physics-handbook/concepts/advanced/kepler-hj.tex

\subsection{The Kepler Problem}

This subsection treats the inverse-square central potential $V(r) = -k/r$ through the Hamilton-- Jacobi formalism, derives conic-section orbits from Jacobi's theorem, and uses action-- angle variables to recover Kepler's third law and the degeneracy that makes bound orbits close.

\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. In spherical coordinates $(r,\theta,\phi)$ the kinetic energy is $T = \tfrac{1}{2}\mu(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\,\dot{\phi}^2)$, and 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'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 $2\mu r^2$:
\[
r^2\left(\pdv{W_{r\theta}}{r}\right)^2 - \frac{2\mu k r}{1} - 2\mu E r^2
= -\left(\pdv{W_{r\theta}}{\theta}\right)^2 - \frac{L_z^2}{\sin^2\theta}.
\]
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's theorem, these equations determine the trajectory without solving any second-- order differential equation.}

\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:
\[
\ell = \frac{L^2}{\mu k},
\qquad
\varepsilon = \sqrt{1 + \frac{2EL^2}{\mu k^2}},
\qquad
\phi_0 = \text{constant}.
\]
The angle $\phi_0$ fixes the orientation of the conic in the orbital plane.}

\pf{Derivation of the orbit from Jacobi'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'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 in the $z$-direction). Differentiating $W_r$ with respect to $L$:
\[
\pdv{W_r}{L} = \int\frac{1}{2\der{W_r}{r}}\cdot\pdv{(\der{W_r}{r})^2}{L}\dd r
= \int\frac{1}{2p_r}\cdot\left(-\frac{2L}{r^2}\right)\dd r
= -\int\frac{L}{r^2 p_r}\,\dd r.
\]
The key observation is that the same integral appears in the relation between $\phi$ and $r$. From Hamilton's equations or from the $\phi$-- part of Jacobi'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 $T = 2\pi\sqrt{\mu a^3/k}$.

\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 $\omega = \sqrt{k/(\mu r^3)}$.

\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 $2\arccos(-1/\varepsilon)$.
\end{enumerate}}

\nt{Action-- angle variables and Kepler'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 $J_r = 2\pi(-L + k\sqrt{\mu/(2|E|)})$. 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'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 $2\pi$, so the orbit closes after exactly one revolution. This is Kepler's third law: the orbital period is determined solely by the energy and is independent of the angular momentum.}

\nt{Comparison with the Binet equation}{
The Binet equation is an alternative derivation of Kepler orbits that begins with Newton's second law and the substitution $u(\phi) = 1/r(\phi)$. The radial equation of motion in polar coordinates is
\[
\mu(\ddot{r} - r\dot{\phi}^2) = -\frac{k}{r^2}.
\]
With the angular momentum $L = \mu r^2\dot{\phi}$, eliminate $\dot{\phi}$ in favor of $L$:
\[
\ddot{r} - \frac{L^2}{\mu^2 r^3} = -\frac{k}{\mu r^2}.
\]
Write $r = 1/u(\phi)$ and use the chain rule to convert time derivatives into $\phi$ derivatives. Since $\dot{\phi} = L/(\mu r^2) = L u^2/\mu$:
\[
\dot{r} = \dv{r}{\phi}\dot{\phi} = -\dv{u}{\phi}\cdot\frac{1}{u^2}\cdot\frac{L u^2}{\mu}
= -\frac{L}{\mu}\dv{u}{\phi}.
\]
Differentiate once more:
\[
\ddot{r} = \dv{}{t}\!\left(-\frac{L}{\mu}\dv{u}{\phi}\right)
= -\frac{L}{\mu}\dv[2]{u}{\phi}\dot{\phi}
= -\frac{L^2}{\mu^2}\dv[2]{u}{\phi}.
\]
Substitute into the radial equation:
\[
-\frac{L^2}{\mu^2}\dv[2]{u}{\phi} - \frac{L^2}{\mu^2}\,u = -\frac{k}{\mu},
\]
which rearranges to the Binet equation:
\[
\dv[2]{u}{\phi} + u = \frac{\mu k}{L^2}.
\]
The general solution is
\[
u(\phi) = \frac{\mu k}{L^2} + A\cos(\phi - \phi_0),
\]
where $A$ is determined by the energy. Inverting $r = 1/u$ gives
\[
r(\phi) = \frac{1}{\mu k/L^2 + A\cos(\phi - \phi_0)}
= \frac{L^2/(\mu k)}{1 + \mu k A/L^2\cos(\phi - \phi_0)},
\]
which matches the form $\ell/(1 + \varepsilon\cos(\phi - \phi_0))$ with $\ell = L^2/(\mu k)$ and $\varepsilon = \mu k A/L^2$. This derivation is shorter but requires knowing the $u = 1/r$ substitution. The Hamilton-- Jacobi approach reaches the same result through quadratures without any clever change of variable, demonstrating the power of Jacobi's theorem as a unifying principle. Moreover, the action-- angle formalism provides immediate access to the energy-- period-- semimajor axis relations that the Binet equation leaves as an afterthought. The Binet method also cannot be extended to noncentral potentials or to higher dimensions without substantial modification, while the Hamilton-- Jacobi approach generalizes naturally to any separable system.}

\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'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'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,
\[
k = 7.94\times 10^{44}\,\mathrm{J\!\cdot\!m},
\qquad
E = -2.65\times 10^{33}\,\mathrm{J},
\]
\[
T = 3.16\times 10^7\,\mathrm{s} \approx 1\,\text{year},
\qquad
v = 29.8\,\mathrm{km/s}.
\]