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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.

The relative displacement of a Kepler orbit sweeps equal areas about the attracting focus of a conic in equal times. A short interval near periapsis covers a wide angle at small separation; the same interval near apoapsis covers a narrow angle at large separation. The swept areas are equal even though the travelled distances are different. In particular, the sectors have curved ellipse boundaries: triangles formed by joining their endpoints only approximate the swept areas for short intervals.

Let $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ be the relative specific angular momentum. The areal velocity isA central force exerts zero torque about its centre, so conservation of angular momentum makes this rate constant. This explanation also applies to other central forces; the ellipse shape specifically comes from Newtonian gravity.

In the $a=1$, $e=0.6$, $\mu =1$ example, specific angular momentum is $h=\sqrt{\mu a(1-e^2)}=\sqrt{0.64}=0.8$, giving areal velocity $0.4$. At periapsis and apoapsis, the velocity is perpendicular to the relative displacement, so $h=rv$. The endpoint speeds are consequentlyThe fourfold speed difference is consistent with the fourfold separation difference. Away from these endpoints, $h/r$ gives only the transverse component of velocity; the full speed follows from the vis-viva equation.

To actually place equal-time markers, use Kepler's equation. Write mean anomaly as $\mathcal{M}=nt$, with $t=0$ at periapsis and mean motion $n=\sqrt{\mu /a^3}=1$. The different symbol $\mathcal{M}$ avoids confusing this angle with the mass $M$. Solve for eccentric anomaly $E$ in radians, then compute the relative position:This timing procedure is explained in NASA educational archive, How Orbital Motion is Calculated.

For the first interval $0\le t\le \pi /6$, the final eccentric anomaly is $E\approx 1.041495$. For the equal interval $\pi \le t\le 7\pi /6$ starting at apoapsis, it ends at $E\approx 3.471065$. Both swept areas are $\Delta A=(h/2)\Delta t=0.4\pi /6=\pi /15\approx 0.209440$, one twelfth of the entire ellipse's area $0.8\pi$. The diagram computes its markers from Kepler's equation, making visible how uniform time steps become uneven spatial steps.

Put $\mathcal{G}=G(M_{\star }+M_{\mathrm{pl}})$, $h=\sqrt{\mathcal{G}a(1-e^2)}$, $\theta =\omega +f$ and $L=n_{\mathrm{pl}}t$. Here $h$ is the specific angular momentum, $f$ is the true anomaly and $\omega$ is the planar longitude of periapsis. Differentiating the Kepler orbit equation with $\dot{f}=h/r^2$ givesResolve the velocity along the radial and tangential unit vectors, $\hat{\mathbf{r}}=(\cos \theta ,\sin \theta )$ and $\hat{\mathbf{\theta }}=(-\sin \theta ,\cos \theta )$. The identities $\sin f\cos \theta -\cos f\sin \theta =-\sin \omega$ and $\sin f\sin \theta +\cos f\cos \theta =\cos \omega$ yieldThis is the Kepler velocity hodograph: the velocity runs around a circle whose centre is displaced by the orbital eccentricity.

The centre of mass condition gives the star's barycentric position ${\mathbf{r}}_{\star }=-\mu a_{\mathrm{pl}}(\cos L,\sin L)$. Translating to the star without rotating the axes gives ${\mathbf{r}}^{\prime }=\mathbf{r}-{\mathbf{r}}_{\star }$ and ${\mathbf{v}}^{\prime }=\mathbf{v}-{\dot{\mathbf{r}}}_{\star }$. From Kepler's third law, $n_{\mathrm{pl}}^2{a}_{\mathrm{pl}}^3=\mathcal{G}$, soConsequentlyThe small displacement and the larger reflex velocity have different orders: $|{\mathbf{r}}^{\prime }-\mathbf{r}|/a=\mu \alpha$, but the reflex speed divided by the circular speed is $\mu {\alpha }^{-1/2}$. These are the source of the astrocentric osculating-element oscillations.

For the coordinate ratio, the exact result isFor fixed $e<1$, $r/a\ge 1-e$, so this is $O(\mu \alpha )$ on any angular region with $|\cos \theta |$ bounded away from zero. The printed coordinate claim is not uniform over a full orbit: at $x=0$ its ratio is undefined. The nonsingular statement is $|{\mathbf{r}}^{\prime }-\mathbf{r}|/r\le \mu \alpha /(1-e)$, assuming the pericentre distance remains much larger than the binary separation.

Squaring the two components of the velocity givesAt $\mu =0$, $\kappa =0$ and $\mathcal{G}=G{M}_{\star }$. Sincethis reduces to the vis-viva equation, $v^{\prime 2}=G{M}_{\star }(2/r-1/a)$.

Now let the barycentric Kepler orbit be circular, write $\Delta =L-\theta$ and $\epsilon =\mu {\alpha }^{-1/2}\ll 1$, and put $v_0=\sqrt{\mathcal{G}/a}$. The osculating orbital elements about the star use the gravitational parameter $G{M}_{\star }=(1-\mu )\mathcal{G}$. To the required order, $r^{\prime }=a[1+O(\mu \alpha )]$ and $v^{\prime 2}=v_0^2[1+2\epsilon \cos \Delta +O({\epsilon }^2)]$. The specific orbital energy therefore impliesHere $\mu =o(\epsilon )$ and ${\epsilon }^2=o(\epsilon )$ under the stated hierarchy. Even at a phase where the leading cosine vanishes, the absolute error bound remains valid.

Finally, use the eccentricity vector ${\mathbf{e}}^{\prime }=({\mathbf{v}}^{\prime }\times {\mathbf{h}}^{\prime })/(G{M}_{\star })-{\hat{\mathbf{r}}}^{\prime }$. In radial and tangential components it isThe translation changes the radial basis only by $O(\mu \alpha )$, whereas $v_r^{\prime }=-v_0\epsilon \sin \Delta +O(v_0\mu \alpha )$ and $v_{\theta }^{\prime }=v_0(1+\epsilon \cos \Delta )+O(v_0\mu \alpha )$. ThusFor the original circular Kepler orbit, $\omega$ and $f$ are separately arbitrary; their sum $\theta$ is well defined. Using the eccentricity vector avoids introducing an undefined pericentre for that circular Kepler orbit.

Use rotating barycentric coordinates $\mathbf{R}=(x,y,z)$, rotating velocity $\mathbf{w}$, and $\mathbf{\Omega }=\hat{\mathbf{z}}$. Multiplying the three equations by the corresponding velocity components and adding cancels the Coriolis force terms:Thus the Jacobi constant is $C=2U-|\mathbf{w}{|}^2$. The inertial barycentric velocity is $\mathbf{V}=\mathbf{w}+\mathbf{\Omega }\times \mathbf{R}$. If $H_z=(\mathbf{R}\times \mathbf{V}{)}_z$ and $E_b=|\mathbf{V}{|}^2/2-{\mu }_1/r_1-{\mu }_2/r_2$, thenThe inertial energy $E_b$ and inertial angular momentum $H_z$ need not separately be constant. Their combination is constant because the two gravitational sources rotate steadily at unit angular speed.

For the Tisserand parameter take ${\mu }_2\ll {\mu }_1\simeq 1$, and compare the osculating orbital elements well outside a close encounter, where ${\mu }_2/r_2$ and barycentre-to-primary corrections are negligible. ThenThis requires a circular secondary orbit, the restricted three-body problem approximation, and no dissipative force or other perturber. $T$ is conserved to leading order between encounter episodes; the osculating $T$ during a close encounter need not be constant. The exact invariant is $C$.

For clarity the exact two-body energy relation is $a={\mu }_1/(2{\mu }_1/r_1-v_1^2)$, rather than the printed formula without the numerator ${\mu }_1$. At the leading order ${\mu }_1=1$ used for $T$, the two agree; retaining finite ${\mu }_1$ in just part of that formula is inconsistent.

At a close encounter put the secondary radius and speed equal to one at leading order, and denote the particle's incoming relative speed outside the secondary's strong-deflection region by $u$. Conservation of relative speed in the short gravitational encounter givesAn incoming orbit with $a=1$ has $|{\mathbf{v}}_1|=1$ at $r_1=1$ by the vis-viva equation; escape there requires an outgoing speed exceeding $\sqrt{2}$. The triangle inequality therefore gives the necessary single-encounter escape velocity boundEquality only gives a marginal parabolic limit. This argument deliberately gives a weak necessary bound, not a sufficient scattering criterion.

The planet-encounter relative velocity is related to the Tisserand parameter byHere $v_{1,\mathrm{tangential}}=h_z$ at the encounter radius, and the formula also applies to inclined passages at the planet's orbital plane. For the intended prograde coplanar incoming orbit, $I=0$ and $a=1$, so $u^2=2-2\sqrt{1-e^2}$. Applying the preceding necessary bound givesCoplanar alone also allows $I=\pi$. For that retrograde case $u^2=2+2\sqrt{1-e^2}$, and even a circular incoming orbit can be scattered onto an escaping orbit; the orbital eccentricity restriction therefore presumes prograde motion. Also, retaining the outgoing relative-velocity geometry gives the sharper necessary condition $|{\mathbf{v}}_{1,\mathrm{out}}|\le 1+u$, hence $u>\sqrt{2}-1$ and $e>\sqrt{4\sqrt{2}-5}/2$ for the same prograde incoming orbit. This stronger bound is consistent with, and implies, the weaker bound above.

For the Tisserand periapsis bound, first writeA bound orbit capable of another close encounter must intersect the secondary's circular radius, so $q\le 1\le Q$. For fixed $q<1$, the largest possible $T$ occurs at $I=0$. The remaining expression decreases with $Q\ge 1$, sinceThus every such orbit obeysOn $0<q<1$, $g(q)$ increases monotonically from $2$ to $3$. For $2<T<3$ the least accessible pericentre is the unique root of $T=g(q)$: the limiting orbit is prograde and coplanar, with apocentre just at the planet. Introduce $s=\sqrt{3-T}$. Its tangential speed at this apocentre is $1-s$, and its vis-viva equation gives the particularly well-behaved answerEquivalently, setting $w=\sqrt{2q/(1+q)}$ gives $T=2-w^2+2w$ and $w=1-s$, with $q=w^2/(2-w^2)$. Rationalizing the previous result yieldsThe apparent singularity at $T=\sqrt{8}$ is removable; the first form gives $q_{\min }=(\sqrt{2}-1)/2$ there. The limits are $q_{\min }\to 0$ as $T\downarrow 2$ and $q_{\min }\to 1$ as $T\uparrow 3$. The same lower envelope also excludes a smaller pericentre on an escaping trajectory when $2<T<3$. To see this without introducing an apocentre, use the planet-frame velocity sphere: $u=\sqrt{3-T}<1$ forces the tangential component at an encounter to be at least $w=1-u>0$, so total specific angular momentum obeys $h\ge w$. At pericentre the stellar energy obeys $h^2=2q+2E{q}^2$, while $E=h_z-T/2\le h-T/2$. ThusIf $q<q_{\min }=w^2/(2-w^2)$, then $q^2<w\le h$, and the left side is increasing with $h$. Hence $w^2\le 2q+(2w-T)q^2$. But $2w-T=w^2-2$, and the right side is smaller than $w^2$ for $0\le q<q_{\min }$, a contradiction. This establishes the barrier for all Kepler trajectories that reach the encounter radius, including unbound ones.

For $0<T\le 2$, bound encounter-crossing orbits can approach $q=0$ (take $Q\to 2/T$), so the positive root extrapolated from the displayed expression is not a universal barrier. For $T>3$ there is no trajectory reaching a close encounter in this approximation, since $u^2=3-T$ would be negative. Multiple encounters can approach the derived lower envelope by changing energy, angular momentum and inclination while preserving the Tisserand parameter; the invariant alone does not guarantee a particular encounter history reaches it.

The transverse velocity is $v_{\theta }=h/r$. Subtracting its square from the vis-viva equation, and putting $u=a/r$, givesConsequentlyThe positive branch is outward motion and the negative branch is inward motion. The PDF's unsigned expression gives the radial speed, or the outward branch, rather than the signed radial velocity everywhere. For example, on the inward part of any noncircular elliptic orbit, $\dot{r}<0$.

The same result follows from $\dot{r}=(\mu e/h)\sin f$, with $f$ the true anomaly. The radicand is nonnegative precisely on $1/(1+e)\le u\le 1/(1-e)$; it vanishes at periapsis and apoapsis. For a circular Kepler orbit, $e=0$, $u=1$ and $\dot{r}=0$.

Use the separation vector $\mathbf{r}$ from the star to the comet, and let $m_{\mathrm{red}}=m{M}_{\star }/(m+M_{\star })$. The two-body problem separates into uniform centre of mass motion andTaking a cross product with $\mathbf{r}$ proves conservation of angular momentum; taking the scalar product with $\dot{\mathbf{r}}$ proves conservation of energy. The relative constants areThe total relative angular momentum and energy are $m_{\mathrm{red}}\mathbf{h}$ and $m_{\mathrm{red}}\epsilon$, respectively. The conserved eccentricity vector is $\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}/\mu -\hat{\mathbf{r}}$. Its scalar product with $\hat{\mathbf{r}}$ gives $r=h^2/[\mu (1+e\cos f)]$, with $f$ the true anomaly. The periapsis and apoapsis distances are therefore $h^2/[\mu (1+e)]$ and $h^2/[\mu (1-e)]$. Their sum is twice the semi-major axis, soAt either apsis $v=h/r$; substituting an apsidal radius gives $\epsilon =-\mu /(2a)$. Thus the specific orbital energy implies the vis-viva equationThese are exact relative-coordinate formulas for an elliptic orbit, $0\le e<1$. In the barycentric frame the comet's speed is $M_{\star }/(M_{\star }+m)$ times the relative speed; the printed identification with the comet's stellar orbital speed uses $M_{\star }\gg m$.

The radial velocity of a Kepler orbit follows from the vis-viva equation after subtracting the tangential component:Each radial interval is crossed twice per orbital period, so its fraction of time isThis is the phase-mixed radial probability of a Kepler orbit, normalized to one between $q$ and $Q$. Dividing by the annular area $2\pi rdr$ gives the phase-mixed comet surface density per comet:For $Q\gg a_p>q$,The printed density has the powers of $Q$ and $a_p$ interchanged. Its expression is too large by $Q/a_p$ in this limit. The normalized residence-time derivation fixes the corrected expression, which is also required for the later $Q^{1/2}$ ejection-time scaling. For a population of $N_c$ independent identical comets, multiply ${\Sigma }_c$ by $N_c$ to obtain a number surface density of a disk.

Put ${\mu }_{\star }=G{M}_{\star }$. Conservation of specific orbital energy and specific angular momentum givesThe last comet expression is the vis-viva equation at the planet's orbit. The planet has only tangential velocity, while the comet's tangential component is $v_t=h/a_p$. Consequently its prograde relative velocity is exactly, before planetary deflection,In the near-parabolic encounter limit $Q\gg a_p$, this reduces toThe displayed approximation also requires $Q\gg a_p$. The assumptions $Q\gg q$ and $q\ll a_p$ alone do not guarantee it: an orbit with $Q=a_p$ encounters the planet near apoapsis and has $v_{\mathrm{rel}}\simeq v_p$, not $\sqrt{3}{v}_p$.

If all coplanar near-parabolic planet-crossing comets are admitted, so that $0\le q\le a_p$, the tangential component ranges from $-\sqrt{2}{v}_p$ to $+\sqrt{2}{v}_p$ when retrograde orbits are included. ThusFor prograde orbits alone the upper limit is $\sqrt{3}{v}_p$; for the stated deep-crossing limit $q\ll a_p$, both prograde and retrograde encounters approach $\sqrt{3}{v}_p$. These are the speeds outside the planet's gravitational well. The surface impact speed is $v_{\mathrm{imp}}=\sqrt{v_{\mathrm{rel}}^2+v_{\mathrm{esc},\mathrm{p}}^2}$, where $v_{\mathrm{esc},\mathrm{p}}$ is the planetary escape velocity; under negligible gravitational focusing the two speeds agree.

For a phase-mixed orbit, probability in a short arc is $dt/T=ds/(Tv)$, so the line density on a Kepler orbit is inversely proportional to speed. At true anomaly $\theta$,Using the vis-viva equation, $v^2=(G{M}_{\star }/a)[1+2e\cos \theta ]+O(e^2)$ and hence $v=na[1+e\cos \theta ]+O(e^2)$. ThereforeFor one particle, the normalized line probability is $(2\pi a{)}^{-1}(1-e\cos \theta )+O(e^2)$. This is density per arc length; density per unit polar angle has a further factor $ds/d\theta \simeq a(1-e\cos \theta )$ and is proportional to $1-2e\cos \theta$.

At the encounter radius $r=a_p$, the planet's circular Kepler orbit has speedThe comet's speed follows from the vis-viva equation:Its component parallel to the planet's prograde tangential velocity is fixed by its specific angular momentum,The relative velocity therefore satisfiesand hence

Let $h=r^2\dot{f}$ be the specific angular momentum. The radial Kepler orbit equation has semi-latus rectum $p=h^2/\mu$, so comparison withgives $h^2=\mu a(1-e^2)$ for an elliptic orbit and $h^2=\mu a(e^2-1)$ for a hyperbolic Kepler orbit. At either apsis, $\dot{r}=0$ and $v=h/r$. Substituting the apsidal radius and angular momentum into part (i), or equivalently using the vis-viva equation, givesThus the signs are $C=\mp \mu /(2a)$ in the convention of the question.

The limiting trajectory grazes the planet at periapsis $R_p$. Equating the asymptotic and periapsis values of specific angular momentum and using the vis-viva equation givesHence the comet avoids the planet when

A Hohmann transfer orbit connects two coplanar circular orbits using two tangential, effectively instantaneous velocity changes. Take outward transfer radii $r_2>r_1$, and write ${\mu }_g=GM$ for the central body's gravitational parameter. The intervening ellipse has periapsis $r_1$, apoapsis $r_2$, semimajor axis $a_t=(r_1+r_2)/2$ and orbital eccentricity $e=(r_2-r_1)/(r_1+r_2)$.

The circular speed is $\sqrt{{\mu }_g/r}$. Conservation of orbital energy gives the vis-viva equation $v^2={\mu }_g(2/r-1/a_t)$. Therefore the transfer speeds at the two impulses areThe first prograde impulse raises the far side of the orbit; the second prograde impulse raises the near side to circularize at $r_2$. Their magnitudes areThe transfer takes half an ellipse period. Kepler's third law givesFor inward transfer the same ellipse is traversed from apoapsis to periapsis and both burns are retrograde. A rendezvous also needs the correct initial phase of the target: for outward transfer its initial angular lead is $\pi -\sqrt{{\mu }_g/r_2^3}{t}_t$ modulo $2\pi$. The method assumes a two-body gravitational field, negligible atmospheric drag and burns short compared with an orbit. It is an economical two-impulse transfer between these circles; it is not universally the least-$\Delta v$ transfer if additional impulses and very large intermediate radii are permitted.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

In units where a circular planet has orbital radius and speed one, the particle's relative speed $u$ outside a close encounter satisfies $u^2=3-T$, to leading order in the planet-to-star mass ratio. This follows from the vis-viva equation and $u^2=v^2+1-2h_z$ at the encounter radius. A short conservative encounter changes the direction of this relative velocity while preserving its magnitude.

For a Kepler orbit with semi-major axis $a$, the vis-viva equation isFor the convention that a hyperbolic Kepler orbit has positive semi-major-axis magnitude $a$, its final term instead has a plus sign.