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How far Halley's comet is - Giancoli, p. 153, Pr. 59, 3rd Ed

  1. Jul 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Period of Halley's comet is 76 years. It comes very close to teh surface of the Sun on its closest approach.

    How far out from teh Sun is it at its farthest?


    2. Relevant equations
    - since no semimajor/minor axes data given: assume circular orbit.
    - Earth orbit about Sun is: [tex]1.496 \times {10^{11}}m[/tex]
    - Kepler's 3rd Law:
    [tex]\frac{{{r_{Earth}}^3}}{{{r_{Halley}}^3}} = \frac{{{T_{Earth}}^2}}{{{T_{Halley}}^2}} = {76^2}[/tex]



    3. The attempt at a solution

    Plugging in the above numbers gives me a distance of:
    r[HalleyOrbit] = [tex]2.68 \times {10^{12}}m[/tex]

    Giancoli's back of the book gives:
    r[HalleyOrbit] = [tex]5.4 \times {10^{12}}m[/tex]

    My answer is approximately half of Giancoli's answer. I suspect I oversimplified the problem, but nothing is said of the geometry of Halley's comet to suggest I should use something other than a circle-orbit...although Giancoli says the comet comes "very close" to the Sun.

    Am I missing something?
     
  2. jcsd
  3. Jul 22, 2010 #2
    The assumption you make is not really good. The problem has mentioned both closest and farthest distances, so the orbit should be elliptic instead. For that "it comes very close to the surface of the Sun on its closest approach", we can make a good assumption here: the closest distance d << the farthest distance D, and therefore, d can be ignored. The consequential result is 2a = d+D = D, where a is semi-major axis of the orbit. It's easy to see that D is now twice of your answer, right? :wink:

    Have a look at this to see how the real Halley's orbit looks like: http://en.wikipedia.org/wiki/File:AnimatedOrbitOf1PHalley.gif
     
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