This is just elementary calculus, but it is harder than I thought at first glance. A friend told me that there is no formula expressing the altitude of a falling body if you is the inverse square law for the force, rather than making the approximation of a constant gravitational field. As far as I am able to tell, he is right. One can however compute some things.

We study the motion of a mass , moving along an axis, whose position is denoted by . At the time , the mass is at rest in the position . There is a mass , at the position , that does not move and attracts by a force given by the inverse square law:

being the gravitational constant.

Let us write the conservation of the mechanical energy for the mass : at any time ,

We consider a falling body, therefore , and thus

This yields

and therefore

with

To find as a function of , we compute

We have

and therefore

We make the change of variable , which gives

and

We obtain

and finally

We now integrate between and :

This gives us the duration of the fall as a function of the position of the falling mass .

We can find another formula for the antiderivative. Let us make the change of variable . Then

and

being taken between and . We obtain

and therefore

which is the answer that can be found in Wikipedia. Taking , we obtain the free-fall time , the time it takes for the falling mass to reach the center of attraction:

How does we find these changes of variable (this is the only tricky part in these calculations)? Well, if we set , the points , with in , are on the circle of radius centered at . Now if is a parametric representation of this circle, the change of variable allows us to write

In the first change of variable, we used a parametrization of the circle by rational functions:

In the second, we used a parametrization by trigonometric functions:

Both are natural choices (the second maybe more so than the first), and both produce a simple result.

If we set

we obtain

and thus

Indeed, is continuous on , differentiable on , and

Therefore is decreasing, and thus a bijection from to . Its inverse is continuous on and differentiable on .

On the other hand, if we set

we obtain

and thus

The function is continuous and differentiable on and

Its is therefore a diffeomorphism from to . Let us denote by the inverse . It satisfies the differential equation

with . In particular, . This, taken with Equation , implies that

for close to . We have approximately

whith

The right-hand side is the result that we would obtain for a constant gravitational field .

We have

Since

we have

and therefore

Therefore, when tends to ,

Using the differential equation, we also find

which is what we would obtain by formally differentiating the equivalent.

The free-fall time can be rewritten

with the accelaration of gravity at the altitude . How many time does it take to fall from the altitude of the Moon? We take to be the radius of the Earth, and is the distance form the Earth to the Moon. Wikipedia gives

kilometers, kilometers and meters per second squared. We find roughly hours, which is to say four days and twenty hours.

The Wikipedia article I linked above gives two references for the free fall:

1. From Moon-fall to motions under inverse square laws, by S. K. Foong, European Journal of Physics, 2008 29: 987–1003;

2. Radial motion of Two mutually attracting particles, by Carl E. Mungan, The Physics Teacher, 2009, 47: 502-507.

I skimmed the first article (unfortunately it is behind a paywall, but I could get access with my university subscription). The formula for and the numerical value for the free fall time from the orbit of the moon agree with what I found. There is also a series expansion, and probably other stuff. I will read in detail later.