Celestia Forums

I used the equations for the force due to gravity, and circular motion equations to derive the following for the distance of Lagrange 1:

(d^5) - 2a(d^4) + (a^2)(d^3) - (1-k)(a^3)(d^2) + 2(a^4)d - (a^5) = 0

Where d is the distance from the primary to Lagrange 1, a is the separation of the primary and secondary objects and k is the mass of the secondary divided by the mass of the primary.

Now this looks suspiciously similar to a binomial expansion to me, and while I am perfectly capable of using the Newton-Raphson process to find a numerical solution to the equation, I would like to know if there is an algebraic solution.

If you could outline the technique to find such a solution, then maybe I could also apply it to L2 and L3.

The solution of the quintic is kids play. I'd tell you the general solution, but I don't want to spoil the fun

There's no simple, exact algebraic method to find the position of the colinear Lagrangian equilibrium points - there seems to be a host of slightly different formulations of the equilibria, but they all need to be iterated out to find a solution.

For an approximate solution as the mass ratio goes towards zero, L1 and L2 move closer to the Hill radius, and L3 moves closer to the orbital radius of the secondary mass.

Grant