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Free Fall Timescale

  1. Oct 31, 2009 #1
    I've seen this equation in a book once but I can't seem to find a good explanation of how it was derived.


    [tex]\ t_{ff}=\sqrt{\frac{3\pi}{32g\overline{\rho}}}[/tex]
     
    Last edited: Oct 31, 2009
  2. jcsd
  3. Nov 1, 2009 #2

    Doc Al

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    See: http://en.wikipedia.org/wiki/Free-fall_time" [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Nov 1, 2009 #3
    In the book I have it derives it from the acceleration due to gravity equations. There is only one part about it I dont get though.

    [tex]-\frac{GM}{r^{2}} [/tex] Then it says to do this.

    [tex]\int^{R}_{t=0} -\frac{GM}{r^{2}}dr [/tex]

    The boundary conditions are r=R and when t=0, I don't understand what to do for the definite integral. I can integrate but I don't know what to plug in. The question you're suppose to get is different than the one I originally posted.

    [tex]\frac{1}{2}\sqrt{\frac{R^{3}}{GM}} [/tex]
     
  5. Nov 1, 2009 #4

    Doc Al

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    It's a bit of a calculus problem. You are trying to find the time it takes for the outer shell to collapse under the influence of the gravity of the mass below it. You have the acceleration of gravity (thus d2r/dt2) as a function of r:
    Given that, you need to solve for the time it takes to go from r = R to r = 0, starting from rest.
     
  6. Nov 1, 2009 #5
    [tex]\int^{R}_{0} -\frac{GM}{r^{2}}dr [/tex]

    Like this? If I integrate this I get stumped at this,

    [tex] -GM\left[-\frac{1}{r} \right]\right|^{R}_{0}[/tex]

    What do I plug in for r?
     
  7. Nov 1, 2009 #6

    Doc Al

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    No, you need to solve this equation:

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

    One way to solve it is in stages, starting with the trick of using a = dv/dt = v dv/dr.
     
  8. Nov 1, 2009 #7
    Sorry I have to ask but, how do I solve it?
     
  9. Nov 1, 2009 #8

    Doc Al

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    It's a bit of a pain. I'll start you off:

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

    First solve that for v in terms of r. Then use v = dr/dt, and solve that to get the time.
     
  10. Nov 1, 2009 #9
    I've got that

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

    [tex]v=\frac{\sqrt{GM}}{r}[/tex]

    [tex]\frac{dv}{dr}=-\frac{\sqrt{GM}}{r}[/tex]

    [tex]\frac{dr}{dt}=\frac{\sqrt{GM}}{r}[/tex]

    How do I get time?
     
  11. Nov 2, 2009 #10
    Did I do something wrong?
     
  12. Nov 2, 2009 #11

    Doc Al

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    How did you go from the first equation to the second? Please explain each step so I can see what you're trying to do.
     
  13. Nov 2, 2009 #12
    [tex]\frac{dv}{dt}=-\frac{GM}{r^{2}}[/tex]

    [tex]\int -\frac{GM}{r^{2}}dr [/tex]

    [tex]=v=\frac{\sqrt{GM}}{r}[/tex]

    [tex]\frac{dv}{dr}=-\frac{GM}{r^{2}}\cdot\frac{r}{\sqrt{GM}}[/tex]

    [tex]\frac{dv}{dr}=-\frac{\sqrt{GM}}{r^{2}}[/tex]
     
  14. Nov 3, 2009 #13

    Doc Al

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    Sorry, but I'm still struggling to understand what you are doing here.
    This is the starting point. OK.

    Not sure what you mean here. Are you taking the integral of both sides with respect to dr? (Where's the equation?)

    How did you get here?
     
  15. Nov 3, 2009 #14
    I took the integral of [tex]\frac{dv}{dt}[/tex] shouldn't that give me v? Maybe that's were I'm confused. [tex]\int-\frac{GM}{r^{2}}dr=\frac{GM}{r}[/tex].

    It looks like [tex]\frac{dr}{dr}=GMLog(r)[/tex] to get that I did:

    [tex]\int -\frac{GM}{r^{2}}dr=\frac{GM}{r}[/tex]

    [tex]\int\frac{GM}{r}dr=GMLog(r)[/tex]

    So [tex]v=\frac{GM}{r}[/tex],
     
  16. Nov 3, 2009 #15

    Doc Al

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    If you were integrating with respect to t that would be true, but you're integrating with respect to r.

    [tex]\frac{dv}{dt} dr = v dv [/tex]

    See post #8.
     
  17. Nov 3, 2009 #16
    So you're saying that I need to integrate with respect to r? How do I do that?

    [tex]-\frac{GM}{r^{2}}\cdot GMlog(r)[/tex]

    Is that right? Does [tex]dv=GMlog(r)[/tex]?
     
  18. Nov 3, 2009 #17

    Doc Al

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    From an earlier post:

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

    Integrate both sides with respect to dr:

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

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

    (This is just the beginning. This is a several step calculus problem.)
     
  19. Nov 3, 2009 #18
    I will try to solve it but now I'm completely confused as to what I'm trying to do. Could you restate the problem?
     
  20. Nov 3, 2009 #19

    Redbelly98

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    The question is: How long does it take the radius to collapse from a value of R to a value of zero?

    So, ultimately you are looking for a time as the answer.
     
  21. Nov 4, 2009 #20

    Doc Al

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    To add to Redbelly98's statement, take a look at post #4.

    Think of it like this: Imagine you are standing on a planet that has just collapsed, so nothing is holding you up. Assume that all the mass has collapsed to a point. How long would it take for you to fall (you're in free fall, thus the name) from your initial position on the surface at r = R to the center of the collapsed planet at r = 0?
     
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