1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrangian points anyone?

  1. Dec 2, 2007 #1
    "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


    User Avatar

    Staff: Mentor

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


    User Avatar
    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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."
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?