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With constant mean motions and $P_c/P_b=(1+\delta )(p+q)/p$, mean conjunctions advance by $\Delta \Lambda =2\pi /(P_c/P_b-1)$ each synodic period. Tracking one branch every $q$ conjunctions and subtracting $2\pi p$ givesThe usual formula $-2\pi p\delta /(q{P}_b)$ keeps only first order in $\delta$. A labelled branch takes $q{P}_b(1+\delta )/(p|\delta |)$ to turn once, whereas the full set of $q$ interchangeable directions repeats after $P_b(1+\delta )/(p|\delta |)$. Finite orbital eccentricity introduces true-conjunction corrections.

A first-order resonant orbital perturbation changes the semi-major axis and therefore the mean motion. Integrating the sinusoidal mean-motion variation over the near-resonant transit-timing superperiod gives a mean-longitude amplitude proportional to ${\mu }_{2}e/{\Delta }^2$. Converting phase to transit time gives the displayed amplitude ratio. It assumes small perturbations, circulating resonant argument, effectively fixed free orbital eccentricity, and dominance of this enhanced term over direct phase and forced-eccentricity contributions. The formal divergence as $\Delta \to 0$ signals failure of this expansion, not infinitely large transit times.

Take an illustrative elliptic Kepler orbit with semi-major axis $a=1$, orbital eccentricity $e=0.6$ and gravitational parameter $\mu =1$, in consistent dimensionless units. Its semi-minor axis is $b=a\sqrt{1-e^2}=0.8$, mean motion is $n=\sqrt{\mu /a^3}=1$, and specific angular momentum magnitude is $h=\sqrt{\mu a(1-e^2)}=0.8$. About the attracting focus,The two sectors below each span $\Delta t=0.5$, giving $\Delta A=0.2$. Near periapsis the short radius sweeps a wide angle; near apoapsis the long radius sweeps a narrow angle. Constant areal velocity therefore does not require constant angular velocity or constant speed.

To compute the sectors without mistaking equal angles for equal times, parametrize the ellipse by eccentric anomaly $E$:This follows by integrating $\frac{1}{2}(xdy-ydx)$. Kepler's equation gives $E-e\sin E=M=n(t-t_{\mathrm{p}})$, with mean anomaly $M$ measured from periapsis time $t_{\mathrm{p}}$. Consequently $\Delta A=abn\Delta t/2=h\Delta t/2$. The displayed sectors use mean anomaly intervals $[-0.25,0.25]$ and $[\pi -0.25,\pi +0.25]$, rather than equal eccentric anomaly increments.

For positive locally constant $A,C$, supplement damped first-order resonant complex eccentricity by $\dot{a}/a=-2(p+1)Ce\sin \varphi -4A/5$. Balancing the two dissipative rates gives $e_f^2=2/[5(p+1)]$, $\sin {\varphi }_f=-A{e}_f/C$ and $B{x}_f=-(C/e_f)\cos {\varphi }_f$. Two fixed points exist when $A<C\sqrt{5(p+1)/2}$. The negative-cosine branch has $x_f>0$ and is stable in this leading model: a phase increase raises the positive resonant torque, raises semi-major axis, lowers mean motion, and reverses the phase drift. The positive-cosine branch has the opposite feedback.

For a coupled verification put $z=u+iv$. The characteristic polynomial at either fixed point is ${\lambda }^3+2A{\lambda }^2+[A^2+B^2{x}_f^2-2B(p+1)C{u}_f]\lambda -4AB(p+1)C{u}_f$. For $u_f<0$ its positive coefficients satisfy the cubic Routh-Hurwitz stability criterion, since the remaining inequality reduces to $2A(A^2+B^2{x}_f^2)>0$. For $u_f>0$ its negative constant term forces a positive real root. At the equality threshold the equilibria merge and an eigenvalue is zero. This is local stability for the frozen smooth-coefficient approximation, not a conclusion for arbitrary rapidly varying dissipation laws.

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 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.

A lunar phase is the appearance of the sunlit portion of the Moon as seen from Earth. Outside a lunar eclipse, the distant Sun illuminates approximately one hemisphere of the Moon, but an observer sees a changing overlap of that hemisphere with the hemisphere facing Earth. The lunar phase therefore changes even though synchronous rotation keeps approximately the same near side of the Moon facing Earth. ^[ref].

The cycle runs through new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, last quarter and waning crescent. “Waxing” means that the lunar illuminated fraction is increasing; “waning” means it is decreasing. A crescent has less than half its apparent disk sunlit, and a gibbous lunar phase has more than half. First and last quarter each show half a disk: “quarter” refers to progress around the cycle, rather than a quarter of the disk being lit. ^[ref].

An instructive circular orbit model puts Earth at the origin, the Sun far away along $+x$, and the Moon at $R(\cos \theta ,\sin \theta )$. The sunward position $x>0$, $\theta =0$, gives new Moon; the opposite position $x<0$, $\theta =\pi$, gives full Moon. Counterclockwise motion gives first quarter at $\theta =\pi /2$ and last quarter at $\theta =3\pi /2$. The Sun's direction is fixed during this snapshot of the cycle. Every small globe in the orbital view is illuminated on its $+x$ half; the adjacent apparent disks are separate views from Earth, rather than globes viewed from above.

The disk views choose north up, with the upward direction perpendicular to the orbital plane. In this convention waxing lunar phases are bright on the right and waning lunar phases on the left. Local horizon orientation and viewing from different locations on Earth can rotate the apparent disk; its lit edge always points toward the projected direction of the Sun. The lunar illuminated fraction is unaffected by this rotation.

The useful angle is the phase angle $\alpha =\angle SME$, with $S$ the Sun, $M$ the Moon and $E$ the observer at Earth's center: its vertex is at the Moon. Full Moon has $\alpha =0$, quarter has $\alpha =\pi /2$, and new Moon has $\alpha =\pi$ in the ideal model. By contrast, the elongation $ϵ=\angle SEM$ has its vertex at Earth. For a distant Sun, its rays are parallel, soHere $k$ is the lunar illuminated fraction: a fraction of projected disk area, not a fraction of the Moon's total surface area or its received radiant flux relative to full Moon. The phase angle and unsigned elongation each lie between $0$ and $\pi$, so either alone gives the illuminated amount without distinguishing waxing from waning. The orbital direction resolves that ambiguity.

The lunar illuminated fraction has an elegant geometric derivation. Model the Moon as a sphere of unit radius and view it along the $z$ direction. Take the Sun's unit direction to be $(\sin \alpha ,0,\cos \alpha )$. A visible surface point above apparent disk coordinates $(x,y)$ has $z=\sqrt{1-x^2-y^2}$. Direct illumination requires the outward normal's dot product with the sunward direction to be positive. The day-night terminator therefore obeys $x\sin \alpha +z\cos \alpha =0$. At a fixed $y$, set $q=\sqrt{1-y^2}$; the visible day-night terminator is $x_t=-q\cos \alpha$, and the illuminated strip extends from $x_t$ to the right edge $q$. Integrating its width givesThe integral is the area of half a unit circle. Thus a three-dimensional visibility problem reduces to a circle's area and a cosine. The derivation assumes parallel sunlight, a distant observer and a sphere, and ignores a lunar eclipse. Real observed brightness additionally depends on surface scattering of light and observing geometry, so $k=1/2$ does not promise half the full-Moon radiant flux.

For a worked example, elongation $ϵ=60^{\circ }$ gives phase angle $\alpha =120^{\circ }$ and $k=(1+\cos 120^{\circ })/2=1/4$: a crescent with $25\%$ of its apparent disk sunlit. At elongation $120^{\circ }$, the phase angle is $60^{\circ }$ and $k=3/4$: a gibbous lunar phase. At $90^{\circ }$, $k=1/2$. In each case the same $k$ occurs once while waxing and once while waning in the ideal cycle.

Ordinary lunar phases are not caused by Earth's shadow. A lunar eclipse occurs only when the Moon actually enters that shadow near full Moon. The orbital inclination of the Moon is about $5^{\circ }$ relative to Earth's orbital plane around the Sun; most full Moons pass above or below the shadow. This is why full Moon does not imply a lunar eclipse every month. ^[ref].

The phase cycle is a synodic month, about $29.5$ days, whereas the Moon's orbital period relative to distant stars is a sidereal month, about $27.3$ days. During one sidereal month, Earth advances around the Sun, so the Moon must move farther to repeat the sunward alignment. For coplanar circular orbits traversed in the same direction, subtract their mean motions: $n_{\mathrm{syn}}=n_{\mathrm{sid}}-n_{\mathrm{year}}$, with $n=2\pi /P$. HenceUsing $P_{\mathrm{sid}}=27.321661\mathrm{days}$ and $P_{\mathrm{year}}=365.256\mathrm{days}$ gives $P_{\mathrm{syn}}=29.530591\mathrm{days}$, consistent with the mean synodic month of $29.53059$ days. After one sidereal month, the sunward direction has advanced about $26.93^{\circ }$; catching up takes about $2.21$ more days. These are mean-period estimates: orbital eccentricity and changing geometry make individual synodic months vary. ^[ref]. ^[ref].

The lunar phase is also a practical clock for observing. Full Moon is opposite the Sun and rises roughly at sunset; first quarter is about a right angle from the Sun and is convenient in the evening, while last quarter is convenient before dawn. These are rough guides, not exact rise-time predictions for every place and season. ^[ref]. Historically, repeated lunar phases provided the month in lunar calendars, while half-lit disks gave Aristarchus of Samos a way to infer astronomical distances. ^[ref].

The orbital period is the time required to complete one orbit. It is related to mean motion by $T=2\pi /n$.

Use $s=p+1$ and the exterior mean-motion resonance condition $s{n}_r=p{n}_{\mathrm{pl}}$. Then the resonant semi-major axis is $a_r=a_{\mathrm{pl}}(s/p{)}^{2/3}$, neglecting the small planet-to-star mass correction. For the geometric construction, take $a_{\mathrm{pl}}=n_{\mathrm{pl}}=1$, $p=4$, $e=0.2$, $\varpi =\pi /4$ and start the particle at pericentre. Solve Kepler's equation $M=E-e\sin E$ with $M=(4/5)t$. Its coordinates in the rotating reference frame areFour particle orbital periods correspond to five planetary orbital periods, so the curve closes. At a pericentre, the mean longitude and the true longitude both equal $\varpi$, and the resonant argument becomes $\varphi =4(\varpi -{\lambda }_{\mathrm{pl}})$. Thus the four pericentre directions are $\varphi /4$ modulo $\pi /2$. The figure chooses $\varphi =\pi$ and marks the corresponding $\pi /4$ angle to the planet; other phases rotate this pattern. Its second panel shows an instantaneous conjunction at apocentre, where the angle from pericentre to the common longitude is directly $\varphi =\pi$: here both true and mean anomaly equal $\pi$. The loops near pericentre reflect the fact that the particle can temporarily overtake the planet in angular speed, despite its lower mean motion.

For the evolution equations, $\lambda$ must be the mean longitude. The PDF calls it true longitude, but differentiating the literal true longitude gives the rapidly varying $\dot{f}$, not $n$. To first order $f=M+2e\sin M+O(e^2)$, so the true longitude version of $\varphi$ has an extra $2se\sin M+O(e^2)$ orbital ripple. The stated Lagrange planetary equations and slow complex eccentricity equation are interpreted consistently as orbit-averaged equations in mean longitude. The pericentre geometry of an exterior resonance above is unchanged by this distinction.

Let $F_r,F_{\theta }$ be perturbing forces, and let $m$ be the particle's mass. The angular momentum about the star satisfiesThe central gravitational force and the radial perturbation have zero torque. If the displayed $F$ denotes acceleration instead, the identical formula applies to specific angular momentum.

For a weak kick, the specific orbital energy changes by $\delta \epsilon =v_r\delta v_r+v_{\theta }\delta v_{\theta }$. A planet leading the particle pulls it forwards, while a trailing planet pulls it backwards. For a nearly circular exterior Kepler orbit, a forward tangential kick raises angular momentum and semi-major axis, lowers mean motion, and makes the next conjunction occur sooner in the synodic cycle. Its longitude retreats relative to the unperturbed conjunction sequence. A backward kick produces the opposite shift. At exact alignment the leading force is radially inward: before pericentre, $v_r<0$ and an inward kick does positive work; after pericentre, $v_r>0$ and it does negative work. These likewise raise or lower semi-major axis. At an apsis, a purely radial infinitesimal kick has no first-order energy change but changes the eccentricity vector, so angular momentum alone does not determine every timing change.

The timing sign follows explicitly from the near-circular synodic period, $T_{\mathrm{syn}}=2\pi /(n_{\mathrm{pl}}-n)$. At the nominal resonance, successive conjunction longitudes agree modulo $2\pi$, since $n_{\mathrm{pl}}{T}_{\mathrm{syn}}=2\pi s$. A persistent small change givesThis is the accumulated mean timing shift; an actual impulsive encounter can also change the instantaneous orbital phase and longitude of periapsis.

Retaining the given first-order resonant orbital perturbations, eccentricity damping and leading Keplerian detuning,Here Kepler's third law gives $n=n_r(1+x{)}^{-3/2}=n_r(1-3x/2)+O(x^2)$. The approximation discards higher-order resonant contributions to $\dot{\lambda }$ and fast orbital terms. Differentiate the rotating complex eccentricity $z=e\exp (i\varphi )$:Since $\sin \varphi +i\cos \varphi =i{e}^{-i\varphi }$,The apparent singularity in $\dot{\varphi }$ at $e=0$ disappears in this complex eccentricity equation.

Take $A>0$, and the conventional positive resonant coefficient $C>0$. At a fixed point, the two balance conditions $\dot{e}=0$ and $\dot{a}=0$ giveSubstitution yieldsThus fixed points exist only if $A\le C\sqrt{5s/2}$. With strict inequality, put $S=\sqrt{(5/2)s{C}^2-A^2}$ and ${\psi }_f=\arctan (A/S)\in (0,\pi /2)$. Thenrespectively. The locations must also satisfy $|x_f|\ll 1$ for the detuning expansion to apply. At equality the phases merge at $3\pi /2$ and $x_f=0$. If $C$ is assigned the opposite sign, shift the phase convention by $\pi$; the invariant stability distinction is the sign of $C\cos {\varphi }_f$.

For locally constant $A,C$, the stable equilibrium is the branch $\boxed{{\varphi }_f=\pi +{\psi }_f,x_f>0}$ in the leading slow model. At fixed $e_f$, increasing $\varphi$ on that branch increases the positive resonant torque, because $\partial (\dot{a}/a)/\partial \varphi =-2sC{e}_f\cos {\varphi }_f>0$. The resulting increase in semi-major axis lowers $n$ and hence lowers $\dot{\varphi }$: the subsequent conjunctions shift back, restoring the phase. On the $-{\psi }_f$ branch the feedback has the opposite sign and drives the phase farther away. Eccentricity damping removes the oscillatory disturbance on the restoring branch.

One can verify that this feedback stabilizes the coupled variables rather than just the phase. Treat $A,C$ as locally constant, and write $z=u+iv$. Then $\dot{u}=-Au+Bxv$, $\dot{v}=-C-Av-Bxu$ and $\dot{x}\simeq -2sCv-4A/5$. At a fixed point $Bxv=Au$. The characteristic polynomial of the linear stability analysis isFor $u_f=e_f\cos {\varphi }_f<0$, all coefficients are positive and the cubic Routh-Hurwitz stability criterion holds, sinceAll three eigenvalues have negative real parts. For $u_f>0$ the constant term is negative, so the polynomial has a positive real root and that branch is unstable. The merged case has a zero eigenvalue. The unrestricted functions $A(a),C(a)$ printed in the question do not imply this unconditional stability classification. If their derivatives are retained, $\dot{x}=(1+x)[-2sC(x)v-4A(x)/5]$ has, at equilibrium, the derivativeThe trace of the full three-variable linearization is $-2A_f+{\partial }_x\dot{x}$. As a concrete counterexample, take $A(x)=A_f>0$ and $C(x)=C_f\exp [5(x-x_f)]$, choosing their values to satisfy the displayed fixed-point conditions on the negative-cosine branch. Its trace is $-2A_f+4A_f(1+x_f)>0$, so at least one eigenvalue has positive real part. Both coefficients are smooth. Thus the restoring conjunction feedback identifies the stable branch of the usual locally constant model, but additional assumptions on coefficient variation are necessary to assert stability for every allowed function.

Let $\mu =M_p/M_{\star }$, $q=\sqrt{1-\beta }$, and restrict first to $0\le \beta <1$. The radiation-pressure coefficient reduces the grain's central gravitational parameter to $G{M}_{\star }{q}^2$, whereas the planet's is approximately $G{M}_{\star }$. Their mean motions obey $n/n_p=q(1+ϵ{)}^{-3/2}$. Setting this ratio to $j/(j+1)$ gives the radiation-pressure shift of a mean-motion resonance:It lies interior to the planet exactly when $q(j+1)<j$, or $j>q/(1-q)$. Equality puts the frequency commensurability at the planet's radius. For $\beta =0$ every such commensurability is exterior.

There is a frame qualification in the printed encounter formula. The circular inertial speed of the grain at $a_p(1+{ϵ}_1)$ is $v_1=q{v}_p(1+{ϵ}_1{)}^{-1/2}$. Subtracting the planet's inertial velocity at conjunction gives the local planet-relative inertial speedThis is the velocity used in the question's two-body gravitational assist calculation. In a frame actually rotating about the star at $n_p$, the grain's tangential velocity is instead $v_1-n_p{a}_p(1+{ϵ}_1)$, with magnitude $v_p[1+{ϵ}_1-q(1-{ϵ}_1/2)]$. In particular its nonradiating shear is $3v_p{ϵ}_1/2$, rather than $v_p{ϵ}_1/2$. Thus the requested expression uses a local translating, inertial encounter convention despite the source's rotating-frame wording. The following derivation consistently uses that convention, and its coefficients are an encounter estimate rather than an exact rotating-frame result.

Take impact parameter $b\simeq a_p{ϵ}_1$ and a small gravitational scattering angle $\theta$. Expanding the supplied hyperbolic scattering relation gives $\theta \simeq 2G{M}_p/(b{v}_{\infty }^2)$. Initially the relative tangential velocity is $-v_{\infty }$; after the deflection it has tangential component $-v_{\infty }\cos \theta$. The relative speed is unchanged in the local hyperbolic Kepler orbit, but the stellar-frame kinetic energy changes. Taking the dot product with the planet's velocity givesThe specific orbital energy of the grain is $-G{M}_{\star }(1-\beta )/(2a)$. Comparing it immediately before and after the encounter at the same position givesConsequentlyThe encounter increases the grain's semi-major axis within this geometry. These steps assume ${ϵ}_1\ll 1$, $\theta \ll 1$, and an orbital kick small enough for the subsequent linearization.

Choose the initial conjunction longitude as zero. The synodic period is $2\pi /(n_p-n)$, and the planet's longitude accumulated by the next conjunction isThese are unwrapped longitudes, counting complete revolutions. The second uses the new mean orbital period while neglecting the encounter's immediate phase offset and the difference between true and mean conjunction. A comparison reduced modulo $2\pi$ could not implement the stated one-turn sensitivity test.

For $\beta =0$, $v_{\infty }\simeq v_p{ϵ}_1/2$, so $\delta ϵ\simeq 32{\mu }^2{ϵ}_1^{-5}$ and ${\lambda }_c\simeq 4\pi /(3ϵ)$. Conjunction-longitude sensitivity to an encounter givesRequiring this to exceed $2\pi$ gives ${ϵ}_1^7<64{\mu }^2/3$. At a high-index first-order mean-motion resonance, ${ϵ}_j=(1+1/j{)}^{2/3}-1\simeq 2/(3j)$, henceThe interpretation of the resonance-overlap encounter estimate is loss of coherent encounter phases: a small kick changes the next synodic encounter time by more than one planetary revolution. Resonant protection by repeatedly meeting at controlled orbital phases can then fail, allowing irregular kicks and diffusion. This sensitivity criterion motivates an unstable region; it is not a proof that every orbit or resonant phase in it must escape.

For a quantitative sketch with radiation, put $h=1-q(1-ϵ/2)$ and $d=1-q(1+ϵ{)}^{-3/2}$. Differentiating the conjunction formula and substituting the kick givesEvaluate this at the positive ${ϵ}_j$ and take $\mathcal{S}=1$ as the critical-index curve. At small $\beta$, increased encounter speed weakens planetary scattering, moving the threshold to larger $j$. As $\beta$ rises further, the radiation-pressure shift of a mean-motion resonance brings each commensurability closer to the planet, strengthening the close encounters. The curve turns over and approaches the exterior-limit boundary $j=q/(1-q)$. Larger planetary mass produces stronger kicks and a lower critical index. These competing effects are shown below; the curves retain the factors in $\mathcal{S}$, so their $\beta =0$ intercepts differ slightly from the leading high-$j$ formula. The boundary for disappearing exterior commensurabilities is geometric; divergence of the encounter approximation near $ϵ=0$ should not be read as a precise physical scattering rate.

Write $R_p$ for the physical planet radius and $R_c=\eta R_{H1}$ for the debris-cloud radius; the source uses $R_1$ for both even though their stated size orderings are opposite. A central transit duration of a small planet on a nearly circular Kepler orbit is $D_p\simeq 2R_{\star }/(n_1{a}_1)=P_1{R}_{\star }/(\pi a_1)$. Kepler's third law, with $M_{\star }=4\pi {\rho }_{\star }{R}_{\star }^3/3$, givesA noncentral planetary exoplanet transit is shorter, so within the small-radius and nearly circular approximations, a duration requiringis too long for the planet alone and favors extended debris. For a central cloud transit with $R_c\gg R_{\star }$, the path length is approximately $2R_c$. Using the Hill radius $R_{H1}=a_1(M_1/(3M_{\star }){)}^{1/3}$ gives $D_1/P_1\simeq R_c/(\pi a_1)$, hence the Hill-cloud transit mass estimateThis equality assumes a central cloud chord. An orbit described merely as nearly edge-on need not have central cloud crossings. If its projected cloud impact distance is $s$, the measured half-chord is $\sqrt{R_c^2-s^2}$, so the boxed mass is multiplied by $[1-(s/R_c{)}^2{]}^{-3/2}$. It is a lower estimate when that geometry is unknown.

For the dynamical calculation use ${\mu }_2=M_2/M_{\star }$, $A=G{M}_2/a_2$, and $\mathcal{R}=Af(\alpha )e_1\cos {\varphi }_1$. The derivatives in the supplied Lagrange planetary equations hold the other osculating orbital elements fixed:Since $A/(n_1{a}_1^2)=n_1{\mu }_2\alpha$, the physical mean-longitude equation for a first-order resonant term isThe source's epoch notation needs care when $n_1$ varies. Literally differentiating ${\lambda }_1=n_1(t)t+{ϵ}_1(t)$ adds $t{\dot{n}}_1$; one cannot retain that definition and also identify the printed ${\dot{ϵ}}_1$ with ${\dot{\lambda }}_1-n_1$. A consistent version writes ${\lambda }_1={\int }_0^t{n}_1(s)ds+{ϵ}_{\mathrm{int}}$ and assigns the displayed correction to ${\dot{ϵ}}_{\mathrm{int}}$. Equivalently, the canonical mean-longitude equation for a first-order resonant term follows from the disturbing function directly. The result for $g$ is the intended physical equation under that convention.

Away from resonant-argument libration, neglect changes in $e_1,\alpha ,{\varpi }_1$ on one slow cycle to this order. With $n_0=2\pi /P_1$ the reference mean motion, the signed slow frequency isThis is the near-resonant transit-timing superperiod. Choose ${\varphi }_1(0)=0$. Differentiating $n_1\propto a_1^{-3/2}$ with the semi-major axis equation givesLet $C=3j{\mu }_2{n}_0^2\alpha f{e}_1$ and $n_s=n_1(0)$. Direct integration giveswith zero initial mean longitude. The constant and linear terms are absorbed into the fitted transit ephemeris: the measured mean frequency is $\bar{n}=n_s-C/\omega$. This distinction avoids mistaking an arbitrary initial frequency for the long-term fitted period.

For $|\Delta |\ll 1$, the double-integrated frequency term dominates the direct $g$ term provided $g/f$ stays finite. Relative to the fitted linear ephemeris, ${\lambda }_1(t_k)=2\pi k$ givesHere $P_1$ denotes the fitted mean period at the retained order. The signed-frequency convention fixes the phase sign. Replacing $\omega$ by $2\pi /P_t>0$ changes the sign of the signed $A_t$ when needed; a nonnegative amplitude hasUsing a signed $P_t=2\pi /\omega$ instead gives the source's signed proportionality $A_t/P_t\simeq 3{\mu }_2\alpha f{e}_1/(2\pi \Delta )$. The leading implicit transit equation can equally use $P_{1}k$ inside the sine, since the difference is higher order in the perturbation. This is the eccentricity-enhanced near-resonant transit-timing variation; its derivation assumes small longitude oscillations, ${\mu }_2{e}_1\ll {\Delta }^2$ at fixed resonance coefficients, and the effectively fixed eccentricity used in this truncated model. Dominance over forced-eccentricity timing terms typically also requires $e_1\gg |\Delta |$ at fixed $j$. A full transit calculation also includes forced orbital eccentricity and the conversion from mean longitude to the actual sky-crossing angle. Those are lower order in this eccentricity-enhanced regime, rather than identically absent; broader near-resonant formulas are derived by Lithwick, Xie and Wu.

Inside a mean-motion resonance, ${\varphi }_1$ undergoes resonant-argument libration and its evolution must be solved together with $a_1,e_1,{\varpi }_1$. The constant circulation frequency $\omega$ and the associated ${\Delta }^{-2}$ expansion then fail. At approximately fixed eccentricity, the resonant resonant pendulum approximation has libration frequency of order $j{n}_0\sqrt{{\mu }_2\alpha |f|e_1}$, up to numerical factors. The transit-timing variation follows that libration period and depends on its libration amplitude of a resonant argument. Exactly at a resonant equilibrium there need be no libration signal; near a separatrix the period grows, and eccentricity dynamics can modify the simple estimate. Resonant amplitudes remain controlled by the bounded resonant motion rather than diverging as the circulating detuning tends to zero.

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.

In the weak, non-crossing, leading-eccentricity model the exterior resonant coefficient has positive sign: write its disturbing function as $R_{\mathrm{res}}=B_q{e}_c^q\cos {\varphi }_{bc}$, with $B_q>0$ for the orders considered. The physical feedback can be seen for first order. Before an outer conjunction near apoapsis, the faster inner planet pulls the outer one backwards; after passing it pulls forwards. On the outward side, the nearer, stronger earlier pull wins, reducing the outer semi-major axis and increasing its mean motion. Thus ${\varphi }_{bc}<\pi$ is driven upward. On the inward side, the later forward pull wins and drives ${\varphi }_{bc}>\pi$ downward. The paired impulses cancel at the symmetric centre.

For higher order, sum the feedback over all $q$ conjunction phases. The first surviving angular Fourier harmonic is $\sin {\varphi }_{bc}$; the lower Fourier harmonics cancel over the equally spaced directions. A local Hamiltonian reduction with canonical momentum $J$ hasThe negative curvature of the Kepler orbit Hamiltonian gives the displayed sign. Linear stability analysis at $\pi$ gives a restoring acceleration. The symmetric libration centres of an eccentricity resonance are thereforeThe corresponding conjunction phases relative to periapsis are $\{\pi \}$, $\{\pi /2,3\pi /2\}$ and $\{\pi /3,\pi ,5\pi /3\}$. They suppress repeatedly close passages and provide resonance protection.

These are expected centres of the specified leading-harmonic model, not centres fixed solely by the integer order for arbitrary orbital eccentricity and masses. Asymmetric resonant-argument libration can occur when higher Fourier harmonics matter. For example, if $R=A\cos \varphi +B\cos 2\varphi$, $A>0$, $B>A/4$, the symmetric $\pi$ equilibrium loses stability and stable centres satisfy $\cos \varphi =-A/(4B)$. Thus further dynamical information is needed beyond the PDF's unrestricted eccentric orbit to assert unique centres. Exterior asymmetric branches are documented by Winter and Murray, “Resonance and chaos. II. Exterior resonances and asymmetric libration”.