We began by reading Richard Montgomery's Article "A New Solution to The Three Body Problem."  Our goal was to follow the origins of the solution, and then try to replicate it in an animation with Maple VI.  What would distinguish this animation from other animations from the internet was that with a Maple Application, one could input the initial values for the position and velocities of each of the bodies involved.

Newton Said

Kepler's Adaptation of this for the 2-Body Problem

x- is the vector from the object sitting at the origin to the other body;
-k is the constant that involved the masses and the gravitational constant
gm1m2
acceration=force
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Calculus Based Physics Tell Us...

The derivative of the position function gives the velocity function, and the derivative of the velocity function gives us the acceleration function.  Thus, given F=ma for two masses, going from the acceleration to the velocity and finally finding the position function is challenging, BUT feasible.

The CHALLENGE OF THE N-BODY PROBLEM rests in the fact that the position of one mass is the consequence of the affects of the SUM of the other masses (2 or more) pulling on it.  To date, this complication has kept the N-Body problem unsolved, having no general solution.  However, some scientists have given and proven special cases of the 3 body problem, but only after given very specific initial conditions.

• planar (2-dimensional) each mass has 2 coordinates (x1, y1) for mass 1, and (x2, y2) for mass 2. (gives 4 second order equations to solve, or 8 first order equations to solve)
• 3-D- each mass has 3 coordinates: mass 1 (x1,y1,z1), mass 2 (x1, y2, z2), giving 6 second order equations to solve, or 12 first order equations to solve.

• planar (2-diminsional) each mass has 2 coordinates (x1, y1) for mass 1, and (x2, y2) for mass 2, and (x3,y3) for mass 3 (gives 6 second order equations to solve, or 12 first order equations to solve)
• 3-diminsional- each mass has 3 coordinates: mass 1 (x1,y1,z1), mass 2 (x1, y2, z2), and mass 3 (x3,y3,z3) giving 9 second order equations to solve, or 18 first order equations to solve.

And our intial conditions included an assigned value for m1,m2,m3, and the initial coordinates for the starting positions of each mass.

• 1765
• masses remain collinear throughout the rotation
• masses may vary
• one fixed mass in the middle, while mass 2 and mass 3 revolve around mass 1
• the ratios of the distances of of the masses from the middle mass remain a constant throughout the rotation
• never stable

1772

mass 1, mass 2 and mass 3 are all equal in magnitude

each mass placed on the vertices of an equilateral triangle

center of mass is at the centroid of this equilateral triangle

each mass has the same total rotational velocity

shape of the equilateral triangle is maintained

stable when one of the three masses is much greater than the other

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Montgomery's Solution

• 1993
• all masses are equal
• more complex movements of masses such that the velocities are continually fluctuating
• stable

What makes our animations different?
They are done using Maple VI such that one can enter the initial values of the position coordinates and the vertical and horizontal components of the velocities of each element.

Determining the Initial Values for the Euler and LaGrange Solutions:

First, we determined the initial positions of the masses on a unit circle.  Here are the initial coordinates of all three masses

Also, we could differentiate x1(t) to get   and  .Now, we can finally plug all this information into the summation equation  .  Note the denominator change from something extremely long and ugly to something quite simple.  We obtained this by looking at out unit circle, and determining the distance form mass 1 to mass 2 through the properties of isosceles triangles and the Pythagorean theorem.   Now we are at the point where we can just “plug and chug” by using substations for all of the x1, x2, and x3, values, as well as our x’’ value.  Finally, we get and equation in terms of noting but trigonometry functions, r, and alpha.  Through some computations, one can simplify this mess to find that alpha in terms of r such that  .  In this situation, m1=m2=m3.  Therefore, now we have all the initial conditions of the masses and the initial positions.  Now we need to apply alpha to the horizontal and vertical components of velocities for each of the three masses.  This is what we came up with:

Now that we had the initial conditions and the 9 equations from before, we could build a working program to animate this situation.

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Animations

The Awesome "BurtleBurtle" Site:  This site contains tons of awesome animations of the figure eight, and replications of planetary orbits.

UCSC Contributions: Filled with Java Application Animations of the Figure-Eight, the Lagrange, and more!

Maple Student Center Webpage

Maple Commands for Our Results:  These commands are applicable to Maple 6 software only.  This link includes commands only.  You must copy and paste them onto a Maple template in order to see the resulting animations.