(a) Express in terms of and . · OlympiadBench #985

Physics · mechanics

Problem

(a) Express α\alpha in terms of M1M_{1} and M2M_{2}. 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, M1M_{1} and M2M_{2}. 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 M1M_{1} and M2M_{2} in terms of the gravitational constant GG and the position vectors r1→\overrightarrow{r_{1}} and r2→\overrightarrow{r_{2}} which point from the origin to M1M_{1} and M2M_{2}, respectively. Context answer: d2dt2r1→=GM2∣r2→−r1→∣3(r2→−r1→)d2dt2r2→=GM1∣r1→−r2→∣3(r1→−r2→)\begin{aligned} & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\ & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right) \end{aligned} Context question: (b) Find the period TT and angular frequency ω\omega of the orbital motion. Context answer: ω=G(M1+M2)∣r2→−r1→∣3\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}} , T=2π∣r2→−r1→∣3G(M1+M2)T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}} Extra Supplementary Reading Materials: Circular restricted three-body problem Let us transform to a non-inertial frame rotating with angular velocity ω⃗=(0,0,ω)\vec{\omega}=(0,0, \omega) about an axis normal to the orbital plane of masses M1M_{1} and M2M_{2}, with the origin at their center of mass. In this frame, M1M_{1} and M2M_{2} are stationary at the Cartesian coordinates (−αR,0,0)(-\alpha R, 0,0) and ((1−α)R,0,0)((1-\alpha) R, 0,0) respectively. The third mass mm is not stationary in this frame; in this non-inertial frame its position is r⃗(t)=(x(t),y(t),0)\vec{r}(t)=(x(t), y(t), 0). The masses satisfy M1,M2≫mM_{1}, M_{2} \gg m. Consider mm to be so small that it does not affect the motion of M1M_{1} or M2M_{2}.

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