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Apocentre and Eccentricity of a Comet's Orbit

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  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    If a comet passes close to the Earth with a velocity of 38km/s, how far is the apocentre of its orbit from the Sun? What is the eccentricity of its orbit?


    2. Relevant equations
    Vis-Viva Equation:

    (V)[itex]_{ecc}[/itex][itex]^{2}[/itex] = GM [ [itex]\frac{2}{R}[/itex] - [itex]\frac{1}{a}[/itex] ]

    Apocentre distance:

    a(1 + e)


    3. The attempt at a solution

    I found the semi-major axis using the vis-viva equation, but I'm really not sure what to do next,I think I should be trying to find where the radial velocity is zero, but I don't seem to have enough information.
     
  2. jcsd
  3. Feb 4, 2012 #2
    I think I just realized what I need to do. Do I just need to plug the apocentre distance into the vis-viva equation and solve for e?

    EDIT: nevermind, still stuck, just tried this and I got a negative eccentricity.
     
    Last edited: Feb 4, 2012
  4. Feb 5, 2012 #3

    gneill

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    Staff: Mentor

    Is the above the entire problem statement, word for word?

    As it is, the problem statement doesn't provide enough information to determine the precise shape of the comet's orbit. The semimajor axis is accessible via the vis-viva equation as you've already determined. What value did you get? What does that say about the general overall shape of the orbit? Can you at least put bounds on the possible values of the eccentricity?
     
  5. Feb 5, 2012 #4
    Yeah, that's all the problem statement says, word for word.

    Anyway, I got 4.02*10[itex]^{11}[/itex] metres as the semimajor axis. Given that it's a comet, I'm at least assuming that the orbit is highly eccentric, less than 1, since it asks for the apocentre, but greater than, probably 0.9, but without more information it's difficult to get more precise.
     
  6. Feb 5, 2012 #5

    gneill

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    Staff: Mentor

    Yup. 4.02x1011m is about 2.7 AU. 2a about 5.4 AU. Mars' orbit is about 1.5 AU, Jupiter's about 9.5 AU. Looks more like a wayward asteroid than a comet. The aphelion could be as far out as nearly 5.4 AU (assuming the comet is to survive its passage around the Sun), and could be a lot closer to the Earth's 1 AU if the eccentricity is closer to zero.
     
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