Figure 1: three bodies orbiting each other

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1 PHAS2443 Practical Mathematics II Orbiting Bodies: The Dance of the Stars Introduction When one thinks of planetary motion a similar image of our solar system comes up where one massive body is being orbited by many smaller bodies. In a case like the one described an assumption would be that the smaller masses do not interact with each other making the model simpler and a solution can be found easily. When we start thinking about different scenarios, like a system of bodies all of similar mass, then the assumption made fails and the model remains complex and a solution is hard to attain. Orbital patterns that arise out of this system tend to be more exotic then the simple model described before. Figure 1: three bodies orbiting each other This investigation will look at n-body systems of similar masses and learn what kind of orbits happen in these systems and how to attain the best solution to the problem. Investigations of the solutions to the n-body orbits will use notations and coding that used on mathematica. Thus enabling the user to follow my method and check the solutions themselves. The reader will find derivations of equations and code working in my project notes. Orbits: To know what is happening in the system of planets you need to know what forces are acting within the system, also you need the positions and velocity of all the planets that affect each other. Using Newton s laws you can deduce the forces of the system. Each planet feels the sum of all the forces that is acting on it... m r = m m rij rji 3 (eq.1) i j ( r ).( ) 2 ij rji rij r ji i ij i j

2 Initial conditions r[0] = { x [0], y [0], z [0]]} = Initial Positions i i i i ri[0] = { xi[0], y [0], zi[0]} = Initial Velocity { m } = masses i G = Gravitational Constant i Eq.1 is a generalized equation for the force of a planet, of mass m i, feels. This equation by itself will not tell you very much about system because for a solution to become attainable you will need to know, starting positions of the planets, the initial velocity, the masses of each planet and the gravitational constant. r is a vector which lie in the (x, y, z) planes also it is a function of time, making time a parameter for the solutions. Solving the equation analytically is found to be impossible thus we can use mathematica to solve this equation numerically using the function NDSolve. I m assuming that all the particles move in the same plane and all constants like masses are all unity because this will make the equations a lot easier to solve and mathematica will work quicker in finding a solution. This will become important when dealing with high number of particles. Using identical masses means that the system becomes more understandable and easier to control. Figure 2 Figure 3 2

3 In both cases they have the same initial conditions of position and velocity but figure 1 has constants {m1=100, m2=1, m3=13, Gg=10} while all constants in figure 2 are in unity. A more extreme example would be figure 4, where the planets are not in a stable orbit and just looks like a mess. (See notes for details of figures). Figure 4 Errors: When using mathematica to numerical solve the equations there will be a slight error in the results. Although this is an unnoticeable error at first but after an amount of time these effects can become disastrous and may look like nothing from your original plot. An example is to see how this error affects the solution of a simple orbit of 2 particles. I have chosen my initial conditions as follows; r10 = {1, 0, 0} r20 = {-1, 0, 0} r1v = {0, 0.5, 0} r2v = {0, -0.5, 0} Where r10 and r20 are my initial positions, r1v and r2v are my initial velocities. These initial conditions were chosen so the masses have symmetry around the origin making it easier to solve. Solutions; 3

4 p1 = ParametricPlot[Evaluate[{x1[t], y1[t]} /. asol], {t, 0, 4[Pi]}] Figure 5 p3 = ParametricPlot[{Sin[t\2], Cos[t\2]}, {t, 0, 4 [Pi]}] Figure 6 Figure 5 shows the trajectory of mass one, while figure 6 shows the analytic solution to the problem. What you can see is that both masses travel on the same circle, also looking at both plots you would say they are pretty much identical. When the amount of time to pass is increased for both plots, p3 = ParametricPlot[{Sin[t\2], Cos[t\2]}, {t, 0, 300[Pi]}]

5 p1 = ParametricPlot[Evaluate[{x1[t], y1[t]} /. asol], {t, 0, 300[Pi]}] What happened above is that the analytic solution (p3) has maintained its accuracy (Would be surprising if it didn t) but the numerical solution has shifted. This is because the numerical solution can only hold a certain amount of precision for each number that it finds. You can watch how the error evolves by taking away the two graphs; if the numerical solution is 100% accurate then the difference should be zero. What figure 7 (below) shows, is the trajectory of the error on the masses. Meaning that at a point in time the error on the position of a mass is equal to distance of the point on figure 7 to the origin. From figure 7 you can see that after 100 units of time that the error is of the order of (10^-5). This of only a simple system, as soon as you increase the number of masses the error will increase with it. When dealing with high a number of masses you will need to increase the accuracy of the solutions or the solutions will only be acceptable for sort periods of time. er = Flatten[Evaluate[Thread[{x1[t], y1[t]}-n[{cos[t\2], Sin[t\2]}]] /. asol]] ParametricPlot[Flatten[Evaluate[Thread[{x1[t], y1[t]} - N[{Cos[t\2], Sin[t\2]}]] /. asol]], {t, 0, 100}] Figure 7 By using the option WorkingPrecision, you can set the amount of digits that mathematica can hold. When working out solutions mathematica can only solve in a limited amount of time, the amount of times that can be solved depends on the number of calculations that are being dealt with. The max amount of time that the solution will run for will decrease if the amount of masses that you use increase. Also if you increase the working precision that mathematica uses then this also dramatically decreases the max time. If number of masses is high you can always decrease the working precision but this will increase the error on the plot. A right balance is required to find a suitable solution. For instance for the solution above 5

6 This is the solution for the 2 particle system described above. As you can see the max time is up to but at that point the errors would be bigger, as shown in figure 8. In this case the error is now of order 10^ If the working precision is increased, Figure 8 As you can see the max time has decreased dramatically and the amount of time taken to solve has increased dramatically. Using NDSolve on mathematica is not the only way in finding a solution of eq1; other algorithmic methods can be used to solve this equation, like the Runge-Kutta method (Method which integrates, numerically, differential equations using steps at the midpoint of an interval). 6

7 Solutions: There are many different types of solutions out there and this will only be covering a few of them to see generally how planets behave. We have already seen the 2 body problem solved above, when we increase the number of bodies to 3 (keeping the masses identical), a variety of solutions come out depending on the initial conditions. If we use simple initial conditions; r10 = {0, 1, 0} r20 = {-1, 0, 0} r30 = {1, 0, 0} r1v = {0.5, 0, 0} r2v = {0, 0.5, 0} r3v = {0, -0.5, 0} Since there are 3 planets there will be 6 forces since every planet interacts with n-1 planets. This means that for 4 planets there will be 12 forces and for 5 there will be 20 forces. You only need half the amount of these forces as the other half re just the negatives of the other forces. Eq2 are the forces that one planet feels with another. Eq2 Eq3 Eq3 are the equations that need to be solved with initial conditions above to find a plot. 7

8 Orbit 3 will find the solutions of the positions of each planet with a parameter of time, in the form of an interpolating function. Above are the solutions, to find the speed at any point in time all that needs to be done is to differentiate the equations respect to time. The above function will plot the results. Figure 9 Figure 9 shows chaotic behavior in the beginning but after some time one planet flies off (blue line) and the other 2 planets (red, black) form a stable orbit together traveling in a chain. Finding the velocities is just as easy as shown below, 8

9 Figure 10 Figure 10 shows that the velocities agree with figure 9 where the circular portions of the figure show that the planets form a stable orbit near the end, while the blue line will hardly ever be in a stable orbit (check notes for animations). Consider 1 mass now not moving in the centre of the other 2 masses. The initial conditions I chose for this are below, r1v = {0, 0, 0} r2v = {0, 0.5, 0} r3v = {0, -0.5, 0} r10 = {0, 0, 0} r20 = {-1, 0, 0} r30 = {1, 0, 0} 9

10 Following the same method as before in calculating the solution, Figure 11 Figure 12 Figure 11 shows a very stable orbit but you can see the errors occur again and when both planets go near the centre you can see some spikes in the path. Figure 12 shows the max time step available and you can see there is a bigger error as the 2 orbiting planets come nearer to the 3 rd. This could be because as one planet comes nearer to the other two it will have a stronger interaction therefore will need a higher numerical precision which mathematica initially doesn t provide. 10

11 Figure 13 Figure 13 shows the solution with a higher working precision then before, as you can see this has gotten rid of the inaccuracy nearer the centre. The faster the initial speeds and you will find that that the particle paths will become more circular and overlap more, make them to fast and they will break out of orbit. If the speeds slow down the elliptical paths they follow will become more stretched and thinner. To find a system where all 3 planets form a circular orbit with one which is locked in the centre. You can take inspiration from the previous example because as the masses have faster initial velocities the shape becomes more circular and they overlap and pass. So finding this overlap a module function can be devised to print out all the plots with different velocities until one overlaps. Using this function you can make a do loop and solve graphically. As you can see from figure 14 they masses move as described above. 11

12 Figure 14 Another case of interest is that if 2 masses are in a stable orbit then the third mass passes through. Use initial conditions defined below; r10 = {1, 0, 0} r20 = {-1, 0, 0} r30 = {0, 100, 0} r1v = {0, 0.5, 0} r2v = {0, -0.5, 0} r3v = {0, -10, 0} In this case a 3 rd particle is rushing in at a much faster rate then the other to. Solving for these initial conditions you will get; Figure 15 12

13 Figure 15 shows the 3 rd mass unhindered by the other 2 and goes straight through while the 2 left break away forming another stable of a chain. Solving for a slower speed might change the outcome by having the 3 rd planet more time to interact with the more stable pair. Using the new conditions below will show a new solution (figure 16). r10 = {1, 0, 0} r20 = {-1, 0, 0} r30 = {0, 50, 0} r1v = {0, 0.5, 0} r2v = {0, -0.5, 0} r3v = {0, -5, 0} Figure 16 Figure 16 shows that now the 3 rd planet interacts more with the other 2 destroying the stability. If we have even more slower; r10 = {1, 0, 0} r20 = {-1, 0, 0} r30 = {0, 25, 0} r1v = {0, 0.5, 0} r2v = {0, -0.5, 0} r3v = {0, -1, 0} 13

14 Then solutions will be; Figure 17 What you can see from figure 17 is that the 3 rd planet comes smashing in it seems to take the place of a stable planet and forms what I had in figure 15. Conclusion: What you can tell from all these diagrams that there are many possible solutions since there is a large amount of combinations of initial values, most of them being of no interest, you have to pick and chose what you think is important. This is just a few of the possible solutions. I have only dealt with 3 bodies mainly, when looking at more and more masses you will see that you begin to have more elaborate and exotic orbits. Mathematica can generalize the method I used to make a function which can deal with any amount of bodies. Solutions can vary from having a very stable orbit to a mess. There are many solutions that seem very chaotic but after some time there will be brief moment of order or they will all break apart leaving smaller clumps of planets which will be stable (figure 9). The problem with having no analytic solution to these problems is that you only get a vague idea of what they do, which is right in the short term but you can never be to sure after longer periods of time if the solution is right. To find any of these exotic orbits in space will be challenge because either a stable orbit will not last long for large groups or you need very specific initial conditions which the smallest disturbance will ruin the stability. Finding accurate solutions would be very useful in the field of astronomy since in the centre of our galaxy many stars are close together forming intricate orbits. Solving for such scales is impossible since the large number of planets will large amounts of computing power, exceeding the most powerful computers. 14