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Angular Momentum of an asteroid

  1. Jan 11, 2006 #1
    I understand the concept of angular momenum, but I do not understand how to use it in this problem:
    An asteroid of mass 1.0 * 10^5 kg, traveling at a speed of 30 km/s relative to the Earth, hits the Earth at the equator. It hits the Earth tangentially and in the direction of Earth's rotation. Use angular momentum to estimate the fractional change in the angular speed of the Earth as a result of the collision?

    First off, what is fractional change?
    I set up a conservation of momenum problem, momentum of Asteroid + momentum of Earth= momentum of Asteroid prime + momentum of Earth prime.
     
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  3. Jan 11, 2006 #2

    Tide

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    The fractional change in angular momentum is

    [tex]\frac {\Delta L}{L}[/tex]
     
  4. Jan 11, 2006 #3
    This is my first post. I don't usually plea for help like this but I have been stuck for some time? Thanks Tide. How do I find the momentum prime of the Earth? I don't understand how exactly the asteroid affects it.
     
    Last edited: Jan 11, 2006
  5. Jan 11, 2006 #4
    there's an inelastic collision when the asteroid hits the earth
     
  6. Jan 11, 2006 #5
    So I set up a conservation of momentum and a conservation of energy situation. How do I find the radius of the asteroid? Don't I need the radius and/or the initial angular velocity of the asteroid?
     
  7. Jan 11, 2006 #6
    Ie(We-Weprime)=Ia(Waprime-Wa)
    I cannot seem to find the velocity without the moment of inertia of the asteroid. How do I do this without knowing the radius of the asteroid?
     
  8. Jan 11, 2006 #7
    I think that momentum is the same units, linear or angular. This is a tough problem. I would add the linear momentum of the asteroid to the angular momentum of the earth since it hits tangentially, and check to see if the added mass of the asteroid is significant enough to change the I of the earth.
     
  9. Jan 11, 2006 #8
    another definition for angular momentum is mass*velocity*radius
     
  10. Jan 11, 2006 #9

    Tide

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    Take the asteroid to be small compared with the Earth so its moment of inertia is just that of a point particle - or just write its angular momentum as [itex]L_a = m_a v_a R_e[/itex] which is correct at the time of impact given the asteroid makes a tangential hit.
     
  11. Jan 11, 2006 #10
    Thanks. I have never seen that form of angular momentum before and was unaware that the asteroid would take on the Earth's radius.
     
  12. Jan 11, 2006 #11

    Tide

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    I am sure you have. That is how angular momentum is usually introduced in physics courses:

    [tex]\vec L = \vec r \times \vec p[/tex]

    If you're taking an engineering course, however, they may introduce it in terms of moments of inertia.
     
  13. Jan 11, 2006 #12
    Actually, our course(not engineering, just high school Physics) introduced it as the rotational analog to angular momentum, with the moment of inertia being analagous to mass in linear momentum. I understand how that form is derived though.
     
  14. Jan 11, 2006 #13

    Tide

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    Oh, I see. I assumed it was a college course.

    In that case, you can use a moment of inertia approach in which case you'll approximate the asteroid as a point "revolving" about the Earth's axis so [itex]I_a = m_a R_e^2[/itex]. The appropriate angular velocity is then found from [itex]v = \omega R_e[/itex]
     
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