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The Classical Analysis of the Two Body Problem 

Consider two particles, one of mass m and the other of a mass so large it can be considered infinite compared to the mass of the first particle. The center of mass of the system is at the center of the massive particle. Let r be the separation distance of the two particles. Let the force between them be given as
where Q is the product of the force constant and the charges of the two particles.
The angular momentum is:
The term (L/m) is a crucial parameter and it will be denoted as λ, Thus ω=λ/r².
The total energy E of the system is
Since rω is equal to λ/r
Now note that
Therefore
This latter formula will be used later.
Now let u=1/r so
Then furthermore
From the initial dynamic equation
The ratio Q/m occurs often enough to justify it having its own symbol, say q.
Thus
Therefore
By defining w equal to (u−q/λ²) the above equation reduces to
which has the solution
This latter solution can be put into the form
where ε turns out to be the eccentricity of the elliptical orbit;.
By a suitable choice of the polar coordinate system; θ_{0} can be made equal to zero. Thus the solution for the system is
The next step is to derive the velocity v, where
From the solution for r
The tangential velocity is rω, but rω is equal to λ/r. Thus
Combining the two terms for v² gives
The probability density per unit path length is proportional to
This gives a probability density function with one maximum and one minimum.
(To be continued.)
The law of Kepler that the radial line from the Sun to a planet sweeps out a constant amount of area per unit time is easily derived. The area swept out in a unit of time is as shown below.
The area of the triangle is ½r(rω) which is equal to ½λ, a constant.
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