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Radial and Angular Velocities of a Mass in an Elliptical Orbit

  1. Dec 2, 2013 #1
    Given:
    According to my reference (An Introduction to Modern Astrophysics), I should start with this equation:

    [itex]r = \frac{a\left(1 - e^2\right)}{1+ e \cos\left(\theta\right)}[/itex]

    And by using Kepler's Second Law, I should be able to derive the expression for [itex]\mathbf{v}_r[/itex] and [itex]\mathbf{v}_\theta[/itex].

    Attempt:

    Since [itex]\mathbf{v}_r = \frac{dr}{dt}\mathbf{\hat{r}}[/itex], then,

    [itex]\frac{dr}{d\theta}\frac{d\theta}{dt} = \frac{d}{d\theta}
    \left[\frac{a\left(1 - e^2\right)}{1+ e \cos\left(\theta\right)}\right]\frac{d\theta}{dt} \mathbf{\hat{r}}[/itex], and

    [itex]\frac{d}{d\theta}
    \left[\frac{a\left(1 - e^2\right)}{1+ e \cos\left(\theta\right)}\right] = \frac{ae\sin\left(\theta\right)\left(1 - e^2\right)}{\left[1+ e \cos\left(\theta\right)\right]^{2}}[/itex].

    According to Kepler's Second Law,

    [itex]\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt} \\
    \frac{1}{2}\frac{L}{\mu} = \frac{1}{2}r^2\frac{d\theta}{dt} \\
    \frac{L}{\mu r^2} = \frac{d\theta}{dt} \\
    \frac{d\theta}{dt} = \frac{L}{\mu r}\frac{1}{r} = \frac{v}{r} = \frac{2\pi}{P}\\[/itex]

    Thus,

    [itex]\mathbf{v}_r = \frac{2\pi ae\sin\left(\theta\right)\left(1 - e^2\right)}{P\left[1+ e \cos\left(\theta\right)\right]^{2}} \mathbf{\hat{r}}.[/itex]

    The same goes for [itex]\mathbf{v}_\theta[/itex],

    [itex]\mathbf{v}_\theta = r \frac{d\theta}{dt} \mathbf{\hat{\theta}} = \frac{2\pi a\left(1 - e^2\right)}{P\left[1+ e \cos\left(\theta\right)\right]} \mathbf{\hat{\theta}}.[/itex]

    My final answers should only be functions of [itex]P, a, e,[/itex] and [itex]\theta[/itex] only. Please check if my calculations are wrong.

    Now, I need to square both and add them up to be able to acquire this,

    [itex]v = G\left(m_1 + m_2\right)\left(\frac{2}{r} - \frac{1}{a}\right)[/itex]

    Squaring and adding the two velocities,

    [itex]v_r^2 = \frac{4\pi^2 a^2e^2\sin^2\left(\theta\right)\left(1 - e^2\right)^2}{P^2\left[1+ e \cos\left(\theta\right)\right]^{4}}[/itex]

    [itex]v_\theta^2 = \frac{4\pi^2 a^2\left(1 - e^2\right)^2}{P^2\left[1+ e \cos\left(\theta\right)\right]^2}[/itex]

    Their sum should be something like this,

    [itex]v_r^2 + v_\theta^2 = \frac{4 \pi^2 a^2 \left(1 - e^2\right)^2}{P^2\left[1+ e \cos\left(\theta\right)\right]^2}\left\{\frac{e^2\sin^2\left(\theta\right)}{\left[1+ e \cos\left(\theta\right)\right]^2} + 1\right\}[/itex]

    If I multiply [itex]\frac{a}{a}[/itex], I will be able to bring out the gravitational constant [itex]G[/itex],

    [itex]\frac{G\left(m_1 + m_2\right)\left(1 - e^2\right)^2}{a\left[1+ e \cos\left(\theta\right)\right]^2} \left\{\frac{e^2\sin^2\left(\theta\right)}{\left[1+ e \cos\left(\theta\right)\right]^2} + 1\right\}[/itex].

    This is where I get stuck. I tried combining the two latter terms and having a trigonometric identity of [itex]e^2\sin^2\left(\theta\right) + e^2\cos^2\left(\theta\right) = e^2[/itex], but it seems I don't get anywhere doing it. Could you help me or give me tips on what should I do?

    Thank you in advance.
     
  2. jcsd
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