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Gravitational force on a particle from line of mass

  1. Feb 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Mass M is distributed uniformly along a line of length 2L. A particle of mass m is at a point that is a distance a above the center of the line on it's perpendicular bisector. For the gravitational force that the line exerts on the particle. calculate the components perpendicular and parallel to the line. Does your result reduce to the correct expression as a becomes very large?



    2. Relevant equations
    [itex] F=\frac{GMm}{r^2} [/itex]

    [itex] \lambda = M/L [/itex]

    3. The attempt at a solution

    So I've been trying to brush up on some first year stuff as I'm transferring to a four year next semester. I'm getting an answer that is too big by a factor of two, not sure why.

    Solving for the perpendicular component of force. I have

    [tex] \frac {aGmM}{L}\int_{-L}^{L} \frac{dx}{(a^2+x^2)^{3/2}} [/tex]

    this gives me [tex] \frac{2GmM}{a(L^2+a^2)^{1/2}} [/tex] two times to big.

    Where am I making the answer twice as big?
     
  2. jcsd
  3. Feb 25, 2015 #2

    Simon Bridge

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    How do you know it is twice as big as the correct answer?
    How did you do the integration?

    What should the expression turn into as a gets very large?
     
  4. Feb 25, 2015 #3

    ehild

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    The rod is of 2L length.
     
  5. Feb 25, 2015 #4

    Simon Bridge

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    <slap face> I better get some sleep... :(
     
  6. Feb 25, 2015 #5
    Haha, that makes sense. Thanks!
     
    Last edited: Feb 25, 2015
  7. Feb 25, 2015 #6
    Yeah, when I posted this I was pretty much a few hours after my bed time as well. Still not sure I would of caught this mistake though, haha. Thanks for your replies.
     
  8. Feb 27, 2015 #7

    Simon Bridge

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    Usually the rod is length L and you do the integration from -L/2 to L/2 or from 0 to L depending how the coordinates were set up.
    After a while i=t gets to be a reflex t just say L/M for density ... so naturally profs will throw that particular curve...
     
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