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Inverse Cubed Force

  1. Nov 14, 2015 #1
    Question: A body of mass m is moving in a repulsive inverse cubed force given by
    F = K/r^3 where K > 0​
    show the path r(θ) of the body given by
    1/r = A cos[β(θ-θο)]
    Find the values of constants A and β in terms of E, L.

    Work done so far:
    we did a similar problem with inverse square so I am attempting to modify that for use with cubes

    so far I have
    F = K/r^2+K/r^3
    V(r) = ∫ F(r)dr
    V(r) = K/r + K/2r^2
    dr/dθ = r^2/L√[2u(E-L^2/2ur^2-K/r - K/2r^2)]

    distributing and moving L
    dθ=dr/r^2 1/√[2uE/L^2 - 1/r^2 - uK/L^2r^2 - 2uK/L^2r]^-½
    substituting 1/r for ω and dr/r^2 for dω
    dθ=dω1/√[2uE/L^2 - ω^2 - uKω^2/L^2 - 2uKω/L^2]^-½
    from here our professor wants us to reduce this to dθ=dω/√(a^2+x^2) so we can use trig sub we are running into issues and dont really know how to move forward. any help would be useful.

    Thank you in advance
     
  2. jcsd
  3. Nov 14, 2015 #2
    Your question says that the force is inverse cube. In your work, you have two forces, an inverse square, + an inverse cube. Why did you do that?
     
  4. Nov 14, 2015 #3
    I was trying to apply Newton's theorem of revolving orbits and I believe I just went completely off from where I should have been heading
     
  5. Nov 14, 2015 #4

    mfb

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    2016 Award

    Staff: Mentor

    The inverse square term should not be there.
     
  6. Nov 14, 2015 #5
    ok so without that it becomes

    F = K/r^3
    V(r) = ∫ F(r)dr
    V(r) = K/2r^2
    dr/dθ = r^2/L√[2u(E-L^2/2ur^2- K/2r^2)]

    distributing and moving L​
    dθ=dr/r^2 1/√[2uE/L^2 - 1/r^2 - uK/L^2r^2]^-½

    substituting 1/r for ω and dr/r^2 for dω
    dθ=dω1/√[2uE/L^2 - ω^2 - uKω^2/L^2]^-½
    consolidating
    dθ=dω1/√[2uE/L^2 - (1+ uK/L^2)ω^2]^-½
    I still do not where to go from here however

    again thank you for any help provided
     
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