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Speed and Gravity

  1. Jul 20, 2006 #1
    we know from the equation of motion that


    where a is the acceleration
    for the gravity field we have


    So we get




    If we integrate one time from R1 to R2 shouldnt we get the speed?
    Because from the equation og energy we get a different result


    So where am i wrong?
    Last edited: Jul 20, 2006
  2. jcsd
  3. Jul 20, 2006 #2
  4. Jul 20, 2006 #3
    I forgot a 3 and i have put it back
    From the equation




    [tex]\int_a^b r^2dr\frac{dr}{dt}=\int_0^t GMdt[/tex]

    where a=R1 and b=R2


    Last edited: Jul 21, 2006
  5. Jul 21, 2006 #4


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    You just arbitrarily stick a dr and dt in? Where did they come from?

    ?? The integral of a product is not just the product of the integrals.
    [tex]\int_{R_1}^{R_2} r^2 dr= \frac{R_2^3- R_1^3}{3}[/tex]
    [tex]\int \frac{d^2r}{dt^2} dt= \frac{dr}{dt}[/tex]
    but that is NOT what you have.

    You might want to try "quadrature":
    [tex]\frac{d^2r}{dt^2}= -\frac{GM}{r^2}[/tex]
    Let [itex]v= \frac{dr}{dt}[/itex]. Then
    [tex]\frac{d^2r}{dt^2}= \frac{dv}{dt}= \frac{dv}{dr}\frac{dr}{dt}= v\frac{dv}{dr}= -\frac{GM}{r^2}[/tex]
    [tex]v dv= -\frac{GM}{r^2}dr[/tex]
    [tex]\frac{1}{2}v^2= \frac{GM}{r}+ C[/tex]
    [tex]\frac{1}{2}v^2-\frac{GM}{r}= c[/tex]
    Which is "conservation of energy": [itex]\frac{1}{2}v^2[/itex] is kinetic energy and [itex]-\frac{GM}{r}[/itex] is potential energy.
  6. Jul 21, 2006 #5
    Ok bu why with a direct double integral we dont get the same?
    why we have to go to the speed and then to find a displacement as a function of R?
    Why is wrong???

    The dr/dt was a mistake which i corrected
    the question is real and is not a trap
    Last edited: Jul 21, 2006
  7. Jul 21, 2006 #6
    There is no "direct" double integral. You can always write position as

    \vec{r} = \int\int \ddot{r}(t)dt dt

    (with suitable limits)

    but the two dt's have to be dealt with carefully. I don't see how you can get to the position from the acceleration directly. But if you want to get to velocity then note that [itex]vdv = a ds[/itex].
  8. Jul 22, 2006 #7
    It seems as thoughy you treated you assumed

    [tex] \int \frac{d^2r}{dt^2} = \frac{dr}{dt} \int \frac{dr}{dt} [/tex]

    which isn't true. The quadrature method suggested above will give a correct general solution, although it's hideous. r(t) = A*(to - t)^2/3 is a solution as well, although it won't fit arbitrary conditions. The general solution requires integrating

    [tex]\frac{1}{2}v^2-\frac{GM}{r}= c[/tex]

    which, when done by MATLAB gives a horrendous trascendental equation I couldn't begin to invert for r(t). Good luck with that.
  9. Jul 23, 2006 #8
    Haha yeah :biggrin:

    I'm no expert but isn't there any perturbation method available for a solution (just a wild guess...lets see how wild it is).
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