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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(r4/λ²)

Therefore

(dr/dθ)² = (r4/λ²)[2(E+Q/r)/m − λ²/r²)]
which reduces to
(dr/dθ)² = r4(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

Then furthermore

(d²u/dθ²) = d/dt(du/dθ)/(dθ/dt) = (d/dt(−(dr/dt)/λ)/ω
(d²u/dθ²) = −[(d²r/dt²)/λ]/(λ/r²)
(d²u/dθ²) = −(d²r/dt²)/(u²λ²)

From the initial dynamic equation

(d²r/dt²) = −Q/(mr²) + rω² = −(Q/m)/r² + rλ²/r4
= −(Q/m)/r² + λ²/r³

The ratio Q/m occurs often enough to justify it having its own symbol, say q.

Thus

(d²r/dt²) = −qu² + λ²u³

Therefore

(d²u/dθ²) = − (d²r/dt²)/(u²λ²) = q/λ² − u

By defining w equal to (u−q/λ²) the above equation reduces to

(d²w/dθ²) + w = 0

which has the solution

w = W*cos(θ−θ0)
where W is
a constant
and hence
u = W*cos(θ−θ0) + q/λ²
and thus
r = 1/(W*cos(θ−θ0) + q/λ²)

This latter solution can be put into the form

r = (λ²/q)/(1+ε·*cos(θ−θ0))

where ε turns out to be the eccentricity of the elliptical orbit;.

By a suitable choice of the polar coordinate system; θ0 can be made equal to zero. Thus the solution for the system is

r(θ) = (λ²/q)/(1+ε·cos(θ))

The next step is to derive the velocity v, where

v² = (dr/dt)² + (rω)²

From the solution for r

(dr/dt) = [(λ²/q)ε·sin(θ)/(1+ε·cos(θ))²]ω
(dr/dt) = [(λ²/q)ε·sin(θ)/(1+ε·cos(θ))²](λ/r²)
(dr/dt) = (q/λ)ε·sin(θ)
and
(dr/dt)² = (q/λ)²ε²sin²(θ)

The tangential velocity is rω, but rω is equal to λ/r. Thus

(rω)² = λ²(q/λ²)²(1+ε·cos(θ))²
(rω)² = (q/λ)²(1+ε·cos(θ))²
(rω)² = (q/λ)²(1+2ε·cos(θ)+ε²cos²(θ))

Combining the two terms for v² gives

v² = (q/λ)²(1+2ε·cos(θ)+ε²)

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.

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