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Ghost Force(?) & Angular Momentum

  1. Feb 16, 2007 #1
    From Kepler's second law as well as conservation of angular momentum, we know that earth (any many other celestial objects) move faster when they get near the sun during their orbit and get slower when they are far away. This is a "change" in the angular speed so the derivative of it is a constant(if it's not a function). But this change must bring an angular acceleration α which in return brings us a torque due to the fact that

    Torque= I x α = r x F

    What is this force and where does it come from?
    Sun's gravity is a central force and can act only parallel to the displacement vector from sun-to-earth thus it can't have any torque effect at all.

    What force cause this acceleration of earth during these phases?
     
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  3. Feb 16, 2007 #2

    ZapperZ

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    There is no "force". This is the result of the conservation of angular momentum. It is the same effect that you get when you are spinning and bring your arms in. There's no external force or torque of any kind there.

    Zz.
     
  4. Feb 16, 2007 #3
    How would you explain the mathematical consequence then? a change in the angular speed thus an angular acceleration YET no force and no torque??
     
  5. Feb 16, 2007 #4

    ZapperZ

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    Mathematical consequence? What mathmatical consequence?

    The conservation laws of our universe are not mathematically derived. It has no consequences to mathematics. The mathematical description of the conservation of angular momentum is pretty obvious, which I don't think is what you're asking.

    Zz.
     
  6. Feb 16, 2007 #5
    Thank you :)
    What mathematics say that if there's a change in angular speed there MUST be an acceleration. If there's an acceleration there MUST be a force.
    Yet you are saying that "that must not be" that's not how universe works.
    Am I following you?
     
  7. Feb 16, 2007 #6

    ZapperZ

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    Notice that when I wrote my response, I wrote the word "force" in quotes. I assumed that you know about external forces and torques that are missing from the system. This is the reason why the angular momentum of the system is conserved.

    There is also something that you are missing. In classical mechanics, the "conservation laws" actually is the most fundamental aspect of the dynamics of any system. The "forces" are actually not that fundamental, and in fact, in the Lagrangian/Hamiltonian mechanics, forces don't even exist! This is because ALL you are actually detected are the two fundamental variables : the cannonical momentum and the cannonical coordinate positions.

    So I don't quite understand the "obsession" with "forces" here in a system that really do not need such a thing when the application of a conservation law is sufficient and, in fact, simpler.

    Zz.
     
  8. Feb 16, 2007 #7
    It's not an obsession, if you're to think about it in classical terms you're prone to find a force to cause such acceleration.
    But I guess this is where classical physics ends and "beyond" begins.
    I'll think about a "forceless" universe and read Lagrangian/Hamiltonian mechanics.

    Thank you, you opened my mind.
     
  9. Feb 16, 2007 #8
    Classical physics does not end at angular momentum (in this context). Far from it. Perhaps this can contribute to your understanding of angular momentum.
     
  10. Feb 16, 2007 #9
    What happens is that the moment of inertia changes. And when that happens, [tex]\tau = I\alpha[/tex] is no longer good enough. We need to consider the more fundamental relationship [tex]\tau = \frac{dL}{dt}[/tex]. Since the angular momentum is given by [tex]L = I\omega[/tex], we can see that, in cases where I changes, torque is given by [tex]\tau = I\alpha + \frac{dI}{dt}\omega[/tex]. This means that there can be an angular acceleration without any external torque. You just need I to vary in time while the object is rotating; and, that's exactly what's happening here.
     
  11. Feb 16, 2007 #10
    Ah, the perfect explanation! Since [itex]I=r^2m[/itex], zero external torque (in Parlyne's equation) implies [itex]\alpha = - \frac {2 \omega} {r} \frac {d r} {d t} [/itex], or in other words, unless there is a torque the planet has to decelerate (in terms of angular velocity) while-ever it is getting further from the sun ([itex]\frac {d r} {d t}[/itex] is +ve), and conversely a skater must spin faster if she draws in her arms.
     
    Last edited: Feb 16, 2007
  12. Feb 21, 2007 #11
    Crystal clear... thank you!
     
  13. Feb 21, 2007 #12

    russ_watters

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    Why can't we just say that the force here is gravity? If the orbit isn't circular, the object isn't traveling tangentiallly to the gravitational potential, so there is a force component in the direction of motion.
     
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