(c) By considering the centrifugal acceleration ω2r and Coriolis acceleration −2ω×v, find the acceleration dt2d2r of the third mass in terms of the coordinates and constants given, including ρ1 and ρ2.
Given constants:
3. The circular restricted three-body problem
In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.
Two-body problem
Let's start with the motion of two masses, M1 and M2. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
Context question:
(a) Express the equations of motion of M1 and M2 in terms of the gravitational constant G and the position vectors r1 and r2 which point from the origin to M1 and M2, respectively.
Context answer:
dt2d2r1=r2−r13GM2(r2−r1)dt2d2r2=r1−r23GM1(r1−r2)
Context question:
(b) Find the period T and angular frequency ω of the orbital motion.
Context answer:
ω=∣r2−r1∣3G(M1+M2) , T=2πG(M1+M2)∣r2−r1∣3
Extra Supplementary Reading Materials:
Circular restricted three-body problem
Let us transform to a non-inertial frame rotating with angular velocity ω=(0,0,ω) about an axis normal to the orbital plane of masses M1 and M2, with the origin at their center of mass. In this frame, M1 and M2 are stationary at the Cartesian coordinates (−αR,0,0) and ((1−α)R,0,0) respectively. The third mass m is not stationary in this frame; in this non-inertial frame its position is r(t)=(x(t),y(t),0).
The masses satisfy M1,M2≫m. Consider m to be so small that it does not affect the motion of M1 or M2.
Context question:
(a) Express α in terms of M1 and M2.
Context answer:
α=M1+M2M2
Context question:
(b) Let ρ1(t) and ρ2(t) be the distances from m to M1 and M2 respectively. Express ρ1(t) and ρ2(t) in terms of the coordinates and constants given.
Context answer:
ρ1(t)=(x(t)+αR)2+(y(t))2 , ρ2(t)=(x(t)−(1−α)R)2+(y(t))2