Editing SSC/Dynamics/FreeFall (section)

===Single Particle in a Point-Mass Potential===

Suppose we examine the free-fall of a single (massless) particle, located a distance <math>~|\vec{r}|</math> from an immovable point-like object of mass, <math>~M</math>.  The particle will feel a distance-dependent acceleration due to a gradient in the gravitational potential of the form,
<div align="center">
<math>~\frac{d\Phi}{dr} = \frac{GM}{r^2} \, ,</math>
</div>
and the Euler equation, as just derived, serves to describe the particle's governing equation of motion, namely,
<div align="center">
<math>~\ddot{r} =  - \frac{GM}{r^2} \, ,</math>
</div>
where we have used dots to denote differentiation with respect to time (see also equation 1 from LMS65, [[#Free-Fall_Collapse|reprinted above]]).  If we multiply this equation through by <math>~2\dot{r} = 2dr/dt</math>, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2\dot{r} \frac{d\dot{r}}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{2GM}{r^2} \cdot \frac{dr}{dt} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ d(\dot{r}^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2GM \cdot d(r^{-1}) \, ,</math>
  </td>
</tr>
</table>
</div>
which integrates once to give,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\dot{r}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2GM}{r} - k \, , </math>
  </td>
</tr>
</table>
</div>
<span id="RoleOfIntegrationConstant">where, as an integration constant, <math>~k</math> is independent of time.  </span>

<table border="1" width="75%" align="center" cellpadding="10">
<tr><th align="center">
Role of Integration Constant
</th></tr>
<tr><td align="left">
Within the context of this particular physical problem, the constant, <math>~k</math>, should be used to specify the initial velocity, <math>~v_i</math>, of the particle that begins its collapse from the radial position, <math>~r_i</math>.  Specifically,
<div align="center">
<math>~k = \frac{2GM}{r_i} - v_i^2 \, .</math>
</div>
Without this explicit specification, it should nevertheless be clear that, in order to ensure that <math>~\dot{r}^2</math> is positive &#8212; and, hence, <math>~\dot{r}</math> is real &#8212; the constant must be restricted to values,
<div align="center">
<math>~k \leq \frac{2GM}{r_i}  \, .</math>
</div>

</td></tr>
</table>

Taking the square root of both sides of our derived "kinetic energy" equation, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dr}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pm \biggl[ \frac{2GM}{r} - k \biggr]^{1/2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ dt </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \pm \biggl[ \frac{2GM}{r} - k \biggr]^{-1/2} dr \, .</math>
  </td>
</tr>
</table>
</div>

As has been shown by [http://adsabs.harvard.edu/abs/1934QJMat...5...73M McCrea &amp; Milne (1934)], this function can be integrated in closed form to give an analytic prescription for <math>~t(r)</math>.  Equation (17) in the McCrea &amp; Milne (1934) paper presents the function to be integrated as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~t</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int\limits_0^\theta \frac{\theta^{1/2} d\theta}{(\alpha + A\theta)^{1/2}} \, ,</math>
  </td>
</tr>
</table>
</div>
which can be straightforwardly obtained from our expression after adopting the positive root of the "kinetic energy" equation and setting,
<div align="center">
<math>~r \rightarrow \theta \, ,</math> 
&nbsp; &nbsp; &nbsp; 
<math>~2GM \rightarrow \alpha \, ,</math> 
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
<math>~k \rightarrow A' = -A \, .</math> 
</div>
The results obtained assuming three different ranges/values for the constant, <math>~k</math>, are presented at the end of &sect;4 of [http://adsabs.harvard.edu/abs/1934QJMat...5...73M McCrea &amp; Milne (1934)] and are reprinted here in an effort to fully acknowledge this early contribution.
<div align="center" id="McCreaMilne1934Solution">
<table border="1" align="center" cellpadding="5">
<tr>
  <th align="center" colspan="1">
W. H. McCrea and E. A. Milne

[http://adsabs.harvard.edu/abs/1934QJMat...5...73M (1934) ''Quarterly Journal of Mathematics Oxford'', 5, 73]
  </th>
</tr>
<tr>
  <th align="center" colspan="1">
[[File:McCreaMilne1934Solution.png|500px|McCrea &amp; Milne (1934) Time Solution]]
  </th>
</tr>
</table>
</div>

In the following subsections, we will rederive these algebraic, <math>~t(r)</math> solutions in the context of three separate, physically  interesting scenarios, all of which involve infall, so we will adopt the velocity root having only the negative sign. 

====Falling from rest at a finite distance &hellip;====
In this case, we set <math>~v_i = 0</math> in the definition of <math>~k</math>, so,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dr}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- ~\biggl[\frac{2GM}{r} - \frac{2GM}{r_i}\biggr]^{1/2} =
\biggl(\frac{2GM}{r_i}\biggr)^{1/2} \biggl[\frac{r_i}{r}-1  \biggr]^{1/2}
\, , </math>
  </td>
</tr>
</table>
</div>

and the relevant expression to be integrated is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~dt </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl(\frac{2GM}{r_i} \biggr)^{-1/2} \biggl[ \biggl( \frac{r_i}{r} \biggr) - 1 \biggr]^{-1/2} dr  \, .</math>
  </td>
</tr>
</table>
</div>

Following [http://adsabs.harvard.edu/abs/1965ApJ...142.1431L LMS65], we see that this equation can be straightforwardly integrated by first making the substitution,
<div align="center">
<math>~\cos^2\zeta \equiv \frac{r}{r_i}  \, ,</math>
</div>
which also means,
<div align="center">
<math>~dr = - 2r_i \sin\zeta \cos\zeta d\zeta \, .</math>
</div>

The relevant integral is, therefore,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\int_0^t dt </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~+ \biggl(\frac{2r_i^3}{GM} \biggr)^{1/2} \int_0^\zeta \cos^2\zeta d\zeta \, ,</math>
  </td>
</tr>
</table>
</div>
where the limits of integration have been set to ensure that <math>~r/r_i = 1</math> at time <math>~t=0</math>.  After integration, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ t </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl(\frac{2r_i^3}{GM} \biggr)^{1/2} \biggl[ \frac{\zeta}{2} + \frac{1}{4}\sin(2\zeta) \biggr]  \, .</math>
  </td>
</tr>
</table>
</div>
The physically relevant portion of this formally periodic solution is the interval in time from when <math>~r/r_i = 1 ~ (\zeta = 0)</math> to when <math>~r/r_i \rightarrow 0</math> for the first time <math>~(\zeta = \pi/2)</math>.  The particle's free-fall comes to an end at the time associated with <math>~\zeta = \pi/2</math>, that is, at the so-called "free-fall time,"
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tau_\mathrm{ff} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ \biggl(\frac{2r_i^3}{GM} \biggr)^{1/2} \biggl[ \frac{\zeta}{2} + \frac{1}{4}\sin(2\zeta) \biggr]_{\zeta=\pi/2}  
= \biggl(\frac{\pi^2 r_i^3}{8GM} \biggr)^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>

<span id="Parametric">In summary,</span> then, the solution, <math>~r(t)</math>, to this simplified but dynamically relevant problem is provided by the following pair of analytically prescribable parametric relations (see also equations 2 &amp; 3 from LMS65, [[#Free-Fall_Collapse|reprinted above]]):

<table border="1" cellpadding="10" align="center">
<tr><th align="center">
Parametric &nbsp;<math>~r(t)</math>&nbsp; Solution
</th></tr>
<tr><td align="center">

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \frac{r}{r_i} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \cos^2\zeta </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \frac{t}{\tau_\mathrm{ff}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2}{\pi} \biggl[ \zeta + \frac{1}{2} \sin(2\zeta) \biggr] </math>
  </td>
</tr>
</table>
</td></tr>
</table>
We note, as well, that the radially directed velocity is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~v_r = \frac{dr}{dt} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl(\frac{2GM}{r_i} \biggr)^{1/2} \biggl[ \frac{1}{\cos^2\zeta}  - 1 \biggr]^{1/2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl(\frac{2GM}{r_i} \biggr)^{1/2} \tan\zeta \, , </math>
  </td>
</tr>
</table>
</div>
which formally becomes infinite in magnitude when <math>~\zeta \rightarrow \pi/2</math>, that is, when <math>~t \rightarrow \tau_\mathrm{ff}</math>.

It is worth demonstrating that the parametric solution derived here is identical to the solution published by [http://adsabs.harvard.edu/abs/1934QJMat...5...73M McCrea &amp; Milne (1934)] &#8212;  [[#McCreaMilne1934Solution|reprinted above]] &#8212; for the case, <math>~k > 0</math>, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~t</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{\theta^{1/2} (\alpha - A' \theta)^{1/2}}{A'} + \frac{\alpha}{(A')^{3/2}} \sin^{-1} 
\biggl( \frac{A' \theta}{\alpha} \biggr)^{1/2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\alpha}{(A')^{3/2}} \biggl[ - \biggl(\frac{A' \theta}{\alpha} \biggr)^{1/2} \biggl(1 - \frac{A' \theta}{\alpha} \biggr)^{1/2} + \sin^{-1} 
\biggl( \frac{A' \theta}{\alpha} \biggr)^{1/2} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
After reversing the substitutions detailed above, that is, after setting,
<div align="center">
<math>~\theta \rightarrow r \, ,</math> 
&nbsp; &nbsp; &nbsp; 
<math>~ \alpha \rightarrow 2GM \, ,</math> 
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
<math>~A' \rightarrow k   \, ,</math> 
</div>
and remembering that, for this particular model example, we have set <math>~v_i = 0 ~\Rightarrow ~ k = 2GM/r_i</math>, the key dimensionless ratio in the McCrea &amp; Milne expression becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl(\frac{A' \theta}{\alpha} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{r}{r_i} = \cos^2\zeta \, ,</math>
  </td>
</tr>
</table>
</div>
and the pre-factor on the righthand side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[\frac{\alpha^2}{(A')^3} \biggr]^{1/2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{r_i^3}{2GM} \biggr]^{1/2} = \frac{2}{\pi} \tau_\mathrm{ff} \, .</math>
  </td>
</tr>
</table>
</div>

Hence, the McCrea &amp; Milne solution becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{t}{\tau_\mathrm{ff}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{\pi} \biggl[  -\cos\zeta \sin\zeta + \sin^{-1} (\cos\zeta)
\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{\pi} \biggl[  -\frac{1}{2}\sin(2\zeta) + \biggl(\frac{\pi}{2} - \zeta \biggr)
\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 1 - \frac{t}{\tau_\mathrm{ff}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{\pi} \biggl[\zeta  +\frac{1}{2}\sin(2\zeta) \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
We see that, by shifting the defined zero-point in time in this last expression such that <math>~t \rightarrow (\tau_\mathrm{ff} - t)</math>, which also reverses the ''sign'' on time, we have exact agreement between the solution that we have derived &#8212; designed to match the one published by {{ LMS65 }} &#8212; and the result for <math>k > 0</math> that was published by McCrea &amp; Milne in 1934.

====Falling from rest at infinity &hellip;====
In this case, we set <math>~k= 0</math>, so the relevant expression to be integrated is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~dt </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl[ \frac{2GM}{r} \biggr]^{-1/2} dr = - (2GM)^{-1/2} r^{1/2} dr \, .</math>
  </td>
</tr>
</table>
</div>
Upon integration, this gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~t + C_0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -  \frac{2}{3}(2GM)^{-1/2} r^{3/2} \, ,</math>
  </td>
</tr>
</table>
</div>
where, <math>~C_0</math> is an integration constant.  In this case, it is useful to simply let <math>~t=0</math> mark the time at which <math>~r = 0</math> &#8212; hence, also, <math>~C_0 = 0</math> &#8212; so at all earlier times (<math>~t</math> intrinsically negative) we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~- t </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{2r^3}{9GM} \biggr)^{1/2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ r </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{9}{2} \cdot GMt^2 \biggr)^{1/3} \, .</math>
  </td>
</tr>
</table>
</div>


It is straightforward to demonstrate that this derived solution is identical to the solution published by [http://adsabs.harvard.edu/abs/1934QJMat...5...73M McCrea &amp; Milne (1934)] &#8212;  [[#McCreaMilne1934Solution|reprinted above]] &#8212; for the case, <math>~k = 0</math>, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~t</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2}{3} \frac{\theta^{3/2}}{\alpha^{1/2}} 
= \biggl( \frac{4\theta^3}{9\alpha} \biggr)^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>
After reversing the substitutions detailed above, that is, after setting,
<div align="center">
<math>~\theta \rightarrow r \, ,</math> 
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
<math>~ \alpha \rightarrow 2GM \, ,</math> 
</div>
the McCrea &amp; Milne solution becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~t </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{2r^3}{9GM} \biggr)^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>
Aside from reversing the ''sign'' on time, we have exact agreement between the solution that we have derived and the result for <math>~k = 0</math> that was published by McCrea &amp; Milne in 1934.


====Falling from a finite distance with an initially nonzero velocity &hellip;====
Here, we examine the case in which <math>~0 < r_i < \infty</math> and <math>~0 < v_i^2 < GM/r_i</math>, in which case, the constant <math>~k</math> is a nonzero, positive number.  The relevant expression to be integrated is,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ dt</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - k^{-1/2}\biggl[ \frac{a}{r} - 1 \biggr]^{-1/2} dr \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~ a \equiv \frac{2GM}{k} \, .</math>
</div>
Using [http://integrals.wolfram.com/index.jsp?expr=-%28a%2Fx-1%29%5E%28-1%2F2%29&random=false Wolfram Mathematica's online integrator], we find,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~- \int \biggl[ \frac{a}{r} - 1 \biggr]^{-1/2} dr</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r ( ar^{-1} -1 )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(2r-a)(ar^{-1} - 1)^{1/2}}{2(r-a)} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

Hence, we find,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~k^{1/2}(t + C_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r ( ar^{-1} -1 )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(2r-a)(ar^{-1} - 1)^{1/2}}{2(r-a)} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r ( ar^{-1} -1 )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(ar^{-1}-2)(ar^{-1} - 1)^{1/2}}{2(ar^{-1}-1)} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r ( ar^{-1} -1 )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(ar^{-1}-2)}{2(ar^{-1}-1)^{1/2}} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r k^{-1/2}( akr^{-1} -k )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(akr^{-1}-2k)}{2k^{1/2}(akr^{-1}-k)^{1/2}} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

Let's determine the constant, <math>~C_0</math>.  When <math>~t = 0</math>, we can write,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[akr^{-1} - k]_{t=0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2GM}{r_i} - \biggl[\frac{2GM}{r_i} - v_i^2  \biggr] = v_i^2
\, .</math>
  </td>
</tr>
</table>
</div>
Hence,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r_i k^{-1}v_i + \biggl(\frac{a}{2k^{1/2}} \biggr) \tan^{-1} \biggl[ \frac{(v_i^2-k)}{2k^{1/2}v_i} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
GMk^{-3/2} \biggl\{
2[\eta(1-\eta)]^{1/2} + \tan^{-1} \biggl[ \biggl( \eta - \frac{1}{2} \biggr) [ \eta(1-\eta)]^{-1/2} \biggr]
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
where, in this last expression,
<div align="center">
<math>\eta \equiv \frac{v_i^2 r_i}{2GM} \, .</math>
</div>
(This last expression needs to be checked for errors, as it has been rather hastily derived.)

====Relationship to Kepler's 3<sup>rd</sup> Law====

[[File:CommentButton02.png|right|100px|Note from J. E. Tohline:  I was first made aware of this relationship, as a graduate student, while listening to John Faulkner give a lecture on the free-fall problem to a class of undergraduates at UCSC.]]It is useful to note a relationship between [http://astro.physics.uiowa.edu/ITU/glossary/keplers-third-law/ Kepler's 3<sup>rd</sup> law] and the free-fall problem, as introduced [[#Falling_from_rest_at_a_finite_distance_.E2.80.A6|here in the context of the motion of a single (massless) particle that falls from rest]] toward a point-like object of mass, <math>~M</math>.  According to Kepler's 3<sup>rd</sup> law, when a massless particle orbits a point-like object of mass, <math>~M</math>, the particle's orbital period, <math>~P_\mathrm{orb}</math>, is related to the semi-major axis, <math>~a_\mathrm{orb}</math>, of its elliptical orbit via the algebraic expression,
<div align="center">
<math>~P^2_\mathrm{orb} = \frac{4\pi a_\mathrm{orb}^3}{GM} \, .</math>
</div>
This relation works for orbits of any eccentricity, 
<div align="center">
<math>~e \equiv \biggl[ 1 - \frac{b_\mathrm{orb}^2}{a_\mathrm{orb}^2} \biggr]^{1/2} \, ,</math>
</div>
where, <math>~b_\mathrm{orb}</math> is the semi-minor axis of the orbit, the extremes being:  <math>~e = 0 ~(b_\mathrm{orb} = a_\mathrm{orb})</math> for a circular orbit, and <math>~e = 1 ~(b_\mathrm{orb} = 0)</math> for a purely radially directed (in-fall) trajectory.  

In the context of our current discussion, it should be clear that a particle that "free-falls" from rest at an initial distance, <math>~r_i</math>, from a point mass object will follow a trajectory synonymous with a Keplerian orbit having eccentricity, <math>~e=1</math>.  The particle's initial position coincides with the apo-center of this orbit and the point mass object is located at the peri-center (as well as at one focus) of the orbit, so the semi-major axis is <math>~a_\mathrm{orb} = r_i/2</math>.  We also recognize that the particle will move from the apo-center to the peri-center of its orbit &#8212; completing its "free-fall" onto the point-mass object &#8212; in a time, <math>~\tau = P_\mathrm{orb}/2</math>.  From Kepler's 3<sup>rd</sup> law, we therefore deduce that,
<div align="center">
<math>~\tau = \frac{1}{2} \biggl[ \frac{4\pi (r_i/2)^3}{GM} \biggr]^{1/2} = \biggl( \frac{\pi r_i^3}{8GM} \biggr)^{1/2} \, ,</math>
</div>
which precisely matches the free-fall time, <math>~\tau_\mathrm{ff}</math>, derived above.