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Is the second derivative of a circle related to an orbiting object?

  1. Apr 7, 2013 #1
    Okay!

    Earlier today I was thinking about potential energy and how it is related to an orbiting object, O, around a centre, C, from which force emanates if the object O is traveling at radius r from this centre, we conclude that the force given by the change in direction must be equal to the force pulling O towards the field's centre. Great, thanks Captain Obvious. So, I was thinking that the force created by the change in direction of O about C is proportional to the change in the derivative of the function of the orbit.

    Since the equation of a circle is (r2 - x2)1/2 it's derivative is equal to x/(r2 - x2)1/2, which derives to produce r2/(r2 - x2)3/2

    Which is odd considering it's intuitive to think that whatever change in change in direction around C would be constant. Imagine swinging a bottle around your body, it doesn't pull with more force halfway through the swing.

    I thought that perhaps this was due to it being a function of the x axis, so I tried it with parametric equations and got

    y=sint
    x=cost

    dy/dt/dx/dt = cost/-sint = -1/tant = -cot(t) using -cot(t) as the new function for the derivative (y) y=-cot(t), I get -csc(t)/-sin(t), which still isn't constant.

    I really don't understand how the second derivative of the direction of a particle around a circle can't be constant.

    Can anyone clear this up for me?
     
  2. jcsd
  3. Apr 7, 2013 #2

    mfb

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    The parametric approach works, but the absolute change in motion is ##\ddot{x}^2+\ddot{y}^2##. You just calculated the acceleration in y-direction - this cannot be constant (it oscillates in y-direction!).
     
  4. Apr 7, 2013 #3

    Filip Larsen

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    Gold Member

    I'm not sure why you think that the direction of the acceleration in a uniform circular motion would be constant, because it is not. Differentiate the parametric equations twice with respect to time (that is, find d2x/dt2 and d2y/dt2) and you find that the acceleration is anti-parallel to the radius vector.

    Perhaps you have confused direction with magnitude? The magnitude of the acceleration in a uniform circular motion is constant.
     
  5. Apr 7, 2013 #4
    That's exactly what I seem to have done, you see I thought that the change in change of the direction of the object's momentum would be proportional to it's velocity. I now understand how the individual second derivatives of x(t) and y(t) respectively determine the rate of change of rate with respect to each axis, which when added together to form x"(t) + y"(t) (as mfb mentioned) do equal a constant, as far as I can tell.
     
  6. Apr 8, 2013 #5

    sophiecentaur

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    The problem is much easier if you use polar coordinates. The radial acceleration is constant and the angular velocity is constant in a circular orbit.
     
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