Physics

Physics · mechanics

Problem

(e) Hence, write down an expression of the motion of mm which is a constant. 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}. Context question: (a) Express α\alpha in terms of M1M_{1} and M2M_{2}. Context answer: α=M2M1+M2\alpha=\frac{M_{2}}{M_{1}+M_{2}} Context question: (b) Let ρ1(t)\rho_{1}(t) and ρ2(t)\rho_{2}(t) be the distances from mm to M1M_{1} and M2M_{2} respectively. Express ρ1(t)\rho_{1}(t) and ρ2(t)\rho_{2}(t) in terms of the coordinates and constants given. Context answer: ρ1(t)=(x(t)+αR)2+(y(t))2\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}} , ρ2(t)=(x(t)−(1−α)R)2+(y(t))2\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}} Context question: (c) By considering the centrifugal acceleration ω2r⃗\omega^{2} \vec{r} and Coriolis acceleration −2ω×-2 \omega \times v⃗\vec{v}, find the acceleration d2dt2r⃗\frac{d^{2}}{d t^{2}} \vec{r} of the third mass in terms of the coordinates and constants given, including ρ1\rho_{1} and ρ2\rho_{2}. Context answer: d2dt2r⃗=−GM1r⃗−r1⃗ρ13−GM2r⃗−r2⃗ρ23+ω2r⃗−2ω×r⃗\frac{d^{2}}{d t^{2}} \vec{r}=-G M_{1} \frac{\vec{r}-\vec{r_{1}}}{\rho_{1}^{3}}-G M_{2} \frac{\vec{r}-\vec{r_{2}}}{\rho_{2}^{3}}+\omega^{2} \vec{r}-2 \omega \times \vec{r} Context question: (d) Express d2xdt2\frac{d^{2} x}{d t^{2}} and d2ydt2\frac{d^{2} y}{d t^{2}} in terms of UU, where U=−GM1ρ1−GM2ρ2−ω22(x2+y2)U=-\frac{G M_{1}}{\rho_{1}}-\frac{G M_{2}}{\rho_{2}}-\frac{\omega^{2}}{2}\left(x^{2}+y^{2}\right). Context answer: x¨=2ωy˙−∂U∂x\ddot{x} =2 \omega \dot{y}-\frac{\partial U}{\partial x} , y¨=−2ωx˙−∂U∂y\ddot{y} =-2 \omega \dot{x}-\frac{\partial U}{\partial y}

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