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Falling Particle From Space

  1. Dec 4, 2004 #1
    I had this question in class the other day and need some help:

    A particle falls to Earth starting from rest at a great height. Neglecting air resistance, show that the particle requires about 9/11 of the total time of fall to traverse the first half of the distance.

    Anything would be appreciated.

  2. jcsd
  3. Dec 4, 2004 #2


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    That is a very hard thing to show... because it is not correct.

    The equation of motion for a falling body, starting from height D is:

    [tex] h = D - \frac 1 2 g t^2 [/tex]

    so the time to reach the ground is

    [tex] t_0 = \sqrt { \frac { 2 D} g } [/tex]

    The time to reach the half way point is:

    [tex] \frac D 2 = D - \frac 1 2 g t^2 [/tex]


    [tex] t_{\frac D 2} = \sqrt {\frac D g }[/tex]


    [tex] \frac {t_{\frac D 2}} {t_0} = \frac 1 {\sqrt 2} = .707 [/tex]

    this is not [itex] \frac 9 {11} [/itex]
    Last edited: Dec 4, 2004
  4. Dec 4, 2004 #3


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    When the problem states "from a great height" I interpret that as meaning the inverse square behavior of gravity must be taken into account. Ultimately, this means comparing

    [tex]\int_{1/2}^{1} \frac {dr}{\sqrt{ \frac {1}{r} -1}}[/tex]


    [tex]\int_{0}^{1} \frac {dr}{\sqrt{ \frac {1}{r} -1}}[/tex]

    which works out to about 9/11.
  5. Dec 4, 2004 #4


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    That makes sense.
  6. Dec 4, 2004 #5
    Thank you both.
    I was pursuing Integral's method at first before running into the same problem. I agree with Tide's reasoning but I don't follow how you got to that integral, and even moreso how I would solve that integral. Could you help with that?
  7. Dec 4, 2004 #6


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    Energy will be conserved so we can relate speed with position:

    [tex]v = \sqrt { 2GM \left( \frac {1}{r} - \frac {1}{r_0} \right)[/tex]

    But [itex]v = \frac {dr}{dt}[/itex] so we can integrate the equation and find

    [tex]t = \frac {1}{\sqrt {2GM}} \int_{r}^{r_0} \frac {dr'}{\sqrt { \frac {1}{r'} - \frac {1}{r_0}}}[/tex]

    Use [itex]r = 0[/itex] to get the full time to fall and [itex]r = r_0/2[/itex] to get the time to fall halfway. It is helpful to use a new variable of integration given by [itex]r/r_0[/itex]. The rest of the problem reduces to getting approximations for the integrals though the integral from 0 to 1 reduces to something called a Beta Function (with a little effort you can cast it into a form that you can look up in integral tables) and the integral from 1/2 to 1 is reduces to an Incomplete Beta Function and you might have to use a numerical approxiimation to evaluate it. Some graphing calculators will do them both numerically for you.
  8. Dec 6, 2004 #7
    Ohhh, ok. Thank you Tide, much appreciated. I just missed that leap from setting up the integral to what you posted I guess. Thanks everyone.

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