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Adiabatic invariance of an orbital action preserves action values during slow regular changes, but frequencies $\partial H/\partial J_i$ can vary. Closure requires rational frequency ratios, which need not persist. A slowly added harmonic component in a Kepler orbit potential produces generic apsidal precession.

Time averaging over a Kepler orbit gives $⟨\cos f⟩=-e$, as follows from $dt=(1-e\cos E)dE/n$ and $\cos f=(\cos E-e)/(1-e\cos E)$. Consequently slow isotropic stellar mass loss preserves the secular orbital eccentricity. Conserved specific angular momentum then gives $\mu a=\mathrm{constant}$, so pericentre, apocentre and semimajor axis expand inversely with the remaining gravitational parameter.

An aligned eccentric ring consists of nested Kepler orbits with nearly common longitude of periapsis. Its width and surface density of a disk depend on both the distribution of semi-major axis values and the radial gradient of forced eccentricity.

In the isolated model, conservation of energy and angular momentum conservation reduce the exterior test particle's motion toHere $\epsilon$ is specific orbital energy and $h$ is specific angular momentum. The effective potential is ${\Phi }_{\mathrm{eff}}=h^2/(2r^2)-\mu /r$. A zero radial velocity is a radial turning point, and need not mean zero total speed: at the periapsis or apoapsis of a nonradial Kepler orbit, tangential velocity remains. Thus the vertical gap between $\epsilon$ and $-\mu /r$ represents total kinetic energy per unit mass, whereas the gap above ${\Phi }_{\mathrm{eff}}$ represents its radial part.

For $h>0$, the orbital eccentricity obeys $e^2=1+2\epsilon h^2/{\mu }^2$. Negative $\epsilon$ gives an elliptic orbit, including a circular orbit at its minimum allowed energy; zero $\epsilon$ gives a parabolic trajectory; positive $\epsilon$ gives a hyperbolic Kepler orbit. The effective potential tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why escape velocity depends on energy even though angular momentum changes the path. The $h=0$ case is radial motion, rather than a nondegenerate parabola or hyperbola; inward radial motion can end in collision.

For a nearly radial Kepler orbit, $q\simeq h^2/(2GM)$, so modest changes in specific angular momentum can strongly change pericentre. A planetary torque acting for a reference orbital time gives $\delta h\sim (m_p/M)\sqrt{GM{a}_p}$. Coherent kicks change $q$ by order itself after $N\sim (M/m_p)\sqrt{q/a_p}$ orbital-scale encounters, whereas uncorrelated kicks require an order-$N^2$ random-walk count. Both estimates are conditional on encounter geometry: a close encounter with a smaller impact parameter can give a much larger kick.

A circular distant barycentric Kepler orbit around a circular star–planet binary need not have constant star-centred osculating orbital elements. Let $\mu$ be the planet's fraction of the binary mass, $\alpha =a_{\mathrm{pl}}/a\ll 1$, and $\epsilon =\mu /\sqrt{\alpha }\ll 1$. Subtracting the stellar reflex velocity produces radial and tangential perturbations $(-v_0\epsilon \sin \Delta ,v_0\epsilon \cos \Delta )$, where $\Delta$ is the planet's longitude minus the particle's longitude. The vis-viva equation gives the displayed semi-major axis variation. The eccentricity vector components follow from $e_r=r{v}_{\theta }^2/(G{M}_{\star })-1$ and $e_{\theta }=-r{v}_r{v}_{\theta }/(G{M}_{\star })$. The errors are $O(\mu +{\epsilon }^2)$; the spatial translation itself is only $O(\mu \alpha )$. These are changes of instantaneous orbital description, even when the assumed barycentric trajectory is fixed.

For bound old Kepler orbits, $E<0$, $\lambda <0$ and $\bar{E}>0$ give a positive radius, positive isochrone scale $b$, and positive transformed mass. The shifted energy is $\bar{E}-k=\bar{E}/\lambda <0$. For unbound old orbits with $E>0$, $1/2<\lambda \le 1$ gives an attractive unbound isochrone branch. The case $\lambda =0$ is handled by the separate Kepler–harmonic radial duality.

In the flat-surface, constant-gravitational acceleration model, a coasting projectile conserves horizontal velocity. If it arrives at the top of its trajectory with horizontal speed $v_o$, its earlier velocity at angle $\theta$ above the horizontal has vertical component $v_o\tan \theta$. Its remaining height gain is therefore $v_o^2{\tan }^2\theta /(2g)$. The displayed burnout height follows. This is a local projectile motion approximation, not an exact Kepler orbit. Computing $v_o=\sqrt{g(R_E+h_o)}$ assumes constant gravitational magnitude also at the orbit; computing $v_o=\sqrt{g_0{R}_E^2/(R_E+h_o)}$ instead assumes an inverse-square law with surface value $g_0$. Those assumptions must be distinguished.

The centre of mass of an astronomical system. In an isolated two-body problem, both bodies follow orbits around their shared astronomical barycenter; the Kepler orbit of their relative displacement has a semi-major axis equal to the sum of their individual semi-major axes.

A gradual rotation of the direction of periapsis within the orbital plane. A closed Kepler orbit has fixed argument of periapsis $\omega$; relativistic gravity can make $\omega$ advance between successive orbits. The angle change and binary-pulsar orbital decay describe different changes of the orbit.

The geometric light-travel contribution from the pulsar's position within the binary orbit. To construct an eccentric Kepler orbit, use eccentric anomaly $E$, orbital eccentricity $e$, mean anomaly $M=2\pi (t-T_0)/P_b$, and argument of periapsis $\omega$. Here $T_0$ is passage through periapsis. Coordinates in the orbital plane are $X=a_1(\cos E-e)$ and $Y=a_1\sqrt{1-e^2}\sin E$. Rotate through $\omega$ and project by $\sin i$ to obtainThe first equation is Kepler's equation: equal time intervals give equal increments of $M$, not of $E$. Since $1-e\cos E>0$ for $0\le e<1$, each $M$ has a unique $E$ on the corresponding unwrapped orbit. A constant delay is absorbed into the arbitrary arrival-time origin. The circular limit is ${\Delta }_R=x\sin (M+\omega )$, so its amplitude measures $x$. For $e>0$, the delay's offset and amplitude also depend on $\omega$; its maximum is not generally $x$.

For grains of fixed diameter $D$ released instantaneously at $r_0$, the drift law $\dot{r}=-\gamma /(rD)$ integrates to the displayed size-radius map. The time to radius $r<r_0$ is $t_r(D)=D(r_0^2-r^2)/(2\gamma )$. Grains therefore arrive in increasing diameter order. This model of Poynting–Robertson drag assumes nearly circular Kepler orbits, fixed positive $\gamma$, and no grain loss or diameter changes before arrival. At a positive time, the map is monotone in $D$ and allows the initial power-law size distribution to be transformed into a spatial distribution or an arrival-time distribution.

For a central gravitational potential and nonzero specific angular momentum $L$, the positive $L^2/(2r^2)$ contribution to the radial effective potential can prevent the orbit reaching the centre at finite energy. It dominates the attractive $-\mu /r$ term at small $r$ in a Kepler orbit. A more singular attractive term may overcome it: for $\Phi =-\mu /r-\mu a/r^2$, the net inverse-square coefficient is $(L^2-2\mu a)/2$, so a central barrier exists only when that coefficient is positive.

An orbit about a planet. Stellar tides, radiation pressure and other perturbations can limit its stability; a sufficiently small semi-major axis compared with the Hill sphere permits a local approximately planet-centred Kepler orbit.

In the gravitational potential $-\mu (1/r+a/r^2)$, a regular noncircular bound orbit requires $L^2>2\mu a$ and has $u=\mu (L^2-2\mu a{)}^{-1}[1+e\cos \nu (\psi -{\psi }_0)]$ with $0<e<1$. Each radial period advances the angle by $2\pi /\nu$. The full position and velocity return after finitely many radial periods precisely when $\nu$ is rational. Irrational $\nu$ gives a dense rosette on the two-frequency invariant torus. Rationality permits a fifth orbit label fixing the apsidal phase on that resonant angular momentum family, but does not give global superintegrability for all nearby angular momenta. When $a=0$, every regular bound Kepler orbit is closed and the Laplace-Runge-Lenz vector supplies the globally defined additional integral.

Conservation of angular momentum says that total angular momentum stays fixed when the net external torque about the chosen fixed inertial origin vanishes. Individual angular momenta and angular velocities can change. The radius change at constant angular momentum below makes this distinction visible: pulling two rotating point masses inward lowers the moment of inertia, raises the angular velocity, and supplies work that increases the kinetic energy.

Follow the same constant-mass point masses throughout, in an inertial frame, with all positions measured from a fixed origin $O$. If the internal pair forces are opposite and central, meaning directed along each pair's separation, thenThus zero net external torque throughout an interval implies constant total angular momentum throughout that interval. A vanishing external torque component, such as ${\tau }_z^{\mathrm{ext}}=0$, conserves the corresponding $L_z$ even when the other components change. This is the balance law for a fixed collection; angular-momentum flux must also be accounted for when matter crosses its boundary. ^[ref]

The central qualification is essential. Write ${\mathbf{F}}_{ij}$ for the internal force on member $i$ from member $j$. Under Newton's third law, ${\mathbf{F}}_{ji}=-{\mathbf{F}}_{ij}$, but the sum of their torques isThis vanishes when ${\mathbf{F}}_{ij}$ is parallel to ${\mathbf{r}}_i-{\mathbf{r}}_j$. Equal-and-opposite forces alone do not suffice: two opposite transverse forces with separated lines of action form a couple with nonzero torque. Cancellation of the internal pair torques, together with Newton's second law, proves the total balance above.

For one point mass under a central force about $O$, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ gives ${\mathbf{\tau }}_O=\mathbf{r}\times \mathbf{F}=0$. The force may still change the speed and kinetic energy; its zero torque preserves angular momentum. With nonzero ${\mathbf{L}}_O$, $\mathbf{r}\cdot {\mathbf{L}}_O=0$ keeps the motion in a fixed plane. Choosing the $z$ axis perpendicular to that plane, $r^2\dot{\theta }=L_z/m$ governs its angular progress. This removes an unknown from the central-force radial turning point calculation and helps reduce the Kepler orbit to a radial problem.

There is a historical geometric version of this result. In Book I, Section II, Proposition I of the Principia, Newton proves that motion under a central force toward a fixed centre stays in a plane and sweeps equal areas in equal times, using successive central impulses and equal triangle areas. ^[ref] In the modern notation above, constant areal velocity is $\dot{A}=L_z/(2m)$: the old geometric proof and the angular momentum calculation express the same invariant.

For bound Kepler orbits, the Delaunay variables obey $0<G<L$, $-G<H<G$, and three periodic angles. A canonical transformation preserves phase space volume. At fixed $L$, integrating $H$ supplies the measure $2GdG$. Since $G=L\sqrt{1-e^2}$, the fraction with orbital eccentricity less than $e$ is $L^{-2}{\int }_{L\sqrt{1-e^2}}^{L}2GdG=e^2$. Thus the probability density function is $2e$, independently of any normalizable energy-only phase-space distribution function. A Boltzmann distribution is unnecessary. Even the time-dependent density $F(E(L))[1+\eta \cos (l-n(L)t)]$, $|\eta |<1$, has this same marginal because integrating the mean anomaly removes its oscillatory part; an eccentricity marginal alone does not imply equilibrium.

For a bound noncollision Kepler orbit with gravitational parameter $\mu$, the canonical actions per unit mass are $\mathcal{L}=\sqrt{\mu a}$, $\mathcal{G}=\mathcal{L}\sqrt{1-e^2}$ and $\mathcal{H}=\mathcal{G}\cos i$. Their angles are mean anomaly $\ell$, argument of periapsis $\omega$ and longitude of ascending node $\Omega$. The Hamiltonian is $-{\mu }^2/(2{\mathcal{L}}^2)$, so only $\ell$ advances in the unperturbed problem. In the prograde spherical-action chart, $J_r=\mathcal{L}-\mathcal{G}$, $J_{\theta }=\mathcal{G}-\mathcal{H}$, $J_{\varphi }=\mathcal{H}$; the type-two canonical transformation is generated by $F_2=(\mathcal{L}-\mathcal{G})w_r+(\mathcal{G}-\mathcal{H})w_{\theta }+\mathcal{H}{w}_{\varphi }$. Retrograde spherical actions have $J_{\theta }=\mathcal{G}-|\mathcal{H}|$, so the absolute value cannot be dropped globally. Circular and coplanar limits are singular angle charts, not singular orbits.

Doppler spectroscopy detects an exoplanet through the host star's changing Doppler shift. Both bodies orbit their shared astronomical barycenter: the star undergoes stellar reflex motion even when the exoplanet itself is unseen. Its velocity along the line of sight shifts its spectral lines; at speeds small compared with the speed of light, $\Delta \lambda /\lambda \simeq v_r/c$, with positive $v_r$ for recession. Repeated measurements can fit a Kepler orbit, rather than merely detect one shifted spectrum. Starspots and other changes in stellar spectral lines must be distinguished from actual orbital motion.

For one companion in an isolated two-body problem, the stellar radial-velocity semi-amplitude isHere $G$ is the gravitational constant, $P$ the orbital period, $e$ the orbital eccentricity, and $M_{\star },M_p$ the star and exoplanet masses. The orbital inclination $i$ is measured relative to the sky plane: $i=0$ is face-on and $i=90^{\circ }$ is an edge-on orbit. The exact binary mass function measured from the fitted orbit isWith known $M_{\star }$ and $M_p\ll M_{\star }$, this is usually reported as $M_p\sin i$. Taking $\sin i=1$ gives the minimum companion mass; a smaller orbital inclination allows a larger mass. A detected exoplanet transit constrains $i$ close to $90^{\circ }$ for a compact planet orbiting well outside the star's surface, substantially resolving this ambiguity. Combining Doppler spectroscopy and exoplanet transit photometry then supplies a mass and radius for exoplanet interior studies.

For a computed circular example, let $q=M_p/M_{\star }=0.001$ and choose a relative orbital speed $V_{\mathrm{rel}}=100\mathrm{km}{\mathrm{s}}^{-1}$: this is the speed of the planet relative to the star, not its speed about the astronomical barycenter. Conservation of momentum givesThe barycentric velocities point in opposite directions, so their speeds add to $V_{\mathrm{rel}}$. Here $e=0$ and $K=V_{\star }\sin i$: $K=99.9001\mathrm{m}{\mathrm{s}}^{-1}$ at $i=90^{\circ }$, $49.9500\mathrm{m}{\mathrm{s}}^{-1}$ at $i=30^{\circ }$, and zero at $i=0$. The star still orbits in the face-on case, but its orbital velocity has no line-of-sight component. The plotted circular signal is $v_r(t)-\gamma =K\sin (2\pi t/P)$ for a chosen phase origin and constant system velocity $\gamma$.

For two circular Kepler orbits with mutual inclination $I_m$, independently average the stellar quadrupole disturbing function over both mean anomalies. A circular ring with unit normal $\mathbf{k}$ has $⟨{\hat{r}}_i{\hat{r}}_j⟩=({\delta }_{ij}-k_i{k}_j)/2$. Contracting the two ring tensors gives $⟨{\cos }^2\psi ⟩=[1+(\mathbf{k}\cdot {\mathbf{k}}_p{)}^2]/4$, proving the displayed exact angular factor. For small orbital inclinations relative to a common plane, $I_m^2=I^2+I_p^2-2I{I}_p\cos ({\Omega }_p-\Omega )+O(I^4)$. Thus the secular angular factor is $1/4-3(I^2+I_p^2)/8+3I{I}_p\cos ({\Omega }_p-\Omega )/4$. This potential generates the slow stellar contribution to satellite nodal precession.