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Scattering Cross-Section from a Central Force

  1. Nov 23, 2014 #1
    1. The problem statement, all variables and given/known data
    An object from space (like an asteroid) approaches Earth. A collision will occur if the scattering cross-section is less than π*Re2. If the distance of closest approach is much greater than Re, no collision would occur. Find an implicit expression for the cross-section in terms of g, vo, and Re. Show explicitly that the cross-section approaches the minimum value for large vo, and find the appropriate dimensionless parameter.

    g: acceleration due to Earth's gravity
    vo: velocity of incoming object
    Re: radius of Earth

    2. Relevant equations
    Rutherford scattering formula:
    dσ/dΩ = k2/16E2*1/sin4(θ/2)

    Total energy:
    E = ½μr'2 - k/r + L2/2μr2

    3. The attempt at a solution
    I knew to use the Rutherford scattering formula because the scattering cross-section arises from an inverse square law (the gravitational force.) By doing some manipulation of the force constant:
    GMem ≈ g*m
    Me + m ≈ Me ⇒ μ ≈ m
    and approximating E by the kinetic energy T at very large r:
    E ≈ ½mvo2
    I arrive at
    dσ/dΩ = g^2/4vo2*1/sin4(θ/2)

    This is an expression including g, and vo... but where can we relate Earth's radius, Re?

    It seems like the differential cross section will approach Re for large vo, but I can't seem to express that mathematically. I'll need help understanding that first part. I could then use hints for finding the impact parameter per the last part of the question.
    I'm quite confused at this point.
     
  2. jcsd
  3. Nov 24, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    The object will hit earth if its minimal distance is below the radius of earth. For a fixed initial velocity, this corresponds to a certain impact parameter you can calculate.
     
  4. Nov 24, 2014 #3
    Do I calculate the impact parameter in terms of vo using equations of orbital motion?
     
  5. Nov 24, 2014 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    What else?
    If you don't find a direct formula, you can also derive it based on energy and angular momentum conservation.
     
  6. Nov 24, 2014 #5
    Resolved. I derived it the second way. Thanks
     
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