The Kepler Problem
- Classical Mechanics 2.4
- Inverse-Square Force
- Energy and Eccentricity
- Elliptic Orbits
- Motion in Time
- Laplace-Runge-Lenz Vector
- Degeneracy
Inverse-Square Force
The Kepler problem is the central force problem with
\[ f(r)=-\frac{k}{r^2}, \qquad V(r)=-\frac{k}{r}, \]
where $k>0$. For gravitation,
\[ k=Gm_1m_2, \]
and the mass in the one-body problem is the reduced mass
\[ \mu=\frac{m_1m_2}{m_1+m_2}. \]
Binet’s equation is
\[ \odvn{2}{u}{\theta}+u = -\frac{\mu}{\ell^2u^2}f\left(\frac{1}{u}\right). \]
Since
\[ f\left(\frac{1}{u}\right)=-ku^2, \]
we obtain
\[ \odvn{2}{u}{\theta}+u=\frac{\mu k}{\ell^2}. \]
The solution is
\[ u=\frac{\mu k}{\ell^2}\left[1+e\cos(\theta-\theta_0)\right]. \]
Thus
\[ r=\frac{p}{1+e\cos(\theta-\theta_0)}, \qquad p=\frac{\ell^2}{\mu k}. \]
This is the polar equation of a conic section with one focus at the force center.
Energy and Eccentricity
The eccentricity is determined by the energy and angular momentum. Using
\[ E = \frac{\ell^2}{2\mu}\left[ \left(\odv{u}{\theta}\right)^2+u^2 \right]-ku, \]
and substituting the conic solution gives
\[ e^2=1+\frac{2E\ell^2}{\mu k^2}. \]
Therefore,
| energy | eccentricity | orbit |
|---|---|---|
| $E<0$ | $0\le e<1$ | ellipse |
| $E=0$ | $e=1$ | parabola |
| $E>0$ | $e>1$ | hyperbola |
The circular orbit is the special case $e=0$. Then
\[ E=-\frac{\mu k^2}{2\ell^2}. \]
This agrees with the circular orbit condition
\[ r_0=\frac{\ell^2}{\mu k}. \]
Elliptic Orbits
For an elliptic orbit, let $a$ be the semimajor axis. The turning points occur when $\dot{r}=0$, so they are the roots of
\[ E = \frac{\ell^2}{2\mu r^2}-\frac{k}{r}. \]
The sum of the two apsidal distances is $2a$, and this gives
\[ a=-\frac{k}{2E}. \]
Thus, for a Kepler ellipse, the energy depends only on the semimajor axis:
\[ E=-\frac{k}{2a}. \]
The angular momentum is related to $a$ and $e$ by
\[ \ell^2=\mu ka(1-e^2). \]
Hence the orbit may also be written as
\[ r=\frac{a(1-e^2)}{1+e\cos(\theta-\theta_0)}. \]
The nearest and farthest distances are
\[ r_{\min}=a(1-e), \qquad r_{\max}=a(1+e). \]
The inverse-square force is therefore special not only because the orbit is closed, but also because the size of an ellipse is fixed by $E$ alone while its shape is fixed by $\ell$ through $e$.
Motion in Time
The orbit equation gives $r$ as a function of $\theta$. To determine the position at a given time, we must use the areal velocity:
\[ \odv{A}{t}=\frac{\ell}{2\mu}. \]
For elliptic motion, introduce the eccentric anomaly $\psi$ by
\[ r=a(1-e\cos\psi). \]
The Cartesian parametrization of the ellipse, with the origin at the focus and periapsis on the positive $x$-axis, is
\[ x=a(\cos\psi-e), \qquad y=a\sqrt{1-e^2}\sin\psi. \]
The period follows from the fact that the areal velocity is constant. The area of the ellipse is
\[ \pi ab =\pi a^2\sqrt{1-e^2}. \]
Since
\[ \ell^2=\mu ka(1-e^2), \]
we get
\[ \tau =2\pi\sqrt{\frac{\mu a^3}{k}}. \]
Equivalently,
\[ \tau^2=\frac{4\pi^2\mu}{k}a^3. \]
For the gravitational two-body problem, $k=Gm_1m_2$ and $\mu=m_1m_2/(m_1+m_2)$, so
\[ \tau^2=\frac{4\pi^2}{G(m_1+m_2)}a^3. \]
If one mass is much larger than the other, this reduces to the usual form of Kepler’s third law.
Let
\[ \omega=\frac{2\pi}{\tau} =\sqrt{\frac{k}{\mu a^3}}. \]
Integrating the time equation gives Kepler’s equation:
\[ \omega(t-t_p)=\psi-e\sin\psi, \]
where $t_p$ is the time of periapsis passage. The left side is the mean anomaly, $\psi$ is the eccentric anomaly, and the polar angle $\theta$ is the true anomaly. The relation between $\theta$ and $\psi$ is
\[ \cos\theta =\frac{\cos\psi-e}{1-e\cos\psi}, \]
or equivalently,
\[ \tan\frac{\theta}{2} =\sqrt{\frac{1+e}{1-e}}\tan\frac{\psi}{2}. \]
Thus the practical time problem is reduced to solving the transcendental equation
\[ \psi-e\sin\psi=M, \]
where $M=\omega(t-t_p)$. This equation has no elementary inverse in general, so numerical methods are required for high-precision orbital prediction.
For parabolic motion, $e=1$, and the orbit can be parametrized by
\[ x=\tan\frac{\theta}{2}. \]
The time from periapsis is then proportional to
\[ x+\frac{x^3}{3}. \]
Thus the parabolic case is algebraic rather than transcendental, but inversion still requires solving a cubic equation.
Laplace-Runge-Lenz Vector
The Kepler problem has one more simple conserved vector. Let
\[ \b{p}=\mu\dot{\b{r}}, \qquad \b{L}=\b{r}\times\b{p}. \]
For the inverse-square force, define
\[ \b{A}=\b{p}\times\b{L}-\mu k\hat{\b{r}}. \]
which is called the Laplace-Runge-Lenz(LRL) vector. We show that $\b{A}$ is conserved. Newton’s equation is
\[ \dot{\b{p}}=-\frac{k}{r^2}\hat{\b{r}}. \]
Since $\b{L}$ is conserved,
\[ \odv{}{t}(\b{p}\times\b{L}) =\dot{\b{p}}\times\b{L}. \]
Using $\b{L}=\b{r}\times\b{p}$ and the triple product identity,
\[ \begin{align*} \dot{\b{p}}\times\b{L} &= -\frac{k}{r^3}\b{r}\times(\b{r}\times\b{p}) \nl &= -\frac{k}{r^3}\left[\b{r}(\b{r}\cdot\b{p})-r^2\b{p}\right] \nl &= \mu k\odv{}{t}\left(\frac{\b{r}}{r}\right). \end{align*} \]
Therefore
\[ \odv{\b{A}}{t}=0. \]
The vector $\b{A}$ is perpendicular to $\b{L}$:
\[ \b{A}\cdot\b{L}=0. \]
Hence it lies in the orbital plane. Taking the dot product with $\b{r}$ gives
\[ \begin{align*} \b{A}\cdot\b{r} &= \b{r}\cdot(\b{p}\times\b{L})-\mu kr \nl &= \b{L}\cdot(\b{r}\times\b{p})-\mu kr \nl &= \ell^2-\mu kr. \end{align*} \]
If $\theta$ is measured from the direction of $\b{A}$, then
\[ Ar\cos\theta=\ell^2-\mu kr. \]
Thus
\[ r=\frac{\ell^2/(\mu k)}{1+\dps \frac{A}{\mu k}\cos\theta}. \]
Comparing with the conic equation, we identify
\[ e=\frac{A}{\mu k}. \]
Therefore $\b{A}$ points toward periapsis and its magnitude determines the eccentricity.
The constants are not independent. One relation is
\[ \b{A}\cdot\b{L}=0. \]
The other is obtained from the magnitude:
\[ A^2=\mu^2k^2+2\mu E\ell^2. \]
This is equivalent to the eccentricity formula.
Degeneracy
For a general central force, $E$ and $\b{L}$ are the natural conserved quantities. They determine the radial bounds and the plane of motion, but they do not usually determine a closed curve. The radial and angular motions generally have incommensurable periods, so the orbit does not close.
The Kepler problem is exceptional. The Laplace-Runge-Lenz vector fixes the direction of the apsidal line and supplies an additional algebraic constant of motion. This is the mechanical origin of the degeneracy of Kepler orbits: the radial period and angular period match so that every bounded orbit is closed.
The harmonic oscillator has a similar closed-orbit property, but its extra conserved quantity is better expressed as a second-rank tensor. These two cases are precisely the two possibilities singled out by Bertrand’s theorem.
