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Math Help - dynamics

  1. #1
    Newbie
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    dynamics

    *theta=(-)
    A particle of mass m moves under an attractive central force mk/r^a where r is the radial distance from the force centre O.
    assuming that the radial and transverse components of accleration in polar coordinated (r, (-)) are (r''-r*(-)'^2) and (2r'*(-)' + r*(-)'') respectively, show that the differential equation for the orbit is,
    d^2u/d(-)^2 = k/h^2 * u^(a-2)
    where u=1/r and h=r^2 * (-)

    sorry about the notations, i wasn't sure how to present the equation on this.

    I am having trouble with this and a similar question. I am not sure how to derive the differential equation for the radial and transverse components.
    cheers
    p.s i hope this is the right sub-forum to post in
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  2. #2
    MHF Contributor
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    Nov 2008
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    France
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    Hi

    The 2 equations are

    (1) m \left(\ddot{r}-r \dot{\theta}^2\right) = -\frac{mk}{r^a}

    (2) 2 \dot{r} \dot{\theta} + r \ddot{\theta} = 0

    Then

    \frac{du}{d\theta} = \frac{du}{dt} \cdot \frac{dt}{d\theta} = \frac{d \left(\frac{1}{r}\right)}{dt} \cdot \frac{1}{\dot{\theta}} =  -\frac{\dot{r}}{r^2} \cdot \frac{1}{\dot{\theta}} = -\frac{\dot{r}}{r^2  \dot{\theta}}


    \frac{d^2u}{d\theta^2} = \frac{d}{d\theta} \left(\frac{du}{d\theta}\right) = \frac{d}{d\theta} \left(-\frac{\dot{r}}{r^2 \dot{\theta}}\right) = \frac{d}{dt} \left(-\frac{\dot{r}}{r^2 \dot{\theta}}\right) \cdot \frac{dt}{d\theta} = -\frac{\ddot{r}r^2 \dot{\theta}-\dot{r} \left(2r\dot{r} \dot{\theta}+r^2 \ddot{\theta}\right)}{r^4 \dot{\theta}^2} \cdot \frac{1}{\dot{\theta}}

    From equation (2) we know that the parenthesis is equal to 0

    From equation (1) we can substitute \ddot{r} = r \dot{\theta}^2 - \frac{k}{r^a}

    Therefore

    \frac{d^2u}{d\theta^2} = -\frac{\ddot{r}r^2 \dot{\theta}}{r^4 \dot{\theta}^3} = -\frac{\ddot{r}}{r^2 \dot{\theta}^2} = -\frac{r \dot{\theta}^2 - \frac{k}{r^a}}{r^2 \dot{\theta}^2} = -\frac{1}{r}+\frac{k}{r^{a+2} \dot{\theta}^2} = -u + \frac{k}{r^4 \dot{\theta}^2} \cdot u^{a-2}



    This is not exactly what you need to find so I may have missed something ...
    Last edited by running-gag; May 3rd 2010 at 01:28 PM. Reason: Correction
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