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

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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 (77.69 KiB) Viewed 12 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 M, and M₂ (F), and the relative position of the third mass, m, (i=73-R). b. (3 marks) Now insert the solutions for the first two masses, i.e., R = constant and F=a cos (St)* + asin(tt)ÿ. where we have chosen their and i axes to be aligned with the orbit initially. Show that the resulting Lagrangian is equivalent to GM₂m GM₂m 16-12-1 + where ₁ = M₂/(M₁ + M₂) and ₂-M₁7/(M₁ + M₂). c. (3 marks) This is more conveniently expressed in the rotating frame in which M₁ and M₂ lie fixed along the z-axis. Show that the resulting Lagrangian is equivalent to + amlift + một (52 xã) tầm + GM₂m GM₂m √laM₂/(M₁ + M₂) -₂1² + y² √aM₁/(M₂+ M₂) + y₂]² + y² 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-mli (²/2) V(i) = -2m2² (1² GM₂m GM₂m √laM₂/(M₂+ M₂) - y² + y² √ [aM₂/(M₂+ M₂) + 3²₂]² + y² e. (3 marks) There are 5 "Lagrange" points, points in for which VV = 0. Make a contour plot of V() 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.