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Mathematics LibreTexts

Section 6.4 Answers

  • Page ID
    28917
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    1. If \(e=1\), then \(Y^{2}=\rho (\rho -2X)\); if \(e\neq 1\left(X+\frac{e\rho}{1-e^{2}} \right)^{2}+\frac{Y^{2}}{1-e^{2}}=\frac{\rho ^{2}}{(1-e^{2})^{2}}\) if; \(e<1\) let \(X_{0}=-\frac{e\rho }{1-e^{2}},\: a=\frac{\rho }{1-e^{2}},\: b=\frac{\rho}{1-e^{2}}\)

    2. Let \(h=r_{0}^{2}\theta _{0}'\); then \(\rho = \frac{h^{2}}{k},\:e=\left[\left(\frac{\rho }{r_{0}}-1 \right)^{2}+\left(\frac{\rho r_{0}'}{h} \right)^{2} \right]^{1/2}\). If \(e=0\), then \(\theta _{0}\) is undefined, but also irrelevant if \(e\neq 0\) then \(\phi = \theta _{0}-\alpha\), where \(-\pi\leq\alpha <\pi\cos\alpha =\frac{1}{e}\left(\frac{\rho }{r_{0}}-1 \right)\) and \(\sin\alpha =\frac{pr_{0}'}{eh}\).

    3. 

    1. \(e=\frac{\gamma _{2}-\gamma _{1}}{\gamma _{1}+\gamma _{2}}\)
    2. \(r_{0}=R\gamma _{1},\: r_{0}'=0\), \(\theta _{0}\) arbitrary, \(\theta _{0}'=\left[\frac{2g\gamma _{2}}{R\gamma _{1}^{3}(\gamma _{1}+\gamma _{2})} \right]^{1/2}\)

    4. \(f(r)=-mh^{2}\left(\frac{6c}{r^{4}}+\frac{1}{r^{3}} \right)\)

    5. \(f(r)=-\frac{mh^{2}(\gamma ^{2}+1)}{r^{3}}\)

    6. 

    1. \(\frac{d^{2}u}{d\theta ^{2}}+\left(1-\frac{k}{h^{2}} \right)\: u=0,\: u(\theta _{0} )=\frac{1}{r_{0}},\:\frac{du(\theta _{0})}{d\theta }=-\frac{r_{0}'}{h}\)
    2. with \(\gamma = \left| 1-\frac{k}{h^{2}} \right| ^{1/2}\)
      1. \(r=r_{0}\left(\cosh\gamma (\theta - \theta_{0})-\frac{r_{0}r_{0}'}{\gamma h}\sinh\gamma (\theta -\theta_{0}) \right)^{-1}\)
      2. \(r=r_{0}\left(1-\frac{r_{0}r_{0}'}{h}(\theta -\theta _{0}) \right)^{-1}\)
      3. \(r=r_{0}\left(\cos\gamma (\theta -\theta_{0})-\frac{r_{0}r_{0}'}{\gamma h}\sin\gamma (\theta-\theta_{0}) \right)^{-1}\)