- 3 15 Marks The Restricted 3 Body Problem Part I We All Now Know How To Solve The 2 Body Problem By Reducing It To 1 (282.92 KiB) Viewed 10 times
3. [15 marks]: The Restricted 3-Body Problem, Part I. We all now know how to solve the 2- body problem by reducing it to a 1-body problem. However, the same reduction cannot be profitably made for the general 3-body problem, i.e., three independent, gravitating masses. This results in a variety of interesting and unexpected behaviors as soon as more masses are included. We will begin setting this problem up here in the limit that the third body is a test particle, i.e., its mass is insufficient to influence the motion of the other two; this is called the "restricted 3-body problem". We will further simplify this by assuming the two primary masses are in a circular orbit. The two primary masses are M₁ and M₂, their semi-major axis is a, and the magnitude of their orbital angular frequency is = √G(M₁ + M₂)/a³. You may assume that all of the dynamics takes place in the same plane. a. (3 marks) To begin, write down the general 3-body Lagrangian and make the conversion to the center-of-mass frame for M₁ and M₂. That is, express the Lagrangian in terms of the position of the center of mass (R), the separation between M₁ and M₂ (7), and the relative position of the third mass, m, (7 – 73 – R). b. (3 marks) Now insert the
solutions for the first two masses, i.e., R and 7 – a cos(Nt)à - a sin(t) j where we have chosen the and y axes to be aligned with the orbit initially. Show that the resulting Lagrangian is equivalent to 5² - constant GMAR 72 7 where = M₂/M₁M₂) and 72M₁7/(M, - M₂).
the position of the center of mass (R), the separation between M₁ and M₂ (F), and the relative position of the third mass, m, (ỹ = 7³ – R). b. (3 marks) Now insert the
solutions for the first two masses, i.e., R = constant Fa cos(nt) + asin(t)ý, where we have chosen the 2 and ŷ axes to be aligned with the orbit initially. Show that the resulting Lagrangian is equivalent to and 1 L = -m|y² + 2 L= ST + mi (822 xã) tam X + GM₂m GM₂m GM₂m F1-9 F2-y where = M₂/(M₂+ M₂) and 2-M₁/(M₁ + M₂). c. (3 marks) This is more conveniently expressed in the rotating frame in which M, and M₂ lie fixed along the z-axis. Show that the resulting Lagrangian is equivalent to V(7) = — -m£?²³|71|² E VaM2/(M+M2) M + + GM₂m 2 [aM₁/(M₁ + M₂)→+ v²]² + v,² where the position of m is '. d. (3 marks) Write down the conserved energy and show that the potential experienced in the rotating frame by m is (defined as E-m7²/2) GMm ✓/\aM₂/(Mx + M₂) – 9;]² +4² " \aM₁/(M₂ + M₂) + %,]² + 7,² e. (3 marks) There are 5 "Lagrange" points, points in y' for which VI´ – 0. Make a contour plot of V(7) assuming M₁ - 2M₂ and visually identify these five points. Ostensibly, each of these points are equilibrium points. In Part II we will look at which are stable.