\subsection{The Kepler Problem}

This subsection treats the inverse-square central potential $V(r) = -k/r$ through the Hamilton--Jacobi formalism. The gravitational potential $V(r) = -GMm/r = -k/r$ originates from Unit 3 m3-3 (conservative forces and potential energy), and the familiar circular-orbit speed and energy of Unit 6 m6-5 emerge here as special cases: setting $\varepsilon = 0$ in the general conic orbit reproduces $v = \sqrt{GM/r}$ and $E = -GMm/(2r)$.

To appreciate why the Hamilton--Jacobi method is ideally suited to this problem, compare it with the traditional Binet-equation approach. The Binet equation reduces Newton\normalsize{}'s second law in polar coordinates to a linear differential equation for $u(\phi) = 1/r(\phi)$:
\[
\dv[2]{u}{\phi} + u = \frac{\mu k}{L^2}.
\]
This derivation requires knowing in advance to make the substitution $u = 1/r$, a clever trick with no obvious physical motivation. By contrast, the Hamilton--Jacobi method obtains the orbit equation purely from separation constants. The relation between $r$ and $\phi$ follows from Jacobi\normalsize{}'s theorem $\pdv{W}{L} = \beta_L$ as a straightforward quadrature, with no inspired change of variable. The orbit emerges from the geometry of phase space rather than from algebraic guesswork. Moreover, action--angle variables immediately yield Kepler\normalsize{}'s third law and the dynamical degeneracy that makes all bound orbits close, results the Binet equation leaves for separate energy-integral calculations.

\dfn{Kepler Hamiltonian}{
Consider the central potential $V(r) = -k/r$ where $k = GM\mu$ with $G$ the gravitational constant, $M$ the mass of the central body, and $\mu$ the reduced mass of the two--body system. From A.02, the scale factors for spherical coordinates $(r,\theta,\phi)$ are $h_r = 1$, $h_\theta = r$, $h_\phi = r\sin\theta$, giving the kinetic energy $T = \tfrac{1}{2}\mu(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\,\dot{\phi}^2)$. The canonical momenta are $p_r = \mu\dot{r}$, $p_\theta = \mu r^2\dot{\theta}$, $p_\phi = \mu r^2\sin^2\theta\,\dot{\phi}$. The Legendre transform yields the Hamiltonian:
\[
\mcH = \frac{p_r^2}{2\mu} + \frac{p_\theta^2}{2\mu r^2} + \frac{p_\phi^2}{2\mu r^2\sin^2\theta} - \frac{k}{r}.
\]
The Hamilton--Jacobi equation for the principal function $\mcS(r,\theta,\phi,t)$ is
\[
\frac{1}{2\mu}\left(\pdv{\mcS}{r}\right)^2 
+ \frac{1}{2\mu r^2}\left(\pdv{\mcS}{\theta}\right)^2 
+ \frac{1}{2\mu r^2\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 
- \frac{k}{r} + \pdv{\mcS}{t} = 0.
\]
Because $\pdv{\mcH}{t} = 0$ the Hamiltonian is time--independent and energy $E = \mcH$ is conserved.}

\nt{Two--body reduction and reduced mass}{
A system of two bodies with masses $M$ and $m$ interacting through a central potential depends only on the distance between them. Introducing the relative coordinate $\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2$ and the center of mass $\mathbf{R} = (M\mathbf{r}_1 + m\mathbf{r}_2)/(M+m)$, the Lagrangian splits into the free motion of the center of mass and the relative motion with the reduced mass $\mu = Mm/(M+m)$. The relative Hamiltonian has exactly the form of the Kepler Hamiltonian above, with the potential $V(r) = -GMm/r = -k/r$ and $k = GMm = G(M+m)\mu$. In many astrophysical situations $M \gg m$ so that $\mu \approx m$ and the central body is effectively fixed. This reduction is what justifies treating the Hamiltonian as a one--body problem.}

\thm{Separated Hamilton--Jacobi equations for the Kepler problem}{
With the separation ansatz
\[
\mcS(r,\theta,\phi,t) = W_r(r) + W_\theta(\theta) + L_z\phi - Et,
\]
the Hamilton--Jacobi equation breaks into three ordinary differential equations. The azimuthal equation is
\[
\pdv{\mcS}{\phi} = L_z,
\]
a constant. The polar angular equation is
\[
\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = L^2,
\]
where $L$ is the total--angular--momentum separation constant. The radial equation is
\[
\left(\der{W_r}{r}\right)^2 = 2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}.
\]
The three constants of motion $E$, $L$, and $L_z$ provide the complete integral required by Jacobi\normalsize{}'s theorem.}

\pf{Derivation of the separated equations from the full HJ PDE}{
Begin by eliminating the time dependence. Because the Hamiltonian does not depend explicitly on time, set $\mcS(r,\theta,\phi,t) = W(r,\theta,\phi) - Et$. The time derivative contributes $-E$ and the Hamilton--Jacobi equation becomes
\[
\frac{1}{2\mu}\left(\pdv{W}{r}\right)^2 
+ \frac{1}{2\mu r^2}\left(\pdv{W}{\theta}\right)^2 
+ \frac{1}{2\mu r^2\sin^2\theta}\left(\pdv{W}{\phi}\right)^2 
- \frac{k}{r} = E.
\]
The azimuthal angle $\phi$ is absent from the potential, so $\phi$ is a cyclic coordinate. Write $W = W_{r\theta}(r,\theta) + W_\phi(\phi)$, and because the $\phi$--term appears only through $\pdv{W}{\phi}$ it must be a constant:
\[
\pdv{W}{\phi} = L_z.
\]
This is the canonical momentum conjugate to $\phi$ and equals the $z$\-component of the total angular momentum. Substitute $L_z^2$ for $\left(\pdv{W}{\phi}\right)^2$ and rearrange the remaining equation so that the angular terms are separated from the radial terms:
\[
\frac{1}{2\mu}\left(\pdv{W_{r\theta}}{r}\right)^2 - \frac{k}{r} - E
= -\frac{1}{2\mu r^2}\left[\left(\pdv{W_{r\theta}}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta}\right].
\]
Multiply by $2\mu r^2$:
\[
r^2\left(\pdv{W_{r\theta}}{r}\right)^2 - 2\mu k r - 2\mu E r^2
= -\left(\pdv{W_{r\theta}}{\theta}\right)^2 - \frac{L_z^2}{\sin^2\theta}.
\]
The left side depends only on $r$ and the right side depends only on $\theta$. Each must therefore equal a constant, which we call the separation constant $L^2$ because it will be identified with the square of the total angular momentum:
\[
\left(\pdv{W_{r\theta}}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = L^2,
\]
\[
r^2\left(\pdv{W_{r\theta}}{r}\right)^2 - 2\mu k r - 2\mu E r^2 = -L^2.
\]
Assume additive separation $W_{r\theta}(r,\theta) = W_r(r) + W_\theta(\theta)$. The $\theta$--equation becomes
\[
\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = L^2,
\]
and the $r$\-equation becomes
\[
\left(\der{W_r}{r}\right)^2 - \frac{2\mu k}{r} - 2\mu E = -\frac{L^2}{r^2},
\]
which rearranges to
\[
\left(\der{W_r}{r}\right)^2 = 2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}.
\]
The three separation constants are $E$ (total energy), $L$ (total angular momentum magnitude), and $L_z$ (angular momentum $z$\-component). Together with Jacobi\normalsize{}'s theorem, these equations determine the trajectory without solving any second--order differential equation.}

\nt{Centrifugal barrier and effective turning points}{
In the radial equation $\left(\der{W_r}{r}\right)^2 = 2\mu E + 2\mu k/r - L^2/r^2$, the term $L^2/(2\mu r^2)$ acts as a repulsive centrifugal barrier. It can be viewed as part of an effective potential
\[
V_{\mathrm{eff}}(r) = -\frac{k}{r} + \frac{L^2}{2\mu r^2},
\]
so that the radial kinetic energy reads $\tfrac{1}{2}\mu\dot{r}^2 = E - V_{\mathrm{eff}}(r)$. At small $r$ the $1/r^2$ centrifugal term grows faster than the attractive $1/r$ gravitational term, pushing the particle away from the origin even when gravity tries to pull it inward. The turning points of the motion occur where $p_r = 0$, i.e., where $E = V_{\mathrm{eff}}(r)$. For bound orbits ($E < 0$) the equation $2\mu E + 2\mu k/r - L^2/r^2 = 0$ has two positive roots for $1/r$, corresponding to $r_{\min}$ (periapsis) and $r_{\max}$ (apoapsis). The existence of $r_{\min} > 0$ is guaranteed by the centrifugal barrier: whenever $L \neq 0$ the barrier creates a local minimum in $V_{\mathrm{eff}}(r)$ and prevents the particle from reaching the center. This is why all nonradial orbits have a well--defined periapsis and never collide with the center of force.}

\thm{Orbit equation for the Kepler problem}{
The trajectory $r(\phi)$ of a particle moving in the potential $V(r) = -k/r$ is a conic section:
\[
r(\phi) = \frac{\ell}{1 + \varepsilon\cos(\phi - \phi_0)},
\]
where the semi--latus rectum $\ell$ and the eccentricity $\varepsilon$ are determined by the constants of motion:
\[
\ell = \frac{L^2}{\mu k},
\qquad
\varepsilon = \sqrt{1 + \frac{2EL^2}{\mu k^2}},
\qquad
\phi_0 = \text{constant}.
\]
Geometrically, the focus of the conic sits at the origin, which is the center of force. The parameter $\ell$ measures the width of the conic at the point $\phi = \phi_0 + \pi/2$ measured in radians from periapsis. The angle $\phi_0$ gives the angular position of periapsis, the point of closest approach, measured in radians from the reference azimuthal axis. The sign of the total energy selects the conic type: $E < 0$ gives an ellipse (bound orbit), $E = 0$ gives a parabola (marginal escape trajectory), and $E > 0$ gives a hyperbola (unbound scattering orbit).}

\pf{Derivation of the orbit from Jacobi\normalsize{}'s theorem}{
Because the motion takes place in a fixed plane (the plane normal to the angular momentum vector), we may choose the orbital plane as $\theta = \pi/2$. In this plane $\sin\theta = 1$ and the radial momentum equals $p_r = \der{W_r}{r}$. The azimuthal momentum is $p_\phi = L_z = L$ (by choosing the $z$\-axis normal to the orbital plane, the total angular momentum lies along $z$). Jacobi\normalsize{}'s theorem states that the derivatives of the characteristic function $W$ with respect to the separation constants are themselves constants determined by the initial conditions:
\[
\pdv{W}{E} = \beta_E,
\qquad
\pdv{W}{L} = \beta_L,
\qquad
\pdv{W}{L_z} = \beta_{L_z}.
\]
The condition $\pdv{W}{L} = \beta_L$ connects the azimuthal angle with the radial coordinate. We have $W = W_r(r) + W_\theta(\theta) + L_z\phi$. At $\theta = \pi/2$ the polar part of the angular integral is at its turning point and contributes no net change to the derivative. The dependence of $W$ on $L$ enters through $W_r$, where $L$ appears in the effective--potential term $-L^2/r^2$, and through the azimuthal term via $L_z = L$ (since for planar motion all angular momentum lies along the $z$-axis). We evaluate $\pdv{W_r}{L}$ explicitly. Write
\[
W_r(r) = \int \sqrt{2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}}\,\dd r.
\]
The integrand depends on the parameter $L$ through the term $-L^2/r^2$. By the Leibniz integral rule, differentiating with respect to a parameter inside the integral gives
\[
\pdv{W_r}{L} = \int \pdv{}{L}\left[\sqrt{2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}}\right]\dd r.
\]
Differentiating the square root:
\[
\pdv{W_r}{L} = \int\frac{1}{2\sqrt{2\mu E + 2\mu k/r - L^2/r^2}}\cdot\left(-\frac{2L}{r^2}\right)\dd r
= -\int\frac{L}{r^2 p_r}\,\dd r.
\]
The key observation is that the same integral appears in the relation between $\phi$ and $r$. From Hamilton\normalsize{}'s equations or from the $\phi$--part of Jacobi\normalsize{}'s theorem, the azimuthal advance per radial step is
\[
\mathrm{d}\phi = \frac{p_\phi}{\mu r^2}\frac{\dd t}{1}
= \frac{L}{\mu r^2}\frac{\dd r}{p_r/\mu}
= \frac{L}{r^2 p_r}\,\dd r.
\]
Integrating from the initial condition $(r_0,\phi_0)$ to the arbitrary point $(r,\phi)$:
\[
\phi - \phi_0 = \int_{r_0}^{r}\frac{L}{r^2 p_r}\,\dd r.
\]
Comparing the two expressions, the derivative $\pdv{W_r}{L}$ is minus the same integral that gives $\phi - \phi_0$. The condition $\pdv{W}{L} = \beta_L$ therefore relates the azimuthal angle to a constant that sets the orientation of the orbital axis. Evaluating the integral explicitly, write the radial momentum as
\[
p_r = \sqrt{2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}}
= \frac{1}{r}\sqrt{2\mu E r^2 + 2\mu k r - L^2}.
\]
Substitute $u = 1/r$, so $\dd r = -\mathrm{d}\,u/u^2$ and $r^2 = 1/u^2$:
\[
\phi - \phi_0 = \int\frac{L\,(-\mathrm{d}\,u/u^2)}{(1/u^2)\sqrt{2\mu E/u^2 + 2\mu k/u - L^2}}
= -\int\frac{L\,\mathrm{d}\,u}{\sqrt{2\mu E + 2\mu k u - L^2 u^2}}.
\]
Complete the square in the denominator:
\[
2\mu E + 2\mu k u - L^2 u^2
= -L^2\left(u^2 - \frac{2\mu k}{L^2}u - \frac{2\mu E}{L^2}\right)
= -L^2\left[\left(u - \frac{\mu k}{L^2}\right)^2 - \frac{\mu^2 k^2 + 2\mu E L^2}{L^4}\right].
\]
Define $\ell = L^2/(\mu k)$ and $\varepsilon = \sqrt{1 + 2EL^2/(\mu k^2)}$. Then
\[
\frac{\mu^2 k^2 + 2\mu E L^2}{L^4} = \frac{\mu^2 k^2}{L^4}\left(1 + \frac{2EL^2}{\mu k^2}\right)
= \frac{1}{\ell^2}\varepsilon^2.
\]
The integral becomes
\[
\phi - \phi_0 = -\frac{1}{\varepsilon}\arccos\!\left(\frac{\mu k/L^2 - u}{\varepsilon/\ell}\right)
= \arccos\!\left(\frac{\ell/r - 1}{\varepsilon}\right),
\]
up to an integration constant absorbed into $\phi_0$. Inverting this relation:
\[
\cos(\phi - \phi_0) = \frac{\ell/r - 1}{\varepsilon}
= \frac{\ell - r}{\varepsilon r}.
\]
Rearrange:
\[
\varepsilon r\cos(\phi - \phi_0) = \ell - r,
\qquad
r\bigl(1 + \varepsilon\cos(\phi - \phi_0)\bigr) = \ell,
\]
\[
r(\phi) = \frac{\ell}{1 + \varepsilon\cos(\phi - \phi_0)}.
\]
This is the standard polar equation of a conic section with focus at the origin. The parameters $\ell$ and $\varepsilon$ follow from matching the effective--energy expression. The radial turning points occur when $p_r = 0$:
\[
2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2} = 0,
\qquad
r^2 + \frac{\mu k}{\mu E}\,r - \frac{L^2}{2\mu E} = 0.
\]
Solving for the roots gives $r_{\min,\max}$, which for bound orbits are the perihelion and aphelion distances. The difference $r_{\max} - r_{\min} = 2\ell\varepsilon/(1-\varepsilon^2)$ for bound orbits matches the major axis of the ellipse. Matching the conic parameters to the physical constants gives $\ell = L^2/(\mu k)$ and $\varepsilon = \sqrt{1 + 2EL^2/(\mu k^2)}$.}

\mprop{Classification of conic--section orbits by eccentricity}{
The eccentricity $\varepsilon = \sqrt{1 + 2EL^2/(\mu k^2)}$ determines the shape of the orbit $r(\phi) = \ell/(1 + \varepsilon\cos(\phi - \phi_0))$. The orbit is

\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item An ellipse when $\varepsilon < 1$. This corresponds to $-k^2\mu/(2L^2) < E < 0$ and $L \neq 0$. The orbit is bound and closed, with semimajor axis $a = \ell/(1 - \varepsilon^2) = -k/(2E)$ and semiminor axis $b = a\sqrt{1 - \varepsilon^2} = L/\sqrt{2\mu|E|}$. The period of one complete revolution is $T = 2\pi\sqrt{\mu a^3/k}$. The periapsis distance is $r_{\min} = \ell/(1+\varepsilon)$ at angle $\phi = \phi_0$ (measured in radians), and the apoapsis is $r_{\max} = \ell/(1-\varepsilon)$ at $\phi = \phi_0 + \pi$.

\item A circle when $\varepsilon = 0$, which occurs at the special energy $E = -k^2\mu/(2L^2)$. The distance $r = \ell$ is constant throughout the motion, and the motion reduces to uniform circular motion with angular speed $\omega = \sqrt{k/(\mu r^3)}$. This matches the circular-orbit results of Unit 6 m6-5.

\item A parabola when $\varepsilon = 1$, corresponding to the critical energy $E = 0$. The trajectory is unbound, and the particle arrives from infinity, swings by the central mass once, and returns to infinity with zero residual speed.

\item A hyperbola when $\varepsilon > 1$, corresponding to $E > 0$. The trajectory is unbound with positive energy, approaching from infinity with a nonzero residual speed after the encounter. The angle between the two asymptotes of the hyperbola is $2\arccos(-1/\varepsilon)$, measured in radians.
\end{enumerate}}

\nt{Action--angle variables and Kepler\normalsize{}'s third law}{
The three independent action variables for the Kepler problem are computed by integrating the appropriate momenta over one complete cycle of each coordinate. For the azimuthal coordinate, $J_\phi = \oint p_\phi\,\mathrm{d}\phi = 2\pi L_z$. For the polar coordinate, $J_\theta = \oint p_\theta\,\mathrm{d}\theta = 2\pi(L - |L_z|)$. For the radial coordinate, the integral $J_r = \oint p_r\,\dd r$ requires careful evaluation between the two radial turning points for bound orbits ($E < 0$). The result is $J_r = 2\pi(-L + k\sqrt{\mu/(2|E|)})$.

Each action variable carries a distinct physical meaning. The azimuthal action $J_\phi = 2\pi L_z$ measures the conserved rotation about the vertical axis: it is the circulation of angular momentum along the $\phi$ direction and sets the rate of azimuthal precession. The polar action $J_\theta = 2\pi(L - |L_z|)$ measures the inclination of the orbital plane: when the orbit is equatorial ($L = |L_z|$) we have $J_\theta = 0$, and larger values of $L - |L_z|$ correspond to a more tilted orbit with greater polar oscillation between $\theta_{\min}$ and $\pi - \theta_{\min}$. The radial action $J_r = 2\pi(-L + k\sqrt{\mu/(2|E|)})$ measures the extent of the radial excursion between periapsis and apoapsis: for a circular orbit $J_r = 0$ (no radial oscillation), and for highly eccentric orbits $J_r$ grows as the particle swings farther from the center.

Adding all three actions eliminates the angular--momentum dependence:
\[
J_{\mathrm{tot}} = J_r + J_\theta + J_\phi 
= 2\pi k\sqrt{\frac{\mu}{2|E|}}.
\]
Inverting this expression gives $|E| = 2\pi^2\mu k^2/J_{\mathrm{tot}}^2$, which expresses the energy in terms of the single total action $J_{\mathrm{tot}}$. Because $E$ depends on $J_{\mathrm{tot}} = J_r + J_\theta + J_\phi$ through a sum, the three frequency derivatives $\pdv{E}{J_r}$, $\pdv{E}{J_\theta}$, and $\pdv{E}{J_\phi}$ are all equal. Equal frequencies mean the radial period equals the angular period, so every bound orbit closes on itself. This degeneracy is the deep mathematical origin of Kepler\normalsize{}'s third law.}

\ex{Action--angle derivation of $E(J_{\mathrm{tot}})$ and the degenerate frequencies}{
We evaluate the three action variables for the Kepler problem explicitly.

\textbf{Azimuthal action.} The momentum conjugate to $\phi$ is $p_\phi = L_z$, a constant. Integrating over one full revolution:
\[
J_\phi = \oint p_\phi\,\mathrm{d}\phi = \int_{0}^{2\pi} L_z\,\mathrm{d}\phi = 2\pi L_z.
\]

\textbf{Polar action.} The polar momentum is $p_\theta = \sqrt{L^2 - L_z^2/\sin^2\theta}$. The turning points satisfy $\sin\theta_{\min} = |L_z|/L$ and $\sin\theta_{\max} = |L_z|/L$ with $\theta_{\max} = \pi - \theta_{\min}$. The integral over one oscillation is
\[
J_\theta = 2\int_{\theta_{\min}}^{\pi-\theta_{\min}}\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\mathrm{d}\theta.
\]
The substitution $u = \cos\theta$ converts the integrand to $\sqrt{L^2 - L_z^2/(1-u^2)}$, and the integral evaluates to
\[
J_\theta = 2\pi\bigl(L - |L_z|\bigr).
\]

\textbf{Radial action.} The radial momentum is $p_r = \pm\sqrt{2\mu E + 2\mu k/r - L^2/r^2}$. For bound orbits ($E < 0$) write $|E| = -E$. The turning points are the roots of $2\mu|E|r^2 - 2\mu kr - L^2 = 0$, which are
\[
r_{\pm} = \frac{\mu k \pm L\sqrt{\mu^2 k^2 - 2\mu|E|L^2}}{2\mu|E|}.
\]
The radial action integral is
\[
J_r = 2\int_{r_-}^{r_+}\sqrt{2\mu|E| + \frac{2\mu k}{r} - \frac{L^2}{r^2}}\;\frac{\dd r}{r^2/r^2}.
\]
The standard contour--integration or elliptic--integral evaluation gives
\[
J_r = 2\pi\!\left(-L + k\sqrt{\frac{\mu}{2|E|}}\right).
\]

\textbf{Total action and energy.} Adding the three actions:
\[
J_{\mathrm{tot}} = J_r + J_\theta + J_\phi
= 2\pi\!\left(-L + k\sqrt{\frac{\mu}{2|E|}}\right) + 2\pi(L - |L_z|) + 2\pi L_z
= 2\pi k\sqrt{\frac{\mu}{2|E|}}.
\]
The angular--momentum terms $L$ and $|L_z|$ cancel exactly. Invert the total--action relation to obtain the energy:
\[
\sqrt{\frac{\mu}{2|E|}} = \frac{J_{\mathrm{tot}}}{2\pi k},
\qquad
\frac{2|E|}{\mu} = \frac{4\pi^2 k^2}{J_{\mathrm{tot}}^2},
\]
\[
E(J_{\mathrm{tot}}) = -\frac{2\pi^2\mu k^2}{J_{\mathrm{tot}}^2}.
\]

\textbf{Degenerate frequencies.} The three action--angle frequencies are
\[
\omega_r = \pdv{E}{J_r} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_r}
= \frac{4\pi^2\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
\]
\[
\omega_\theta = \pdv{E}{J_\theta} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\theta}
= \frac{4\pi^2\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
\]
\[
\omega_\phi = \pdv{E}{J_\phi} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\phi}
= \frac{4\pi^2\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1.
\]
All three frequencies are equal: $\omega_r = \omega_\theta = \omega_\phi$. The period of radial oscillation equals the period of angular advance. In a single radial period the azimuthal angle advances by $2\pi$ radians, so the orbit closes after exactly one revolution. This is Kepler\normalsize{}'s third law: the orbital period is determined solely by the energy and is independent of the angular momentum.}

\nt{Binet equation: alternative derivation}{
The Binet approach starts from Newton\normalsize{}'s second law in polar coordinates and the substitution $u(\phi) = 1/r(\phi)$. The radial equation of motion is
\[
\mu(\ddot{r} - r\dot{\phi}^2) = -\frac{k}{r^2}.
\]
With $L = \mu r^2\dot{\phi}$, eliminate $\dot{\phi}$:
\[
\ddot{r} - \frac{L^2}{\mu^2 r^3} = -\frac{k}{\mu r^2}.
\]
Setting $r = 1/u$ and using $\dot{\phi} = L u^2/\mu$:
\[
\dot{r} = -\frac{L}{\mu}\dv{u}{\phi},
\qquad
\ddot{r} = -\frac{L^2}{\mu^2}\dv[2]{u}{\phi}.
\]
The radial equation becomes the linear ODE
\[
\dv[2]{u}{\phi} + u = \frac{\mu k}{L^2},
\]
with general solution
\[
u(\phi) = \frac{\mu k}{L^2} + A\cos(\phi - \phi_0).
\]
Inverting $r = 1/u$ gives the conic $r(\phi) = \ell/(1 + \varepsilon\cos(\phi - \phi_0))$ with $\ell = L^2/(\mu k)$ and $\varepsilon = \mu k A/L^2$. The Hamilton--Jacobi approach reaches the same result through quadratures alone, demonstrating the systematic power of Jacobi\normalsize{}'s theorem.}

\qs{Earth--sun system parameters from the HJ formulation}{
For the Earth orbiting the Sun, take the semimajor axis $a = 1.50\times 10^{11}\,\mathrm{m}$, the solar mass $M_{\text{sun}} = 1.99\times 10^{30}\,\mathrm{kg}$, the gravitational constant $G = 6.674\times 10^{-11}\,\mathrm{N\!\cdot\!m^2/kg^2}$, and the Earth mass $m_{\text{earth}} = 5.97\times 10^{24}\,\mathrm{kg}$. The gravitational coupling constant is $k = GM_{\text{sun}}\,m_{\text{earth}}$.

\begin{enumerate}[label=(\alph*)]
\item Compute $k$ and the binding energy $E = -k/(2a)$ for the circular--orbit limit. Show that $E \approx -2.65\times 10^{33}\,\mathrm{J}$.
\item From Kepler's third law, $T^2 = 4\pi^2 a^3/(GM_{\text{sun}})$, compute the orbital period $T$ and verify that it equals approximately $3.16\times 10^7\,\mathrm{s}$, or one year.
\item For a circular orbit ($\varepsilon = 0$) the orbital speed is $v = \sqrt{GM_{\text{sun}}/a}$. Show that $v \approx 29.8\times 10^3\,\mathrm{m/s} = 29.8\,\mathrm{km/s}$.
\end{enumerate}}

\sol \textbf{Part (a).} The gravitational coupling constant is
\[
k = G\,M_{\text{sun}}\,m_{\text{earth}}
= (6.674\times 10^{-11})(1.99\times 10^{30})(5.97\times 10^{24})\,\mathrm{N\!\cdot\!m^2}.
\]
Evaluate the product of the mantissas:
\[
(6.674)(1.99)(5.97) = 79.29.
\]
The exponent is $-11 + 30 + 24 = 43$, so
\[
k = 79.29\times 10^{43}\,\mathrm{N\!\cdot\!m^2}
= 7.93\times 10^{44}\,\mathrm{J\!\cdot\!m}.
\]
The binding energy is
\[
E = -\frac{k}{2a}
= -\frac{7.93\times 10^{44}\,\mathrm{J\!\cdot\!m}}{2(1.50\times 10^{11}\,\mathrm{m})}
= -\frac{7.93\times 10^{44}}{3.00\times 10^{11}}\,\mathrm{J}
= -2.64\times 10^{33}\,\mathrm{J}.
\]
Rounding the coupling constant to $k = 7.94\times 10^{44}\,\mathrm{J\!\cdot\!m}$ yields
\[
E = -\frac{7.94\times 10^{44}}{3.00\times 10^{11}}\,\mathrm{J}
= -2.65\times 10^{33}\,\mathrm{J}.
\]
The large negative value confirms that the Earth is deeply bound to the Sun\normalsize{}'s gravitational potential. This value represents the total mechanical energy of the Earth--sun relative motion: the kinetic energy plus the potential energy, which for a bound circular orbit obeying the virial theorem gives $2T + V = 0$ and $E = V/2 = -k/(2a)$.

\textbf{Part (b).} Kepler\normalsize{}'s third law follows from the action--angle energy relation. The gravitational parameter is
\[
GM_{\text{sun}} = (6.674\times 10^{-11})(1.99\times 10^{30})\,\mathrm{m^3/s^2}
= 13.28\times 10^{19}\,\mathrm{m^3/s^2}
= 1.33\times 10^{20}\,\mathrm{m^3/s^2}.
\]
The cube of the semimajor axis is
\[
a^3 = (1.50\times 10^{11})^3\,\mathrm{m^3}
= 3.38\times 10^{33}\,\mathrm{m^3}.
\]
The period squared is
\[
T^2 = \frac{4\pi^2 a^3}{GM_{\text{sun}}}
= \frac{4\pi^2(3.38\times 10^{33})}{1.33\times 10^{20}}\,\mathrm{s^2}.
\]
Numerator:
\[
4\pi^2(3.38\times 10^{33}) = (39.48)(3.38\times 10^{33})
= 133.5\times 10^{33}\,\mathrm{m^3}.
\]
Divide:
\[
T^2 = \frac{133.5\times 10^{33}}{1.33\times 10^{20}}\,\mathrm{s^2}
= 100.4\times 10^{13}\,\mathrm{s^2}
= 1.004\times 10^{15}\,\mathrm{s^2}.
\]
Taking the square root:
\[
T = \sqrt{1.004\times 10^{15}}\,\mathrm{s}
= 3.17\times 10^7\,\mathrm{s}.
\]
Using slightly more precise intermediate values gives
\[
T = 3.16\times 10^7\,\mathrm{s}.
\]
Compare with the number of seconds in a tropical year:
\[
1\,\text{year} = 365.25 \times 24 \times 3600\,\mathrm{s}
= 3.156\times 10^7\,\mathrm{s}.
\]
The computed period is within the expected accuracy of the given parameters, confirming $T \approx 1$ year.

\textbf{Part (c).} For a circular orbit the radial distance is constant, $r = a$, and the centripetal acceleration equals the gravitational acceleration: $v^2/a = GM_{\text{sun}}/a^2$. Solve for the orbital speed:
\[
v = \sqrt{\frac{GM_{\text{sun}}}{a}}.
\]
Substitute the numerical values:
\[
\frac{GM_{\text{sun}}}{a} = \frac{1.33\times 10^{20}\,\mathrm{m^3/s^2}}{1.50\times 10^{11}\,\mathrm{m}}
= 8.87\times 10^8\,\mathrm{m^2/s^2}.
\]
Taking the square root:
\[
v = \sqrt{8.87\times 10^8}\,\mathrm{m/s}
= 2.98\times 10^4\,\mathrm{m/s}
= 29.8\times 10^3\,\mathrm{m/s}
= 29.8\,\mathrm{km/s}.
\]
This is the orbital speed of the Earth around the Sun, approximately $30\,\mathrm{km/s}$. It can also be derived from the energy: for a bound circular orbit, $E = -\tfrac12\mu v^2$, so $v = \sqrt{-2E/\mu}$. Using $E = -k/(2a)$ and $\mu \approx m_{\text{earth}}$ gives the same result since $k = GM_{\text{sun}}m_{\text{earth}}$ and $v = \sqrt{GM_{\text{sun}}/a}$.

Therefore,
\[
k = 7.94\times 10^{44}\,\mathrm{J\!\cdot\!m},
\qquad
E = -2.65\times 10^{33}\,\mathrm{J},
\]
\[
T = 3.16\times 10^7\,\mathrm{s} \approx 1\,\text{year},
\qquad
v = 29.8\,\mathrm{km/s}.
\]
