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Minimum distance from a repulsive central force

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves under action of a repulsive central force Fr=Cr-3 with constant C greater than 0. At a very large distance from the centre of the force, the partcle has kinetic energy K and its impact parameter is b. Use conservation of energy and angular momentum to show that the closest m comes to the centre of force is rmin=[tex]\sqrt{b^2+ C/2K}[/tex]


    2. Relevant equations
    L=mbv0
    E=[tex]\frac{1}{2}[/tex]mvr2+L2/2mr2+V(r)



    3. The attempt at a solution
    So far, I said that since the object is far away its total energy is K
    I think V(r) = -3C/r2, but I'm not sure.
    And at rmin, E==[tex]\frac{1}{2}[/tex]mvrmin2+(mbv)2/2mr2-3C/rmin2

    I don't know what to do next.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 22, 2009 #2
    PHY2333...Assignment is due tomorrow... Still can't get this either. Where do you get -3C/r^2 for V(r)? Would it not be -0.5C/r^2?
     
  4. Oct 22, 2009 #3
    Yeah, I still don't know how to get it.
     
  5. Oct 22, 2009 #4
    oops wrong topic
     
    Last edited: Oct 22, 2009
  6. Oct 22, 2009 #5
    Take Force integrate from rmin to b, as when r is infinity, it will approximately equal to be, because it is essentially uneffected by the force anyway. this integration will give you the answer. I know its not using angular momentum, but I don't care at this point.
     
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