I've been trying to derive the full solution to the problem of free fall when the distances considered are large (and hence, the acceleration is no constant, but a function of height). This isn't homework, I was just curious.(adsbygoogle = window.adsbygoogle || []).push({});

The problem is innocently simple, but yields a differential equation which I cannot solve (and it's in principle solvable).

Free fall in a gravitational field

Consider a large mass [itex]M[/itex], which we place at the origin of a 1D coordinate system. We then drop a test mass [itex]m[/itex] from a certain height [tex]r=h_0[/itex], from rest. The problem is to calculate its position as a function of time, [tex]r(t)[/itex]. We assume that [itex]M \gg m[/itex] so the larger mass remains fixed at the origin and does not move.

An example of this would be a high-orbit satellite free-falling onto the Earth in a radial trajectory.

My attempt at a solution

The physics of the problem is very straightforward. At a distance [itex]r[/itex] from the origin (where the larger mass is located), the local acceleration due to the gravitational force is given by

[tex]g(r) = -\frac{G M}{r^2}[/tex]

And so the (differential) equation of motion of the test mass is

[tex]\frac{d^2r}{dt^2} = -\frac{G M}{r^2}[/tex]

or, put simply,

[tex]r^2 \frac{d^2r}{dt^2} = -\mu[/tex]

where [itex]\mu \equiv GM[/itex] is the standard gravitational parameter of the system. This second-order ODE is not linear, so its solution might be complicated, but it's not impossible.

We define [itex]v \equiv dr/dt[/itex] so that [itex]d^2r/dt^2 = dv/dt[/itex], and using the chain rule we write the ODE as:

[tex]r^2 \frac{dv}{dr} \frac{dr}{dt} = -\mu[/tex]

or, since [itex]dr/dt = v[/itex],

[tex]v r^2 \frac{dv}{dr} = -\mu[/tex]

In this form, the ODE is separable:

[tex]v dv = -\frac{\mu}{r^2} dr [/tex]

Integrating, we get

[tex]\frac{v^2}{2} = \frac{\mu}{r} + C [/tex]

where [itex]C[/itex] is an integration constant. Since we assume the object is dropped from rest from a height [itex]h_0[/itex], we can set [itex]v(r=h_0)=0[/itex] to obtain [tex]C[/itex]:

[tex]0 = \frac{\mu}{h_0} + C [/tex]

and hence

[tex]C = -\frac{\mu}{h_0}[/tex]

The solution of the ODE for the velocity is therefore (edit: I just found out that this can be obtained very easily by an energy consideration):

[tex]\frac{v^2}{2} = 2\mu \left( \frac{1}{r} - \frac{1}{h_0} \right) [/tex]

which gives the velocity as a (ugly) function of height:

[tex]v(r) = -\sqrt{2\mu \left( \frac{1}{r} - \frac{1}{h_0} \right)} [/tex]

(we take the negative root because we expect the velocity to be negative as the particle falls). The problem I encounter is when I try to integrate this equation:

[tex]-\int \frac{dr}{\sqrt{2\mu \left( \frac{1}{r} - \frac{1}{h_0} \right)}} = \int dt [/tex]

The right-hand side is trivial, and we can wright the left-hand side as follows:

[tex] -\frac{1}{\sqrt{2\mu}} \int \frac{dr}{\sqrt{\frac{\displaystyle 1}{r} - A}} = t + C [/tex]

where [itex]A \equiv 1/h_0[/itex] is a positive constant.

Here's where I'm stuck. I've no clue how to solve that integral. I gave it to Maxima, and I get a pretty ugly solution. Can anyone help? is there another way of solving the original differential equation?

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# Free fall (full solution)

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