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Gravitational attraction of a hemisphere at its pole

  1. Jul 3, 2012 #1
    Find the gravitational attraction of a solid hemisphere of radius a anddensity1 on a unit point mass placed at its pole.

    My attempt:

    Obviously I figured that spherical coordinates would work nicely for this problem.

    I decided to invert the hemisphere thus placing the pole on the origin. Then I decided to split the upside down quarter disc into two integrals. The first integral was then for a cone shaped figure (r1 in the picture i made).

    I set it up as follows

    [itex] G \int_{\vartheta=0}^{2\pi} \int_{
    \varphi=0}^{\pi/3}\int_{\rho=0}^{a/sin(\phi)} sin(\phi)cos(\phi)d\phi d\rho d\theta[/itex]

    Plz let me know if i have made any mistakes with the above integrand (pi/3 is where a cone will intersect a circle of equal radius).

    However I could not manage to think up the equation (in spherical coordinates) for the lower circular bound. I considered changing coordinate systems for the outer spherical part but setting up the equation for force of gravity might have been tricky. Any help will be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Jul 3, 2012 #2

    Simon Bridge

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    well it would be given as the field due to the total mass within a gaussian surface through the point as if all that mass were concentrated at the center of mass.

    Note: you can make your life easier by exploiting the symmetry even further - can you find the field at a perpendicular distance r from a disk radius R and thickness dz?
     
  4. Jul 3, 2012 #3
    Yes I could but were looking for the gravitational attraction on the surface of the pole since that is where the unit mass is placed, or am i misinterpreting you?
     
  5. Jul 3, 2012 #4

    Simon Bridge

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    You sort-of have ... if you orient the hemesphere so it's axis is the z axis, and slice it into disks thickness dz, then the biggest disk (the base) will be distance R from the test mass.

    I just have a feeling you've not got one of the limits right ... one of the advantages of the slice method is that you can conceptually check each stage. Still - run with what you are comfy with ;)
     
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