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For a parabolic trajectory with periapsis radius $r_p$, put $D=\tan (f/2)$, where $f$ is true anomaly. The Barker equation is $t-t_p=\sqrt{2r_p^3/(GM)}(D+D^3/3)$. Since $r=r_p(1+D^2)$, it is equivalent to the radius-time relation for a parabolic Kepler orbit. The sign of $D$ distinguishes approach from recession; $t_p$ is the periapsis time.

For two uniform angular motions with $P_b<P_c$, an inner exoplanet transit at zero and next outer exoplanet transit at $\tau$, the next conjunction solves $(n_b-n_c)t+n_c\tau =2\pi$. Hence its longitude modulo $2\pi$ isFor an eccentric outer body this is only an approximation. Let $f_L,M_L$ be its line-of-sight anomalies and ${\Delta }_e(t)=f(t)-M(t)-(f_L-M_L)$. Exact true alignment instead requires $(n_b-n_c)t+n_c\tau -{\Delta }_e(t)=2\pi$. The correction follows directly from uniform mean anomaly and nonuniform true anomaly.

An exoplanet passes in front of its host star as seen by an observer and blocks part of the stellar flux. Exoplanet transit photometry infers geometry and radii from the resulting light curve. A true alignment involves the true anomaly, not generally the mean anomaly.

View a circular orbit along its perpendicular spin axis. Let $f$ be the orbital angle measured from a fixed direction, choosing an arbitrary zero for the circular limit of true anomaly, and let $\theta$ be the direction of an outward arrow painted on the body, measured from the same fixed direction. The direction from the body to its companion is $f+\pi$. Thus an inward-facing arrow satisfiesMore generally synchronous rotation keeps $\theta -f$ constant; its particular value selects which surface direction faces the companion. Over one orbital period $T$, both $f$ and the unwrapped $\theta$ increase by $2\pi$. This is the short geometric explanation for one turn per orbit. For a nonspinning body, $\dot{\theta }=0$ instead, and the companion direction sweeps through all surface longitudes. On an eccentric orbit, mean $1:1$ spin does not force $\theta -f$ to stay exactly constant: true anomaly advances nonuniformly, producing the longitudinal component of libration.

For perturbing acceleration $R\hat{\mathbf{r}}+T\hat{\mathbf{\theta }}$ on an osculating orbital element system with fixed central parameter $\mu$,To prove it, differentiate the specific orbital energy $-\mu /(2a)$, obtaining $\dot{a}=(2a^2/\mu )(R\dot{r}+Th/r)$. Substitute $\dot{r}=\mu e\sin f/h$, $h/r=\mu (1+e\cos f)/h$ and $h^2=\mu a(1-e^2)$. Here $f$ is the true anomaly; a normal perturbing acceleration does no instantaneous work.

For a nearly radial Kepler orbit with pericentre $q$, specific angular momentum is $h\simeq \sqrt{2GMq}$. Relative to a circular reference orbit of radius $a_p$ and mean motion $n_p$, the instantaneous azimuthal rate changes from faster to slower at $r_x=\sqrt{h/n_p}\simeq a_p(2q/a_p{)}^{1/4}$. In a rotating reference frame the polar angle has a turning point at this radius. The true anomaly there differs from $\pi$ by approximately $2\sqrt{q/r_x}$.

The bound relative Kepler orbit is an ellipse with the attracting centre at one focus of a conic. In polar coordinates, its separation is the displayed function of true anomaly $f$, the angle measured from periapsis. For $a=1$ and $e=0.6$, it becomes $r=0.64/(1+0.6\cos f)$, giving $0.4$ at $f=0$ and $1.6$ at $f=\pi$. Setting orbital eccentricity to zero gives a circular orbit. This specifies the path; Kepler's second law specifies how it is traversed.

Uniform mean anomaly implies a line probability inversely proportional to the orbital speed. For small orbital eccentricity and true anomaly $f$, it is proportional to $1-e\cos f$.

For a quantity on a periodic Kepler orbit, its orbit average is $⟨x⟩=P^{-1}{\int }_0^{P}x(t)dt$. Equal increments of true anomaly are not equal increments of time. Eccentric anomaly gives the weight $dt=(1-e\cos E)dE/n$, which yields $⟨\cos f⟩=-e$. Slow orbital perturbations can be averaged using elements held fixed over a single orbital period.

For isotropic stellar mass loss the eccentricity vector obeys $\dot{\mathbf{e}}=-(\dot{\mu }/\mu )(\mathbf{e}+\hat{\mathbf{r}})$. With true anomaly $f$ and $e>0$, this gives $\dot{e}=-(\dot{\mu }/\mu )(e+\cos f)$ and $\dot{\varpi }=-(\dot{\mu }/\mu )\sin f/e$. The osculating pericentre satisfies $\dot{q}/q=-(\dot{\mu }/\mu )(1-\cos f)/(1+e)$, hence cannot decrease during mass loss and is instantaneously stationary at pericentre. The vector form is nonsingular at a circular orbit.

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.

Put $k=p+1$, $\mu =G{M}_{\star }$ and $n_{\mathrm{pl}}=\sqrt{\mu /a_{\mathrm{pl}}^3}$. To leading order the mean-motion resonance fixes $a=a_{\mathrm{pl}}{k}^{-2/3}$. The specific angular momentum isThe asteroid's instantaneous angular speed is $h/r^2$. Thus the high-eccentricity angular-speed crossover satisfies $h/r_x^2=n_{\mathrm{pl}}$, givingHere $q\ll r_x\ll a$, so a parabolic Kepler orbit gives the local motion accurately. Its polar equation is $r=2q/(1+\cos f)$. Write $f_x=\pi -{ϵ}_x$. Since $1+\cos f_x\simeq {ϵ}_x^2/2$,This is the outward crossing; the inward crossing has true anomaly $-f_x$ modulo $2\pi$.

For the pericentre-to-crossover flight time use $D=\tan (f/2)$. The parabolic Kepler orbit has $r=q(1+D^2)$ and $dt=r^{2}df/h$, whence direct integration gives the Barker equationAt the crossover $D_x\simeq \sqrt{r_x/q}\gg 1$. Consequently the planet's angular displacement isThe error tends to zero as $q/a_{\mathrm{pl}}\to 0$ at fixed $k$; it includes the finite binding energy of the elliptic Kepler orbit as well as the subleading term in the Barker equation.

At the $j$th pericentre passage the asteroid's unwrapped mean longitude is $\lambda =\varpi +2\pi j$. The resonant argument therefore givesThus $\varphi /k$ specifies the planet's direction relative to the asteroid's longitude of pericentre when the asteroid passes pericentre, with $k$ branches separated by $2\pi /k$. It is not the instantaneous true-longitude separation throughout the orbit. At high orbital eccentricity the asteroid quickly sweeps to a direction nearly opposite pericentre, while the planet scarcely moves. During the long outer excursion the asteroid's direction changes slowly and the planet advances substantially. This makes an astronomical conjunction during the outer excursion much more dangerous than an astronomical conjunction close to pericentre.

At the following apocentre, the planet has advanced $\pi /k$. The asteroid's direction in the rotating reference frame is consequentlyThese $k$ apocentre directions are equally spaced. Phase protection of interior integer resonances places the planet in the largest gap between them, rather than at an apocentre direction. For $k=2$, $\varphi =\pi$ puts an apocentre toward the planet, whereas $\varphi =0$ puts the apocentres at $\pm \pi /2$. For $k=3$, $\varphi =0$ puts an apocentre toward the planet, whereas $\varphi =\pi$ puts the apocentres at $\pi /3,\pi ,5\pi /3$. Hence the geometrically favored resonant-argument libration centres in this regime of high orbital eccentricity areThis is an encounter-avoidance argument for likely stability, not a calculation of a resonant Hamiltonian or a claim that every orbit near either centre is stable at arbitrary orbital eccentricity.

The following rotating-frame drawing uses $a_{\mathrm{pl}}=1$, $a=3^{-2/3}$, $q=0.002$ and $\varphi =\pi$. It follows three successive Kepler orbits, which fill one planet orbital period and close the pattern. The blue segment is just the first outward half-orbit. The marked crossover is a maximum of its rotating-frame polar angle, since $d\psi /dt=h/r^2-n_{\mathrm{pl}}$ changes from positive to negative there. The planet stays at $(1,0)$; the three apocentres avoid that direction.

To estimate angular-momentum kicks to nearly radial orbits, note that $q\simeq h^2/(2\mu )$, so an order-one fractional change in $q$ requires $|\Delta h|$ of order $h$. A characteristic distant encounter at radii of order $a_{\mathrm{pl}}$ has tangential acceleration of order $G{M}_{\mathrm{pl}}/a_{\mathrm{pl}}^2$ and duration of order $n_{\mathrm{pl}}^{-1}$. Its torque per unit asteroid mass therefore producesWith the most favorable coherent signs, the crude encounter count isOrder-one factors depend on $p$ and encounter geometry. Uncorrelated signs instead give a random walk count of order $(M_{\star }/M_{\mathrm{pl}}{)}^{2}q/a_{\mathrm{pl}}$ when this is large. A librating phase-protected orbit can suppress the kicks even further.

There is an important limit to calling this a minimum. The question does not specify an encounter impact parameter. For a weak close encounter with impact parameter $b$ and relative speed $u\sim \sqrt{G{M}_{\star }/a_{\mathrm{pl}}}$, the impulse estimate gives $|\delta h|\sim G{M}_{\mathrm{pl}}{a}_{\mathrm{pl}}/(bu)$ and hence $N\sim (M_{\star }/M_{\mathrm{pl}})(b/a_{\mathrm{pl}})\sqrt{q/a_{\mathrm{pl}}}$, capped below by one. In the 2:1 case the Kepler orbit can cross the planet's orbit, so one suitably close encounter can suffice: there is no impact-parameter-independent large lower bound. For $k\ge 3$, $Q\simeq 2a_{\mathrm{pl}}{k}^{-2/3}<a_{\mathrm{pl}}$ bounds the separation below by $a_{\mathrm{pl}}-Q$, and supplies a geometric factor at fixed $k$. The boxed count is the usual orbital-scale, maximally coherent estimate; the assumptions are essential.

The central equation must be read as $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$; its printed left-hand expression lacks $=0$. Initially $\mu$ is constant. Dotting this equation with the velocity givesTaking its cross product with $\mathbf{r}$ gives $d(\mathbf{r}\times \mathbf{v})/dt=0$. In the fixed orbital plane the magnitude is $h=r^2\dot{\theta }$. These are conservation of specific orbital energy and specific angular momentum.

Let $f=\theta -\varpi$. Differentiating the polar equation of the Kepler orbit, with its elements initially constant, givesSubstitution into the specific orbital energy yields

For isotropic stellar mass loss with no recoil or drag, the force is still central, so $\mathbf{h}$ remains exactly constant, although $C$ does not. NowThe same algebraic relation between $C,h,e$ holds for the instantaneous osculating orbital elements. Differentiating it and using $h^2/(\mu r)=1+e\cos f$ givesand, for $e>0$,A particularly useful nonsingular description uses the eccentricity vectorThe vector equation remains meaningful at a circular orbit, where a longitude of pericentre is undefined. Its components along and perpendicular to $\mathbf{e}$ give the scalar formula above and

The osculating pericentre is $q=h^2/[\mu (1+e)]$. ThereforeFor mass loss, $\dot{\mu }<0$, this is nonnegative: pericentre does not decrease. It is positive except at pericentre itself, where $\cos f=1$ and $\dot{q}=0$. Thus a strictly opposite sign at every instant is not literally true; a pericentre passage with nonzero mass-loss rate is a counterexample to the strict wording. The intended monotonicity follows exactly, without requiring slow mass loss.

For adiabatic orbital expansion under isotropic mass loss, average over the unperturbed Kepler orbit and hold $\dot{\mu }/\mu$ constant to leading order during that orbit. If $E$ is the eccentric anomaly and $n=\sqrt{\mu /a^3}$, thenIt follows immediately thatAveraging with uniform true anomaly instead of uniform time would give the wrong result. Since $\mu a(1-e^2)=h^2$, conserved $h$ and constant secular orbital eccentricity giveThe pericentre and apocentre expand in the same secular proportion. These are leading adiabatic invariants, rather than exact invariants for arbitrary time-dependent $\mu$.

For the instantaneous precession fraction write $\theta =f+\varpi$ and $\dot{\theta }=h/r^2$. The adiabatic apsidal condition is controlled byIt is a signed fraction: apsidal advance contributes positively on the outward leg during mass loss, and negatively on the inward leg. Let ${\tau }_{\mu }=|\mu /\dot{\mu }|$. Ordinary orbit averaging requires ${\tau }_{\mu }\gg P$, together with negligible variation of the mass-loss law over one orbit. If the pericentre direction is also to vary negligibly relative to the instantaneous azimuthal motion, the maximum absolute fraction must be small. A precise condition for $0<e<1$ isThe maximizing cosine is $c=(1-\sqrt{1+8e^2})/(2e)$: differentiating the last factor gives $c+2e-e{c}^2=0$. Thus one can evaluate $F(e)$ explicitly by putting $\sqrt{1-c^2}/(1+ec{)}^2$ into the formula. Its useful limits areFor orbital eccentricity of order unity, mass loss much slower than an orbital period suffices; near a circular orbit, keeping the pericentre direction slowly varying additionally requires $n{\tau }_{\mu }e\gg 1$. The latter divergence is a singularity of the pericentre coordinate, not a divergence of the eccentricity vector dynamics. Indeed, at frozen $\dot{\mu }/\mu$, integrating the scalar evolution to first order gives $\delta e=-(\dot{\mu }/\mu )(1-e^2)\sin E/n$ up to an integration constant. Its absolute amplitude stays small for slow mass loss even when the scalar longitude of pericentre becomes unsuitable.

In the center of mass frame, the velocities are ${\mathbf{v}}_1=(M_2/M)\mathbf{v}$ and ${\mathbf{v}}_2=-(M_1/M)\mathbf{v}$. Hence the kinetic energy is $T=\frac{1}{2}\mu v^2$, while the Newtonian gravitational potential energy is $U=-G{M}_1{M}_2/r=-GM\mu /r$. The two-body orbital energy is thereforeThe relative equation of Newtonian gravity is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. It is a central force, so conservation of angular momentum confines the motion to a plane and preserves the specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|=r^2\dot{\theta }$.

For completeness, derive the polar equation of a Kepler orbit. Let $w=1/r$ and use primes for $\theta$ derivatives. Then $\dot{r}=-h{w}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{w}^2(w^{\prime \prime }+w)$. The radial equation gives the Binet equationChoose the angular origin at closest approach. Its solution is $w=(GM/h^2)(1+e\cos \theta )$, soFor the ellipse, the orbital eccentricity satisfies $0\le e<1$. Its extreme separations are $r_p=l/(1+e)$ and $r_a=l/(1-e)$. The semi-major axis is half their sum, giving $a=l/(1-e^2)$ and thus $l=a(1-e^2)$. The true anomaly $\theta$ runs through a full $2\pi$; the endpoints are the same position, and for a circular Kepler orbit the angular origin is arbitrary.

The radial and transverse relative speeds are $\dot{r}=(GM/h)e\sin \theta$ and $r\dot{\theta }=h/r$. Substituting these into the specific orbital energy givesTherefore the conserved orbital energy isThe negative specific orbital energy and fixed angular momentum characterize the bound Kepler orbit; neither should be assumed unchanged through a mass-ejecting explosion.

Initially the circular Kepler orbit has $r=a$ and $v^2=GM/a$. Treat the supernova kick in a binary star as impulsive: the relative position does not change, the companion's velocity is unchanged during the impulse, and the new neutron star receives $\mathbf{u}$ with $|\mathbf{u}|=\alpha v$. Thus ${\mathbf{v}}^{\prime }=\mathbf{v}+\mathbf{u}$, and with $\psi$ the angle between these vectors before the kick,Write $M^{\prime }=M_1^{\prime }+M_2$, $\beta =M^{\prime }/M$ and ${\mu }^{\prime }=M_1^{\prime }{M}_2/M^{\prime }$. The post-explosion specific orbital energy isUsing ${\epsilon }^{\prime }=-G{M}^{\prime }/(2a^{\prime })$ givesHere $a^{\prime }>0$ for a bound Kepler orbit. For a hyperbolic Kepler orbit, this formula uses the signed energy parameter $a^{\prime }<0$, rather than the positive geometric magnitude of the hyperbola's semi-major axis. At the parabolic Kepler orbit boundary, $1/a^{\prime }=0$.

The kick-direction binary survival criterion is bound if $S<2\beta$ and unbound with positive asymptotic speed if $S>2\beta$. To prove the printed sufficient disruption condition, choose a perpendicular kick, $\cos \psi =0$: then $S=1+{\alpha }^2$, so $M^{\prime }<\frac{1}{2}(1+{\alpha }^2)M$ produces positive specific orbital energy. This is sufficient, not the sharp existence threshold. Sincethe complete kick-direction binary survival criterion isThe second inequality follows by taking a kick directly opposite to $\mathbf{v}$. All directions remain bound if $2\beta >(1+\alpha {)}^2$, while every direction has positive escape energy if $2\beta <(1-\alpha {)}^2$. For $\alpha >0$, the marginal direction obeys $\cos \psi =(2\beta -1-{\alpha }^2)/(2\alpha )$ when the right-hand side lies in $[-1,1]$. Equalities give marginal parabolic Kepler orbits for the relevant extreme direction. With zero kick, losing more than half the original total mass unbinds the circular Kepler orbit.

For an escaping pair, the Newtonian gravitational potential energy approaches zero at infinite separation. Conservation of the post-explosion two-body orbital energy then gives $\frac{1}{2}{\mu }^{\prime }{V}^2={\mu }^{\prime }{\epsilon }^{\prime }$, or the asymptotic relative speed of a disrupted binaryThis is the relative recession speed, not the velocity of the new center of mass or either star's individual velocity in the original inertial frame. A marginal parabolic Kepler orbit separates with $V=0$ at infinity.

For a spherical grain of density $\rho$, the radiation-pressure coefficient isThe $r^{-2}$ dependence of radiation pressure cancels that of stellar gravity. Here radiation-pressure efficiency includes absorption and the momentum-transfer part of scattering, averaged over the stellar spectrum. In the geometrical optics regime, $D$ exceeds the important stellar wavelengths, $Q_{\mathrm{pr}}$ is of order unity, and $\beta \propto D^{-1}$: cross-sectional area grows as $D^2$, whereas mass grows as $D^3$. Around the stellar wavelength, Mie scattering can produce a broad maximum and material-dependent structure. In the Rayleigh scattering regime, absorption gives $Q_{\mathrm{abs}}\propto D$, while scattering gives $Q_{\mathrm{sca}}\propto D^4$. Thus small absorbing grains approach an approximately constant $\beta$, whereas nearly transparent grains have $\beta \propto D^3$. A universal fall to zero at small $D$ is therefore inappropriate without specifying the optical properties. The sketch shows both possible small-grain limits; its vertical normalization is illustrative.

Assume release with negligible velocity relative to the planet, negligible planetary gravity after release, and a constant radiation-pressure coefficient. Put $\mu =G{M}_{\star }$ and ${\mu }_d=(1-\beta )\mu$. The inherited speed and specific angular momentum are $v_p=\sqrt{\mu /a_p}$ and $h=a_p{v}_p$. The new specific orbital energy isFor $0\le \beta <1/2$, the dust orbit released from a circular parent ring is an ellipse. Using ${\epsilon }_d=-{\mu }_d/(2a_d)$ and $h^2={\mu }_d{a}_d(1-e_d^2)$ givesFor $\beta >0$, release is at periapsis, since $a_d(1-e_d)=a_p$. At the radiation-pressure blowout threshold, $\beta =1/2$, the orbit is parabolic. For $1/2<\beta <1$ it is hyperbolic: the signed semimajor axis is negative and $e_d>1$. For $\beta \ge 1$ the central force ceases to be attractive, so the elliptical interpretation and the positive-eccentricity formula cannot be extended unchanged.

With ${\theta }_d$ measured from release, it is the true anomaly. The polar equation of a Kepler orbit and conservation of specific angular momentum giveDefine the signed dust-tail angular lag by $\Delta \theta ={\theta }_d-{\theta }_p$, with ${\theta }_p=n_{p}t$ and both angles unwrapped. For fixed ${\theta }_d$, expansion to first order in $\beta$ givesso $\Delta \theta =2\beta (\sin {\theta }_d-{\theta }_d)+O({\beta }^2{\theta }_d)$. The dust trails the planet, hence the signed lag is negative. Its slope is $-4\beta {\sin }^2({\theta }_d/2)$; it decreases monotonically, has horizontal tangents at successive release-direction passages, and has a sinusoidal ripple about $-2\beta {\theta }_d$. At the end of orbit $j$, ${\theta }_d=2\pi j$ and $\Delta \theta \simeq -4\pi \beta j$. At five orbits this is $-0.2\pi$ for $\beta =0.01$ and $-0.8\pi$ for $\beta =0.04$. The upper-$\beta$ sketch remains a first-order approximation; its accumulated error is $O({\beta }^2{\theta }_d)$.

For the subsequent positive-distance statements, set $s=-\Delta \theta ={\theta }_p-{\theta }_d\ge 0$. The mean motion of the grain isA full relative wrap, $s=2\pi$, therefore takes $t_{\mathrm{wrap}}\simeq \pi /(\beta n_p)$ after averaging over the orbital ripple. During this time Poynting–Robertson drag changes the semimajor axis byThe signed change is negative. This estimate treats $a_d\simeq a_p$ as fixed in the rate and computes the drag accumulated over the radiation-pressure wrap time. Since $v_p/c\ll 1$, drag causes little migration over that time. For an arbitrarily tiny $\beta$, however, this small migration can itself alter the relative phase appreciably: neglecting that feedback on the wrap time additionally requires $v_p/c\ll \beta$.

Rapid dust sublimation is a plausible sink near a hot planet: a grain can evaporate long before it completes a relative wrap. Destructive collisions can also remove visible grains. Poynting–Robertson drag alone, with the estimate above, does not explain immediate removal from a short tail. If the destruction time is $t_{\mathrm{loss}}$, a short tail requires roughly $2\beta n_p{t}_{\mathrm{loss}}\ll 2\pi$.

For a steady dust-tail continuity equation, assume a constant injection rate $\dot{N}$, grains with the same fixed $\beta$, negligible release-speed dispersion, and negligible destruction within the segment being calculated. A finite destruction lifetime can terminate that segment, or multiply its density by a survival probability. To first order,To obtain the explicit trigonometric profile, make the additional secular phase approximation for a dust tail, $s\simeq 2\beta {\theta }_d$, discarding the bounded $-2\beta \sin {\theta }_d$ term in the angle-to-age map while retaining the periodic instantaneous drift speed. ThenSince the square is even, the same written profile applies to the signed $\Delta \theta$ if number intervals are interpreted with positive orientation. The result is an approximate phase-substitution profile, not a uniformly valid consequence of $\beta \ll 1$. The consistent first-order description is instead parametric:In particular, immediately after release, $s\sim \beta {\theta }_d^3/3$ and $n\propto s^{-2/3}$, whereas the phase-substitution expression would give $s^{-2}$. Nor does $s\ll 2\pi$ by itself justify neglecting the orbital ripple. The formal enhancements where $\dot{s}=0$ are dust-tail caustics; a spread of radiation-pressure coefficients, finite release velocities, and destruction smooth them. With constant lifetime $\tau$, the parametric profile acquires $\exp [-t({\theta }_d)/\tau ]$. These qualifications state precisely the extra assumptions behind the explicit profile.

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 $M_{\star }\gg m$.

Treat $a,e$ as instantaneous osculating orbital elements of the fixed-central-parameter Kepler orbit. The perturbing work changes the specific orbital energy according toThe specific angular momentum and conic equation giveSubstituting proves the Gauss planetary equation for the semi-major axis:Here $f$ denotes true anomaly, not the fractional luminosity used in Question 1. The equation follows from work alone; a normal perturbing acceleration has no instantaneous contribution because the velocity lies in the orbital plane. For slowly averaged evolution, the perturbation must be weak over one orbital period.

First adopt the uniform-angular-motion approximation implicit in the requested formula. Set the line of sight to longitude zero, the $b$ exoplanet transit to $t=0$, and the next $c$ exoplanet transit to $t=\tau$, with $0\le \tau <P_c$. Write $n_b=2\pi /P_b$, $n_c=2\pi /P_c$, with $P_c>P_b$. The next conjunction satisfiesThe common longitude is $n_{b}t$ modulo $2\pi$. Subtracting one $2\pi$ gives the conjunction longitude from a transit time lagSubstitution of the period ratio yieldsIt is exact in this uniform-angle model; no first-order expansion in $\delta$ is necessary here. For other choices of which exoplanet transit is used, choose the appropriate whole-turn branch.