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Intermediate Mechanics Practice Test Problem

  1. May 7, 2007 #1
    Hello everyone. The following is from the practice test given to me for my intermediate mechanics final. I'm, at the moment, completely lost on what to do. If you have even just a few ideas of what I should be doing they would be apperciated. If you want to give a detailed solution it would be wonderful.

    Problem: The relative coordinate of a system of two masses, m and M, is moving in a circular orbit of radius r_0 because of the mutual gravitational attraction. Suddenly the particle with mass m receives a kick in negative radial direction. As a consequence of this kick, the particles approach each other until the centrifugal force finally repels them again.

    What radial velocity, v_0 as a result of the kick is necessary to have the particles approach each other to a minimum distance of 10% of r_0?
    Is the new trajectory a circle, ellipse, parabola or hyperbola?

    Thank you for your time.
     
  2. jcsd
  3. May 10, 2007 #2
    Hello, this is an interesting question. The following is my opinion only and something detail won't be metioned.

    The system in the question is an isolated two-body one.
    One can apply a coordinate transformation from the usual inertial frame to the relative-motion frame, i.e. observe m on M.
    Acturally, the frame is a non-inertial one.
    But one can make correction: [tex]m\rightarrow\frac{Mm}{M+m}[/tex], which let Newton's law applied well.
    Therefore, on the [tex]M[/tex]-frame, after [tex]V_0[/tex] being added on [tex]m[/tex], the conservation law of angular momentum and mechanical energy should be satisfied.
    From the beginning point (relative radius [tex]r_0[/tex] , tangent velocity [tex]\sqrt{\frac{G(M+m)}{r_0}}[/tex], radial velocity [tex]V_0[/tex]) to the nearest point (relative radius [tex]\frac{r_0}{10}[/tex], (assume) tangent velocity [tex]U[/tex] , no radial velocity):
    Angular momentum conservation:
    [tex]\frac{Mm}{M+m}\left(\sqrt{\frac{G(M+m)}{r_0}}\right)r_0=\frac{Mm}{M+m}U\frac{r_0}{10}\Rightarrow U=10\sqrt{\frac{G(M+m)}{r_0}}[/tex]
    Mechanical energy conservation:
    [tex]-\frac{GMm}{r_0}+\frac{1}{2}\frac{Mm}{M+m}\left(\sqrt{\frac{G(M+m)}{r_0}}^2+V_0^2\right)=-\frac{GMm}{r_0/10}+\frac{1}{2}\frac{Mm}{M+m}U^2\Rightarrow V_0=9\sqrt{\frac{G(M+m)}{r_0}}[/tex]
    The first equation gives [tex]U=10\sqrt{\frac{G(M+m)}{r_0}}[/tex].
    The second equation gives [tex]V_0=9\sqrt{\frac{G(M+m)}{r_0}}[/tex].
    One can check easily that the mechanical energy in the second equation is positive:
    R.H.S.[tex]=-\frac{GMm}{r_0/10}+\frac{1}{2}\frac{Mm}{M+m}U^2=40\frac{GMm}{r_0}[/tex].
    Therefore the trajectory is hyperbola(mechanical energy >0), in the point of view of M .
     
    Last edited: May 10, 2007
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