Dynamics of Planetary Systems

1. The two-body problem

1.2 The shape of the Kepler orbit

• the shape of the orbit is given by

  $$\begin{array}{}\text{(1.29)}& r=\frac{a(1-e^2)}{1+e\cos f}\end{array}$$

  where $f=\phi -\varpi$ is known as the true anomaly

• the closest approach of the two bodies occurs at $f=0$ or azimuth $\phi =\varpi$, the longitude of periapsis is $\varpi$.

1.3 Motion in the Kepler orbit

• the mean motion or mean rate of change of azimuth

  $$\begin{array}{}\text{(1.44)}& n=\frac{2\pi }{P}\end{array}$$

• the mean anomaly

  $$\begin{array}{}\text{(1.45)}& l=2\pi \frac{t-t_0}{P}=n(t-t_0)\end{array}$$

• define the eccentric anomaly by

  $$\begin{array}{}\text{(1.46)}& r=a(1-e\cos u)\end{array}$$

• Kepler's equation

  $$\begin{array}{}\text{(1.49)}& l=u-e\sin u\end{array}$$

1.8 Nearly circular orbits

1.8.2 The epicycle approximation

• the azimuthal frequency is

  $$\begin{array}{}\text{(1.161)}& {\kappa }_{\varphi }(R_g)\equiv \frac{L_z}{R_g^2}={\left(\frac{1}{R}\frac{\partial \mathrm{\Phi }}{\partial R}\right)}_{(R_g,0)}^{1/2}\end{array}$$

  the vertical frequency is

  $$\begin{array}{}\text{(1.165)}& {\kappa }_z(R_g)={\left(\frac{{\partial }^2\mathrm{\Phi }}{\partial z^2}\right)}_{(R_g,0)}^{1/2}\end{array}$$

  the radial or epicycle frequency is

  $$\begin{array}{}\text{(1.168)}& {\kappa }_R(R_g)={(\frac{3L_z^2}{R^4}+\frac{{\partial }^2\mathrm{\Phi }}{\partial R^2})}_{(R_g,0)}^{1/2}={(\frac{3}{R}\frac{\partial \mathrm{\Phi }}{\partial R}+\frac{{\partial }^2\mathrm{\Phi }}{\partial R^2})}_{(R_g,0)}^{1/2}\end{array}$$

  derivation (Lagrange's equation & perturbation method)

  consider an axisymmetric potential $\mathrm{\Phi }(R,z)$ in cylindrical coordinates $(R,\varphi ,z)$, and assume that the potential is symmetric about the equatorial plane $z=0$, so $\mathrm{\Phi }(R,-z)=\mathrm{\Phi }(R,z)$. The equations of motion for a test particle, $\ddot{r}=-\mathrm{\nabla }\mathrm{\Phi }$, can be written:

  $$\begin{array}{}\text{(1.156)}& \ddot{R}-R{\dot{\varphi }}^2=-\frac{\partial \mathrm{\Phi }(R,z)}{\partial R},2\dot{R}\dot{\varphi }+R\ddot{\varphi }=0,\ddot{z}=-\frac{\partial \mathrm{\Phi }(R,z)}{\partial z}\end{array}$$

  the second equation implies,

  $$\begin{array}{}\text{(1.157)}& R^2\dot{\varphi }=\text{constant}=L_z\end{array}$$

  the first equation can be rewritten as

  $$\begin{array}{}\text{(1.158)}& \ddot{R}-\frac{L_z^2}{R^3}=-\frac{\partial \mathrm{\Phi }(R,z)}{\partial R}\end{array}$$

  examine a circular orbit in the equatorial plane, $R(t)\equiv R_g=\text{constant}$, $z(t)=0$. Equation (1.158) yields

  $$\begin{array}{}\text{(1.159)}& L_z^2=R_g^3{\left(\frac{\partial \mathrm{\Phi }}{\partial R}\right)}_{(R_g,0)}\end{array}$$

  1. Equation (1.157) can be solved$$\boxed{{\kappa }_{\varphi }=\dot{\varphi }=\frac{L_z}{R_g^2}={\left(\frac{1}{R}\frac{\partial \mathrm{\Phi }}{\partial R}\right)}_{(R_g,0)}^{1/2}}$$

  Now consider a nearly circular orbit with the same $z$-component of angular momentum $L_z$ as in equation (1.159). Let

  $$\begin{array}{}\text{(1.162)}& x\equiv R-R_g\end{array}$$

  and expand the potential in a Taylor series around $(R,z)=(R_g,0)$:

  $$\begin{array}{}\text{(1.163)}& \mathrm{\Phi }(R,z)={\left(\frac{\partial \mathrm{\Phi }}{\partial R}\right)}_{(R_g,0)}x+\frac{1}{2}{\left(\frac{{\partial }^2\mathrm{\Phi }}{\partial R^2}\right)}_{(R_g,0)}{x}^2+\frac{1}{2}{\left(\frac{{\partial }^2\mathrm{\Phi }}{\partial z^2}\right)}_{(R_g,0)}{z}^2\end{array}$$

  The epicycle approximation neglect all terms in the expansion above that are higher than second order in $x$ and $z$. 2. Substituting the Taylor series (1.163) into the third of equations (1.156) to obtain

  $$\begin{array}{}\text{(1.164)}& \ddot{z}+{\kappa }_z^{2}z=0,\end{array}$$

  where the vertical frequency is

  $$\begin{array}{}\text{(1.165)}& \boxed{{\kappa }_z(R_g)={\left(\frac{{\partial }^2\mathrm{\Phi }}{\partial z^2}\right)}_{(R_g,0)}^{1/2}}\end{array}$$

  3. We next turn to the radial equation of motion (1.158), replacing $\ddot{R}$ by $\ddot{x}$, the potential $\mathrm{\Phi }(R,z)$ by its Taylor expansion (1.163), and $L_z^2/R^3$ by its Taylor expansion $L_z^2/(R_g+x{)}^3=L_z^2/R_g^3-3(L_z^2/R_g^4)x+O(x^2)$. The terms independent of $x$ cancel because of equation (1.159), and discarding all terms that are higher than first order in $x$ or $z$, we obtain$$\begin{array}{}\text{(1.167)}& \ddot{x}+{\kappa }_R^{2}x=0,\end{array}$$where the radial or epicycle frequency is$$\begin{array}{}\text{(1.168)}& \boxed{{\kappa }_R(R_g)={(\frac{3L_z^2}{R^4}+\frac{{\partial }^2\mathrm{\Phi }}{\partial R^2})}_{(R_g,0)}^{1/2}={(\frac{3}{R}\frac{\partial \mathrm{\Phi }}{\partial R}+\frac{{\partial }^2\mathrm{\Phi }}{\partial R^2})}_{(R_g,0)}^{1/2}}\end{array}$$

• the solution of $R$ and $\varphi$:

  $$R(t)=R_g+x_0\cos ({\kappa }_{R}t+\eta )$$$$\begin{array}{}\text{(1.170)}& \varphi (t)={\kappa }_{\varphi }t+{\varphi }_0-\frac{2x_0}{R_g}\frac{{\kappa }_{\varphi }}{{\kappa }_R}\sin ({\kappa }_{R}t+\eta )\end{array}$$

  derivation

  1. It's easy to solve $x$ from equation (1.167)$$\begin{array}{}\text{(1.169)}& x(t)=x_0\cos ({\kappa }_{R}t+\eta ),\end{array}$$with integration constants $x_0\ge 0$ and $\eta$.
     Substituting into relation (1.162), it reads$$\boxed{R(t)=R_g+x_0\cos ({\kappa }_{R}t+\eta )}$$
  2. Solve the azimuthal motion by writing equation (1.157) in the form$$\dot{\varphi }=\frac{L_z}{(R_g+x{)}^2}={\kappa }_{\varphi }(1-\frac{2x}{R_g})+O(x^2),$$dropping terms higher than $O(x)$ and using equation (1.169), we find$$\begin{array}{}\text{(1.170)}& \boxed{\varphi (t)={\kappa }_{\varphi }t+{\varphi }_0-\frac{2x_0}{R_g}\frac{{\kappa }_{\varphi }}{{\kappa }_R}\sin ({\kappa }_{R}t+\eta )}\end{array}$$

  comfine the integration constants

  Comparing the $R(t)$ with the equation (1.46),

  $$\begin{array}{rl}R(t)& =R_g+x_0\cos ({\kappa }_{R}t+\eta )\\ \text{(1.46)}& r& =a(1-e\cos u)\end{array}$$

  we find that $R_g=a$, $x_0=ae$, so we have

  $$\begin{array}{}\text{(a)}& \boxed{\begin{array}{rl}R(t)& =a(1+e\cos ({\kappa }_{R}t+\eta ))\\ \varphi (t)& ={\kappa }_{\varphi }t-\frac{2{\kappa }_{\varphi }e}{{\kappa }_R}\sin ({\kappa }_{R}t+\eta )+{\varphi }_0\end{array}}\end{array}$$