Editing SSC/Dynamics/FreeFall (section)

====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,
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<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,
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<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,
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<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.