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Central force attraction

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    a particle of mass m is attracted to the origin by a force F=-k/r^2

    find the time t for the mass to reach the origin

    2. Relevant equations

    [tex]\Delta[/tex]U= U-U0 = [tex]\int[/tex][tex]\textbf{F}[/tex](r) dr



    3. The attempt at a solution


    I found [tex]\Delta[/tex]U by [tex]\Delta[/tex]U= U-U0 = [tex]\int[/tex]F(r) dr and by following the convention r0=[itex]\infty[/itex] U([itex]\infty[/itex])=0

    so [tex]\Delta[/tex]U = [tex]\frac{-k}{r}[/tex]

    -[tex]\Delta[/tex]U=[tex]\textbf{W}[/tex]

    not sure what to do now but i think im just missing some easy steps. I know I have to equate this with 1/2mv^2 and then probably take a time derivative...
     
  2. jcsd
  3. Feb 9, 2009 #2
    It seems to me that k is not a constant but is a function of mass. If it were otherwise, cricket balls would fall more quickly than cannonballs.
    So if we calculate the work done by the fall and equate that to the kinetic energy gained we should be able to derive an expression for the time.
     
    Last edited: Feb 9, 2009
  4. Feb 9, 2009 #3
    K is given as a constant, analogous to Coulombs or the gravitational constant. The force could be expressed as

    F = -k/r^2 = m[itex]\ddot{x}[/itex]

    and we can see that acceleration is as always inversely proportional to the mass of the particle.

    I believe you need to use U to solve this problem but i'm not sure how
     
  5. Feb 9, 2009 #4
    In that case, the method I already outlined will work but the time taken will be a function of both the mass and the distance.
     
  6. Feb 9, 2009 #5
    ok here's what I did

    integrated F with respect to r and got W

    W=-k/r + c1

    using W = -[tex]\Delta[/tex]U I found U to be k/r

    I then used the expression

    [tex]\textbf{E}[/tex] = [tex]\frac{k}{r1}[/tex] +[tex]\frac{1}{2}[/tex]mv1[tex]^{2}[/tex] = [tex]\frac{k}{r2}[/tex] +[tex]\frac{1}{2}[/tex]mv2[tex]^{2}[/tex]

    and since v1= 0 and r1=0

    [tex]\textbf{E}[/tex] = [tex]\frac{k}{r1}[/tex] = [tex]\frac{1}{2}[/tex]mv2[tex]^{2}[/tex]

    Now I solve for v

    v = [tex]\frac{dL}{dt}[/tex] = [tex]\sqrt{\frac{2k}{mr}}[/tex]

    [itex]\int[/itex]dL = [tex]\int[/tex] [tex]\sqrt{\frac{2k}{mr}}[/tex]dt

    I end up with

    t = [tex]\frac{L}{\sqrt{\frac{2k}{mr}}}[/tex]

    is this the proper expression for time t it takes for the particle to move to the origin?
     
  7. Feb 10, 2009 #6
    Doesn't L = r ?
     
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