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Homework Help: Gravitational force on infinitely long rod

  1. Oct 30, 2007 #1
    A infinitely long rod, [tex]A[/tex], with linear mass density, say [tex]p[/tex], is placed along the z-axis. Another thin rod, [tex]B[/tex], of length [tex]L[/tex] and mass density, say [tex]p_{2}[/tex], is placed orthogonally to [tex]A[/tex] on the x-axis from [tex]x=a[/tex] to [tex]x=a+L[/tex]. What is the gravitational force experienced by rod [tex]B[/tex] from rod [tex]A[/tex]?
    I have solved a similar problem which asks for the gravitational potential given a rod of finite length and a point a perpendicular distance away from the rod's midpoint (thornton and marion problem 5-7). I am trying to extend this but not having much luck. So if we were trying to find the gravitational potential of the end point on rod [tex]B[/tex] closest to [tex]A[/tex] we would get:

    [tex] \phi =-G \int_{A} \frac{p}{r}ds =- p G \int_{A} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z\rightarrow \infty} \int_{-z}^{z} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z \rightarrow \infty} \ln(x+\sqrt{x^2+a^2})|_{-z}^{z}[/tex]

    This already goes to infinity. I was thinking that I could make [tex]\phi[/tex] a function of [tex]a[/tex] and then once I find a function for the gravitational potential, I'll integrate over rod [tex]B[/tex] to get the entire potential and hence the force by [tex]F=-\nabla U = -\nabla (m U ) = -m \nabla U[/tex]

    However, I guess I am stuck at step one since I cannot get the integral to work out. If you have any suggestions, I'd really appreciate the help. Thanks.
  2. jcsd
  3. Oct 30, 2007 #2


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    the gravitational force experienced by rod B from rod A will be a finite force per unit length of rod B. since rod B is infinitely long, it would be an infinite force (acting on an infinite mass, and the attractive acceleration would be finite).

    edit: WOW, this is my PF post #1000! a sorta milestone.
  4. Oct 30, 2007 #3


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    But B isn't infinitely long. You shouldn't be discouraged by infinite potentials if their derivative is finite. Congrats rbj!
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