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Lagrangian points anyone?

  1. Dec 2, 2007 #1
    http://en.wikipedia.org/wiki/Lagrangian_point
    "stationary solutions of the circular restricted three-body problem" So for earth sun nearly circular orbit, an object would be stationary. Quite interesting site.
     
  2. jcsd
  3. Dec 2, 2007 #2

    russ_watters

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    Staff: Mentor

    Yes, I'll have some....
     
  4. Dec 2, 2007 #3

    wolram

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    Gold Member

    It is an interesting subject, for the life of me i can not understand the R3 point, maybe some one will be kind and explain.
     
  5. Dec 2, 2007 #4

    D H

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    Suppose that the only two masses in the universe are the Earth, the Sun, and a tiny little test mass over which we have control. The Earth and Sun are in circular orbits about the Sun-Earth center of mass. Like any of the other Lagrange points, it is most convenient to work in a rotating reference frame whose origin is the Sun-Earth center of mass and rotating such the Sun and Earth have a constant position.

    I'll use subscripts "e" and "s" to denote the Earth and Sun. Put a tiny mass at some distance [itex]R_e-\epsilon_r[/itex] from the Sun but opposite the Earth. We want to set [itex]\epsilon_r[/itex] so that the point orbit will orbit the Sun-Earth center of mass with exactly the same rotation rate as the Earth and Sun orbit about the Sun-Earth center of mass.

    Denoting the ratio of the Earth's mass to the Sun's mass as k, [itex]k=M_e/M_s[/itex], our test point is a distance [itex]R_e(1+k/(1+k)-\epsilon_r/Re)[/tex] from the center of mass. To make Newton's laws work in this rotating frame we need to add a fictitious centripetal acceleration [itex]a_{frame} = R_e(1+k/(1+k)-\epsilon_r/Re)\omega^2[/itex]. The frame rotation rate is determined by Newton's second law,
    [itex]\omega^2 = G(M_s+M_e)/R_e^3 = (1+k)GM_s/R_e^3[/itex]. We want to place the our test mass so that the gravitational attraction to the Sun and to the Earth exactly counterbalances this centripetal acceleration. After a little math, [itex]\epsilon_r \approx 7/12kR_e[/itex]. That is, the point is a little closer to the Sun than is the Earth. However, the Earth orbits the Sun-Earth center of mass at a distance [itex]R_e(1-k/(1+k)) \approx Re(1-k)[/itex], so the L3 point is a little outside the Earth's orbit, but seemingly paradoxically a bit closer to the Sun than is the Earth.

    Which is what the Wiki article says: "L3 in the Sun–Earth system exists on the opposite side of the Sun, a little outside the Earth's orbit but slightly closer to the Sun than the Earth is."
     
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