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Newtonian Gravity- Body 'falling' into the sun integration problems

  1. Apr 11, 2005 #1
    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] \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
    With G and M constant:

    [tex] \int {\frac{GM}{r^2}dr = GM\int{\frac{1}{r^2}}dr[/tex]

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

    Use the power rule for integration to find this integral. There are no square roots involved.
     
  4. Apr 11, 2005 #3
    Appreciated Woozum, but you mis-read my post. Maybe i should clarify further:

    [tex] \dot{r} = \frac {dr}{dt}, [/tex]

    and that i solved the first integral of this second order differential equation, but cannot, in particular, solve this:

    [tex] \int \frac {1}{\sqrt{1/r - 1/R}} dr = -\sqrt{2GM}\int dt, [/tex]

    reminding that 'R' is the initial distance of the body from the sun and that 'r' is variable distance that you integrate to.

    So, any ideas anyone? :smile: :confused:

    Regards

    Romeo
     
  5. Apr 11, 2005 #4

    BobG

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    whozum's right, but, more importantly, how is that going to get your time?

    Your first equation is your acceleration. To get your velocity, your integration should be:

    [tex]\dot{r}=\int{a}\, dt[/tex]

    rdot is your velocity, which you can integrate wrt time to get your position.

    That get's you an initial value that you can solve for time (just in case you didn't know the equation, already).
     
  6. Apr 11, 2005 #5
    BobG, I understand what you're intending to say, but haven't quite got what i'm trying to say:

    I have the second order differential equation [tex] \ddot{r} = -GM /r^2 [/tex] to solve, aiming to find an expression for 'r', which will then simply be be used to find an expression for the time it takes for a body to 'fall' into the sun from a general distance.

    I neglected to show the work involved to solve the latter, but I am pretty certain that the expression for [tex] \dot{r} [/tex] is correct.

    With that in mind, any ideas folks? :smile: :confused:

    Regards

    Romeo
     
  7. Apr 11, 2005 #6
    Honestly dont know what your trying to do, but

    How does [itex] \dot{r} = - {\sqrt{2GM}}{\sqrt{1/r - 1/R}} [/tex] stem from [itex] \ddot{r} = \frac {GM}{r^2} [/tex]

    For constant acceleration,
    [tex] x(t) = x_0 + v_0t + \frac{gt^2}{2} \mbox{ where } g = \frac{GM}{r^2} \mbox{ so then } [/tex]

    [tex] x(t) = \frac{GMt^2}{2r^2}[/tex] with [tex] x_0 = v_0 t = 0 [/tex]

    and [tex] t = \sqrt{\frac{2xr^2}{GM}} [/tex]
     
  8. Apr 11, 2005 #7

    dextercioby

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    Dog Maddit !!! :mad:Do not double post !You got the same problem solved by Arildno in the Diff.Eq.forum.

    Daniel.
     
  9. Apr 12, 2005 #8
    Thank you anyway Whozum. I think maybe I should have stated that there will of course not be constant acceleration, hence your approach is invalid. However, there is now a solution in Diff Equ/ns, so check out what I was trying to do!

    Regards

    Romeo
     
  10. Apr 12, 2005 #9

    arildno

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    Dearly Missed

    A word of thanks could work wonders, you know.
    Not a wholesale devotional of course, but an appreciative nod might be in order..
     
  11. Apr 12, 2005 #10
    Thank you all, apologies for the double post, all is accounted for in the Diff Equ forum. And again, thank you Arildno for the solution.

    Regards

    Romeo
     
  12. Mar 24, 2011 #11
    The problem can also be solved without performing any integration by applying Kepler's laws of planetary motion.

    From Kepler's first law the orbit of any mass around the sun is an ellipse with the sun at one of the two foci. Lets call the semimajor axis of this ellipse 'a' and the semiminor axis of the elipse 'b'.

    An object with zero initial velocity falling into the sun is the limiting case of this orbit as the minor axis of the ellipse (b) goes to 0. The distance f from the centre of the ellipse to the foci is f = (a^2 - b^2) ^(1/2). Obviously as b -> 0, f ->a. So the initial distance R from the object to the sun is R = 2a in this limit.

    Kepler's third law gives the orbital period T in terms of the semimajor axis length.

    T^2 = a^3 (4*Pi^2)/MG

    Bearing in mind that on object falling into the Sun only completes half of its 'orbit' and using the relationship between a and R above you can solve the problem very simply.
     
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