1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar

    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


    User Avatar
    Science Advisor
    Homework Helper

    But B isn't infinitely long. You shouldn't be discouraged by infinite potentials if their derivative is finite. Congrats rbj!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Gravitational force on infinitely long rod