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1-dimensional problem in newtonian gravity- HELP

  1. Apr 11, 2005 #1
    1-dimensional problem in newtonian gravity- HELP!!

    The problem is this:

    Given sun of mass M and a body of mass m (M>>m) a distance r from the sun, find the time for the body to 'fall' into the sun (initially ignoring the radius of the sun).


    Our first equation is therefore [tex] \frac {d^2r}{dt^2} = \ddot{r} = \frac {GM}{r^2} [/tex].

    I am able to integrate this, giving:
    [tex] \dot{r} = - {\sqrt{2GM}}{\sqrt{1/r - 1/R}} [/tex],

    where R is the inital distance of the body from the sun. However, I am unable to integrate this again. I have shoved it into wolfram's integrator for an indicator of what to aim for, but cannot come close.

    Any thoughts would be greatly appreciated.

    Regards

    Romeo
     
  2. jcsd
  3. Apr 11, 2005 #2

    pervect

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    You can think of this as a zero-width elliptical orbit, so you can use Kepler's law relating the orbital period to the length semi-major axis, and then take 1/4 of the orbital period of any orbit with the same semi-major axis.

    If you insist on carrying out the integration, my textbook also suggests the substition

    r = a(1-e*cos(psi))

    in this case I think the eccentricity, e would be 1, so you'd have

    r = a(1 - cos(psi))

    I haven't tried this out, though.
     
  4. Apr 12, 2005 #3
    Thank you pervect, it's an approach I hadn't considered.

    Regards

    Romeo
     
  5. Apr 12, 2005 #4

    arildno

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    Aah, the triple post.
    This has already been solved in Diff-eq.
     
  6. May 7, 2005 #5
    Here is my try at it. Can you check for errors?
    A rock is dropped from [tex] r_0 [/tex] and falls straight down to R.
    How long does it take. Where R is earth radius and [tex]r_0 >> R [/tex].

    The gravity force equation [tex] F= \frac{GMm}{r^2} [/tex] ,

    integrates to a specific energy equation [tex] W= -\frac{u}{r}[/tex].

    The velocity at some intermediate r is given by:

    [tex] V=\sqrt{V_0^2+\frac{2u}{r} -\frac{2u}{r_0}}[/tex] .

    This is a hard differential equation to solve for r(t). I believe I have a form which resembles your integral.

    Change of symbols:

    [tex] A=2u,\ \ \ \ B = -\frac{2u}{r_0},\ \ \ \ V_0=0[/tex]

    [tex] V(r) = \sqrt{\frac{A}{r}+B}= \frac{dr}{dt}[/tex]

    Now the part which I have never trusted.

    [tex] \frac{dt}{dr}=\frac{1}{\sqrt{\frac{A}{r}+B}}[/tex]

    Then [tex] \int_R^r dt= \int_\frac{A}{R}^\frac{A}{r} \frac{1}{\sqrt{\frac{A}{r}+B}}\ \ dr [/tex]

    So [tex] T = \int_\frac{A}{R}^\frac{A}{r} \frac{1}{\sqrt{\frac{A}{r}+B}}\ \ dr [/tex]

    A change of variables [tex] \frac{A}{r}=x,\ \ r=\frac{A}{x},\ \ dr=\frac{-A}{x^2}dx [/tex]

    [tex]
    T = \int_\frac{A}{R}^\frac{A}{r} \frac{\frac{-A}{x^2}}{\sqrt{x+B}}\ \ dx = -A \left[ \frac{-\sqrt{B+x}}{Bx} -\frac{1}{2B} \left( \frac{2}{\sqrt{-B}} \arctan{ \sqrt{\frac{B+x}{-B}}} \right) \right]_\frac{A}{R}^\frac{A}{r} [/tex]

    [tex] = \left[ \frac{-r_0 \sqrt{x-\frac{2u}{r_0}}}{x} -\frac{2r_0}{\sqrt{\frac{2u}{r_0}}} \arctan{\sqrt{\frac{x-\frac{2u}{r_0}}{\frac{2u}{r_0}}}} \right]_\frac{2u}{R}^\frac{2u}{r_0} =
    [/tex]

    [tex] = \frac{-r_0 \sqrt{\frac{2u}{r_0}-\frac{2u}{r_0}}}{\frac{2u}{r_0}} -\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{\frac{2u}{r_0}-\frac{2u}{r_0}}{\frac{2u}{r_0}}}} \
    +\ \frac{r_0 \sqrt{\frac{2u}{R}-\frac{2u}{r_0}}}{\frac{2u}{R}} +\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{\frac{2u}{R}-\frac{2u}{r_0}}{\frac{2u}{r_0}}}}
    [/tex]

    [tex] = \frac{Rr_0}{\sqrt{2u}} \sqrt{\frac{1}{R}-\frac{1}{r_0}}+\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{r_0}{R}-1}}= \sqrt{\frac{Rr_0}{2u}} \sqrt{r_0-R} +\sqrt{\frac{2}{u}} r_0^\frac{3}{2} \arctan{\sqrt{\frac{r_0}{R}-1}} [/tex]
    This of course means you have to work the problem backwards usually as an iteration, but I would use a binary search which I found converges rapidly for the Kepler problem. It should be ok here too.
     
    Last edited: May 7, 2005
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