The position of the Lagrange point L1
Let the distance from the Earth to the center of mass of the Earth-Moon system be r1, and the distance from the Earth to L1 be x. The distance between the Earth and the Moon is a. The mass of the Earth is M and of the Moon m. We assume that a body in L1 has a mass so small that the influence on the orbits of the Earth and the Moon is negligible.
The Earth, the Moon and a body in L1 are assumed to travel in circular orbits around the center of mass. The period of the moon is given by (derivation here):
In L1 the sum of the gravitational forces from the Earth and the Moon should be just sufficient to supply the required centripetal force to make a body in L1 go in a circular orbit with the same period as the Moon. The body should then continue to lie on a line connecting the Earth and the Moon. Newton's second law and gravitational law, and an expression for the centripetal acceleration, give
We divide all terms by GM and multiply all terms by the common denominator. Putting k = m/M, we get:
We know that (see the simulation "Orbit of the Moon")
It is convenient to introduce the dimensionless quantity z = x/a. We divide the equation by a5 and use the expression for r1. We then get the following equation to determine the position of L1:
By studying the graph of the lefthand side for various values of k, we discover that this fifth-order equation has one (and only one) solution in the interval 0 to 1.