The big news for this week is that NASA has discovered running water on Mars. As the liquid water flows across Mars, it absorbs chemicals from Mars's atmosphere, which in turn lower the water's freezing temperature so that it can stay liquid. Understanding the absorbance spectrum of these dissolved chemicals is how NASA was able to determine that the water was there and that it was liquid. As cool as that is, the question that's on everyone's mind is whether this means there could be life on Mars.
This past summer, I read a book called Alone in the Universe, by John Gribbin (not to be confused with the song of the same name). Gribbin asks the same question, but on a larger scale -- are we the only intelligent life in the Galaxy? Before I read the book, I probably would have said no. The Galaxy is a big place, and since we're here, it means that intelligent life isn't impossible. Statistically speaking, then, it should follow that with sheer quantity of stars, some other intelligence should have evolved.
The story that Gribbin presents, however, is much more complicated. He talks about the Drake equation, which I had heard of before, but he investigates factors that I hadn't ever considered before. (Side note: I read the book a while ago, so unfortunately a lot of this is from memory.) For example, he talked about a habitable region in the Galaxy, not only determined by the density of stars, but also by their metallicities. As we've learned so far in Astro 17, the Milky Way has a greater number density of stars towards the center of the Galaxy than it does closer to the outskirts, so this factor alone cuts down pretty heavily on the sheer numbers I had been counting on initially. Among other factors, he also goes on to talk about the set-up of our solar system; I believe he brought up that our planets are spaced out just right and that having a gas giant like Jupiter positioned properly may be vital to protecting rockier planets from constant bombardment from meteors.
He also goes on to talk about Earth's geology (less interesting to me) and the processes that life itself on Earth had to go through to get to the point where I'd be sitting on a futon in the wee hours of the morning writing an increasingly-existential blog post. Gribbin argues (although I don't think it was initially his idea) that there are certain "checkpoints" that life reaches in its evolution, but that at least one of them acts as a bottleneck, restricting the development of intelligent life. The problem is we don't know if we've already gotten lucky and that that bottleneck is behind us, or whether it'll be basically impossible for us to make it to the next stage of our evolution. (Cue suspenseful music.)
Anyway, since this is starting to move away from astronomy and towards biology (I'm a biologist at heart, sorry), I'll wrap up. Basically, Alone in the Universe presented a lot of interesting ideas about the search for extraterrestrial intelligence that I hadn't considered before. Although Gribbin didn't extrapolate for other galaxies, I'm no longer quite so sure what to think about our own Galaxy. I guess being alone in the Galaxy is pretty incredible in that we've made it this far, but definitely sobering to think about just how alone we are. Either way, I think it's safe to bet that the next update on the Mars situation probably won't look like this.
Tuesday, September 29, 2015
Monday, September 28, 2015
Blog #12: William Herschel
William Herschel was born November 15, 1738 in Hanover, although he eventually moved to England and did most of his work there. Over his long career, he catalogued over 800 binary star systems and over 2400 nebulae, discovered a planet and four moons in the solar system, and discovered infrared radiation, as well as making his own telescopes (including one that was the longest in the world for around 50 years) and composing over 20 symphonies.
His work with binary star systems is particularly interesting. Many at the time believed that the perceived motion was due to the Earth's rotation around the Sun, rather than the actual motion of the stars that were being observed. Herschel, however, hypothesized that the stars in binary systems did move, orbiting each other because of their mutual gravitational attraction. He later confirmed his hypothesis and continued to discover binary and multiple star systems.
During Herschel's time, probably the only binary systems that could be detected were visual binaries -- systems in which the stars are far enough apart to be visually distinguished. However, pairs of stars that are closer together can sometimes be detected by measuring the Doppler effect on the system as each star moves closer or farther from our point of observation. These are called spectroscopic binaries. A third type of binary star system is the eclipsing binary. If the plane of the stars' orbit is properly aligned with the observer such that the stars eclipse each other as they orbit, this will be observable and will indicate that the stars form a system.
He worked closely with his sister Caroline, who acted as his secretary, recording his descriptions of the objects he saw in the night sky. They must have made a pretty good team, because they tracked the relative positions of pairs of stars for 25 years carefully enough that they were able to determine that the stars were orbiting each other. In fact, the first recorded orbit of a pair of stars actually wasn't completed until 5 years after Herschel died.
For these and his other discoveries, Herschel became quite well known as an astronomer. After discovering a planet and trying to name it after King George III, Herschel was knighted (although I'm curious as to whether he got that honor before or after the name "Uranus" was picked instead). He died at 83, but his son John continued his work and eventually became a prominent astronomer in his own right.
Sources:
 |
| Herschel's 40-foot telescope |
His work with binary star systems is particularly interesting. Many at the time believed that the perceived motion was due to the Earth's rotation around the Sun, rather than the actual motion of the stars that were being observed. Herschel, however, hypothesized that the stars in binary systems did move, orbiting each other because of their mutual gravitational attraction. He later confirmed his hypothesis and continued to discover binary and multiple star systems.
During Herschel's time, probably the only binary systems that could be detected were visual binaries -- systems in which the stars are far enough apart to be visually distinguished. However, pairs of stars that are closer together can sometimes be detected by measuring the Doppler effect on the system as each star moves closer or farther from our point of observation. These are called spectroscopic binaries. A third type of binary star system is the eclipsing binary. If the plane of the stars' orbit is properly aligned with the observer such that the stars eclipse each other as they orbit, this will be observable and will indicate that the stars form a system.
| Binary star system |
He worked closely with his sister Caroline, who acted as his secretary, recording his descriptions of the objects he saw in the night sky. They must have made a pretty good team, because they tracked the relative positions of pairs of stars for 25 years carefully enough that they were able to determine that the stars were orbiting each other. In fact, the first recorded orbit of a pair of stars actually wasn't completed until 5 years after Herschel died.
For these and his other discoveries, Herschel became quite well known as an astronomer. After discovering a planet and trying to name it after King George III, Herschel was knighted (although I'm curious as to whether he got that honor before or after the name "Uranus" was picked instead). He died at 83, but his son John continued his work and eventually became a prominent astronomer in his own right.
 |
| "That's Sir William Herschel to you!" |
Sources:
http://listverse.com/2011/09/08/top-10-most-important-astronomers/
https://en.wikipedia.org/wiki/William_Herschel
https://en.wikipedia.org/wiki/40-foot_telescope
http://www.space.com/17432-william-herschel.html
https://en.wikipedia.org/wiki/Binary_star
https://upload.wikimedia.org/wikipedia/commons/3/36/William_Herschel01.jpg
https://upload.wikimedia.org/wikipedia/commons/6/62/Herschel_40_foot.jpg
http://www.dailygalaxy.com/.a/6a00d8341bf7f753ef01bb08126160970d-pi
Blog #11: WS4.1, #5
5. Let's think about the time dependence of magnification of a lensing event by considering the lens and the source stars as they move through the Galaxy. The stars have a non-zero relative, projected speed of \(v_{rel}\), which appears as a relative angular motion \(\mu_{rel}\) on the sky. A typical \(\mu_{rel}\) might be 10.5 mas/yr. As a result of this relative proper motion, the projected angular distance between the objects, u, will change with time. We'll call the projected distance of closest approach \(u_0\). It is analogous to an impact parameter.
a) Give an equation for u(t), the instantaneous projected separation as a function of time, in terms of \(u_0, t_0\), and anything else relevant. \(t_0\) is the time at which the closest projected approach occurs, so \(u(t_0) = u_0\). Recall that u(t) is measured in units of the Einstein radius.
a) Give an equation for u(t), the instantaneous projected separation as a function of time, in terms of \(u_0, t_0\), and anything else relevant. \(t_0\) is the time at which the closest projected approach occurs, so \(u(t_0) = u_0\). Recall that u(t) is measured in units of the Einstein radius.
The source star is moving along the arrow labeled \(\mu_{rel}\), and how far it has moved on that line is \(\mu_{rel}\) multiplied by a ∆t. We know that at \(t_0\), the source star is at its minimum distance \(u_0\), so we can use that time point to calibrate the time point we use. By using the right triangle, we can find the value of u(t). \(\theta_E\) is used to normalize \(\mu_{rel}\).
\([u(t)]^2 = [u_0]^2+[\frac{\mu_{rel}(t-t_0)}{\theta_E}]^2\)
\(u(t) = [u_0^2+(\frac{\mu_{rel}(t-t_0)}{\theta_E})^2]^{1/2}\)
b) What might be a good timescale (\(t_E\) in this problem? Rewrite part a in terms of this timescale. Express this timescale algebraically, involving 'typical' values for \(M_L, \pi_{rel}\), and \(\mu_{rel}\). Find a typical \(t_E\) we might encounter for microlensing events in the Milky Way.
A good timescale will simplify the expression and make it easier to tell that the units cancel out, so \(t_E = \frac{\theta_E}{\mu_{rel}}\)
\(u(t) = [u_0^2+(\frac{t-t_0}{t_E})^2]^{1/2}\)
\(t_E = \frac{2}{c\mu_{rel}}(GM_L\pi_{rel})^{1/2}\)
\(t_E = \frac{2[GM_{L(typ)}\pi_{rel(typ)}]^{1/2}}{c\mu_{rel(typ)}}(\frac{\mu_{rel(typ)}}{\mu_{rel}})(\frac{M_L}{M_{L(rel)}})^{1/2}(\frac{\pi_L}{\pi_{L(rel)}})^{1/2}\)
\(\theta_E = (\frac{4GM_L}{c^2}\pi_{rel})^{1/2} = (\frac{M}{10^{11.09}M_{\odot}})^{1/2}(\frac{\pi_{rel}^{-1}}{Gpc})^{-1/2} arcsec\) (with help)
If the typical values are:
\(M_L = 0.3M_{\odot}\)
\(\pi_{rel} = \frac{D_S-D_L}{D_SD_L} = \frac{8kpc-4kpc}{(4kpc)(8kpc)} = 0.125/kpc\)
\(\mu_{rel} = 10.5 mas/yr \rightarrow 10.5*10^{-3}arcsec/yr\)
Then \(\theta_E = (\frac{0.3M_{\odot}}{10^{11.09}M_{\odot}})^{1/2}(\frac{8*10^{-6}Gpc}{Gpc})^{-1/2} arcsec = 5.52*10^{-4}arcsec\)
\(t_E = \frac{\theta_E}{\mu_{rel}} = \frac{5.52*10^{-4}arcsec}{10.5*10^{-3}arcsec/yr} = 0.0525 yr \rightarrow 19 days\)
c) Trace out a microlensing curve. How does changing the the lens mass and \(u_0\) affect the shape of the light curve?
A is the magnification of the source, and t is the time. The peak occurs when the source is as close to being directly in line with the source as possible. As the mass of the lens increases, the Einstein radius increases (since \(\theta_E = (\frac{4GM_L}{c^2}\pi_{rel})^{1/2}\)), so the time that it takes for the source to pass across the Einstein radius increases, and the peak widens. By increasing the \(u_0\), the source will be farther out of line with the lens, and will therefore not be amplified as effectively; thus the peak will fall.
Thursday, September 24, 2015
Blog #10: WS4.1, #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 of "mass" \(m_{\gamma}\) passing near an object of mass \(M_L\); we'll call this object a "lens" (the '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.
To do this, we can consider two different ways of calculating force and set them equal to each other, then solve for the acceleration.
\(F = m_{\gamma}a = \frac{GM_Lm_{\gamma}}{b^2}\)
\(a = \frac{GM_L}{b^2}\)
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.
\(v_{photon} = c\)
\(c = \frac{2b}{∆t}\) (speed = displacement/time)
\(∆t = \frac{2b}{c}\)
c) Solve for the change in velocity, ∆v, in the direction perpendicular to the original photon path over this time of interaction.
We can assume that the photon's gravitational acceleration is constant as it moves past the lens.
\(∆v = ∆ta = \frac{2b}{c}\frac{GM_L}{b^2} = \frac{2GM_L}{cb}\)
d) Solve for the deflection angle (\(\alpha\)) in terms of G, \(M_L\), b, and c using the answers from parts a, b, and c.
\(tan(\alpha) = \frac{∆v}{c}\)
\(tan(\alpha) = \frac{2GM_L}{c^2b}\)
Because of the small angle approximation, \(tan(\alpha) = \alpha\)
\(\alpha = \frac{2GM_L}{c^2b}\)
This result is a factor of 2 smaller than the correct, relativistic result, so \(\alpha = \frac{4GM_L}{c^2b}\)
Monday, September 21, 2015
Blog #9: The View from a Marble
So this post isn't really related to galactic astronomy, but timing, as they say, is everything, and my mom just happened to send me this video a couple hours ago (one of the few times waiting until the last minute has actually helped me). Also shout out to my mom for her impeccable timing and her ability to find awesome videos.
The basic idea of the video is that these guys decided it would be sweet to make a to-scale model of the solar system, since if you search "solar system" in Google Images you'll get super artistic but scientifically bogus depictions of the solar system with all the planets practically touching each other:
The basic idea of the video is that these guys decided it would be sweet to make a to-scale model of the solar system, since if you search "solar system" in Google Images you'll get super artistic but scientifically bogus depictions of the solar system with all the planets practically touching each other:
(This one, on the other hand, is 100% accurate.) As a result, it's really hard for anyone to visualize the solar system in any other way, and most people probably don't even realize that these depictions are so flawed. Even the distance between the Earth and the Moon usually isn't portrayed accurately, and this is what the guys in the video start with.
After demonstrating the scale distance between the Earth and the Moon using a marble to model the Earth and thus reigniting everyone's appreciation for the Moon landings, these dudes decide to take it to the next level and model the entire solar system based on Earth-as-marble. If they had gone off of my Google Image results of the solar system's scale, they probably could've done the whole thing in my dorm room...but to do it to the actual scale, they need seven whole miles of open space. Based on our humble abode being the size of a marble. Seven miles.
So they go out to the middle of nowhere and drive around a bunch to create these to-scale orbits. One of the moments that stood out to me was around 2:50, when they're setting up Mars. It's this tiny marble 269m away from the center, and the narrator says, "We've got a couple of robots rolling around on that one." Again, it renewed my sense of awe at our space program and the fact that we were able to launch multiple probes that far, to land on that small of a target. 'Murica.
Another moment that really got to me was around 4:08 in the video, where they show the orbit of Neptune, and they have to zoom out drastically before you can see any of the other orbits. I know this whole post has basically been me saying, "WOW the solar system is so big" but it is really really big. I'm just bummed that they didn't include Pluto -- and I'm sure I'm not the only one -- but if they had, they would have needed an extra 4-5 miles to include Pluto's entire orbit.
Anyway, their finished product is really beautiful. Although I understand why the insane scale is often neglected in depictions of the solar system, it's really important to be reminded how tiny we are once in a while. I think it helps us to appreciate what we have already done as well as how much is left to be accomplished -- plus it is definitely humbling. After all, if all we have is the view from a marble, it's incredible to imagine what else could be out there.
Blog #8: WS3.1, #7
7. The mass of the Milky Way, measured using stars and probes of the gas in the Galaxy, yield a Milky Way mass of around \(10^{10}-10^{11}M_\odot\). The puzzle of the apparent mass discrepancy with the actual mass has confounded scientists since the 1930s with no clear resolution. They dubbed this mysterious and invisible mass 'dark matter.' Many theories attempt to explain the nature of dark matter. A notable one was the MACHO -- MAssive Compact Halo Object. The idea is that the invisible mass of the Galaxy is entirely due to a great number of non self-luminous individual astrophysical bodies such as brown dwarfs (tiny, failed stars), white dwarfs (the core of a star left over after the star dies), very low-mass red dwarfs (small, faint stars), or even black holes! All of these objects are very hard to detect by conventional techniques. In this problem, we'll test the validity of the MACHO theory of dark matter.
a) If MACHOs are composed entirely of brown dwarfs, and a typical brown dwarf has mass 0.05 \(M_\odot\), what must the number density of brown dwarfs in the Galaxy be to account for the mass deficit between the stellar and total mass of the Galaxy? Assume the MACHOs are distributed uniformly in a sphere with radius 10kpc.
Comparing the stellar mass of the Galaxy with the total mass of the Galaxy
\(M_{stellar} = 10^{10}M_\odot\)
\(M_{tot} = 3*10^{12}M_\odot\)
shows a deficit of \(2.99*10^{12}M_\odot\), which needs to be made up with brown dwarfs.
\(\frac{2.99*10^{12}M_\odot}{\frac{4}{3} \pi (10kpc)^{3}} \frac{BD}{0.05M_\odot} = 1.43*10^{10}BD/kpc^3 \rightarrow 14BD/pc^3\)
b) What does the number density imply about the distance to the nearest MACHO?
Equating a volume of a sphere unknown radius to the volume that should contain one brown dwarf: \(\frac{4}{3}\pi r^3 = \frac{1}{14}pc^3\)
\(r = (\frac{3}{56\pi})^{1/3} = 0.38pc\) is the implied distance to the nearest MACHO.
0.38pc is a much smaller distance than our distance to the nearest brown dwarf in reality, Luhman 16, which is 2pc away. Therefore, this theory is probably incorrect.
c) If MACHOs were made of nachos, how many nachos would be required to populate the galaxy for dark matter to be a solved problem? What would be our distance to the nearest nacho?
We'll assume an average nacho mass of 10g, and we need to use these nachos make up the deficit of \(2.99*10^{12}M_\odot\) or \(2.99*10^{45}g\).
\(\frac{2.99*10^{45}g}{10g} = 2.99*10^{44}\) nachos would be needed
\(\frac{2.99*10^{44}nachos}{\frac{4}{3}\pi (10kpc)^3} = 7.14*10^{40}nachos/kpc^3 \rightarrow 7.14*10^{31}nachos/pc^3\)
\(\frac{4}{3}\pi r^3 = \frac{1}{7.14*10^{31}}pc^3\)
\(r=(\frac{1}{7.14*10^{31}}\frac{3}{4\pi})^{1/3} = 1.5*10^{-11}pc\) would be the distance to the nearest nacho...close, but not nearly close enough for Chris Pratt.
via GIPHY
a) If MACHOs are composed entirely of brown dwarfs, and a typical brown dwarf has mass 0.05 \(M_\odot\), what must the number density of brown dwarfs in the Galaxy be to account for the mass deficit between the stellar and total mass of the Galaxy? Assume the MACHOs are distributed uniformly in a sphere with radius 10kpc.
Comparing the stellar mass of the Galaxy with the total mass of the Galaxy
\(M_{stellar} = 10^{10}M_\odot\)
\(M_{tot} = 3*10^{12}M_\odot\)
shows a deficit of \(2.99*10^{12}M_\odot\), which needs to be made up with brown dwarfs.
\(\frac{2.99*10^{12}M_\odot}{\frac{4}{3} \pi (10kpc)^{3}} \frac{BD}{0.05M_\odot} = 1.43*10^{10}BD/kpc^3 \rightarrow 14BD/pc^3\)
b) What does the number density imply about the distance to the nearest MACHO?
Equating a volume of a sphere unknown radius to the volume that should contain one brown dwarf: \(\frac{4}{3}\pi r^3 = \frac{1}{14}pc^3\)
\(r = (\frac{3}{56\pi})^{1/3} = 0.38pc\) is the implied distance to the nearest MACHO.
0.38pc is a much smaller distance than our distance to the nearest brown dwarf in reality, Luhman 16, which is 2pc away. Therefore, this theory is probably incorrect.
c) If MACHOs were made of nachos, how many nachos would be required to populate the galaxy for dark matter to be a solved problem? What would be our distance to the nearest nacho?
We'll assume an average nacho mass of 10g, and we need to use these nachos make up the deficit of \(2.99*10^{12}M_\odot\) or \(2.99*10^{45}g\).
\(\frac{2.99*10^{45}g}{10g} = 2.99*10^{44}\) nachos would be needed
\(\frac{2.99*10^{44}nachos}{\frac{4}{3}\pi (10kpc)^3} = 7.14*10^{40}nachos/kpc^3 \rightarrow 7.14*10^{31}nachos/pc^3\)
\(\frac{4}{3}\pi r^3 = \frac{1}{7.14*10^{31}}pc^3\)
\(r=(\frac{1}{7.14*10^{31}}\frac{3}{4\pi})^{1/3} = 1.5*10^{-11}pc\) would be the distance to the nearest nacho...close, but not nearly close enough for Chris Pratt.
via GIPHY
Blog #7: WS3.1, #6
6. Now consider a spherically symmetric galaxy with a density profile:
\[\rho (r) = \frac{C}{4\pi r^2(r+a)^2}\] where a and C are constants.
a) What is M(<r) for this galaxy in terms of C and a?
The mass would be the integral of the density function to the appropriate radius, so \(M(<r) = \int_{0}^{r} \rho (r')dV\)
\(dV = 4\pi r'^2 dr'\) because you're integrating over a spherical volume.
\(M(<r) = \int_{0}^{r} \frac{C}{4\pi r'^2(r'+a)^2}4\pi r'^2dr'\)
\(M(<r) = \int_{0}^{r} \frac{C}{(r'+a)^2}dr'\)
\(M(<r) = \int_{0}^{r} C(r'+a)^{-2}\)
\(M(<r) = \frac{-C}{r'+a}]_{0}^{r}\)
After plugging in the upper and lower limits for r', his simplifies to \(\frac{Cr}{a(r+a)}\).
b) The total mass of the system is the limit of M(<r) as \(\rightarrow \infty\); calculate this total mass.
\(\lim_{x\rightarrow\infty}\frac{Cr}{a(r+a)}\)
As r approaches infinity, it will become much larger than a, so (r+a) can therefore be approximated as simply r; this allows simplification to the solution \(\frac{C}{a}\).
c) If we write the total mass as \(M_{tot}\), rewrite M(<r) and \(\rho (r)\) to eliminate C in favor of \(M_{tot}\).
The total mass includes all mass regardless of distance from the center of the galaxy, so it should be the mass as r approaches infinity (the answer to part b): \(M_{tot} = \frac{C}{a}\).
\(C = aM_{tot}\)
\(\rho (r) = \frac{aM_{tot}}{4\pi r^2(r+a)^2}\)
d) What is the rotation curve for the galaxy? The circular velocity should go to a constant as \(r\rightarrow 0\), i.e. at radii r<<a, this galaxy appears to have a flat rotation curve. If we write this constant velocity as \(v_0\), write M(<r) and \(\rho (r)\) in terms of \(v_0\), a, and r. What is \(M_{tot}\) in terms of \(v_0\) and a?
Rotation curve: \(v(r) = (\frac{GM(<r)}{r})^{1/2}\) (from problem 3c)
Constant velocity: \(v_0 = (\frac{GM_{tot}}{a})^{1/2}\)
\(M_{tot} = \frac{v_0^2a}{G}\)
Since mass and radius scale linearly, we can use the ratio \(\frac{M(<r)}{M_{tot}} = \frac{r}{r+a}\)
\(M(<r) = \frac{r}{r+a}M_{tot}\)
\(M(<r) = \frac{r}{r+a}\frac{v_0^2a}{G} = \frac{v_0^2ra}{G(r+a)}\)
\(\rho(r) = \frac{a^2v_0^2}{4\pi r^2G(r+a)^2}\) based on the equation from part c.
e) If we consider this as a model of the Milky Way, what is a in kpc if \(M_{tot} = 10^{12}M_\odot\) and \(v_0\) = 240km/s?
\(v_0 = 240km/s \rightarrow 240*10^3km/s\)
\(M_{tot} = 10^{12}*10^{33}g = 10^{45}g \rightarrow 10^{42}kg\)
\(G = 6.67*10^{-11}m^3/kgs^2\)
\(240*10^3m/s = (\frac{(6.67*10^{-11}m^3/kgs^2)(10^{42}kg)}{a})^{1/2}\)
\(a = 1.16*10^{21}m \rightarrow 37.7kpc\)
\[\rho (r) = \frac{C}{4\pi r^2(r+a)^2}\] where a and C are constants.
a) What is M(<r) for this galaxy in terms of C and a?
The mass would be the integral of the density function to the appropriate radius, so \(M(<r) = \int_{0}^{r} \rho (r')dV\)
\(dV = 4\pi r'^2 dr'\) because you're integrating over a spherical volume.
\(M(<r) = \int_{0}^{r} \frac{C}{4\pi r'^2(r'+a)^2}4\pi r'^2dr'\)
\(M(<r) = \int_{0}^{r} \frac{C}{(r'+a)^2}dr'\)
\(M(<r) = \int_{0}^{r} C(r'+a)^{-2}\)
\(M(<r) = \frac{-C}{r'+a}]_{0}^{r}\)
After plugging in the upper and lower limits for r', his simplifies to \(\frac{Cr}{a(r+a)}\).
b) The total mass of the system is the limit of M(<r) as \(\rightarrow \infty\); calculate this total mass.
\(\lim_{x\rightarrow\infty}\frac{Cr}{a(r+a)}\)
As r approaches infinity, it will become much larger than a, so (r+a) can therefore be approximated as simply r; this allows simplification to the solution \(\frac{C}{a}\).
c) If we write the total mass as \(M_{tot}\), rewrite M(<r) and \(\rho (r)\) to eliminate C in favor of \(M_{tot}\).
The total mass includes all mass regardless of distance from the center of the galaxy, so it should be the mass as r approaches infinity (the answer to part b): \(M_{tot} = \frac{C}{a}\).
\(C = aM_{tot}\)
\(\rho (r) = \frac{aM_{tot}}{4\pi r^2(r+a)^2}\)
d) What is the rotation curve for the galaxy? The circular velocity should go to a constant as \(r\rightarrow 0\), i.e. at radii r<<a, this galaxy appears to have a flat rotation curve. If we write this constant velocity as \(v_0\), write M(<r) and \(\rho (r)\) in terms of \(v_0\), a, and r. What is \(M_{tot}\) in terms of \(v_0\) and a?
Rotation curve: \(v(r) = (\frac{GM(<r)}{r})^{1/2}\) (from problem 3c)
Constant velocity: \(v_0 = (\frac{GM_{tot}}{a})^{1/2}\)
\(M_{tot} = \frac{v_0^2a}{G}\)
Since mass and radius scale linearly, we can use the ratio \(\frac{M(<r)}{M_{tot}} = \frac{r}{r+a}\)
\(M(<r) = \frac{r}{r+a}M_{tot}\)
\(M(<r) = \frac{r}{r+a}\frac{v_0^2a}{G} = \frac{v_0^2ra}{G(r+a)}\)
\(\rho(r) = \frac{a^2v_0^2}{4\pi r^2G(r+a)^2}\) based on the equation from part c.
e) If we consider this as a model of the Milky Way, what is a in kpc if \(M_{tot} = 10^{12}M_\odot\) and \(v_0\) = 240km/s?
\(v_0 = 240km/s \rightarrow 240*10^3km/s\)
\(M_{tot} = 10^{12}*10^{33}g = 10^{45}g \rightarrow 10^{42}kg\)
\(G = 6.67*10^{-11}m^3/kgs^2\)
\(240*10^3m/s = (\frac{(6.67*10^{-11}m^3/kgs^2)(10^{42}kg)}{a})^{1/2}\)
\(a = 1.16*10^{21}m \rightarrow 37.7kpc\)
Blog #6: WS3.1, #3
3. We will approximate the shape of our Galaxy as 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 can be expressed in terms of the orbital frequency \(\Omega \equiv 2\pi /P\) (i.e. orbits/time) and the distance from the center \(GM_{tot} = r^3\Omega ^2\).
Kepler's 3rd states \(P^2 = \frac{4\pi ^2a^3}{GM_{tot}}\) where P is the orbital period, a is the semimajor axis, \(M_{tot}\) is the sum of the masses in the system, and G is the gravitational constant.
\(a \equiv r\) (distance from the object to the galactic center)
\(P = \frac{2\pi}{\Omega}\)
\((\frac{2\pi}{\Omega})^2 = \frac{4\pi ^2r^3}{GM_{tot}}\)
\(GM_{tot} = r^3\Omega ^2\)
b) Assume that the Milky Way has a spherical mass distribution -- this is a good approximation when talking about the total mass distribution. Rewrite the above for an object orbiting a radius r from the center of the galaxy.
Since the mass inside the orbit of the object can be approximated as a point mass at the center of the galaxy, the equation remains essentially the same, but with the mass and radius adjusted for the orbiting object.
\(r^3\Omega ^2 = GM_{enc}\) where \(M_{enc}\) is the mass enclosed within the object's orbit.
c) Next, let's call the velocity of this 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 \(M_{enc}\)), in terms of \(M_{enc}\), G, and r. This is known as the Keplerian rotation curve. It describes the motion of the planets in the solar system, since the Sun has nearly all of the mass.
\(\Omega = \frac{v(r)}{r}\) (orbital velocity is the ratio between linear velocity and radius)
\(r^3(\frac{v(r)}{r})^2 = GM_{enc}\)
\(rv^2(r) = GM_{enc}\)
\(v^2(r) = \frac{GM_{enc}}{r}\)
\(v(r) = (\frac{GM_{enc}}{r})^{1/2}\)
a) Show that Kepler's 3rd can be expressed in terms of the orbital frequency \(\Omega \equiv 2\pi /P\) (i.e. orbits/time) and the distance from the center \(GM_{tot} = r^3\Omega ^2\).
Kepler's 3rd states \(P^2 = \frac{4\pi ^2a^3}{GM_{tot}}\) where P is the orbital period, a is the semimajor axis, \(M_{tot}\) is the sum of the masses in the system, and G is the gravitational constant.
\(a \equiv r\) (distance from the object to the galactic center)
\(P = \frac{2\pi}{\Omega}\)
\((\frac{2\pi}{\Omega})^2 = \frac{4\pi ^2r^3}{GM_{tot}}\)
\(GM_{tot} = r^3\Omega ^2\)
b) Assume that the Milky Way has a spherical mass distribution -- this is a good approximation when talking about the total mass distribution. Rewrite the above for an object orbiting a radius r from the center of the galaxy.
Since the mass inside the orbit of the object can be approximated as a point mass at the center of the galaxy, the equation remains essentially the same, but with the mass and radius adjusted for the orbiting object.
\(r^3\Omega ^2 = GM_{enc}\) where \(M_{enc}\) is the mass enclosed within the object's orbit.
c) Next, let's call the velocity of this 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 \(M_{enc}\)), in terms of \(M_{enc}\), G, and r. This is known as the Keplerian rotation curve. It describes the motion of the planets in the solar system, since the Sun has nearly all of the mass.
\(\Omega = \frac{v(r)}{r}\) (orbital velocity is the ratio between linear velocity and radius)
\(r^3(\frac{v(r)}{r})^2 = GM_{enc}\)
\(rv^2(r) = GM_{enc}\)
\(v^2(r) = \frac{GM_{enc}}{r}\)
\(v(r) = (\frac{GM_{enc}}{r})^{1/2}\)
Monday, September 14, 2015
Blog #5: WS2.1, #3
3. You observe a star and measure its flux to be \(F_\bigstar\). If the luminosity of the star is \(L_\bigstar\),
a) Give an expression for how far away the star is.
\(F_\bigstar = \frac{L_\bigstar}{4\pi r^2}\)
\(r^2 = \frac{L_\bigstar}{4\pi F_\bigstar}\)
\(r=(\frac{L_\bigstar}{4\pi F_\bigstar})^{1/2}\)
b) What is its parallax?
\(\theta = \frac{L}{D}\)
\(\theta = \frac{1AU}{(\frac{L_\bigstar}{4\pi F_\bigstar})^{1/2}}=1AU(\frac{4\pi F_\bigstar}{L_\bigstar})^{1/2}\)
c) If the peak wavelength of its emission is at \(\lambda_0\), what is the star's temperature?
\(\lambda_0 = \frac{b}{T}\)
\(T = \frac{b}{\lambda_0}\)
\(b = 2.9*10^{-3}mK\) (Wien's displacement constant)
d) What is the star's radius, \(R_\bigstar\)?
\(L_\bigstar = \sigma AT^4\)
\(A = \frac{L_\bigstar}{\sigma T^4} = 4\pi R_\bigstar ^2\)
\(R_\bigstar = (\frac{L_\bigstar}{4\pi \sigma T^4})^{1/2}\)
a) Give an expression for how far away the star is.
\(F_\bigstar = \frac{L_\bigstar}{4\pi r^2}\)
\(r^2 = \frac{L_\bigstar}{4\pi F_\bigstar}\)
\(r=(\frac{L_\bigstar}{4\pi F_\bigstar})^{1/2}\)
b) What is its parallax?
\(\theta = \frac{L}{D}\)
\(\theta = \frac{1AU}{(\frac{L_\bigstar}{4\pi F_\bigstar})^{1/2}}=1AU(\frac{4\pi F_\bigstar}{L_\bigstar})^{1/2}\)
c) If the peak wavelength of its emission is at \(\lambda_0\), what is the star's temperature?
\(\lambda_0 = \frac{b}{T}\)
\(T = \frac{b}{\lambda_0}\)
\(b = 2.9*10^{-3}mK\) (Wien's displacement constant)
d) What is the star's radius, \(R_\bigstar\)?
\(L_\bigstar = \sigma AT^4\)
\(A = \frac{L_\bigstar}{\sigma T^4} = 4\pi R_\bigstar ^2\)
\(R_\bigstar = (\frac{L_\bigstar}{4\pi \sigma T^4})^{1/2}\)
Blog #4: WS2.1, #2
2. You are floating in space with Sandra Bullock. It's dark and there is a barely discernible 1W lightbulb 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?
L = \(10^7\) erg/s (luminosity)
a = \(\pi 0.25^2 cm^2 = 0.196cm^2\) (area of your pupil)
\(\lambda\) = 500*\(10^{-7}\)cm (wavelength in the visible light portion of the spectrum)
\(t_{refresh}\) = 0.1s (time it takes for your eyes to absorb the photons)
R = ? cm
\(E_{photon} = \frac{hc}{\lambda}\)
\(F = \frac{10E_{photon}}{at_{ref}} = \frac{L}{A}\)
\(\frac{10hc}{at_{ref}\lambda} = \frac{L}{4\pi R^2}\)
\(R = (\frac{Lat_{ref}\lambda}{40\pi hc})^{1/2}\)
\(R = (\frac{(10^7 erg/s)(0.196cm^2)(0.1s)(500*10^{-7}cm}{(40\pi )(6.6*10^{-27} erg*s)(3*10^{10} cm/s)})^{1/2}\)
R = 2.0*\(10^7\)cm
L = \(10^7\) erg/s (luminosity)
a = \(\pi 0.25^2 cm^2 = 0.196cm^2\) (area of your pupil)
\(\lambda\) = 500*\(10^{-7}\)cm (wavelength in the visible light portion of the spectrum)
\(t_{refresh}\) = 0.1s (time it takes for your eyes to absorb the photons)
R = ? cm
\(E_{photon} = \frac{hc}{\lambda}\)
\(F = \frac{10E_{photon}}{at_{ref}} = \frac{L}{A}\)
\(\frac{10hc}{at_{ref}\lambda} = \frac{L}{4\pi R^2}\)
\(R = (\frac{Lat_{ref}\lambda}{40\pi hc})^{1/2}\)
\(R = (\frac{(10^7 erg/s)(0.196cm^2)(0.1s)(500*10^{-7}cm}{(40\pi )(6.6*10^{-27} erg*s)(3*10^{10} cm/s)})^{1/2}\)
R = 2.0*\(10^7\)cm
Sunday, September 13, 2015
Blog #3: WS1.2, #1
1. Create an illustration of the Milky Way galaxy as viewed from the side and from above. Include the following:
a) Location of the Sun
b) Thin/thick disks, bulge, halo
c) Globular clusters
d) The Small Magellanic Cloud and the Large Magellanic Cloud
e) Sgr A* (black hole)
f) Location of Orion star forming region, and the nearest and furthest known open clusters to the Sun
g) Scale length and scale height
Nearest open cluster: Hyades (153ly)
Furthest open cluster: Berkeley 29 (15,000ly)
Sources:
http://galaxymap.org/drupal/node/171
a) Location of the Sun
b) Thin/thick disks, bulge, halo
c) Globular clusters
d) The Small Magellanic Cloud and the Large Magellanic Cloud
e) Sgr A* (black hole)
f) Location of Orion star forming region, and the nearest and furthest known open clusters to the Sun
g) Scale length and scale height
Nearest open cluster: Hyades (153ly)
Furthest open cluster: Berkeley 29 (15,000ly)
Sources:
http://galaxymap.org/drupal/node/171
http://www.space.com/25584-milky-way-structure-spiral-arms.html
https://en.wikipedia.org/wiki/Orion_Nebula
http://www.atlasoftheuniverse.com/milkyway.html
https://en.wikipedia.org/wiki/Milky_Way
http://www.ess.sunysb.edu/fwalter/AST101/images/mw-side.jpg
http://pages.uoregon.edu/jimbrau/BrauImNew/Chap23/6th/23_10Figure-F.jpg
https://upload.wikimedia.org/wikipedia/commons/thumb/8/8d/Milky-way-edge-on.pdf/page1-945px-Milky-way-edge-on.pdf.jpg
http://astronomy.swin.edu.au/cms/cpg15x/albums/userpics/thickdisks1+0.jpg
http://www2.astro.psu.edu/users/cpalma/astro1h/Images/FG14_09.JPG
https://en.wikipedia.org/wiki/Hyades_(star_cluster)
http://www.univie.ac.at/webda/cgi-bin/ocl_page.cgi?dirname=be029
https://en.wikipedia.org/wiki/Galactic_coordinate_system
https://www.univie.ac.at/webda/cgi-bin/ocl_page.cgi?dirname=mel025
http://www.univie.ac.at/webda/cgi-bin/ocl_page.cgi?dirname=ngc1976
http://www.poyntsource.com/Richard/ngc_5694_escaping_globular.htm
https://en.wikipedia.org/wiki/Small_Magellanic_Cloud
https://en.wikipedia.org/wiki/Large_Magellanic_Cloud
Blog #2: WS1.1, #2-4
3. Estimate the average number density of stars throughout the Milky Way, using the distribution n(r)=k*exp(-r/Rs)
a) Consider that within a 2pc radius of the Sun there are 5 stars.
b) Estimate based on the Galaxy's scale height of 330pc, and assuming an average stellar mass of half the mass of the Sun.
4. Determine the collision rate of the stars using the number density of the stars, the cross-section for a star, and the average velocity of the stars. How many stars will collide every year? Is the Sun likely to collide with another star?
Blog #1: "Hello, Universe!"
My name is Kaelyn and I'm a junior at Harvard College, studying Molecular and Cellular Biology. I'm taking Astronomy 17 because I've always found space really interesting, but have never had the opportunity to take a formal course on it until this semester. If I end up enjoying the course enough, I might pursue a secondary in Astrophysics, but my ultimate goal is to go to medical school. Thanks for checking out my blog!
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