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2-body central force ( gravity ) problem

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A comet is released from rest at a distance r_max from the sun.
    Therefore it has no angular momentum.
    We can assume the sun is a point mass ( i.e. sun's radius is zero ).

    How long does it take for the comet to hit the sun?

    Let m = comet mass
    let M_s = sun's mass
    let G = gravitational constant
    r = distance of comet from the sun, with origin at the sun.

    2. Relevant equations

    t = sqrt( m/2 ) * ∫ dr / sqrt( E - U(r) )

    3. The attempt at a solution

    Since the comet is released from rest, it initially has kinetic energy T = 0 and potential
    U = G*M_s*m/r_max.

    E = T + U = G*M_s*m/r_max.

    U(r) = G*M_s*m/r

    E-U(r) = (G*M_s*m)( 1/r_max - 1/r)

    After grinding a bit I got:

    t = 1/sqrt( 2*G*M_s) ∫ dr/sqrt( 1/r_max - 1/r )

    with the limits of integration being r_max to 0.

    This is where my eyes glazed over, since it's been 25 years since I've been near a calculus course. Any hints about how to proceed?

    This is Taylor, _Classical Mechanics_, problem 8.21, part (c).

    I am NOT in a class. I'm a software grunt trying to educate myself, any help offered would be greatly appreciated. This problem has really had me tearing my hair out for about 3 days.

    Thanks,
    John
    (jobendorfer@cyberoptics.com)
     
  2. jcsd
  3. Mar 15, 2012 #2

    ehild

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    Welcome to PF!

    The potential energy of the comet is negative. U=-G*M_s*m/r_max.

    You see, the kinetic energy is all right now, greater than zero.

    Actually, v=-dr/dt as r decreases with t as the comet approaches the sun.


    There is an easy way if you have problems with an integral. Try to put it into wolframalpha.com. But before that, remove irrelevant constants and denote the necessary ones (r_max) by a simple letter. It is also advisable to change the variable so you get a simpler function. This case you can use u=r/r_max.

    When you want to do the integration yourself, it is a long way... start with substituting sqrt(r_max/r-1)=u.





    ehild
     
    Last edited: Mar 15, 2012
  4. Mar 15, 2012 #3

    ehild

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    ?????

    ehild
     
  5. Mar 15, 2012 #4
    Ehild,

    Thanks for the welcome and for your help. Very much appreciated!

    I've been lurking here off and on for years, I just tend to read and listen a lot more than I write -- for some strange reason. :-)

    J.
     
  6. Mar 15, 2012 #5

    ehild

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    Don't be shy! Come and talk.:smile:

    How do you proceed with that integral?

    ehild
     
  7. Mar 16, 2012 #6
    I tried your suggestion u = r/r_max:

    dr = r_max*du

    Integral[ r_max*du/sqrt( 1/u*r_max - 1/r_max ) ]

    With a bit of algebra:

    r_max^1.5 * Integral [ du / sqrt( 1/u - 1 ) ]

    I fed that integral into Mathematic 6's Integrate function as:

    Integrate[ (1/u - 1)^-0.5, u]

    What I got was:

    (0.666667 (1.- 1. u)^0.5 u Hypergeometric2F1[1.5, 0.5, 2.5,
    u])/(-1. + 1/u)^0.5

    It's an answer, but ... wow. Not even sure what a "Hypergeometric" is.
    So I'm thinking that there's got to be some clever transformation that gets rid of the 1/u.
    Or I've bodged the algebra again. Take your (non-exclusive) pick!

    But I did figure out where I went wrong with the minus signs on the gravity potential, so the day wasn't completely unproductive :-)

    Regards,

    John
     
  8. Mar 16, 2012 #7

    ehild

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    Hi John :smile:

    Sometimes you get the result in a very complicated form with these programs. With Wolframalpha, it is


    http://www3.wolframalpha.com/Calculate/MSP/MSP18861a0f7db9ai44e4be00000h0feidd94e0ec2a?MSPStoreType=image/gif&s=3&w=470&h=97 [Broken]
    http://www.wolframalpha.com/input/?i=int(1/sqrt(1/x-1)+dx)

    You can simplify and it results in something as pi/2 after inserting the boundaries 0 and 1.

    You can use also 1/u as variable ... But try what I suggested sqrt(r_max/r-1)=u. Do not forget to transform dr too.

    There is an other method to get the time: Think of Kepler's Third Law.


    ehild
     
    Last edited by a moderator: May 5, 2017
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