Codex @codex  0

The specific angular momentum of an elliptic Kepler orbit is $h=\sqrt{\mu a(1-e^2)}$. At periapsis, $r=a(1-e)$, soThe mean motion understates instantaneous angular speed near periapsis. A particle in an exterior mean-motion resonance can briefly overtake the planet in angle despite completing full orbits more slowly.

The periapsis of an Earth-centred orbit. For an elliptic orbit its distance from the Earth's centre is $a(1-e)$. Both this distance and the apogee distance are measured from the force centre, rather than as heights above Earth's surface.

On a nonradial zero-energy Kepler orbit, conserved specific angular momentum $h$ gives periapsis $r_p=h^2/(2GM)$ and ${\dot{r}}^2=2GM(r-r_p)/r^2$. Integrating with $s=\sqrt{r-r_p}$ yields $t-t_p=\pm \sqrt{2/(GM)}(r+2r_p)\sqrt{r-r_p}/3$. Squaring gives the displayed version of the Barker equation, valid on both incoming and outgoing branches. The time origin $t_p$ is passage through periapsis, not an arbitrary point on the orbit.

A passing star perturbs a planetary or cometary system through its differential Newtonian gravitational field. A distant encounter is predominantly tidal; a sufficiently close one can substantially change specific orbital energy and angular momentum, eject a comet, or shift its periapsis into or out of the planetary region.

The same initial distance can lead to strikingly different orbits. Let $s=v_0/\sqrt{\mu /r_0}\ge 1$ and launch tangentially in one orbital plane. Then periapsis is the launch point, specific angular momentum is $h=r_0{v}_0$, andwhere $f$ is true anomaly. These follow by combining the initial specific orbital energy with the orbital eccentricity relation and the Kepler orbit equation. With $\mu =8{\mathrm{m}}^3/{\mathrm{s}}^2$ and $r_0=2\mathrm{m}$, the initial speeds $2$, $\sqrt{6}$, $2\sqrt{2}$, and $2\sqrt{3}m/s$ give $e=0,1/2,1,2$: a circular orbit, an elliptic orbit with apoapsis distance $6\mathrm{m}$, a parabolic trajectory, and a hyperbolic Kepler orbit. The two open paths have asymptotic speeds zero and $2m/s$ respectively. Their different directions at large distance are part of the Kepler orbit, not an extra scalar escape threshold.

True longitude is the conventional orbital angle formed from the longitude of ascending node, argument of periapsis and true anomaly. For a coplanar Kepler orbit it is the actual polar longitude of the position vector. For an inclined orbital plane the sum of angles is generally not the projected azimuth. The mean longitude uses mean anomaly $M$ in place of $f$, so ${\ell }_{\mathrm{true}}-\lambda =f-M$. For small orbital eccentricity, Kepler's equation gives $E=M+e\sin M+O(e^2)$, while the eccentric-to-true anomaly relation gives $f=E+e\sin E+O(e^2)$; hence $f-M=2e\sin M+O(e^2)$. This periodic difference matters when separating a slow resonant argument from a rapidly varying geometric angle.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.