The position of the Lagrange point L1

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Let the distance from the Earth to the center of mass of the Earth-Moon system be *r*_{1},
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 *a*^{5}
and use the expression for *r*_{1}. 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.