but we need 10 photons to see the light and end up with \(E = 4x10^{-11}\). Next we must approximate how long it will take that many photons to enter the eye. We will approximate that the speed at which we can process light or the eye's "refresh rate" as 24 hz which converts to 40ms or .04s between. Now we have the conversion for the amount of energy our eye needs to take in per second\[\frac{4x10^{-11}}{.04} = 1x10^{-9} \frac{erg}{s}\]The last step is to find how far away the lightbulb. We can do this by comparing the amount of power our eye is receiving, the amount of power given off by the light bulb, and the sizes of our eye and the light. The amount of light entering our eye will differ proportionally based on how far away from the bulb we are. We approximate the radius of our eye as .5cm which gives us the area as a .25cm^2 circle receiving the light from a spherical bulb. So the equation becomes \[\frac{A_{eye}}{A_{light}} = \frac{E_{eye}}{E_{light}}\] where \[a_{eye} \thickapprox .25 cm^2 \qquad a_{light} = 4 d^2 \qquad E_{eye} = 1x10^{-9} \qquad E_{light} = 1x10^7\] Plugging in a rearranging for d we get \[4d^2 = \frac{1x10^7 \frac {erg}{s} \cdot .25cm^2}{1x10^{-9} \frac {erg}{s}}\] which gives us a final answer of \[d = 250 km\]

Blog #4, Worksheet #2.1, Black Bodies and Flux

3. You observe a star and measure its flux to be \(F_\star\). If the luminosity of the star is \(L_\star\), 

a) Give an expression for how far away the star is. 

Luminosity is a measured constant dependent on the specific star. Flux is the amount of power that is emitted from a certain area of a black body. The two are related with the simple equation \[F = \frac{L}{A}\] where A is the area of a sphere through which the flux is being measured. The area of a sphere is defined as \[A = 4 \pi d^2\] where d is the distance away from the star. Thus to solve for distance we simply rearrange and solve for d and get \[d = (\frac{L_\star}{4\pi F_\star})^{\frac{1}{2}}\]

b) What is its parallax?

From the first worksheet we derived that the parallax is \[P = \frac{1}{D[pc]}\] It is important to note that the distance, D, here is in parsecs. To use our equation from part A we must convert parsecs to a metric unit, centimeters, by multiplying the denominator \[1pc = 3.08 x 10^{18} cm\]We can then use our equation for distance from part A to solve for the parallax of the star.

c) If the peak wave length of its emission is \(\lambda_0\), what is the star's temperature?

We use the Wein Displacement Law to relate peak wavelength to temperature with \[\lambda_0 = \frac{b}{T}\] where b is a constant of proportionality equal to \[b = 2.89 x10^{-3} m \cdot K\] Rearranging to solve for T, we get \[T = \frac{b}{\lambda_0}\]

d) What is the star's radius, \(R_\star\)? 

Luminosity is directly related to both radius and temperature through the equation \[L_\star = 4\pi {R_\star}^2 \sigma T^4\] where \[\sigma = 5.67x10^{-8} W \cdot m^{-2} \cdot K^{-4}\] and is known as the Steffan-Boltzmann constant. We then rearrange to solve for \(R_\star\) and get \[R_\star = (\frac{L_\star}{4\pi \sigma T^4})^{\frac{1}{2}}\]And for a final step we can put in our answer from part C for the temperature and get \[R_\star = (\frac{L_\star {\lambda_0}^4}{4\pi \sigma b^4})^{\frac{1}{2}}\]

Blog post #1, Hello Universe Post

Hello Universe! 

My name is Simon Shuham and I am a junior in Currier House. I study mechanical engineering and am pursuing a secondary in astrophysics as a way to satisfy my interests in aerospace engineering. I am hoping this class will give me the baseline knowledge to begin building vehicles that may someday explore the places we are learning about. 

Blog #3, Worksheet #1.2, Milky Way Map

Create a map of the Milky Way

Our sun is located about 8.5 kiloparsecs or 27,000 light years away from the galactic center. The sun lies on the inner edge of the Orion Arm of the galaxy. Our galaxy is constantly rotating and the sun has an orbit of approximately 240 million years around the Milky Way.

The side view of the galaxy shows that there are a few distinct areas. Two of these are the thin and thick disks of the Milky Way. The disks are defined by their contents. The thin disk, located closer to the galactic center, is mostly made up of gas, dust, and newly formed stars and holds most of the star forming regions of the galaxy. The thin disk is only about 400 light years in height. As you move further away from the center, towards the thick disk, there are better-established stars. The further you move from the center, the older the stars become. Most stars in the thick disk are more than 10 billion years old. The thick disk is about 1,000 light years in height.

The halo is a roughly spherical region surrounding the galactic disks. It contains very little gas or dust and is mostly comprised of older stars, some extending back to 15 billion years old. The halo is home to many star clusters or loose groupings of stars the gravitationally attract one another. Much of the halo is not visible and is approximated based on the positions of various star clusters.

At the center of the galaxy is the nucleus or bulge. The bulge is nearly 10,000 light years across and made up of almost completely very old stars. The stars are most densely packed in the bulge.

Globular clusters are very dense, roughly spherical star formations. Clusters contain stars that formed around the same time and are thus often used to categorize stars and make presumptions about different characteristics of star groupings. Clusters often contain hundreds of thousands to millions of stars. Most of the stars in globular clusters are 10-15 billion years old.

The Small Magellanic Cloud (SMC) is a very small dwarf galaxy located very close to our Milky Way (only about 200,000 light years). It is only 7,000 light years in diameter and thus much smaller than the Milky Way. Some researchers believe the SMC used to be a regular spiral galaxy but was disrupted by the gravitational pull of the Milky Way and formed its current irregular shape.

The Large Magellanic Cloud is another galaxy very close to the Milky Way. The LMC is closer (153,000 light years) and larger (14,000 light years in diameter) than its partner the SMC. There is a bridge of gases across the sky that links the SMC and LMC. Like the SMC, the LMC is part of the local group of galaxies that are within observational distance of the Milky Way allowing us to study it very closely.

Sagittarius A*, abbreviated Sgr A*, is an incredible bright and dense light and radio source at the center of the Milky Way. Because of its properties, researchers believe it is the sight of a super massive black hole. Although the black hole is not visible, by observing the rotation of stars around the center of the Milky Way, there is now strong evidence that large black holes exist at the center of many spiral galaxies like the Milky Way.

The Orion Star Forming Region is a nebula about 1,300 light years away from Earth. It is made up of gas and various gasses that allow it to be a breeding ground for new stars in the galaxy.  The Orion Nebula is one of the closest and brightest nebulae and can be seen with the naked eye from Earth’s surface. The region is one of the most observed and studied entities in the galaxy.

An open cluster is a grouping of stars much more loosely joined the globular clusters. Nevertheless, open clusters are made up of thousands of stars formed from the same region around the same time and thus usually have very similar properties. The closest open cluster to the Sun is Hyades at only 153 light years away. The furthest discovered light cluster from our solar system (as of 2005) is the Berkeley 29 cluster, at over 48,000 light years away.

Work Used

http://www.astro.umass.edu/~myun/teaching/a100_old/images/MWmodelb.jpg

http://www.nasa.gov/mission_pages/sunearth/news/gallery/galaxy-location.html

http://www.universetoday.com/18256/where-is-the-sun/

http://csep10.phys.utk.edu/astr162/lect/milkyway/components.html

http://hubblesite.org/reference_desk/faq/answer.php.id=39&cat=galaxies

http://astronomy.swin.edu.au/cosmos/T/thick+disk

http://earthsky.org/clusters-nebulae-galaxies/the-small-magellanic-cloud

http://imagine.gsfc.nasa.gov/features/cosmic/nearest_galaxy_info.html

http://www.nasa.gov/mission_pages/chandra/multimedia/black-hole-SagittariusA.html

http://www.astronomynotes.com/chapter1/s2.htm

http://www.univie.ac.at/webda/recent_data.html

Blog #2, Worksheet #1.1, Milkomeda Calculations

*Worked with Tim Mcnamara 

2. How long will it take for Andromeda to collide with the Milky Way? The time-scale here is the free fall time, tff. One way of finding this is to assume that Andromeda is on a highly elliptical orbit around the Milky Way (e -> 1). With this assumption we can use Kepler's Third Law where P is the period of the orbit and a is the semi major axis. How does tff relate to the period? Estimate it to an order of magnitude.


We will solve this problem from the Milky Way's frame of reference meaning we will picture the Andromeda Galaxy orbiting the milky way with a highly elliptical shape. The Milky Way will remain stationary and will serve as one of the foci for the Andromeda's orbit. Because it is so highly elliptical, we make the extreme assumption that the Andromeda is more or less moving in a line towards the Milky Way. Since the full period, \(P\), is the time it takes for the Andromeda to get to the Milky Way and then return to its original position, the time it takes to collide with the Milky Way is simply, \(\frac{P}{2}\). The Milky Way and Andromeda have roughly the same mass thus \(M_{Mw}\thickapprox M_{And}\). Thus we can assume Kepler's equation as: \[P^2 = \frac{4\pi^2a^3}{G(2M)}\] If we substitute in \(t_{ff} = \frac{P}{2}\) and simplify we end up with \[t_{ff} = \frac{\pi a^{\frac{3}{2}}}{\sqrt{G(2M)}}\] Substituting in our values for \(M = 10^{12} M_\odot\), \(G = 6.67 x 10^{-11} \frac{N m^2}{kg^2}\), and \(a = 800 kpc\), we end up with \[t_{ff} = 1.1 x 10^{10}\] or 11 billion years!

3. Let’s estimate the average number density of stars throughout the Milky Way, n. First, we need to clarify the distribution of stars. Stars are concentrated in the center of the galaxy, and their density decreases exponentially. 

Part (a): Consider that within a 2 pc radius of the Sun there are five stars. 

We use the given information to first find our proportionality constant that we can then use to find the average distribution throughout the galaxy. We begin by using the fact that the sun is 8 kpc from the center of the universe so \[n(8000) = \frac{stars}{volume} = \frac{5}{\frac{4}{3} \cdot \pi \cdot 2^3} = \frac{15}{32\pi}\]. We also know that \[n(8000) = C \cdot e^{\frac{-8000}{R_s}}\] where \(R_s = 3.5 kpc\). Which gives us constant, C: \[C = 1.47 \frac{stars}{pc^3}\] This is the central density of our galaxy and can be taken as the average number density of the stars throughout our Milky Way. 

Part (b): Use the galaxy's "Scale height" of 330 pc. Use the galaxy's scale lengths as the lengths of the volume within the galaxy containing most of the stars. Assume a typical stellar mass of \(0.5_\odot\). 

To begin we will assume that our galaxy takes the rough shape of a cylinder with the scale lengths given: \[h = 330 pc \qquad r = 3.5 kpc\] We then solve for the volume of the galaxy using \[v = \pi \cdot r^2 \cdot h = 1.27x10^{10} pc^3\] We also know that the total mass of the Milky Way is \(10^10 M_\odot\) and the avg. stellar mass is \(0.5_\odot\) which gives us an approximation of \(2x10^{10}\) stars in the galaxy. We then solve for the density using \[\rho = \frac{m}{v}\] which leads to \[\rho = \frac{2x10^{10}}{1.27x10^{10}}\] which gives us a similar final answer of \[\rho_{avg} = 1.57 \frac{stars}{pc^3}\]

4. Determine the collision rate of the stars using the number density of the stars (n), the cross-section for a star \(\sigma_\divideontimes\), and the average velocity of the Milkomeda's stars as they collide \(\overline{v}\). How many stars will collide every year? Is the Sun safe, or likely to collide with another star?

We begin with finding a good estimate for the cross-sectional area of a star. Because our sun is an average sized star, we will use its radius to approximate a circular cross-section of an average star. The sun has a radius \(R_s \thickapprox 10^9 m\). When converting this to parsecs we end up with \(R_s = 3.24 x 10^{-8} pc\). We then solve for the cross-sectional area \(\sigma_\divideontimes = \pi \cdot {R_s}^2 = 1x10^{-15}pc^2\).

For the next step we must solve for the velocity, \(\overline{v}\), using our information from part two. We know that the time, t, that it will take for stars in the Milkomeda to collide is \(t = 1.1 x 10^{10} years\), and the distance between the two galaxies is 800 kpc. Using \(v = \frac{d}{t}\), we find that \(\overline{v} = 7.2x10^{-5} \frac{pc}{year}\). Generalizing our problem, we can assume that the number of collisions is equal to the number of stars occupying the set volume in a set amount of time. To find the volume occupied we multiply our area, \(\sigma_\divideontimes\) by the distance it will travel in a certain time. Thus \(Volume = \sigma_\divideontimes \cdot \overline{v} \cdot t\). Then our collisions will be this volume multiplied with the density of stars in the Milkomeda, which we found in part 3 (though we must use twice the density to approx. the density of both galaxies) We are left with \(collisions = Volume \cdot n = \sigma_\divideontimes \cdot \overline{v} \cdot t \cdot n\). But we are looking for \(\frac{collisions}{year}\) and thus must divide by time, t, leaving us with \(\frac{collisions}{year} = \sigma_\divideontimes \cdot \overline{v} \cdot 2n\). We then put in our constants \[\sigma_\divideontimes = 1x10^{-15}pc^2 \qquad \overline{v} = 7.2x10^{-5}\frac{pc}{year} \qquad n = 1.47 \frac{stars}{pc^3}\] Which gives us a final answer of \[2.26x10^{-19} \frac{collisions}{year}\]This is an incredibly low number of collisions and thus it would be expected that our sun is relatively safe and would not collide with another star once the two galaxies begin to collide.