Lagrangian points in circular restricted three-body problem

by johnbills111
Tags: circular, lagrangian, points, restricted, threebody
johnbills111 is offline
Feb27-12, 12:32 AM
P: 1
In the circular restricted 3-body problem, if we consider motion confined to the x-y plane and adopt units such that G(m1 + m2) = 1 [m1 and m2 are the masses of the two heavy bodies], the semimajor axis of the relative orbit of the massive bodies = 1, and n = 1 (n is mean motion, 2∏/period), the effective potential associated with Jacobi's constant is U = -1/2(x2 + y2) - μ1/r1 - μ2/r2, where μ[SUP]1 = m1/(m1+m2), μ[SUP]2 = m2/(m1+m2), r12 = (x + μ2)2 + y2 and r22 = (x - μ1)2 + y2. [Note - it doesn't say so in the question, but I assume x and y are in the rotating frame of the two massive bodies, with both bodies along the x axis, and the origin at their centre of mass.]
(a) The Lagrangian points are locations where a test particle could remain stationary in the rotating frame. Show that these are stationary points of U.
(b) Find the positions of the triangular Lagrangian points L4 and L5 by showing that dU/dr1 = 0 and dU/dr2 = 0 [partial derivatives] would give stationary points.

(a) I am unsure of the overall approach to take.
I calculated the partial derivatives of U with respect to x and y. I know that to find stationary points, I set both of these to zero, then solve the two equations simultaneously. But this will be very messy and it doesn't seem like the right approach here.
I also know -[itex]\nabla[/itex]U should equal the force per unit mass on the test particle, which must be zero if the particle is stationary, as is the case at Lagrangian points. So we get 0 = -dU/dx -dU/dy -dU/dz [partial derivatives]. dU/dz = 0, so we get dU/dx = -dU/dy. As above, I know what dU/dx and dU/dy are. But subbing them in gives an extremely messy equation. And anyway, where would I go next?? Would solving the equation for x and y give me sets of coordinates that I could sub into dU/dx and dU/dy to show they would be zero?
(b) I have written x and y as functions of r1 and r2. Then I could sub these into the expression for U and take dU/dr1 and dU/dr2. But these are also very messy...
Phys.Org News Partner Science news on
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior

Register to reply

Related Discussions
Lagrangian problem (rigid body+particle) Advanced Physics Homework 16
Circular restricted three body problem Astrophysics 1
c code, restricted three body Engineering, Comp Sci, & Technology Homework 2
Restricted circular 3 body problem RC3BP Astrophysics 1
gravitational field, two body problem, lagrangian points General Physics 5