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Falling under gravity with decreasing r

  1. Jun 25, 2013 #1
    If you drop an object from a great height with no air resistance. e.g. 1000 km above the moon. How long will it take to hit the surface?

    (GM[itex]_{}1[/itex]M[itex]_{}2[/itex])/r[itex]^{}2[/itex] = M[itex]_{}2[/itex]a


    Hmm this is my first post, not quite getting the symbols.

    "The second derivative of r is equal to a constant divided by r squared"

    So presumably you integrate that nonlinear differential equation. Somehow.

    This is quite an obvious question. It seems like something Isaac Newton would have asked when he first discovered the equations of gravity. A really classic piece of physics. But I can't seem to calculate the answer or find it anywhere.
  2. jcsd
  3. Jun 26, 2013 #2


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    As code: $$\ddot{r}=\frac{-k}{r^2}$$

    Or, in ASCII: d^2/dt^2 r = -k/r^2

    The minus sign gives the usual convention that r is the (positive) radius and k is positive, too.

    With energy conservation, you can calculate the velocity for every point of your motion. The inverse velocity is "time per length" - if you integrate that over the length of the motion, you get the total time.
  4. Jun 27, 2013 #3
    Here is a rule which can be useful sometimes.
    [itex]\ddot{r} = \frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} =\frac{dv}{dr} v [/itex]

    Applying this to this situation:
    [itex]\frac{dv}{dr} v = - \frac{k}{r^2}[/itex]

    [itex]v \mathop{dv}= - \frac{k}{r^2}\mathop{dr}[/itex]

    [itex]\frac{v^2}{2} = \frac{k}{r} + C_1 [/itex]
    We want that at [itex]r_0[/itex], the velocity is zero
    [itex]\frac{v^2}{2} = \frac{k}{r} - \frac{k}{r_0}[/itex]

    Next I used the fact that the velocity of the falling object will be negative.

    [itex]v = \frac{dr}{dt} = - \sqrt{\frac{k}{r} - \frac{k}{r_0}}\;\;\;\; r < r_0 [/itex]

    [itex]-\int_{r_0}^{r_1} \frac{dr}{\sqrt{\frac{k}{r} - \frac{k}{r_0}}} = \int_0^t dt'\;\;\;\; r_1 < r_0 [/itex]
    So the time would be
    [itex]t = -\frac{1}{\sqrt{k}}\int_{r_0}^{r_1}
    \frac{dr}{\sqrt{\frac{1}{r} - \frac{1}{r_0}}} \;\;\;\; r_1 < r_0 [/itex]
    Now comes the task of finding this integral.

    [itex]x = \frac{1}{r} \;\;\;\; dx = -\frac{dr}{r^2}[/itex]

    [itex]dr = -r^2 dx = -\frac{dx}{x^2}[/itex]

    [itex]-\int \frac{dr}{\sqrt{\frac{1}{r} - \frac{1}{r_0}}} = \int \frac{dx}{x^2\sqrt{x - \frac{1}{r_0}}} [/itex]

    Many steps omitted. Integral obtained using Wolfram Alpha.

    [itex]\int \frac{dx}{x^2\sqrt{x - \frac{1}{r_0}}} [/itex][itex]= \frac{r_0
    \sqrt{x - \frac{1}{r_0}} }{ x} +r_0^{3/2}tan^{-1}\left(\sqrt{r_0}\sqrt{x - \frac{1}{r_0}}\right)[/itex]

    substituting back for x:
    [itex] = rr_0\sqrt{\frac{1}{r} - \frac{1}{r_0}} +r_0^{3/2}tan^{-1}\left(\sqrt{r_0}\sqrt{\frac{1}{r} - \frac{1}{r_0}}\right)[/itex]

    note that this will in fact be zero at [itex]r = r_0[/itex] so we have

    [itex]t = \frac{1}{\sqrt{k}} \left[ r_0r_1\sqrt{\frac{1}{r_1} - \frac{1}{r_0}} +r_0^{3/2}tan^{-1}\left(\sqrt{r_0}\sqrt{\frac{1}{r_1} - \frac{1}{r_0}}\right) \right] \;\;\;\; r_1 < r_0[/itex]

    I am not sure that I have not made a mistake. One thing you could do to check this is to compare it to the falling time for constant acceleration and see if it is close for small distances.

    [itex]t = \sqrt{\frac{2(r_0 - r_1)}{g}} = \sqrt{\frac{2r_0^2(r_0 - r_1)}{k}} \;\;\;\; r_1 < r_0[/itex]
  5. Jun 28, 2013 #4
    Sorry, I was missing a [itex]\sqrt{2}[/itex].

    [itex]t = \frac{1}{\sqrt{2k}} \left[ r_0r_1\sqrt{\frac{1}{r_1} - \frac{1}{r_0}} +r_0^{3/2}tan^{-1}\left(\sqrt{r_0}\sqrt{\frac{1}{r_1} - \frac{1}{r_0}}\right) \right] \;\;\;\; r_1 < r_0[/itex]
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