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free fall time needed to double separation


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This isn't homework, I'm reviewing physics after many years of neglect.


I thought this particular problem would be simple - it isn't. 2 particles of mass m<<M fall from rest along the same straight line directly down toward gravitational source M. The particles are initially a distance [latex]2R_i[/latex] apart and their center of mass is initially a distance [latex]r_i>>R_i[/latex] from M. Therefore, just to be clear, the upper particle, particle 2, falls from an initial height of [latex]r_i+R_i[/latex], and the lower particle, particle 1, falls from an initial height of [latex]r_i-R_i[/latex], both falling straight down along the exact same straight line toward M. How much time is needed for their separation to double, to [latex]4R_i[/latex]?


Let the unprimed frame be the inertial frame in which M is stationary at the origin.

Let the primed frame be the non-enertial frame in which the C.M. of the 2 particles is stationary at the origin. Therefore the primed frame falls with the particles. Let us call the primed frame 'the local frame' and the C.M. of the 2 particles the 'local C.M..'


Therefore the tidal force on particle 1 [latex]=-\frac{2GMmR}{r^3}[/latex] implies that the eqn. of motion of particle 1 in the local frame is [latex]\ddot{R}-\frac{2GMR}{r^3}=0[/latex](eqn.1) and the eqn. of motion of the local C.M. in the unprimed frame is [latex]\ddot{r}+\frac{GM}{r^2}=0[/latex](eqn.2).


The soln. to eqn.2 is well known, [latex]t=\frac{r_i^\frac{3}{2}}{\sqrt{2GM}}[\sqrt{\frac{r}{r_i}}\sqrt{1-\frac{r}{r_i}}+cos^{-1}\sqrt{\frac{r}{r_i}}][/latex].


Keep in mind [latex]\frac{\ddot{R}}{R}=\frac{2GM}{r^3}[/latex] and [latex]\dot{r}=-\sqrt{2GM}\sqrt{\frac{1}{r}-\frac{1}{r_i}}[/latex].


Good luck.

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I'm not really sure why you're trying to use that equation for tidal forces. It's just the leading term from a series expansion of the exact force differential between two radially separated points, and is only valid for small separations.


The time for a small mass starting at rest at coordinate [math]r_0[/math] to fall to coordinate [math]r[/math] is given by:


[math]t=\sqrt{\frac{r_0^3}{2GM}} \, \left(\sqrt{{\frac{r}{r_0}\left(1-\frac{r}{r_0}\right)}} + \tan^{-1} \sqrt{\frac{r_0}{r}-1} \,\right)[/math]


Therefore the two particles, after falling for some time, have r-coordinates which can be related to each other:


[math]\sqrt{\frac{(r_i-R)^3}{2GM}} \left(\sqrt{{\frac{r_1}{r_i-R}\left(1-\frac{r_1}{r_i-R}\right)}} + \tan^{-1} \sqrt{\frac{r_i-R}{r_1}-1} \,\right) = \sqrt{\frac{(r_i+R)^3}{2GM}} \left(\sqrt{{\frac{r_2}{r_i+R}\left(1-\frac{r_2}{r_i+R}\right)}} + \tan^{-1} \sqrt{\frac{r_i+R}{r_2}-1} \,\right)[/math]


Now, you could in principle solve this for either [math]r_1[/math] or [math]r_2[/math] given the condition [math]r_2-r_1=4R.[/math] Once you've done that you can just plug the value into the time equation. The above looks incredibly tedious, which is why I won't try it. If you're feeling brave I suppose you could try it yourself.



EDIT: Now that I read your question again, it seems you're more interested in the case where the separation between the particles is small. You could expand that massive equation above into two series and discard all higher order terms in R.

Edited by elfmotat
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Moderator Note


Dekan - please keep your comments on the topic. For your guidance - the reason for using latex is that it is the only real way to show equations of such complexity; it has nothing to do with seeking admiration - it is merely sharing information.


Let's all get back to the physics - no need to comment


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elfmotat, Thanks for the suggestion.


I did try essentially what you suggest, realized how tedious it would be, and gave up.


I only attempted the original problem because I thought it would be straightforward, lol.


I will back-burner this one till a get back to higher-level dynamics. Thanks.

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