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
To find as a function of , we compute
We make the change of variable , which gives
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
being taken between and . We obtain
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
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
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
The right-hand side is the result that we would obtain for a constant gravitational field .
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.