Chicago State University
REU program-Summer 2003
Univ. of Wisconsin - Madison
Madison, WI 53706
Research projects of other REU students
Useful links
My conclusions
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Jeremy Harrison Chicago State University
REU program-Summer 2003 |
Research projects of other REU students Useful links My conclusions |
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In the galaxy, every object has there own velocity of 220 km/s for circular orbits, even gas clouds orbiting the milky way galaxy, (as observed from the center of the galaxy.) However, from our perspective things can be moving at an angle and thus, we can be moving towards some things and away from some things. It's called the red shift! Things moving away from us is shifted to the red, light, specturem and things moving towards us is shifted to the blue, light, specturem. However, these high velocity gass clouds are moving towards us, to galatic center either because they are in a decaying orbit or because they experience a drag force. Prehaps, both!
The bad news is that we can't observe this directly. To do so, would take about a million years or more. For you nonscience guys out there this cloud that we're talking
about is about ... lets say about 30,000 light years (10KPC), now a light year is the distance that light travels in one year is 1629360000 miles (94608000000000 meters)
because of the speed of light is 186,000 miles/h ( 3*10^7 meters/second.) and so the total distance would be 48880800000000 miles. Our space vehicles would take about a trillion-billion years or more to go that far out!
The other thing about these gas clouds orbiting the galaxy is that the galaxy has different gravitational fields at different coordinates of (X, Y, Z). On the earth we live in a constant gravitational field of 9.8 meters/second squared (32 feet/second squared), unless of course we go up as far the space shuttle which has a gravitational acceleration less then 32feet/second squared. You see the equation for calculating the force of gravity is (GMeMo/(R + h)^2), where G is the universal gravitional constant, Mo is Mass of the object, Me is the Mass of the earth, R is the radius of the earth and h is the height above the earth. Now when "h" is small we can leave the h out. Even if your on a big mountain, we can still say, "You're in a gravitiation field of 32 feet/second^2, but when "h" get's so high that you are above the earth then we say that your in a different gravitional field of the earth.
Life would be so much better if things were that easy! On earth, just like in the Galaxy, you have a drag force that complicates things. The drag force is a little bit more complex because it is different for slow moving objects then for fast moving objects! "How so you ask?!" Well, for slow moving object the drag force is directly porpotional to the velocity and for fast moving objects, we have that the drag force is directly porpotional to the velocity squared. In Classical Mechanics, I mainly solved such types of problems anayltically, but such an equation is a 'grasp' NON-LINEAR differential equation. At least when the velocity is squared. My professor back home once told me that nobody really nows how to do non-linear differential equations so what we scientist do is linearize such equations. For example, the pendulum has a theta dependence (which I'll call 'A'), because the differential equation for the pendulum is actually ( d^2A/dt^2 = sinA ), but when A is very small we can just as easily say that the equation is ( d^2A/dt^2 = A ), and making life easier!
Okay so what do we do if we don't want to linearize the pendulum, Well we can go through a very tricky method of changing variables. Using some tricks with trig functiions and tricks with calculus, which is going to take me forever to explain, but is very interesting! I suggest reading "Symon," (it's a tough book, but it has a lot of information in it) or we can use numerical methods. Numerical Methods is the easy way to do differential equations for linear and nonlinear systems. The method we choose was the Runge-Kutta method because it was more accurate then any other method, plus it was the simplest method too. The problem was that I never did any numerical analysis before. To help me me understand how to do numerical analysis I wrote a program about about the acceleration of things here on earth. Now to make things simpler, or as my DEQ professor would say, "Let's make life easier." ( He only said that when he was feeling genourous and he didn't feel genourous a whole lot!) I had a term 'b' represent the mass of the object, the coefficient of drag, Cd, the area and the density of the air. Now on earth the density depends on height only, but in the Galaxy the density is a function of the height and the radius of the Galaxy. Moreover, no human has ever visited the outer reaches of the Galaxy so the best we can do is rely on emperical data, from our instruments here on earth and guessing (to be honest). I solved for the terminal velocity anayltically to be the squareroot of 'g' if 'b' was equal to one! Now I had to write the numerical methods program that would show that. It took me about 3 - 4 days of trail and error to finally get it right! When I got an answer, on the computer, a good scientist will always come back to the problem at hand to check the answer he/she got. Why? Because we are after the truth, whatever that might be! or so I hear! : )
Well, I finally had the numerical methods technique down. Now to write the program. ( dV/dt = g - bV^2) remember that the gravitiational field 'g' is in all three directions in the Galaxy and is different and so is the density. The other equation is ( dx/dt = V ) and these two equation is suppose to represent vectors in all three directions. Just one of these equations actually contain three equations, therefore we actually have six equations to worry about. One for dv/dt, and the other for dx/dt in all three directions. Moreover, unlike for the drag problem on earth I wasn't expecting the numbers to converge. I had to look for something different instead of the numbers converging to one value. The first thing we thought up is the energies. This is a conservative system without drag. So the energy should be constant. Also, momentum is conserved. Our initial thoughts were that the magnitude of the angular momentum should be conserved. That wasn't true but it was actually the Z component of the angular momentum that was conserved. This is because our clouds are cylindrical and the Z component is cylindrically symmetrical. The velocityy we noticed oscilated nicely, so you could represent it by the cosine function. Now with drag turned on we notice that the Z component of the angular momentum decreassed but at some point it leveled out to some value. This surprised us because we thought that the angular momentum should continue to decrease but it didn't. Moreover, over time, with drag the gas clouds became a perfect circle. The theory to explain that is ... Imagine you're in the water moving at some speed and then you enter a whirlpool big enough to trap ships or something like that to pull you down or whatever, at some point you will move at the speed of the whirlpool no matter how hard you try to over come it. The same is true for the gas clouds, it is submegged in a fluid of a circular velocity and it has no choice but to move with that fluid over time.
Here is a graph of how the Magellanic Stream (MS) moves with time. The dots represent 250Myrs. The MS is a cross between an equitorial orbit and a polar orbit.
there is a drag force on this cloud. You can tell because the velocity levels out to some constant non-zero value.
Here are two equitorial orbits. The simplest cases. The first one is without drag and the second one is with standard drag. For a purely equitorial orbit the Z velocity should be zero, but the program keeps
missing up if we didn't give Z velocity some non zero value.
The only position we have is in the Y direction and therefore all the velocity is in the X direction The numbers are the initial conditions.
Vx = V*(y/R).
One problem with the graphs is that we found out that with our program, we had to use very small timesteps, because our answer changed slightly with timesteps. For example, with timestep=.1, we got a nice looking orbit and then with a timestep=.01, we got a slightly different orbit. It's somewhat to be expected since we are using numerical methods to calculate the orbits, Luckily it doesn't change too much. Otherwise, that would indicate something was obvious wrong with our program!
Now that I showed you the equitorial orbits I should also show the polar orbits, again here is a case without drag and a case with drag.
