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Kepler's third law

  1. May 17, 2005 #1
    not sure what to do here..

    Im being asked to compare the periods of 2 different satellites in orbit around a planet.


    the first one is a circular orbit of radius = r

    the second one orbits 1r to the left and 3r to the right around the planet.

    I'll attempt to draw it here :tongue:

    0 is the planet




    (---------0---------)------------------------)
    <---r----><---r----><---------2r------------>



    I understand how the period works in the first circular orbit.. but not the second one. any ideas?
     
  2. jcsd
  3. May 17, 2005 #2

    Doc Al

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    Staff: Mentor

    Start by reviewing what Kepler's 3rd law says.
     
  4. May 17, 2005 #3
    it states that r^3/T^2 = K

    is it just the average radius of the second satellite? (in this case 2r ?)
     
  5. May 17, 2005 #4

    BobG

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    Yes, the mean distance, average radius, or, more commonly, the semi-major axis.

    The shape of the orbit doesn't matter (your second orbit has an eccentricity of .5)
     
  6. May 17, 2005 #5

    DaveC426913

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    So, a satellite in an eccentric orbit will have a period that is equivalent to a circular orbit whose radius is equal to (aphelion minus perihelion) of the eccentric orbit?

    So, if an asteroid happened to be on an orbit that went out as far as Jupiter, and in as far as Mercury, its orbital period would be equivalent to a circular orbit whose radius is (Jupiter's - Mercury's) orbit?

    I'd always wondered that.
     
  7. May 17, 2005 #6

    SpaceTiger

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    A circular orbit whose radius is equal to the semimajor axis of the ellipse. Perihelion is

    [tex]r_{peri}=a(1-e)[/tex]

    where a is the semimajor axis and e is the eccentricity. Aphelion is

    [tex]r_{ap}=a(1+e)[/tex]

    So, the semimajor axis is given not by r_ap - r_peri, but rather:

    [tex]a=\frac{1}{2}(r_{ap}+r_{peri})[/tex]

    This is what scales with period in Kepler's 3rd law:

    [tex]P^2 \propto a^3[/tex]
     
  8. May 17, 2005 #7

    James R

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