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## Main Question or Discussion Point

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.

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

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.

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 (

[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?

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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