Kepler's ODE, Central Coordinates

Let (rho, nu) and beta be as developed above. Then as we have seen,
[1] rho*cos(nu) = e + cos(beta)
[2] rho*sin(nu) = b*sin(beta)
[3] rho ~= 1 + e*cos(beta)
[4] nu-dot = K/rho2 (K a constant)
Now we wish to derive Kepler's ODE in terms of beta-dot and beta alone,
[14] beta-dot = K/b*rho = K/b*(1 + e*cos(beta))

Derivation of [14]

Note [3] gives rho as a function of beta only, so we seek an expression of beta-dot involving beta and rho, equivalent to [4] above.

To begin, we differentiate [1] and [3] above, getting,

[1'] rho-dot*cos(nu) - rho*sin(nu)*nu-dot = - sin(beta)*beta-dot
[3'] rho-dot = - e*sin(beta)*beta-dot
Now eliminate rho-dot in [1'] using [3'], getting,
[8] e*sin(beta)*beta-dot*cos(nu) + rho*sin(nu)*nu-dot = sin(beta)*beta-dot
Collecting beta terms on the left and nu on the right,
[9] (1 - e*cos(nu))*sin(beta)*beta-dot = rho*sin(nu)*nu-dot
Next we replace rho*sin(nu) on the right using [2], and cancel sin(beta) from each side, finding
[10] (1 - e*cos(nu))*beta-dot = b*nu-dot
This is a simple relation between beta-dot and nudot, but we still wish to eliminate nu on the left. So multiply both sides by rho, and again use [1], to find
[11] (rho - e*(e + cos(beta))*beta-dot = rho*b*nu-dot
and next use [3] again to replace e*cos(beta) on the left. After minor simplification, we have,
[12] (1 - e2)*beta-dot = rho*b*nu-dot
in which we again recognize, b2 = 1 - e2, so, cancelling b,
[13] b*beta-dot = rho*nu-dot
This is so simple, there must be a shorter route! In any case, recall that [3] expresses rho as a function of beta, while [4] gives nu-dot as a function of rho, and hence beta, so this is what we are after. So substituting for nu-dot on the right with [4], we have,
[14] beta-dot = K/b*rho = K/b*(1 + e*cos(beta))
using [3] once again. And we are done!
Revised 31 dec 2001 by ralph abraham