限制性三体问题(2): 拉格朗日点与其稳定性分析

上一篇介绍了平面圆形限制性三体问题在旋转坐标系O-xyz中的运动方程:

\ddot{x}-2n\dot{y}=\frac{\partial \Omega}{\partial x} \\\ddot{y}+2n\dot{x}=\frac{\partial \Omega}{\partial y} \\\ddot{z}=\frac{\partial \Omega}{\partial z} \\

其中：

\Omega=\frac{1}{2}n^{2}\left(x^{2}+y^{2}\right)+\frac{1-\mu}{\sqrt{\left(x+\mu \right)^{2}+y^{2}+z^{2}}}+\frac{\mu}{\sqrt{\left(x-1+\mu \right)^{2}+y^{2}+z^{2}}} \\

为等效势, \frac{1}{2}n^{2}\left(x^{2}+y^{2}\right)为离心势.

拉格朗日点

令v_{x}=\dot{x},v_{y}=\dot{y},v_{z}=\dot{z},则运动方程可以写为:

\dot{v}_{x}=2v_{y}+\frac{\partial \Omega}{\partial x} \\\dot{v}_{y}=2v_{x}+\frac{\partial \Omega}{\partial y} \\\dot{v}_{z}=\frac{\partial \Omega}{\partial z} \\

若m_{3}在惯性坐标系中某些点速度为\textbf{v}=\textbf{0},则称这些点为拉格朗日点.此时v_{x}=v_{y}=v_{z}=\dot{v}_{x}=\dot{v}_{y}=\dot{v}_{z}=0, 根据上述运动方程, 则得到:

\frac{\partial \Omega}{\partial x}=\frac{\partial \Omega}{\partial y}=\frac{\partial \Omega}{\partial z}=0 \\

此时该方程得到的解为旋转坐标系O-xyz中的不动点解, 在惯性坐标系中O-\xi\eta\zeta中称为平动点.

将\Omega的表达式代入上述方程, 得:(已假定n=1)

\frac{\partial \Omega}{\partial x}=x-\frac{1-\mu}{\rho_{1}^{3}}(x+\mu)-\frac{\mu}{\rho_{2}^{3}}(x+\mu-1)=0 \\\frac{\partial \Omega}{\partial y}=y\left(1-\frac{1-\mu}{\rho_{1}^{3}}-\frac{\mu}{\rho_{2}^{3}}\right)=0 \\\frac{\partial \Omega}{\partial z}=-z\left(\frac{1-\mu}{\rho_{1}^{3}}+\frac{\mu}{\rho_{2}^{3}}\right)=0 \\

其中

\rho_{1}(x,y,z)=\sqrt{\left(x+\mu \right)^{2}+y^{2}+z^{2}} \\\rho_{2}(x,y,z)=\sqrt{\left(x-1+\mu \right)^{2}+y^{2}+z^{2}} \\

根据\frac{1-\mu}{\rho_{1}^{3}}+\frac{\mu}{\rho_{2}^{3}}>0可知z=0, 故不动点均处于Oxy平面上.

1. 若y=0, 则\rho_{1}(x,0,0)=|x+\mu|,\rho_{2}(x,0,0)=|x-1+\mu|.则得到：

x-\frac{1-\mu}{|x+\mu|^{3}}(x+\mu)-\frac{\mu}{|x-1+\mu|^{3}}(x+\mu-1)=0 \\

(a). 若-\mu<x<1-\mu, 此时不动点x_{1}在m_{1},m_{2}之间, 则:

\rho_{1}=x_{1}+\mu,\rho_{2}=-x_{1}+1-\mu,\rho_{1}+\rho_{2}=1 \\

此时方程为：

1-\mu-\rho_{2}-\frac{1-\mu}{\rho_{1}^{2}}+\frac{\mu}{\rho_{2}^{2}}=0 \\

整理,得：

\left(\frac{\mu}{3}\right)^{\frac{1}{3}}=\rho_{2}\left(\frac{1-\rho_{2}+\frac{1}{3}\rho_{2}^{2}}{1-2\rho_{2}+\rho_{2}^{2}+2\rho_{2}^{3}-\rho_{2}^{4}}\right)^{\frac{1}{3}}=\alpha \\

泰勒展开：(其中\rho_{2}\ll 1)

\frac{1}{1-2\rho_{2}+\rho_{2}^{2}+2\rho_{2}^{3}-\rho_{2}^{4}}=1+2\rho_{2}+3\rho_{2}^{2}+O\left(\rho_{2}^{3}\right) \\

则：

\frac{1-\rho_{2}+\frac{1}{3}\rho_{2}^{2}}{1-2\rho_{2}+\rho_{2}^{2}+2\rho_{2}^{3}-\rho_{2}^{4}}=1+\rho_{2}+\frac{4}{3}\rho_{2}^{2}+O\left(\rho_{2}^{3}\right) \\

则:

\alpha=\rho_{2}\left(1+\rho_{2}+\frac{4}{3}\rho_{2}^{2}+O\left(\rho_{2}^{3}\right)\right)^{\frac{1}{3}}=\rho_{2}+\frac{1}{3}\rho_{2}^{2}+\frac{1}{3}\rho_{2}^{3}+O\left(\rho_{2}^{4}\right) \\

根据级数反演公式,得：

\rho_{2}=\alpha-\frac{1}{3}\alpha^{2}-\frac{1}{9}\alpha^{3}+O\left(\alpha^{4}\right) \\

故此时得到一个近似解:

x_{1}=1-\mu-\left(\frac{\mu}{3}\right)^{\frac{1}{3}}+\frac{1}{3}\left(\frac{\mu}{3}\right)^{\frac{2}{3}}+\frac{1}{9}\left(\frac{\mu}{3}\right)^{\frac{3}{3}}+... \\

该解x_{1}称为拉格朗日L_{1}点;

(b). 若x>1-\mu, 此时不动点x_{2}在m_{2}的右边, 则：

\rho_{1}=x_{2}+\mu, \rho_{2}=x_{2}+\mu-1, \rho_{1}=1+\rho_{2} \\

此时方程为：

1-\mu+\rho_{2}-\frac{1-\mu}{\rho_{1}^{2}}-\frac{\mu}{\rho_{2}^{2}}=0 \\

整理,得：

\left(\frac{\mu}{3}\right)^{\frac{1}{3}}=\rho_{2}\left(\frac{1+\rho_{2}+\frac{1}{3}\rho_{2}^{2}}{1+2\rho_{2}+\rho_{2}^{2}+2\rho_{2}^{3}+\rho_{2}^{4}}\right)^{\frac{1}{3}}=\alpha \\

泰勒展开：(其中\rho_{2}\ll 1)

\frac{1}{1+2\rho_{2}+\rho_{2}^{2}+2\rho_{2}^{3}+\rho_{2}^{4}}=1-2\rho_{2}+3\rho_{2}^{2}+O\left(\rho_{2}^{3}\right) \\

则：

\frac{1+\rho_{2}+\frac{1}{3}\rho_{2}^{2}}{1+2\rho_{2}+\rho_{2}^{2}+2\rho_{2}^{3}+\rho_{2}^{4}}=1-\rho_{2}+\frac{4}{3}\rho_{2}^{2}+O\left(\rho_{2}^{3}\right) \\

则:

\alpha=\rho_{2}\left(1-\rho_{2}+\frac{4}{3}\rho_{2}^{2}+O\left(\rho_{2}^{3}\right)\right)^{\frac{1}{3}}=\rho_{2}-\frac{1}{3}\rho_{2}^{2}+\frac{1}{3}\rho_{2}^{3}+O\left(\rho_{2}^{4}\right) \\

则有：

\rho_{2}=\alpha+\frac{1}{3}\alpha^{2}-\frac{1}{9}\alpha^{3}+O\left(\alpha^{4}\right) \\

故此时得到一个近似解:

x_{2}=1-\mu+\left(\frac{\mu}{3}\right)^{\frac{1}{3}}+\frac{1}{3}\left(\frac{\mu}{3}\right)^{\frac{2}{3}}-\frac{1}{9}\left(\frac{\mu}{3}\right)^{\frac{3}{3}}+... \\

该解x_{2}称为拉格朗日L_{2}点;

(c). 若x<-\mu, 此时不动点x_{3}在m_{1}左侧, 则:

\rho_{1}=-x_{3}-\mu, \rho_{2}=1-x_{3}-\mu, \rho_{1}=\rho_{2}-1 \\

此时方程为：

1-\mu-\rho_{2}+\frac{1-\mu}{(\rho_{2}-1)^{2}}+\frac{\mu}{\rho_{2}^{2}}=0 \\

不妨设\rho_{2}-1=1-\beta, \beta\ll 1, 则上述方程可以整理为：

\mu\left[1+\frac{1}{(1-\beta)^{2}}-\frac{1}{(2-\beta)^{2}}\right]=\frac{1}{(1-\beta)^{2}}-(1-\beta) \\

泰勒展开:

\frac{1}{(1-\beta)^{2}}=1+2\beta+3\beta^{2}+4\beta^{3}+O\left(\beta^{4}\right) \\\frac{1}{(2-\beta)^{2}}=\frac{1}{4}+\frac{1}{4}\beta+\frac{3}{16}\beta^{2}+O\left(\beta^{3}\right) \\

则：

\mu=\left[1+\frac{1}{(1-\beta)^{2}}-\frac{1}{(2-\beta)^{2}}\right]^{-1}\left[\frac{1}{(1-\beta)^{2}}-(1-\beta)\right] \\=\frac{12}{7}\beta\left(1+\beta+\frac{4}{3}\beta^{2}\right)\left(1-\beta-\frac{17}{28}\beta^{2}\right)+O\left(\beta^{6}\right) \\=\frac{12}{7}\left[\beta-\frac{23}{84}\beta^{3}-\frac{163}{84}\beta^{4}-\frac{68}{84}\beta^{5}\right]+O\left(\beta^{6}\right) \\

根据级数反演公式,得:

\beta=\frac{7}{12}\mu+\frac{1127}{20736}\mu^{3}+O\left(\mu^{4}\right) \\

则:

x_{3}=-1-\mu+\beta=-1-\frac{5}{12}\mu+\frac{1127}{20736}\mu^{3}+... \\

该解x_{3}称为拉格朗日L_{3}点.

上述给出L_{1},L_{2},L_{3}的推导过程. 下图给出函数

f(x)=x-\frac{1-\mu}{|x+\mu|^{3}}(x+\mu)-\frac{\mu}{|x-1+\mu|^{3}}(x+\mu-1) \\

的图像,其中\mu=0.01. 可见该函数有三个零点,分别对应于三个拉格朗日点.

1. 若y\neq 0, 则有:

1-\frac{1-\mu}{\rho_{1}^{3}}-\frac{\mu}{\rho_{2}^{3}}=0 \\

而根据:

\frac{\partial \Omega}{\partial x}=x-\frac{1-\mu}{\rho_{1}^{3}}(x+\mu)-\frac{\mu}{\rho_{2}^{3}}(x+\mu-1)=0 \\

得:

\rho_{1}=\sqrt{\left(x+\mu \right)^{2}+y^{2}}=\rho_{2}=\sqrt{\left(x-1+\mu \right)^{2}+y^{2}} \\

解得:

x=-\mu+\frac{1}{2}, y=\pm \frac{\sqrt{3}}{2} \\

该两点称为拉格朗日L_{4},L_{5}点,且与两个主天体m_{1},m_{2}形成等边三角形,故称为三角不动点.

下图给出当\mu=0.1时,五个拉格朗日点的分布图.

拉格朗日点的稳定性

上述根据计算给出五个拉格朗日点, 那么这五个点是否稳定呢？接下来就来考虑这个问题.

设(x_{0},y_{0})为这五个拉格朗日点中任意一点的坐标. 考虑该点附近坐标为(x,y)=(x_{0}+\delta x, y_{0}+\delta y)的一点, 则在该点有:

\frac{\partial \Omega}{\partial x} \approx \frac{\partial^{2}\Omega}{\partial x^{2}}\delta x+\frac{\partial^{2}\Omega}{\partial x \partial y}\delta y \\\frac{\partial \Omega}{\partial y} \approx \frac{\partial^{2}\Omega}{\partial y^{2}}\delta y+\frac{\partial^{2}\Omega}{\partial x \partial y}\delta x \\

则该点的运动方程可以写为:

\frac{d^{2}\delta x}{dt^{2}}-2\frac{d \delta y}{dt}=\frac{\partial^{2}\Omega}{\partial x^{2}}\delta x+\frac{\partial^{2}\Omega}{\partial x \partial y}\delta y \\\frac{d^{2}\delta y}{dt^{2}}+2\frac{d \delta x}{dt}=\frac{\partial^{2}\Omega}{\partial y^{2}}\delta y+\frac{\partial^{2}\Omega}{\partial x \partial y}\delta x \\

1. 若(x_{0},y_{0})为L_{1},L_{2},L_{3}中的某点, 则y_{0}=0. 此时：

\frac{\partial^{2}\Omega}{\partial x^{2}}|_{x=x_{0}}=1+2\gamma \\\frac{\partial^{2}\Omega}{\partial y^{2}}|_{x=x_{0}}=1-\gamma \\\frac{\partial^{2}\Omega}{\partial x \partial y}|_{x=x_{0}}=0 \\

其中:

\gamma=\frac{1-\mu}{\rho_{1}^{3}}+\frac{\mu}{\rho_{2}^{3}} \\

此时运动方程为:

\frac{d^{2}\delta x}{dt^{2}}-2\frac{d \delta y}{dt}=(1+2\gamma)\delta x \\\frac{d^{2}\delta y}{dt^{2}}+2\frac{d \delta x}{dt}=(1-\gamma)\delta y \\

设置检测轨道:\delta x=Ae^{\omega t},\delta y=Be^{\omega t},其中A,B,\omega均为常数.

若\omega为非零的实数,则可以判断该拉格朗日点是不稳定的； 若\omega为纯虚数, 则可以判断该拉格朗日点是稳定的.

将检测轨道代入上述的运动方程, 且消去参数A,B,得到关于\omega的方程:

\omega^{4}+(2-\gamma)\omega^{2}+(1+2\gamma)(1-\gamma)=0 \\

解得:

\omega^{2}=\frac{\gamma-2\pm\sqrt{\gamma(9\gamma-8)}}{2} \\

要想让\omega为纯虚数, 则:

\gamma(9\gamma-8)>0, (1+2\gamma)(1-\gamma)>0 \\

解得: \frac{8}{9}<\gamma<1. 此为稳定判据.

而事实上, 对于L_{1},L_{2},L_{3}点, 有：

\frac{\partial \Omega}{\partial x}=x-\frac{1-\mu}{\rho_{1}^{3}}(x+\mu)-\frac{\mu}{\rho_{2}^{3}}(x+\mu-1)=0 \\

其中\rho_{1}=|x+\mu|,\rho_{2}=|x-1+\mu|.

整理得:

1-\gamma=\frac{\mu(1-\mu)}{x}\frac{\rho_{2}^{3}-\rho_{1}^{3}}{\rho_{1}^{3}\rho_{2}^{3}}<0 \\

故对于L_{1},L_{2},L_{3}点,\gamma>1,即满足稳定判据, 也就是说L_{1},L_{2},L_{3}点是不稳定点.

2. 若(x_{0},y_{0})为L_{4},L_{5}中的某点, 则x_{0}=-\mu+\frac{1}{2}, y_{0}=\pm\frac{\sqrt{3}}{4}. 此时：

\frac{\partial^{2}\Omega}{\partial x^{2}}|_{x_{0},y_{0}}=\frac{3}{4} \\\frac{\partial^{2}\Omega}{\partial y^{2}}|_{x_{0},y_{0}}=\frac{9}{4} \\\frac{\partial^{2}\Omega}{\partial x \partial y}|_{x_{0},y_{0}}=\pm \frac{3\sqrt{3}}{4}(1-2\mu)=a \\

此时运动方程为:

\frac{d^{2}\delta x}{dt^{2}}-2\frac{d \delta y}{dt}=\frac{3}{4}\delta x+a\delta y \\\frac{d^{2}\delta y}{dt^{2}}+2\frac{d \delta x}{dt}=\frac{9}{4}\delta y+a\delta x \\

将检测轨道代入上述的运动方程, 且消去参数A,B,得到关于\omega的方程:

\omega^{4}+\omega^{2}+\frac{27}{4}\mu(1-\mu)=0 \\

解得:

\omega^{2}=\frac{-1\pm\sqrt{1-27\mu(1-\mu)}}{2} \\

要想让\omega为纯虚数, 则:

1-27\mu(1-\mu)>0, \frac{27}{4}\mu(1-\mu)>0 \\

解得: 0<\mu<\frac{1}{2}-\sqrt{\frac{23}{108}}\approx 0.0385.

由上可知,当\mu<0.0385时, 拉格朗日点L_{4},L_{5}比较稳定. 在以太阳、行星为主天体的限制性三体系统内, 稳定的L_{4},L_{5}点附近可能存在小天体,这些小天体被称为特洛伊小天体/特洛伊小行星(Trojans asteroids).

比如在太阳-木星系统中, 木星轨道前后方,也就是木星的L_{4},L_{5}点附近发现有数千颗特洛伊小行星. 在木星轨道前方,特洛伊小行星称为“希腊群”, 在木星轨道后方,特洛伊小行星称为“特洛伊群”.

此外,太阳-火星系统发现有4颗特洛伊小行星,太阳-海王星系统已经发现了17颗特洛伊小行星, 太阳-地球系统在L_{4}点也发现一个特洛伊小天体.

到此, 已简介完平面圆形限制性三体问题的拉格朗日点和其稳定性.