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Kepler's First Law of Planetary Motion

  1. Feb 6, 2012 #1
    I'm reviewing my old calculus textbook and I stumbled upon a proof of Kepler's First Law of Planetary Motion which uses vector valued functions along with all of the operations to demonstrate the material. I understand the math and how to to DO it but what I am struggling with is why.

    It goes from:

    v X h = GM u + c

    and then does the dot product of both sides with r.


    Why is the above equation "dotted" with r.

    This is apparently found in most calculus textbooks but if it is not:
    v is the velocity
    u is the unit vector of direction
    h is a constant vector of rXv
    c is a constant vector of integrating u'
     
  2. jcsd
  3. Feb 6, 2012 #2

    D H

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    The typical approach is to start by showing that the specific angular momentum [itex]\vec h = \vec r \times \vec v[/itex] and the eccentricity vector [itex]\vec e = \vec v \times \vec h / \mu - \hat r[/itex] are constants of motion, where [itex]\mu=GM[/itex].

    That [itex]\vec h[/itex] is constant is a direct consequence of central motion, any kind of central motion. That the eccentricity vector is constant is only true for inverse square law central motion, [itex]\vec a = -\mu/r^2 \hat r[/itex].

    To show that the eccentricity vector is constant, look at the time derivative of [itex]\vec v \times \vec h[/itex]:
    [tex]\begin{aligned}
    \frac{d}{dt}(\vec v \times \vec h)
    &= \frac{d\vec v}{dt}\times h \\
    &= -\frac{\mu}{r^2}\hat r \times \vec h \\
    &= -\frac{\mu h}{r^2} \hat r \times \hat h \\
    &= \mu\dot{\theta}\hat{\theta} \\
    &= \mu\frac{d\hat r}{dt}
    \end{aligned}[/tex]
    The penultimate step relies upon [itex]h=r^2\dot{\theta}[/itex] and [itex]\hat h \times \hat r = \hat{\theta}[/itex]. The final step relies upon the definition of the circular unit vectors [itex]\hat r[/itex] and [itex]\hat{\theta}[/itex].

    From the above, [itex]d/dt(\vec v \times \vec h) = \mu d/dt(\hat r)[/itex], or [itex]d/dt(\vec v \times \vec h - \mu\hat r) = 0[/itex]. Thus [itex]\mu\vec e = \vec v \times \vec h - \mu\hat r[/itex] is a constant of motion.

    Now we're at the point where we can take the inner product with the position vector.
    [tex]\begin{aligned}
    \mu\vec e \cdot \vec r
    &= (\vec v \times \vec h)\cdot \vec r - \mu \hat r \cdot \vec r \\
    &= (\vec r \times \vec v)\cdot \vec h - \mu r \\
    &= h^2 - \mu r
    \end{aligned}[/tex]

    Another way to express the above inner product is in terms of the magnitudes of the two vectors and the angle f between them: [itex]\mu\vec e \cdot \vec r = \mu e r \cos f[/itex]. Equating these approaches yields
    [tex]\mu e r \cos f = h^2 - \mu r[/tex]
    from which
    [tex]r = \frac{h^2/\mu}{1+e\cos f}[/tex]
    Defining [itex]p=h^2/\mu[/itex] yields [itex]r=p/(1+e\cos f)[/itex], which is the equation of a conic section with semi latus rectum p, eccentricity e, and the origin at one of the foci.
     
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