Blog Post 13, Biography of Ejnar Hertzsprung

For this blog post, we had to write a post about a famous astronomer in the form of a short form biography. While I strongly considered investigating the lives of Johannes Kepler or Isaac Newton, having just talked about the astounding impact their thoughts had on society in my fantastic History of Science class, I decided to go for someone whose name I have heard quite a bit, but knew next to nothing about: Ejnar Hertzsprung. Hertzsprung's name is commonly heard connected to the Hertzsprung-Russell Diagram, graphing the relationship between the absolute magnitude or luminosity of a star vs. the star's surface temperature. One example is shown below:

Source: http://www.wwu.edu/skywise/a101_hrdiagram.html

The Hertzsprung-Russell shows that instead of being random, a distinct relationship exists between the temperature and the luminosity of a star, allowing us to group stars into several different groups by their observed characteristics. The majority of stars, including the Sun, appear on the Main Sequence, where stars spend much of their lifetime, but several other groups appear in unique locations on the diagram including giants, supergiants, and white dwarves. A star's location on the Hertzsprung-Russell diagram can be used to infer many properties of the star itself, using only two observed characteristics, temperature and luminosity.

Source: http://www.phys-astro.sonoma.edu/brucemedalists/Hertzsprung/hertzsprung.jpg

Ejnar Hertzsprung was a Danish astronomer and chemist who lived from 1873-1967. Though his name is heard often throughout the astronomy sphere, he worked primarily as a chemist for the first part of his life. He studied chemical engineering at Copenhagen Polytechnic, later continuing his work in chemistry in both St. Petersburg and Leipzig, focusing on the chemistry of photography. In 1901, he returned to Denmark to pursue an independent career in astronomy. For the next few years, Hertzsprung applied his knowledge of photography to investigate the relationship between a star's color and brightness. The relationship that Hertzsprung discovered (shown in the diagram at top) is now used as a basis for spectroscopic parallax, a method for estimating the distance of a star from the Earth.

In 1909, Karl Schwarzschild invited Hertzsprung to work with him in Göttingen, Germany, and the two of them moved later that year to the Potsdam Astrophysical Observatory. Hertzspring worked at the Leiden Observatory in the Netherlands from 1919 to 1944, finally retiring back in Denmark after his final nine years as direction of the observatory. 

Blog Post 12, Free Form Post, Gravitational Slingshots

NASA's New Horizons mission to Pluto still has everyone in awe since its close flyby of Pluto on July 14, 2015. Though New Horizons is now about 0.61 AU (as of 9/28/15) away from Pluto* (more than half the distance from the Earth to the Sun!), the probe continues to send back more and more stunning images like the following: 

(Source: http://pluto.jhuapl.edu/Multimedia/Science-Photos/pics/P_LORRI_FULLFRAME_COLOR.png)

But how did New Horizons manage to make it to Pluto in just nine years? Though it seems like a long time, had the New Horizons team not employed a gravitational slingshot around Jupiter, the trip would have taken an extra five years.

A gravitational slingshot (or gravity assist) is a method of using the orbital momentum of a planet (or other object) to either positively or negatively accelerate the velocity of a spacecraft. How does a gravitational slingshot work? Let's look at the system below, with a spacecraft of initial velocity v approaching a planet of initial velocity U.  As the spacecraft flies closer to the planet, the gravitational attraction between the much larger planet and the much smaller spacecraft draws the spacecraft in very close to the planet. 

(Source: https://en.wikipedia.org/wiki/File:Gravitational_slingshot.svg)

We know that the total energy of the system must be the same before and after the gravity assist. If we consider the system from the planet's point of view, the velocity of the planet is zero (the planet is at rest), and the velocity of the approaching spacecraft is U + v, or the original velocity (v) added to the velocity of the reference frame (U). As the spacecraft leaves the planet, the spacecraft must have the same velocity as when it entered, or U + v, but in the opposite direction as before. Then, to switch back to the reference frame of the observer, we must add the velocity of the reference frame (U), to get a final velocity of 2U + v. 

Obviously, the path of a spacecraft approaching an object will not always enter and exit in parallel paths. As it turns out, the path of the spacecraft will be hyperbolic, and the amount by which the velocity increases is dependent upon the angle at which the spacecraft enters and exits. In general, if the spacecraft's velocity is more in the direction of the planet's orbit after its encounter, the velocity of the spacecraft will increase. In turn, the velocity of the planet will actually slow down the smallest bit, because of Newton's Third Law - every action has an equal and opposite reaction. The momentum gained by the spacecraft is lost by the planet. Check out some examples of spacecrafts entering at different angles: 

Source: http://www.planetary.org/blogs/guest-blogs/2013/20130926-gravity-assist.html

So thank you, gravitational assists, for our newest images of Pluto! For more information, check out the following links: 

http://www.planetary.org/blogs/guest-blogs/2013/20130926-gravity-assist.html

http://physics.stackexchange.com/questions/53050/why-does-gravity-assist-transfer-twice-the-planets-velocity

https://en.wikipedia.org/wiki/Gravity_assist

http://www.askamathematician.com/2010/05/q-how-does-a-gravitational-sling-shot-actually-speed-things-up/

Blog Post 11, Worksheet 4.1, Problem 2

Consider a "lens" star at a distance \(D_L\) from the Earth, which lies near the line of sign to a background "source" star at a distance \(D_S > D_L\), as shown in the figure above. If the source and lens are separated by an angle \(\beta\). The bending of light from the source will create an image that is an angle \(\theta\) away from the lens, and an angle \(\alpha\) from the source. 

(Click on image to see larger version)

(a) Convince yourself that \(\beta = \theta - \alpha '\), where \(\alpha'\), the source-image angular separation, is distinct from \(\alpha\), the deflection angle of the source, in the figure above. This trivial expression is known as the lens equation. 

From the diagram, we can see that \(\alpha' + \beta = \theta\). To check this, think about the scenario in which the telescope, the lens, and the source are all directly lined up. The angle between the source and the lens, \(\beta\), would then be zero. Then \(\theta\), the angle between the image and the lens, is the same as \(\alpha'\), which agrees with the lens equation.


(b) Next, show that \[ \alpha' = (\frac{D_S-D_L}{D_S})\alpha\] and substitute it into the lens equation.

The key to understanding this equation is to realize that all the angles given in the above diagram are extremely small. Both \(\alpha\) and \(\alpha'\) are extremely small, but are not the same. Thus, we can think of two triangles with angles so small that they can be thought of as right triangles, with their hypotenuse and longest leg having approximately the same length:

As shown in the diagram above, both triangles share a sidelength, which we can arbitrarily name length L. Now, we can use trig to show the relationship between the two triangles: \[\text{sin}(\alpha) = \frac{L}{D_S} \approx \alpha'\] \[\text{sin}(\alpha) = \frac{L}{D_S-D_L} \approx \alpha\] If we solve for L in both equations, we get: \[L = \alpha(D_S - D_L) = \alpha'D_S\] Now, we can solve for \(\alpha'\): \[ \alpha' = (\frac{D_S-D_L}{D_S})\alpha\] Substituting this into the lens equation from part (a), we get: \[\beta = \theta - (\frac{D_S-D_L}{D_S})\alpha\]

(c) Now, substitute in the correct expression for \(\alpha\) from Question 1(d) to express \(\beta\) in terms of G, \(M_L\), b, \(D_L, D_S, \theta\), and c. Remember that your result from 1(d) is a factor of two smaller than the correct expression due to our Newtonian approximations. 

The correct expressions for \(\alpha\) from Problem 1d is: \[\alpha = \frac{4M_LG}{bc^2}\] Substituting this into the equation above, we get: \[\beta = \theta - (\frac{D_S-D_L}{D_S})(\frac{4M_LG}{bc^2})\]

(d) Replace b with an expression involving \(\theta\) and \(D_L\). Rearrange algebraically to show that \[\beta = \theta - (\frac{4GM_L}{\theta c^2})(\frac{D_S-D_L}{D_SD_L})\] Take a step back and remind yourself what each variable (\(\beta, \theta\), etc.) means and how they relate to each to each other. 

From Problem 1, we know that b is the distance between the lens and a photon of light. Combining these two images, we see the following setup:

Because theta is extremely small, we can solve for theta using the small angle theorem: \[\text{tan}(\theta) = \frac{b}{D_L} \approx \theta\] \[b = \theta D_L\] Now we can substitute this expression for b into the equation from part (c): \[\beta = \theta - (\frac{D_S-D_L}{D_S})(\frac{4M_LG}{bc^2})\] \[\beta = \theta - (\frac{D_S-D_L}{D_S})(\frac{4M_LG}{\theta D_Lc^2})\] \[\beta = \theta - (\frac{(D_S-D_L)4M_LG}{D_S D_L \theta c^2})\] \[\beta = \theta - (\frac{4GM_L}{\theta c^2})(\frac{D_S-D_L}{D_SD_L})\]

(e) When the lens and source lie along the exact same sight line, \(\beta = 0\). For this case, solve for the special value of \(\theta_E\), known as the Einstein ring radius, in terms of the lens mass, \(M_L\), and the relative parallax, \(\pi_{\text{rel}} = D_L^{-1} - D_S^{-1}\)

If we look more closely at the expression given for \(\pi_{\text{rel}}\), we see that it's very familiar:

\[\pi_{\text{rel}} = D_L^{-1} - D_S^{-1} = \frac{1}{D_L} - \frac{1}{D_S} = \frac{D_S}{D_SD_L} - \frac{D_L}{D_SD_L} = \frac{D_S - D_L}{D_SD_L}\]

\(\pi_{\text{rel}}\) is actually just a quantity already in the equation! With \(\beta = 0\), we are left with \[\theta_E = \pm(\frac{\pi_{\text{rel}}4M_LG}{c^2})^{1/2}\]

(f) For a typical lens of mass 0.3 \(M_\odot\) (an M-dwarf) at a distance \(D_L = 4\) kpc and a typical source at \(D_S = 8\) kpc, what is the sie of \(\theta_E\) in arcseconds?

This part just involves plugging in the given values for the equation we found in part (e) for \(\theta_E\): 

\[\theta_E = \pm(\frac{\pi_{\text{rel}}4M_LG}{c^2})^{1/2}\]

\[\pi_{\text{rel}} = \frac{D_S - D_L}{D_SD_L} = \frac{8 \text{ kpc} - 4 \text{ kpc}}{(8 \text{ kpc})(4 \text { kpc})} = \frac{4 \text{ kpc}}{32 \text{ kpc}^2} = \frac{1}{8 \text{ kpc}}\]

\[\theta_E = (\frac{(\frac{1}{8 \text{ kpc}})\:(4)\:(0.3 M_\odot * \frac{10^{33} \text{ g}}{M_\odot})\:(6.7 * 10^{-8} \text{ cm}^3 \text{ g}^{-1} \text{ s}^{-2})}{(3 * 10^{10} \text{ cm s}^{-1})^2})^{1/2} (\frac{ 4.8 * 10^{-6} \text{ arcseconds}}{ \text{1 radians}})\] \[\theta_E = \pm5.6 * 10^{-4} \text{ arcseconds}\]

(g) In astronomy, it is often very useful to case equations in terms of isolated variables measured in units of their typical order-of-magnitude values, with any power law dependencies explicitly labelled for each variable. There should be only one numerical constant out front bearing the proper units for the quantity of interest. This means calculating out all the remaining constants. In equations written in this way, not only can you see exactly how your quantity varies and by how much with changes in each variable. For example, Kepler's Law II recast in units relevant for a 'typical' stellar/planetary system might read: \[P = 1 \text{ yr } (\frac{a}{1 \text{ AU}})^{3/2} \:(\frac{M_\odot}{M_{\text{tot}}})^{1/2}\] 

You can easily see from this equation that, for a star-planet system whose semimajor axis measures 1 AU and whose stellar mass is 1 \(M_\odot\), the orbital period is 1 yr, as you would expect for the Sun-Earth system. If you wanted to find the orbital period for Mars around the Sun and you know \(a_{\text{Mars}}\) = 1.5 AU, you can readily deduce that its orbital period would be \(1.5^{3/2} * 1 \text{ yr} \approx 1.8 yr\)

Recast the equation you derived in 2(f) for \(\theta_E\) in terms of the typical lens mass and relative parallax. Your answer should appear in the form: \[\theta_E \approx \text{ mas } (\frac{M_{\text{lens}}}{M_{\text{lens (typical)}}})^x(\frac{\pi_{\text{rel}}}{\pi_{\text{rel(typical)}}})^y\] where mas is milli-arcseconds.

We are given that the mass of a normal lens, \(M_L\), is 0.3 \(M_\odot\), and the relative parallax, \(\pi_{\text{ relative}}\) is \(\frac{1}{8 \text{ kpc}}\), as calculated in the previous question. If we separate out the constants from the above equation, we are left with: \[\theta_E = \pm(\pi_{\text{rel}}^{1/2})(M_L)^{1/2}(\frac{4G}{c^2})^{1/2}\] Now, we can rearrange this formula in the form as asked:

\[\theta_E = \pm(\frac{4G}{c^2})^{1/2}(\frac{\pi_{\text{rel}}}{1/8 \text{ kpc}})^{1/2}(\frac{M_L}{0.3 M_\odot})^{1/2}\] We then calculate the final constant to get: \[\theta_E = \pm(5.6 * 10^{-1} \text{ mas})(\frac{\pi_{\text{rel}}}{1/8 \text{ kpc}})^{1/2}(\frac{M_L}{0.3 M_\odot})^{1/2}\]

Blog Post 10, Worksheet 4.1, Problem 1

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 on "mass" m, 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. 

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

The expression for the gravitational force between objects of mass M and m at a distance b from each other is given by the equation: \[F = ma = \frac{GMm}{r^2}\] The two smaller masses cancel, and we are left with: \[a = \frac{GM}{b^2}\]

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

We know that the photon is traveling at a speed c, \(3 * 10^{10} \text{ cm s}^{-1}\). As the formula for velocity, or speed, is given by speed = \(\frac{\text{distance}}{\text{time}}\), the formula for the photon's speed is given by \[c = \frac{2b}{\Delta t}\] Rearranging the equation, we have \[\Delta t = \frac{2b}{c}\]

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

If the acceleration of an object is constant, we know that the change in velocity of that object is given by the equation \(\Delta v = a(\Delta t)\), where a is the object's acceleration. As we know the acceleration of the object from part (a) and the \(\Delta t\) from part (b), we can solve for the \(\Delta v\) by plugging in our previous answers: \[\Delta v = (\frac{GM}{b^2})(\frac{2b}{c})\] \[\Delta v = \frac{2GM}{bc}\]

(d) Now solve for the deflection angle (\(\alpha\)) 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. 

If the think of the photon's original speed as a horizontal vector of length c, and the photon's final velocity as a result of the gravitational acceleration as a vector of length \(v_{\text{final}}\), we know by addition of vectors that the change in velocity, \(\Delta v\), must be the vertical vector as shown in the diagram below:

From this diagram, we know that \[\text{tan}(\alpha) = \frac{\Delta v}{c}\] As \(\alpha\) is extremely small, we approximate \[ \text{tan}(\alpha) = \frac{\Delta v}{c} \approx \alpha\] \[\alpha \approx \frac{(\frac{2M_LG}{bc})}{c}\] \[\alpha = \frac{2M_LG}{bc^2}\]

Blog Post 9, WIMPs

In Worksheet 3.1, Milky Way Rotation, we looked at the reason behind dark matter theories. If we calculate the amount of mass in the Milky Way Galaxy using Kepler’s Third Law and the observed orbital rotation velocity of objects within the galaxy, we find that the mass of the galaxy should be approximately \(10^{12} M_\odot\). However, the stellar mass of the galaxy as observed is approximately \(10^{10} M_\odot\), or only around 1% of the expected mass. The conclusion, therefore, is either that our estimation using Kepler’s 2rd law and Newtonian mechanics is wrong, or there must be some form of matter hidden in the galaxy that we cannot detect, but gravitationally affects all other matter in the galaxy. This “hidden” matter is known as dark matter. For this blog post, I was interested in learning about one of the major theories for what dark matter is – WIMPs.

WIMPs

WIMP stands for weakly interacting massive particles. The idea is that a WIMP is a non-baryonic particle (i.e. not composed of protons or neutrons), with a mass in a range from 1 GeV – 1 TeV, that only interacts through the electroweak interaction. I’ve never taken a course in particle physics, but from my understanding this means that these particles are extremely hard to detect, because they don’t interact with any of the matter we observe.

The theory behind the existence of WIMPs is that in the early universe, there were equal amounts of photons and WIMPs, or particles and antiparticles, and because of the extreme high temperature of the universe, these particles could be created and annihilated freely. As the temperature cooled to a point below the mass of the WIMPs, they could no longer be created, but could still be annihilated by joining with another WIMP. The density of WIMPs dropped until they had a density such that it was extremely unlikely that they would run into another WIMP, and that’s what the density is now.

From what I’ve read so far, the really fascinating part about WIMPs is that they align very coincidentally with the current theory of supersymmetry. Supersymmetry, from my understanding, is the theory that for every known particle in the universe, there is an “antiparticle” or supersymmetric paired particle. For example, the pair to a photon is photino, and the pair of the quark is the squark. Apparently, the predicted density of dark matter needed to fill the matter gap between our observations and our calculations fits the predicted density of the WIMPs that supersymmetry predicts.

As WIMPs supposedly interact through the weak force, and not the electromagnetic force, they have yet to be detected in the form that supersymmetry predicts. There are several types of experiments that are attempting to detect these particles.

One method attempts to directly detect WIMPs. WIMPs interact through the weak force, and are thus not usually detectable, even though there are theoretically several hundred going through one's body at any given moment [Source 2]. However, every now and then a WIMP might interact with an atomic nucleus by "scattering" [Source 1] off of it, causing the nucleus to recoil and possibly vibrate. This scattering is extremely difficult to detect, however, because this interaction is extremely rare and unpredictable.

Source: http://cdms.berkeley.edu/Education/DMpages/science/directDetection.shtml

Another method attempts to indirectly detect WIMPs. By the current theory, as WIMPs have been around for quite a long time, and therefore gravity must have caused some WIMPs must have settled at the core of the Earth and the Sun by now. As the density of the WIMPs at these cores increases, the possibly of these WIMPs colliding and annihilating to form neutrinos increases as well. Thus, scientists are attempting to detect the presence of neutrinos, formed by this annihilation, streaming out from the Earth’s and Sun’s cores.

Neither of these methods have yet to confirm the existence of WIMPs, but as scientists across the globe continue to experiment, they continue to explore and eliminate possibilities for the locations of WIMPs, hopefully drawing closer to their real location (or not, in which case it’s on to another theory).

Sources:

1. https://www.astro.umd.edu/~ssm/darkmatter/WIMPexperiments.html

2. http://www.astro.caltech.edu/~george/ay20/eaa-wimps-machos.pdf

Blog Post 8, Worksheet 3.1, Problem 5

5. M(<r) is related to the mass density \(\rho(r)\) by the integral: \[M(<r) = \int_0^r 4πr'^2\rho(r')dr'\] (Recall that the \(4πr'^2\) comes from the surface area of each spherical shell, and the dr' is the thickness of each thin shell). The fundamental theorem of calculus then implies that \(4πr^2\rho(r) = dM(<r)/dr\). For the case in question 4, what is \(\rho(r)\)? Is the density finite as r --> 0 in the case of a flat rotation curve?

This problem gives the relationship between mass, M(<r), and mass density \(\rho(r)\). In problem 4, we found that the mass, M(<r), is given by: \[ M(<r) = \frac{V_c^2r}{G},\] where \(V_c\) is the constant rotational velocity of an object in the galaxy, r is the distance from the center of the galaxy, and G is the gravitational constant. We now have two equations for M(<r) - the integral from the problem, and our equation from question 4. We can set these two definitions for M(<r) equal to each to solve for \(\rho(r)\). \[ \int_0^r 4πr'^2\rho(r')dr' = \frac{V_c^2r}{G}\] Taking the derivative of each side with respect to r, we get: \[4πr^2\rho(r) = \frac{dM(<r)}{dr} = \frac{V_c^2}{G}\] Rearranging this equality, we get: \[ \rho(r) = \frac{V_c^2}{4πr^2G}\] As a result, the density is not finite as r --> 0 as in the case of a flat rotation curve, because as r goes to zero, \(\rho(r)\) goes to infinity.

Blog Post 7, Worksheet 3.1, Problem 4

4. We actually observe a flat rotation curve in our own Milky Way. This means v(r) is nearly constant for a large range of distances. 

(a) Let's call this constant rotational velocity \( V_c\). 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 \(V_c\), r, and G?

From Problem 3, we know that relationship between velocity, mass enclosed, and distance from the center is expressed by the equation: \[v(r) = (\frac{GM_{\text{enc}}}{r})^{1/2}\] Now, we set this equation for velocity equation to a constant rotational velocity \(V_c\): \[v_C = (\frac{GM_{\text{enc}}}{r})^{1/2}\] From here, we solve for \(M_{\text{enc}}\), as we are searching for the M(<r) as a function of r: \[V_C = (\frac{GM_{\text{enc}}}{r})^{1/2}\] \[ \frac{GM_{\text{enc}}}{r} = V_c^2\] \[M(<r) = \frac{V_c^2r}{G}\] Thus, M(<r) as a function of r in therms of \(V_c\), r, and G is: \[M_{\text{enc}} = \frac{V_c^2r}{G}\]
(b) How does this compare with the picture of the galaxy you drew last week with most of the mass appearing in the bulge?

Unlike the picture we drew last week, this equation shows that the mass enclosed in the galaxy increases proportionally to the distance from the center of the galaxy - that is, when the distance increases, the mass increases. In the picture from last week, this did not appear to be true, as most of the mass appeared to be located in the center of the galaxy.

(c) If the Milky Way rotation curve is observed to be float (\(V_c \approx\) 250 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 that the total mass of stars in the Milky Way, a number you have been given in your first assignment and should commit to memory. 

Using the given values for \(V_c\) (240 km/s) and r (100 kpc), we can plug these values into the equation for \(M_\text{enc}\): \[M_{\text{enc}} = \frac{V_c^2r}{G}\] \[M_{\text{enc}} = \frac{(240 \text{ km/s})^2(100 kpc)}{G}\] \[M_{\text{enc}} = \frac{((240 \text{ km/s})(10^5 \text{ cm/km}))^2(100 \text{ kpc} (\frac{3 * 10^23 \text{ cm}}{100 \text{ kpc}})))}{7 * 10^{-8} \text{ cm}^3 \text{ s}^{-1} \text{ g}^{-1}}\] \[M_\text{ enc} = 2.6 * 10^{45} \text{ g  } (\frac{1 M_\odot}{2 * 10^{33} \text{ g}})\] \[M_\text{ enc} = 1.3 * 10^{12} M_\odot\]
If we compare this value to the stellar mass of the galaxy, \(10^{10} M_\odot\), we see that the stellar mass makes up only around 1% of the matter that should be in our galaxy: \[\frac{10^{10} M_\odot}{1.3 * 10^{12} M_\odot} * 100 \%≈ 1\% \]

Blog Post 6, Worksheet 3.1, Problem 3

3. Last time we approximated the shape of our Galaxy as a cylinder. This time it will be a sphere. If there are no other large-scale forces other than gravity (a good approximation in most galaxies), then an object's orbit around the galactic center will be approximately circular. 

(a) Show that Kepler's 3rd law can be expressed in terms of the orbital frequency \(\Omega \equiv 2π/P\) (i.e. orbits/time) and the distance from the center: \[r^3\Omega^3 = GM_{\text{ tot}}\]
As Kepler's Third Law expresses the relationship between period, P, and semimajor axis, a, we need to solve the given relationship between orbital frequency and period for P in order to be able to make the proper substitution in Kepler's Third Law: \[\Omega = 2π/P\] \[ P = \frac{2π}{\Omega}\] Now, we can plug this definition for P into Kepler's Third Law, as given in Problem 1: \[P^2 = \frac{4π^2a^3}{GM_{\text{tot}}}\] \[(\frac{2π}{\Omega})^2 = \frac{4π^2a^3}{GM_{\text{ tot}}}\] \[ \frac{4π^2}{\Omega^2} = \frac{4π^2a^3}{GM_{\text{tot}}}\] The \(4π^2\) terms cancel, and as we are referring to orbits as cylindrical (as opposed to elliptical), we can replace a, the semimajor axis of an ellipse, with r, the radius of a circle: \[ GM_{\text{tot}} = r^3 \Omega^2\] Thus, Kepler's Third law can be expressed in terms of the orbital frequency and the distance from the center.

(b) Now, assume that the Milky Way has a spherical mass distribution - this is a good approximation when talking about the total mass distribution. Using what you learned from Problem 2, rewrite the above for an object orbiting a radius r from the center of the galaxy.

From the Shell Theorem, we know that an object orbiting a galaxy at a distance r from the center will only experience the gravitational effects from the mass located between the center of the galaxy up to a distance r from the center of the galaxy. As a result, \(M_{\text{ enc}}\) can be written as M(<r), or the mass located in the galaxy up to a distance r from the center, and the equation above can be written as: \[ r^3 \Omega^2 = GM(<r)\]
(c) Next, let's call the velocity of the object at a distance r away from the center, v(r). Use Kepler's Third Law as expressed above to derive v(r) for a mass m if the central mass is concentrated in a single point at the center (with mass \(M_{\text{enc}}\)), in terms of \(M_{\text{enc}}\), G, and r. This is known as the Keplerian rotation curve. As you saw earlier, it describes the motion of the planets in the solar system, since the Sun has nearly all of the mass.

As we are trying to determine the velocity, v, for a mass m in terms of \(M_{\text{enc}}\), G, and r, we can use the standard relationship between linear velocity v and orbital frequency \(\Omega\) to introduce velocity into the equation for Kepler's Third Law: \[ v = \Omega r\] \[\Omega = \frac{v}{r}\] Now we can plug this definition for the orbital frequency into Kepler's Third Law as expressed above: \[ r^3 \Omega^2 = GM(<r)\] \[ r^3 (\frac{v}{r})^2 = GM(<r)\] \[ \frac{r^3}{r^2} v^2 = GM(<r)\] \[ v^2 r = GM(<r)\] \[v^2 = \frac{GM(<r)}{r}\] \[v = (\frac{GM(<r)}{r})^{1/2}\] Thus, the relationship between v and \(M_{\text{enc}}\), G, and r is given by: \[v = (\frac{GM_{\text{enc}}}{r})^{1/2}\]

Blog Post 5, Worksheet 2.1, Problem 4

4. A supernova goes off and you can barely detect it with your eyes. Astronomers tell you that supernovae have a luminosity of \(10^{42}\) ergs/s; what is the distance of the supernova?
Assume the supernova emits most of its energy at the peak of the eye's sensitivity and that it explodes isotropically. 

From problem 3, we have know the relationship between the luminosity of an object, the flux observed, and the distance between the object and the observer to be: \[ d = (\frac{L_*}{4πF})^{1/2},\] where d is the distance, L is the luminosity, and F is the flux observed.

The problem gives the luminosity of the supernova. In order to solve for the distance, we need to determine the observed flux.

From problem 1, we know that the human eye must receive ~ 10 photons every .1 seconds in order to register the presence of a light. We can call this: \[ N_{\text{photons}} = 100 \text{ photons / s}\] We are also given that the energy of a photon is E = \(h\nu\), for a frequency \(\nu\), and Planck's constant h = 6.6 * \(10^{-27}\) erg*s.

The minimum energy we can receive from the supernova and still register it with our eye is thus: \[ N_{\text{photons}}h\nu,\]with units of \[(\frac{\text{photons}}{\text{s}})(\frac{\text{erg * s *} \frac{1}{\text{s}}}{\text{photon}}) = \frac{\text{erg}}{\text{s}} = \frac{\text{energy}}{{\text{time}}}\]

The quantity we are looking for, flux, needs to have units of energy per time per area, or \[ \frac{\text{energy}}{\text{time * area}}\] Thus,  \[ N_{\text{photons}}h\nu = FA\]  \[F = \frac{N_{\text{photons}}h\nu}{A}\], where F is the flux observed, \(\nu\) is the frequency of the emitted radiation, h is Planck's constant as above, and A is the area over the the flux is being observed - in this case, the area of a human pupil.

We don't know the frequency of emission from the supernova, but we do know the wavelengths at which the human eye can register light. Using the relationship c = \(\lambda\nu\), where c is the speed of light, \(\lambda\) is the wavelength of the light, and \(\nu\) is the frequency of the light, we can replace \(\nu\) in the equation for \(\frac{c}{\lambda}\): \[F = \frac{N_{\text{photons}}h(\frac{c}{\lambda})}{A}\] \[F = \frac{N_{\text{photons}}hc}{A\lambda}\] We can now plug this expression for the flux into our previous equation for distance: \[ d = (\frac{L_*}{4πF})^{1/2}\] \[ d = (\frac{L_*}{4π\frac{N_{\text{photons}}hc}{A\lambda}})^{1/2}\] \[ d = (\frac{L_*A\lambda}{4π N_{\text{photons}}hc})^{1/2}\]
Now, we plug in the known value for each quantity, approximating the area of the pupil as a circle using a standard pupil radius of 1 mm, and an average wavelength of 500 nm:
\[ d = (\frac{(10^{42} \frac{\text{erg}}{\text{s}})(π (0.1 \text{ cm})^2)(500 \text{ nm} (\frac{1 \text{ cm}}{10^7 \text{ nm}}))}{4π (100 \text{ s}^{-1})(6.67 * 10^{-27} \text{  erg * s})(3 * 10^{10} \frac{\text{ cm}}{\text{ s}})})^{1/2}\]
\[ d = (\frac{(10^{42})(3 (0.1)^2)(5 *10^{-5})}{10 (100)(7 * 10^{-27})(3 * 10^{10})})^{1/2} \text{ cm} = 3 * 10^{24} \text{ cm}\]
\[ d = 3 * 10^{24} \text{ cm} (\frac{1 \text{ pc}}{3 * 10^{18} \text{ cm}})\] \[ d = 9 * 10^5 \text{pc}\] The supernova is \( 9 * 10^5\) pc away.

Blog Post 4, Worksheet 2.1, Problem 3

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

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

The flux of an object is the total energy that crosses a unit area per unit time, and has a unit of energy per area per time. The luminosity of an object is a measure of the object's intrinsic brightness, and has units of energy per time. The relationship between flux and luminosity is expressed in the following equation: \[ F = \frac{L}{4πd^2},\] where F is the flux, L is the luminosity, and d is the distance between the object and the location at which the flux is being observed. From this equation, we can solve for d, how far away the star is.
\[ F_* = \frac{L_*}{4πd^2}\] \[  4πd^2 = \frac{L_*}{F_*}\] \[  d^2 = \frac{L_*}{4πF_*}\] \[  d = (\frac{L_*}{4πF_*})^{1/2}\]
(b) What is its parallax?

The parallax of an object is the angle the object appears to move against more distant, "fixed" stars as the Earth moves a distance of 1 AU in its orbit.

The relationship between the parallax, given in arcseconds, and the distance, given in parsecs, is shown by the following equation: \[ \text{parallax [arcsecond]} = (\frac{1}{\text{distance [pc]}})\]
Using this equation, the parallax of this star is given by: \[p = (\frac{L_*}{4πF_*})^{-1/2},\] where p is equal to the parallax, \( F_*\) is the observed flux of the star, and \( L_*\) is the star's luminosity.

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

In question 1c on Worksheet 2.1, we derived an equation for the wavelength \( \lambda_{\text{max}}\) corresponding to the peak of the intensity distribution of a blackbody at a given temperature T: \[ \lambda_{\text{max}} = \frac{hc}{5kT},\] where h is Planck's constant, 6.6 * \(10^-27\) ergs*s, c is the constant for the speed of light, k is the Boltzmann's constant 1.4 * \(10^{-16} \text{ ergs K}^{-1}\), and T is the temperature of the blackbody in Kelvin.

Using this relationship, if the peak wavelength of the star's emission is at \( \lambda_0\), the relationship between the this wavelength and the star's temperature can be solved for temperature:
\[ \lambda_{\text{max}} = \frac{hc}{5kT}\] \[ 5kT = \frac{hc}{\lambda_0}\] \[ T = \frac{hc}{5k\lambda_0}\]
(d) What is the star's radius, \( R_*\)?

To solve this problem, we look first to the information that we already know.

- The flux emitted from a black body as a function of temperature is given by the equation: \[ F(T) = \sigma T^4,\] with \( \sigma = \frac{2πk^4}{c^2h^3}\int_{0}^{\infty}\frac{u^3}{e^u-1}du\), \(u = \frac{h\nu}{kT}\), and \( \nu\) as the frequency of emission. The units of flux are energy per time per area of the object, or \( \frac{\text{energy}}{\text{time * area}}\).

- The luminosity of the star is given as \(L_*\), and has units of \(\frac{\text{energy}}{\text{time}}\).

- The temperature of a black body as a function of wavelength, as shown in part (c), is given by: \[ T = \frac{hc}{5k\lambda_0}\]
By looking at the units for flux and luminosity, we can see that the relationship between flux and luminosity is given by: \[ F = \frac{L}{A},\] with F = flux, L = luminosity, and A = area. By assuming that the star is spherical, we can plug in the formula for the surface area of a sphere, SA = \(4πr^2\), for the area of the object: \[ A = \frac{L}{F} = 4πr^2,\] with r as the radius of the sphere. From this, we can solve for the radius of the object: \[r^2 = \frac{L}{4πF}\] \[r = (\frac{L}{4πF})^{1/2}\] Now, we can plug in the values known for this star to get the radius of the star, \(R_*\):
\[R_* = (\frac{L_*}{4πF_*})^{1/2}\] \[R_* = (\frac{L_*}{4π\sigma T^4})^{1/2}\] \[R_* = (\frac{L_*}{4π\sigma (\frac{hc}{5k\lambda_0})^4})^{1/2}\]

Blog Post 3, Worksheet 1.2: Drawing Our Galaxy

Side view of the Milky Way Galaxy: 

Top view of the Milky Way Galaxy:

The Milky Way is a barred spiral galaxy, with four main arms spiraling outwards from a center bar structure. These arms are formed of several components, including stars, gas, and dust.

a) Location of the Sun

The Sun (and therefore the Earth as well) is located approximately 8 kpc from the galactic center on what is known as the Orion Arm, a smaller section of the galaxy located between two of the main spiral arms (the Carina-Sagittarius Arm and the Perseus Arm, not labeled).

b) Thin disk

This is a relatively thin disk structure surrounding the center of the galaxy that contains stars as well as gas and dust. The stars within the thin disk make up the majority of stars within the galaxy, and are relatively young compared to stars in the thick disk.

c) Thick disk

A much larger disk of stars surrounding the center of the galaxy containing significantly older stars than the thin disk.

d) Bulge 

This “bulge” at the center of the Milky Way is tightly packed with stars, gas, and dust, and is roughly 3 kpc across.

e) Halo

The halo of the Milky Way contains sparsely distributed old stars and globular clusters (see below). The halo is thought to be as wide as 40 kpc across. 

f) Globular clusters

Globular clusters are dense clusters of old stars located within the galaxy’s halo. They are much older and denser than open clusters, which are located within the disk(s).

g) Small Magellanic Cloud (SMC)

The Small Magellanic Cloud, or SMC, is a dwarf galaxy with a central bar structure similar to the Milky Way, but with an irregular shape. It is located approximately 61 kpc from the Milky Way. The SMC and LMC (see below) are thought to possibly be orbiting the Milky Way due to the gravitational attraction.  The SMC is much smaller than the Milky Way, with a mass of approximately 7 billion \( M_{\odot}\) (as compared to the Milky Way’s 10¹² \( M_{\odot}\)), and has a diameter of around 2.1 kpc (as compared to the Milky Way’s diameter of roughtly 30.5 kpc).

h) Large Magellanic Cloud (LMC)

The Large Magellanic Cloud, or LMC, is also an irregular dwarf galaxy approximately 50 kpc from the Milky Way that possibly orbits the Milky Way due to the gravitational attraction between the two systems. The LMC is much smaller than the Milky Way, with only around a hundredth of the Milky Way’s mass and a diameter of around 4.3 kpc  

i) Sgr A*

Sagittarius A* is a small and very bright source of radiation that is believed to be a supermassive black hole at the very center of the Milky Way. When viewed from telescopes on Earth, it appears as part of the Sagittarius constellation, and is surrounded by a spiral structure known as Sagittarius A West.

j) Location of Orion star forming region

The Orion Nebula is a region in which stars are formed by a gravitational collapse of clouds containing clouds of gas and dust. This star forming region is located approximately 412 pc from the Earth, and has a mass of approximately 2000 \( M_{\odot}\).

k) Nearest known open cluster to the Sun

Open clusters are groups of relatively young stars that formed from the same star-forming cloud, and due to their proximity are still loosely bound by gravity. They are generally found in the galactic disk(s), as opposed to globular clusters, which are found in the halo. The nearest known open cluster to the Sun is known as Hyades, at a distance of approximately 47 pc from the Sun.

l) Furthest known open cluster to the Sun

The furthest known open cluster from the Sun is known as Berkeley 29, at a distance of about 15 kpc away from the Sun. This is roughly twice the distance from the Sun to the galactic center.

m) Scale length

The scale length of a galaxy is the length over which the number density of stars drops by a factor of e, or approximately 2.7. The Milky Way Galaxy has a scale length of 3.5 kpc. The distance from the galactic center to the Sun is approximately twice the scale length of the galaxy.

n) Scale height

The scale height of a galaxy is the perpendicular height over which the density of stars drops by a factor of e, or approximately 2.7. The Milky Way has a scale height of approximately 330 pc.