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Kepler's 2nd across a system of planets

  1. Oct 18, 2015 #1
    An interesting question, which I have just seen for the first time...
    Does Kepler's 2nd mean that, for instance, both Pluto and Mercury sweep out an equal area over 1 hour? My gut reaction, without calculating anything, is "yes".
     
  2. jcsd
  3. Oct 18, 2015 #2

    marcus

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    Actually not.
    The law is about one planetary orbit. It doesn't equate sweep-rates of different planets.
    You could imagine a highly elliptical comet orbit with perihelion equal to Mercury's and aphelion equal to Plutos. (nearest and farthest points in orbit)

    At farthest, the comet would be sweeping slower than Pluto, because destined to fall in towards Sun.
    At nearest it would be sweeping faster than Mercury because destined to swing out away from Sun.
    The comet's two rates would be equal (and slower than Pluto's but faster than Mercury's).

    So Pluto > Mercury.

    Also think of the area sweep rate as expressing angular momentum of unit mass in the given orbit.
    Pluto has more angular momentum (per unit mass)
     
    Last edited: Oct 18, 2015
  4. Oct 18, 2015 #3

    Janus

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    This can be mathematically calculated as the Areal velocity for the object.
    for circular orbits, this is found by:
    [tex]A = \sqrt{GMa}[/tex]

    where a is the semi-major radius of the orbit (mean orbital distance)

    for an elliptical orbit, it is found by:
    [tex] \sqrt{GMa \frac{1+e}{1-e}}[/tex]

    Where e is the eccentricity of the orbit.

    Note that as the mean orbital distance goes up, so does the Areal velocity (However, since it also goes up with eccentricity, it would be possible for an object with a high eccentricity to have a greater Areal velocity than a object with a higher mean orbital distance but a lower eccentricity.)
     
  5. Oct 18, 2015 #4
    Yup, makes perfect sense.
     
  6. Oct 18, 2015 #5
    Indeed.
     
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