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Calculating Fgrav at a non-point mass

  1. Nov 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Hello! I signed up about week ago, in case i needed some help in physics, and today i do. I Hope someone can explain this to me a little better.

    Anyways, yesterday we started the chapter on gravitation. Our professor layed out eight steps we should follow to calculate the Fgrav at a non-point mass. They were,

    1. Draw a diagram
    2. Pick a small chunk (dm)
    3. Draw a line from dm to the 2nd mass
    4. Draw a force vector starting at the 2nd mass
    5. Calculate dF created between dm and the 2nd mass
    6. Check symmetry
    7. Calculate remaining components of dF
    8. Integrate


    2. Relevant equations

    He gave us the following:

    [tex]dF\ =\ G\frac{dm M_{2}}{r^{2}}[/tex]

    Which he broke down into components,

    [tex]dF_{x}\ =\ G\frac{dm M_{2}}{r^{2}}cos\theta[/tex]


    [tex]F_{y}\ =\ 0[/tex] by symmetry.

    3. The attempt at a solution

    When i went home, i proceeded to find dFx
    [tex]dm\ =\lambda dl[/tex]

    [tex]\lambda\ =\frac{M}{L}[/tex]

    [tex]r\ =\sqrt{(R^{2}+l^{2})}[/tex]

    [tex]cos\theta\ =\frac{R}{\sqrt{(R^{2}+l^{2})}}[/tex]

    [tex]dF_{x}\ =\ G\frac{\lambda dl M_{2}}{(R^{2}+l^{2})}\frac{R}{\sqrt{(R^{2}+l^{2})}}[/tex]

    [tex]F_{x}\ = \int dF_{x}\ =2\int^{\frac{L}{2}}_{0} \ G\frac{\lambda dl M_{2}}{(R^{2}+l^{2})}\frac{R}{\sqrt{(R^{2}+l^{2})}}[/tex]

    [tex]F_{x}\ = 2 \ G\lambda M_{2}R\int^{\frac{L}{2}}_{0}\frac{dl} {(R^{2}+l^{2})^{\frac{3}{2}}} [/tex]

    Now to finish this up, it seems to me like trig substitution. Is that the correct way to proceed with he integral?
    My main question that i dont understand is, What does he mean "by symmetry"? I first assumed he was going to use [tex]sin\theta[/tex] for Fy but he didnt. Why is Fy=0?
    Thanks for any help you can give me about the symmetry part.

    Attached Files:

  2. jcsd
  3. Nov 26, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Since M2 is symmetrically located with respect to your extended mass, the y-components of force from the upper half will be exactly equal and opposite to the y-components from the lower half. Thus, "by symmetry", you know they will cancel out. Of course you can certainly work it out using sinθ, but you'll get the same answer.
  4. Nov 26, 2008 #3
    Oh, i see that. Thanks for that clarification!
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