Lagrangian points anyone?

http://en.wikipedia.org/wiki/Lagrangian_point" [Broken]
"stationary solutions of the circular restricted three-body problem" So for earth sun nearly circular orbit, an object would be stationary. Quite interesting site.

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russ_watters
Mentor
Yes, I'll have some....

wolram
Gold Member
Dearly Missed
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.

D H
Staff Emeritus
I'll use subscripts "e" and "s" to denote the Earth and Sun. Put a tiny mass at some distance $R_e-\epsilon_r$ from the Sun but opposite the Earth. We want to set $\epsilon_r$ 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, $k=M_e/M_s$, our test point is a distance $R_e(1+k/(1+k)-\epsilon_r/Re)[/tex] from the center of mass. To make Newton's laws work in this rotating frame we need to add a fictitious centripetal acceleration [itex]a_{frame} = R_e(1+k/(1+k)-\epsilon_r/Re)\omega^2$. The frame rotation rate is determined by Newton's second law,
$\omega^2 = G(M_s+M_e)/R_e^3 = (1+k)GM_s/R_e^3$. 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, $\epsilon_r \approx 7/12kR_e$. 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 $R_e(1-k/(1+k)) \approx Re(1-k)$, 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.