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Free fall far away from Earth (integral substitution problem)

  1. Feb 14, 2008 #1
    [SOLVED] Free fall far away from Earth (integral substitution problem)

    1. The problem statement, all variables and given/known data


    [tex]v(x) = -v_1\sqrt{\left(\frac{R}{x} - \frac{R}{h}\right)}[/tex]

    Find the time t.

    2. Relevant equations

    Listed above where [tex]v_1 , R , h[/tex] are all constant.

    3. The attempt at a solution

    [tex]v(x) = \frac{dx}{dt}[/tex]

    [tex]dt = \frac{dx}{\left[-v_1\sqrt{\left(\frac{R}{x} - \frac{R}{h}\right)}\right]}[/tex]

    [tex]t = -\frac{1}{v_1}\int\frac{dx}{\left[\sqrt{\left(\frac{R}{x} - \frac{R}{h}\right)}\right]}[/tex]

    I'm stuck on this integral but i'm convinced it's some kind of trigonometric substitution although i can't figure out which one. I've spent hours searching the internet for help and couldn't find anything so any help would be appreciated.

    Perhaps if someone could at least point out which trigonometric substitution (if any at all) i should be trying to work with i'd be extremely grateful.

    Thanks in advance.
  2. jcsd
  3. Feb 14, 2008 #2

    Shooting Star

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    R/x - R/h =R(h-x)/hx.

    The integrand becomes [1/(sqrt R)]*sqrt(hx)dx/sqrt(h-x). Put x = h*(sin^2 u).

    It is now solved very easily. In the num, there is only sin^2u*du.
  4. Feb 14, 2008 #3
    Thanks a lot that seemed to solve it for me ^_^
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