Lagrangian points in circular restricted three-body problem

[Deleted member 396972]

QUESTION:
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), μ2 = 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.

MY PROGRESS:
(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 -$\nabla$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...</sup>

 

[AndyW]

Hello, thank you for your question. I would approach this problem by first understanding the physical significance of the Lagrangian points in the circular restricted 3-body problem. These points represent locations in the x-y plane where a test particle can remain stationary relative to the two massive bodies, due to the combined gravitational forces acting on it.

To show that these points are stationary points of the effective potential U, we can use the definition of a stationary point, which is where the gradient of U is equal to zero. In this case, we have a two-dimensional system, so we can use the partial derivatives with respect to x and y.

Starting with the partial derivative with respect to x, we have:
dU/dx = -x - μ1(x+μ2)/r1^3 - μ2(x-μ1)/r2^3

Setting this equal to zero and simplifying, we get:
x = -μ2(r1^3 - r2^3)/(r1^3 + r2^3)

Similarly, for the partial derivative with respect to y, we have:
dU/dy = -y - μ1y/r1^3 - μ2y/r2^3

Setting this equal to zero and simplifying, we get:
y = 0

These two equations represent the coordinates of the stationary points in terms of the distances r1 and r2. To find the actual coordinates, we can substitute these expressions for x and y into the equations for r1 and r2, which are:
r1 = √[(x+μ2)^2 + y^2]
r2 = √[(x-μ1)^2 + y^2]

Solving for r1 and r2, we get:
r1 = √[(-μ2)^2 + 0^2] = μ2
r2 = √[(-μ1)^2 + 0^2] = μ1

Substituting these values back into the expressions for x and y, we get the coordinates of the stationary points:
x = -μ2(r1^3 - r2^3)/(r1^3 + r2^3) = -μ2(μ2^3 - μ1^3)/(μ2^3 + μ1^3)
y = 0

These coordinates correspond to the Lagrangian points L1 and L2,