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Simple gravitational force problem (Gregory Classical Mechanics)

  1. Mar 3, 2013 #1
    1. The problem statement, all variables and given/known data

    This is question 3.7 from Gregory's Classical Mechanics textbook.

    A symmetric sphere of radius a and mass M has its center a distance b from an infinite plane containing a uniform distribution of mass ## \sigma ## per unit area. Find the gravitational force exerted on the sphere


    2. Relevant equations



    3. The attempt at a solution

    From what I understand, a solid sphere can be represented as if it were a single particle of mass M concentrated at its center of mass, call this point ##S##. So, the way I approached the problem was summing (integrating) up the forces exerted on this point by each infinitesimally small area ##dxdy## on the plane.

    The mass of each infinitesimal on the plane is ##m = \sigma dA##. I also let ##\theta## represent the angle between SB (where B is the straight line distance from the point S) and the line drawn from S to the infinitesimal. Then, the equation I got was:

    ## F = MG \int _A \sigma \cos (\theta) / R^2 dA ##
    ## F = MG\sigma \int _A R\cos (\theta) / R^3 dA ##
    ## F= MG\sigma \int _A b/R^3 dA ##
    ## F = MGb\sigma \int _{-\infty}^\infty \int _{-\infty}^\infty 1/(x^2 + y^2 + b^2)^{3/2} dxdy ##

    If anyone could let me know if I have set this up correctly (I get the feeling I have not), and how to approach this problem, I'd be really grateful. Thanks!
     
  2. jcsd
  3. Mar 3, 2013 #2
    The reduction of the problem to that of a point mass is correct.

    Then, as far as I can see, you integrate the parallel component of the force, ignoring the perpendicular component. I think you owe us an explanation here.
     
  4. Mar 3, 2013 #3
    So from what I understand, if we drop a straight line from the center of the sphere S to where it meets the plane, say that point is O, we'll have a line SO perpendicular to the plane of length ##b##. Then if we drop another straight line from S to a particular point on the plane, say X, then SOX becomes a right triangle where the angle created by the lines SO and SX is ##\theta##. Also:

    ##|SO| = b##
    ##|OX| = R \sin(\theta)##
    ##|SX| = R##

    Then, I supposed by symmetry, that all the component forces by each point ##X## in the plane will sum up to a single perpendicular force pointing in the direction of ##SO##. So, if each point ##X## attracts S in its direction with a force of magnitude

    ## F' = m(\sigma dA)G/R^2 ##

    Then its force in the direction of SO ought to be ##F = F'\cos (\theta)##. That's how I derived my integral in the form of

    ##F = mG \int _A \sigma \cos (\theta)/R^2 dA ##

    Please help me understand where I've made my error! Thanks again!
     
  5. Mar 3, 2013 #4
    There is no error in your derivation. It is entirely correct. You just need to finish integration.
     
  6. Mar 3, 2013 #5
    Hmm, then maybe I am having trouble with simplifying the integral. What is the most efficient way to do it? What I thought to do is this:

    ##F = MG\sigma \int _A \cos (\theta) / R^2 dA ##
    ##F = MG\sigma \int _A R\cos (\theta) /R^3 dA ##
    ##F = MG\sigma \int _A b / R^3 dA##

    After this, I assumed by the pythagorean theorem that ##R = \sqrt{ x^2 + y^2 + b^2 }##, but then I end up with what looks like a nasty integral:

    ##F = MG\sigma b \int _{-\infty}^{\infty} \int _{-\infty}^{\infty} 1/ (\sqrt{x^2 + y^2 + b^2})^3 dxdy##

    Is there another way to approach this, or am I simplifying the integral wrongly? Thanks!
     
  7. Mar 3, 2013 #6
    Ah never mind, I've figured it out. The better way is to use polar coordinates for the area of the plane. The integral simplifies to

    ## F = MGb\sigma \int _{0}^{2\pi} \int _{0}^{\infty} r/(r^2+b^2)^{3/2}drd\phi ##

    Thanks for the help.
     
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