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Net gravitational force inside a shell

  1. Jul 28, 2014 #1
    When I was in introductory physics I remember being told that the net force (gravitational for example) inside a shell is always 0. I always felt that, intuitively, this would only be true at the center of the ring. Not sure what made me think of it today, by I decided to sit down and do the calculation.

    The vector formulation for a force of gravity

    [itex]F=\frac{Gm_1m_2}{|r|^2}\frac{r}{|r|}[/itex]

    m1 will be the mass of the particle inside the shell, m2 will be the mass of the shell. Inside a shell this becomes

    [itex]F=\frac{Gm_1m_2}{|r|^2}<cos(θ),sin(θ)>[/itex]

    [itex]dF=\frac{Gm_1dm_2}{|r|^2}<cos(θ),sin(θ)>[/itex]

    where

    [itex]dm_2=\frac{m_2}{2π}dθ[/itex]

    If r is changing as a function of θ, we can write it in this form

    [itex]dF=\frac{Gm_1m_2}{2π}<\frac{cos(θ)}{r(θ)^2},\frac{sin(θ)}{r(θ)^2}>dθ[/itex]

    For the r(θ), I made up a simple function that satisfied my question. [itex]r(θ)=2+sin(θ-π/2)[/itex] will put my particle somewhere in the right section (as long as it's not centered) of my oval shaped ring. (the biggest radius from my particle would be 3, the smallest would be 1). Finally we get

    [itex]F=\frac{Gm_1m_2}{2π}∫<\frac{cos(θ)}{(2+sin(θ-π/2))^2},\frac{sin(θ)}{(2+sin(θ-π/2))^2}>dθ[/itex]

    The integral is from 0 to 2π (not sure how to do a definite). After doing the integrals in my calculator I get roughly

    [itex]F=\frac{Gm_1m_2}{2π}<1.21,0>[/itex]

    There would be a pull to the right. So unless the particle was exactly centered, it would never just float, there would always be a force on it. Anyways, just thought I'd share. Why are we told to assume the net force at every point inside the shell is 0?
     
  2. jcsd
  3. Jul 28, 2014 #2
  4. Jul 28, 2014 #3
    It's the same in principle, isn't it? I just found it easier to make up an equation where r is a function of θ. If you were to take a point inside a spherical shell that wasn't at the center, wouldn't the result for net force be non-zero?
     
  5. Jul 28, 2014 #4
    It would be zero in a spherically symmetric shell. Try it out, and then check the wikipedia I posted which has the derivation.

    Intuitively you could think that in any direction you will be closer to a small amount of mass and further from a large amount of mass such that the force from each is equal in magnitude and opposite in direction (they cancel out). If the shell isnt spherically symmetric then this balance wont necessarily happen.
     
  6. Jul 28, 2014 #5
    Ahh I see. I'm not quite sure how to develop that equation for r, but assume it has something to do with a coordinate transformation. I will go work on that, thanks ModusPwnd.
     
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