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Time for freefall in a varying gravitational field?

  1. Jan 21, 2009 #1
    I have been trying to calculate the time it takes for an object of negligible mass to free fall towards an object with a large mass, taking into account that the gravitational acceleration experienced by the small object increases as it moves closer to the big object.
    The first thing I tried was setting up a diffeq:

    n''=GM/(h-n)^2
    Where G is the gravitational constant, M is the mass of the object, h is the initial height, n is the distance traveled, and ' denotes a derivative. After trying for a few days to solve this equation for n(as a function of t, time), I gave up and tried to solve it using change in kinetic energy, which also didn't work. Any ideas? Thanks.
     
  2. jcsd
  3. Jan 21, 2009 #2

    turin

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    This problem has been discussed somewhere on this forum, but I don't remember the title of the thread. Anyway, the differential equation (not exactly yours, but basically the same idea) has been solved. I'll give you a hint:

    [tex]
    \frac{d^2x}{dt^2}=v_x\frac{dv_x}{dx}
    [/tex]

    You should prove this to yourself. (because it is actually not always true, so you should understand why it is true for your particular problem)
     
  4. Jan 22, 2009 #3
    I'm not sure I understand why that is true. Is v(x) velocity with respect to x, where x is distance traveled? Is v(x) just an arbitrary name for a function which fits the relationship?
     
  5. Jan 22, 2009 #4

    turin

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    Yes, v_x (not v(x), BTW) is the velocity component in the x direction:

    [tex]
    v_x\equiv{}\frac{dx}{dt}
    [/tex]

    You can replace x with whatever coordinate you want, but wierder coordinates will require more care in determining when this relationship is true. Are you familiar with the chain rule?
     
  6. Jan 22, 2009 #5
    sure, f(g(x))'=f'(g(x))g'(x)
    Ok, so far I am following you. The expression rewrites the second derivative of position(acceleration) as the first derivative of position(velocity) times the derivative of velocity with respect to position. Using gravitational potential energy I have figured out the equation for velocity with respect to space, and I know how to take the derivative of that. That leaves just a single order diffeq. A single order diffeq of the form a'=a*k should have a general solution of the form e^(tk), so then acceleration should equal ce^kt where k is the derivative of velocity with respect to position right?
     
  7. Jan 24, 2009 #6

    turin

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    Are you sure it has that form?
     
  8. Jan 27, 2009 #7
    Hi, Fr33Fa11!
    Your attempt to use change in kinetic energy should take into accout the change in potential energy, equal to GMm/(h - n) - GMm/h. Basing on it you can find the velocity dependence on distance: V = f(n). The time of flight can be found by integration of the expession dn/f(n).
     
    Last edited: Jan 27, 2009
  9. Feb 9, 2009 #8
    I've solved it, and I have calculated the constants that come out of the problem. Do you know if this has been done(it would be really cool if it hasn't)
     
  10. Feb 10, 2009 #9

    turin

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    I thought we already said that. Sorry, no Nobel prize today.
     
  11. Feb 10, 2009 #10
    I was expecting that something like this would have been done before. Do you have a link to the paper/post where it was solved, I want to see if they have a faster way of solving it, because my work is kind of messy, as is the answer.
     
  12. Feb 10, 2009 #11

    D H

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    This exact topic was discussed in the threads [thread=243444]Two particles and gravity; non-constant acceleration[/thread] and [thread=246833]Time for two bodies to collide under gravity[/thread].
     
  13. Feb 10, 2009 #12
    They did not arrive at a fully solved solution. I'm also trying to solve it for any number of dimensions.(Where GM/r^(n-1)) is the formula for gravitational acceleration with any number of dimensions.
     
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