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Computing semimajor axis from ecentricity and period

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Compute the semimajor axes "a" of Halley's Comet.
    Given:
    orbital eccentricity e = 0.9673
    period P =76 days (2.39674E9 seconds)
    Gravitational Constant G = 6.67428E-11
    solar mass M = 1.9891E30 Kg.

    Also used in equations are:
    L = angular momentum of center of mass
    [tex]\mu[/tex] = reduced mass
    r = distance from focus to comet mass
    b= semiminor axis of elipse.

    2. Relevant equations
    I used:

    [tex]
    \frac{dA}{dt} = \frac{L}{2\mu} = \frac{A}{P}
    [/tex]

    [tex] A=\pi a b [/tex]

    [tex] b^2=a^2(1-e^2) [/tex]

    [tex] L=\mu\sqrt{GMa(1-e^2)}[/tex] from text

    3. The attempt at a solution

    [tex] L=\frac{2\pi \mu a^2 (1-e^2)}{P} = \mu \sqrt{GMa(1-e^2}[/tex]

    [tex] a = \sqrt[1/3]{\frac{GMP^2}{4\pi^2 (1-e^2)}} [/tex]

    Unfortunately "a" turns out to be 6.6964E12 meters = 44.76 AU .
    Since the data is for Halley's Comet, a should be 17.8 AU

    Where did I go wrong
     
  2. jcsd
  3. Oct 3, 2009 #2

    gabbagabbahey

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    Looks like you've got an extra factor of [itex](1-\epsilon^2)^{-1/3}[/itex] in your final result...I think the [itex]1-\epsilon^2[/itex] you get from [itex]b^2=a^2(1-\epsilon^2)[/itex] should cancel with the [itex]1-\epsilon^2[/itex] you get from [itex]L=\mu\sqrt{GMa(1-\epsilon^2)}[/itex] when you substitute everything in properly...
     
  4. Oct 3, 2009 #3

    D H

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    Staff Emeritus
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    The eccentricity is not involved in Kepler's third law, suggesting you made an error. And you did. What is the area of an ellipse?
     
  5. Oct 4, 2009 #4
    OK Stupid error. Its not the first time. After getting that straightened out, on rereading the problem I wonder if it is possible to solve for the semimajor axis "a" with only the eccentricity, period and gravitational constant.(ie no solar mass) . Any comments?

    Thanks for your help

    Gary
     
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