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Homework Help: Speed and Gravity

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

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

    where a is the acceleration
    for the gravity field we have

    [tex]a=\frac{GM}{r^2}[/tex]

    So we get
    [tex]\frac{d^2r}{dt^2}=\frac{GM}{r^2}[/tex]

    [tex]\frac{(R1^3-R2^3)}{3}\frac{dr}{dt}=GMt[/tex]

    [tex]\frac{dr}{dt}=\frac{3GMt}{R1^3-R2^3}[/tex]

    [tex]u=\sqrt{\frac{3GMt}{R1^3-R2^3}}[/tex]

    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

    [tex]u=\sqrt{2GM}\sqrt{\frac{R1-R2}{R1*R2}}[/tex]

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

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

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

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

    where a=R1 and b=R2

    [tex]\frac{(R1^3-R2^3)}{3}\frac{dr}{dt}=GMt[/tex]

    [tex]\frac{dr}{dt}=\frac{3GMt}{R1^3-R2^3}[/tex]
     
    Last edited: Jul 21, 2006
  5. Jul 21, 2006 #4

    HallsofIvy

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    Science Advisor

    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]
    and
    [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

    [tex]
    \vec{r} = \int\int \ddot{r}(t)dt dt
    [/tex]

    (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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