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The Classical Analysis of the Two Body Problem

Classical Analysis of a 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

−Q/r²
and the potential
energy is
− Q/r

where Q is the product of the force constant and the charges of the two particles.

The angular momentum is:

L = mr²ω
and thus
ω = (L/m)/r²

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

E = ½m(dr/dt)² + ½m(rω)² −Q/r

Since rω is equal to λ/r

(dr/dt)² = 2(E+Q/r)/m − λ²/r²

Now note that

(dr/dθ) = (dr/dt)/(dθ/dt)
but (dθ/dt)=ω=λ/r² so
(dr/dθ) = (dr/dt)(r²/λ)
and hence
(dr/dθ)² = (dr/dt)²(r^{4}/λ²)

Therefore

(dr/dθ)² = (r^{4}/λ²)[2(E+Q/r)/m − λ²/r²)]
which reduces to
(dr/dθ)² = r^{4}(E+Q/r)/(mλ²) − r²

This latter formula will be used later.

Now let u=1/r so

(du/dθ) = −(1/r²)(dr/dθ)
and since (dr/dθ=(dr/dt)(r²/λ)
(du/dθ) = −(1/r²)(dr/dt)/λ

The probability density per unit path length is proportional to

1/|v| = (λ/q)/(1+2ε·cos(θ)+ε²)^{½}

This gives a probability density function with one maximum and one minimum.

(To be continued.)

Kepler's Law

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.