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The argument of periapsis is the angle in the orbital plane from the ascending node to periapsis, measured in the direction of motion. It differs from the longitude of periapsis $\varpi =\Omega +\omega$. For an inclined orbit this sum is a conventional longitude, not generally the azimuth of the projected periapsis. The argument is undefined for a circular orbit and the node-based description is singular when the orbit and reference plane coincide.

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 $x$.

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].

For two nearly edge-on orbits with small orbital inclination difference $\Delta I$ and small longitude of ascending node difference $\Delta \Omega$, expanding the dot product of their normals gives $I_m^2=(\Delta I{)}^2+(\Delta \Omega {)}^2+O({\delta }^4)$, with angles in radians and all offsets from edge-on of order $\delta$. Thus equality of observed sky-plane inclinations alone does not imply aligned orbital planes.

In a right-handed reference frame, the rotation matrix from an orbital plane frame whose first axis points along the ascending node is $R_z(\Omega )R_x(I)$. Its columns are the orbital frame's orthonormal basis expressed in reference coordinates. The unit vector normal is $(\sin \Omega \sin I,-\cos \Omega \sin I,\cos I)$; its dot product with another such normal gives the cosine of their mutual inclination.

For a bound Kepler orbit, the semimajor axis is half the long diameter of the ellipse; if $r_p,r_{mathrma}$ are its apsidal distances, $a=(r_p+r_{mathrma})/2$. The orbital eccentricity is $e=(r_{mathrma}-r_p)/(r_{mathrma}+r_p)$, with $0\le e<1$ and semiminor axis $b=a\sqrt{1-e^2}$. The gravitating centre is a focus, displaced by $ae$ from the ellipse's geometric centre.

Choose a reference plane, a positive normal and a fixed reference direction in that plane. The ascending node is where the orbit crosses toward positive normal displacement. The longitude of ascending node $\Omega$ is the reference-plane angle from the fixed direction to that node. The argument of periapsis $\omega$ is the angle from the ascending node to periapsis, measured in the orbital plane in the direction of motion. The orbital inclination $i$ is the angle between the reference normal and the orbital angular-momentum normal, with $0\le i\le \pi$. The diagram separates the ellipse's shape from these orientation angles.

Let $\mu =GM$ and use actions per unit test-particle mass. In the prograde spherical-action chart underlying the stated Hamiltonian, $J_{\varphi }=L_z$ and $J_{\theta }=L-L_z$, so $J_{\theta }+J_{\varphi }=L$. The radial action then satisfies $J_r+L=\sqrt{\mu a}$, consistently with $H_K=-\mu /(2a)$. DefineIf $(w_r,w_{\theta },w_{\varphi })$ are the old angles, a type-two generating function isDifferentiation with respect to the old angles recovers $J_r,J_{\theta },J_{\varphi }$, and differentiation with respect to the new actions gives the canonical transformationIts triangular integer action matrix has determinant one, so it preserves the angle periodicities. In the new coordinates,These are the Delaunay variables:With the usual choices of angle origins, $w_1=\ell$ is mean anomaly, $w_2=\omega$ is argument of periapsis, and $w_3=\Omega$ is longitude of ascending node. The generator $\mathcal{G}=L$ rotates the orbit within its plane, whereas $\mathcal{H}=L_z$ rotates that plane about the reference normal; this explains the angular identification. The fast angle obeys Kepler's equation $\ell =\mathcal{E}-e\sin \mathcal{E}$, where $\mathcal{E}$ is eccentric anomaly.

For retrograde orbits, signed $J_{\varphi }=L_z$ gives $J_{\theta }=L-|L_z|$ and the spherical Hamiltonian uses $J_r+J_{\theta }+|J_{\varphi }|$. The stated plus-sign form is therefore a prograde chart, not a global signed-action formula. The Delaunay variables still use $\mathcal{H}=L_z$; on the retrograde branch the generating function is $(\mathcal{L}-\mathcal{G})w_r+(\mathcal{G}+\mathcal{H})w_{\theta }+\mathcal{H}{w}_{\varphi }$. At $e=0$, periapsis is undefined, and at $i=0$ or $\pi$ the node is undefined: these are angle-chart degeneracies, not physical singularities.

For the perturbed planet orbit, the radial and vertical frequencies mean small oscillations about a circular equatorial orbit of cylindrical radius $R$. Write $r=\sqrt{R^2+z^2}$ and retain the stated quadrupole potential. On the equator,At fixed axial specific angular momentum $L_z$, the radial effective potential is $\varphi (R,0)+L_z^2/(2R^2)$. Its second derivative at circular balance $L_z=R^2{\Omega }_c$ gives the radial epicyclic frequencyReflection symmetry decouples vertical motion to first order. Expanding the potential in $z$ givesso the vertical epicyclic frequency isThe circular orbit is radially stable only for $R^2>3{\alpha }^2$. For $\alpha \ne 0$ in that domain, $\nu >{\Omega }_c>\kappa$. Vertical node crossings repeat at frequency $\nu$ while azimuth advances at ${\Omega }_c$, so the line of nodes has nodal precession rate ${\Omega }_c-\nu <0$. Radial periapsis passages repeat at $\kappa$, giving apsidal precession rate ${\Omega }_c-\kappa >0$. To leading quadrupole order, with $n=\sqrt{\mu /R^3}$,This is oblate-quadrupole epicyclic precession. Here $\varpi$ denotes the nearly equatorial longitude of periapsis, not $\omega$ alone.

The precession signs above have a small-eccentricity, small-inclination domain. They are not a universal claim for every inclined Kepler ellipse. To see the distinction with the same Delaunay variables, average the perturbation over an unperturbed orbit. Since $z/r=\sin i\sin (\omega +f)$ and $dt=r^{2}df/\mathcal{G}$, the term linear in $e\cos f$ integrates to zero and the ${\sin }^2(\omega +f)$ average in the $r^{-3}$ integral is $1/2$. ThusDifferentiating at fixed other actions yields the first-order secular ratesTheir nearly equatorial sum reproduces the prograde $\dot{\varpi }$. For a polar orbit, however, the node is fixed and $\dot{\omega }<0$. This supplies a concrete counterexample to extending the requested forward periapsis precession beyond the intended epicyclic regime.

For a circular orbit the origin of mean anomaly is arbitrary because there is no distinguished pericentre. Choose $M=0$ at the ascending node, so $M$ is also the argument of latitude. If a different periapsis convention is used, replace $M$ below by $M+\omega$. The orbital-frame rotation from inclination and node givesIn particular the crossing at $M=0$ has increasing $z$ for prograde $n>0$ and $0<I<\pi$, confirming the ascending node convention.

Put $ϵ=a/a_{\mathrm{pl}}$ and $c=\cos \psi =\hat{\mathbf{r}}\cdot {\hat{\mathbf{r}}}_{\star }$. The binomial series expansion of the direct disturbing function isThe indirect disturbing function subtracts $G{M}_{\star }ac/a_{\mathrm{pl}}^2$, cancelling the dipole term. The constant $G{M}_{\star }/a_{\mathrm{pl}}$ has no spatial derivative and does not perturb the particle. After omitting that constant, the leading stellar quadrupole disturbing function isThus the PDF's equality is understood modulo a dynamically irrelevant additive constant and higher multipoles.

Define $\ell =M+\Omega$ and let ${\ell }_{\star }$ be the star's orbital longitude as seen from the planet. Its orbital plane has $I_{\mathrm{pl}},{\Omega }_{\mathrm{pl}}$. The vector from planet to star is the negative of the planet's heliocentric vector: if ${\ell }_{\mathrm{pl}}$ describes that latter vector, use ${\ell }_{\star }={\ell }_{\mathrm{pl}}+\pi$. This sign matters in $\cos \psi$ but drops out of ${\cos }^2\psi$. Expanding the two rotated unit vectors givesFor example, the particle's horizontal correction is $-(I^2/2)\sin (\ell -\Omega )(-\sin \Omega ,\cos \Omega )$, while its vertical component is $I\sin (\ell -\Omega )$; their dot products with the corresponding stellar components produce these terms. No cubic terms occur because simultaneous reversal of both small orbital inclinations leaves the dot product unchanged.

The two mean longitudes are uniform for the circular Kepler orbits, so orbit averaging is uniform in both $\ell$ and ${\ell }_{\star }$. Set $c_0=\cos (\ell -{\ell }_{\star })$ and $c_2=\cos \psi -c_0$ through quadratic order. The needed averages follow by integrating sine and cosine products over $[0,2\pi {]}^2$:For the last identity, expand $c_0=\cos \ell \cos {\ell }_{\star }+\sin \ell \sin {\ell }_{\star }$ and factor the two integrations, each using $⟨{\sin }^2\ell ⟩=⟨{\cos }^2\ell ⟩=1/2$. Squaring $c_0+c_2$ now yieldswhere the error includes all fourth-order combinations of the two orbital inclinations. ThereforeMultiplying this by $G{M}_{\star }{a}^2/a_{\mathrm{pl}}^3$ gives the requested secular disturbing function. Equivalently, the exact angular factor for two circular rings is $(3{\cos }^2{I}_m-1)/8$, where $I_m$ is their mutual inclination; its small-angle expansion gives the same result. This is the double-averaged circular tidal potential.

For aligned equatorial and planetary orbital planes, set $I_{\mathrm{pl}}=0$. The secular stellar quadrupole disturbing function then has derivative ${\partial }_I{\overline{\mathcal{R}}}_{\star }=-3G{M}_{\star }{a}^{2}I/(4a_{\mathrm{pl}}^3)$. The supplied Lagrange planetary equations for nodal precession givesThe planetary quadrupole coefficient $J_2>0$ contributes ${\partial }_I{\overline{\mathcal{R}}}_{J_2}=-3G{M}_{\mathrm{pl}}{J}_2{R}_{\mathrm{pl}}^{2}I/(2a^3)$, henceBoth are retrograde nodal precession rates. Equating them cancels $n$ and gives the Laplace radiusThe planetary quadrupole dominates inside $a_L$ and the stellar tide outside it. These are small-inclination rates for planet-bound Kepler orbits in the hierarchical regime $a\ll a_{\mathrm{pl}}$. At exactly $I=0$, the node itself is undefined; the rates are the limiting precession frequencies of an infinitesimal tilt.

Take the sky frame and disk frame to be right-handed orthonormal bases, with positive angular displacement measured from $\hat{X}$ toward $\hat{Y}$. In particular, positive $Z$ is toward the observer, so material crossing the sky plane with positive $\dot{Z}$ is at the ascending node. Write $c_d=\cos {\Omega }_d$, $s_d=\sin {\Omega }_d$, $C_d=\cos I_d$, and $S_d=\sin I_d$. The disk frame's first unit vector lies along its ascending node, and its second has positive $Z$ component. The orbital-frame rotation from inclination and node therefore givesEach column is a disk-frame unit vector in sky coordinates, so $\mathbf{X}=T\mathbf{x}$ and the inverse is $\mathbf{x}=T^{\mathsf{T}}\mathbf{X}$. At the disk's ascending node, the $Z$ component of its tangential velocity is positive, confirming the sign of the tilt.

Let $T_p=R_Z({\Omega }_p)R_X(I_p)$ define a provisional planet frame whose first axis is the planet's sky-plane ascending node. Its normal is ${\hat{z}}_p=(\sin {\Omega }_p\sin I_p,-\cos {\Omega }_p\sin I_p,\cos I_p)$. Expressing that normal in the disk orthonormal basis, with $\Delta \Omega ={\Omega }_p-{\Omega }_d$, givesThe last component is the dot product of the two orbital normals, hence the cosine of the mutual inclination. The mutual ascending node points along ${\hat{z}}_d\times {\hat{z}}_p$: motion there has positive disk-normal component. SetThen $\hat{x}\prime =(A\hat{x}+B\hat{y})/\sqrt{A^2+B^2}$, and the quadrant-correct longitude of ascending node is ${\Omega }_m=atan2(B,A)$. ThusThe tangent alone cannot fix the quadrant, and ${\Omega }_m$ is undefined for exactly coincident orbital planes.

For the two rotation routes, define $\psi$ as the oriented angle, within the planet's orbital plane, from its sky-plane ascending node to its disk-plane ascending node. If ${\hat{x}}_p,{\hat{y}}_p$ are the first two columns of $T_p$, take $\psi =atan2(\hat{x}\prime \cdot {\hat{y}}_p,\hat{x}\prime \cdot {\hat{x}}_p)$. Both routes describe the same planet orthonormal basis:The left route goes through the disk frame; the right goes through the sky-node planet frame. Comparing their third columns gives the two boxed relations, while comparison of first columns fixes the extra in-plane angle.

Put $I_d=\pi /2+{\delta }_d$ and $I_p=\pi /2+{\delta }_p$. A second-order Taylor expansion gives $\cos I_m=1-[({\delta }_p-{\delta }_d{)}^2+\Delta {\Omega }^2]/2+O({\delta }^4)$, and therefore the mutual inclination of nearly edge-on orbits satisfiesAll expansion angles are in radians. For plotting in degrees this is $I_m\simeq \sqrt{(I_p-87^{\circ }{)}^2+(4^{\circ }{)}^2}$. The given planet has exact $I_m\simeq 4.119^{\circ }$ and ${\Omega }_m\simeq 76.044^{\circ }$. The exact minimum occurs at $I_p=atan2(\sin I_d\cos \Delta \Omega ,\cos I_d)\simeq 86.993^{\circ }$, with $I_{m,\min }\simeq 3.995^{\circ }$. At $I_p=85^{\circ }$ and $95^{\circ }$, the second-order values are ${\sqrt{20}}^{\circ }\simeq 4.472^{\circ }$ and ${\sqrt{80}}^{\circ }\simeq 8.944^{\circ }$.

For the disk evolution, neglect disk self-gravity, collisions that align the planes, and back-reaction on the planet. In the initial disk plane, use the complex inclination $w=I{e}^{i\Omega }$ for a ring and $w_p=I_m{e}^{i{\Omega }_m}$ for the planet. Linear Laplace-Lagrange secular theory has forced inclination $w_p$ and solution $w(a,t)=w_p[1-e^{-i\nu (a)t}]$ for initially flat rings. For an inner circular perturber, the Laplace coefficient formula is $\nu =n(M_p/M_{\star })\alpha b_{3/2}^{(1)}(\alpha )/4$, with $\alpha =a_p/a$. Its quadrupole approximation uses $b_{3/2}^{(1)}\simeq 3\alpha$, yieldingIn the $(I\cos \Omega ,I\sin \Omega )$ plot referenced to the initial disk, every ring runs clockwise on the same circle centered at $w_p$, starting at the origin. Inner rings advance more quickly. If the axes are instead referenced to the planet's plane, the relative tilt is $w-w_p=-w_p{e}^{-i\nu t}$: these circles are centered at the origin, and the mutual inclination is constant. Both conventions describe the same nodal precession; they should not be mixed.

At an intermediate epoch, rings with $\nu t\gtrsim 1$ have tilted substantially while outer rings with $\nu t\ll 1$ remain near the original disk plane. This radial variation is a planet-induced debris-disk warp. The characteristic affected radius grows like $t^{2/7}$. Conservative differential nodal precession preserves each ring's tilt relative to the planet, so it does not by itself align every orbit; an unresolved inner region can acquire a mean plane near the planetary plane through phase mixing. The secular interpretation is also described in Wyatt's planetary dynamics lectures.

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.

Let ${\mu }_p=G{M}_p$, $A=\beta G{M}_{\star }/c$ and $\kappa =A/a_p^2$. A tightly bound circumplanetary orbit lies well inside the Hill sphere, so $a/a_p\ll 1$ and its orbital period is short compared with the planet's year. Treat the stellar flux and stellar direction $\hat{\mathbf{R}}$ as constant during one dust orbit. With stellar-frame velocity ${\mathbf{V}}_p+\mathbf{u}$, the velocity-dependent acceleration isThe terms independent of $\mathbf{u}$, including the leading static radiation pressure, do no net work on an unperturbed closed circular orbit. HenceFor a coplanar circular orbit, $u^2={\mu }_p/a$ and the component along the stellar direction has mean square $u^2/2$. Thus $\dot{a}=2a^2{\dot{\epsilon }}_p/{\mu }_p$ yields the Poynting–Robertson decay of a circumplanetary orbitThe coefficient three assumes coplanarity, which the PDF does not state. For orbital normal $\hat{\mathbf{k}}$, the general circular covariance is $⟨u_i{u}_j⟩=(u^2/2)({\delta }_{ij}-k_i{k}_j)$, givingAn orbital plane normal to the stellar radial direction has coefficient two during that orbit, a counterexample to an orientation-independent coefficient three. Averaging also over the planet's circular stellar orbit, with fixed dust-plane orbital inclination $I$ to it, gives coefficient $3-\frac{1}{2}{\sin }^{2}I$. The conservative forces must be weak enough for the assumed approximately circular, planet-bound orbit to persist; the small planet-to-star mass ratio alone does not ensure this for arbitrary grain $\beta$.

A simultaneous transit in a resonant chain requires both conjunction patterns to admit the observer's longitude at the same time. In the leading mean-conjunction geometry put $x=\Lambda -{\varpi }_c$. The necessary phase compatibility at symmetric centres isLet $d=\gcd (q,n)$. Eliminating $x$, and using that $q\ell -nk$ ranges over multiples of $d$, gives the exact modular compatibility criterionHere $\gcd$ denotes the greatest common divisor. For the orders $q,n\in \{1,2,3\}$ above, the two phase sets intersect only whenThe common directions are the two quadratures relative to periapsis. In reduced resonances this requires $p$ and $m$ odd; otherwise the apparent second-order ratios reduce to different resonance orders. There is no additional restriction on those odd values from angular compatibility alone. A shared direction and suitable temporal phase can be chosen, and the rational period ratios allow its recurrence.

This is a potential low-amplitude configuration, not a guarantee of an observed triple exoplanet transit. The observer must lie in the common orbital plane and near an allowed direction. Finite stellar radii, nonzero libration amplitudes of a resonant argument, apsidal motion and asymmetric centres broaden or change the possibilities. At finite $e_c$, true simultaneous alignment obeys the more precise equations $(p+q)y-px={\varphi }_{bc}$ and $(m+n)x-my={\varphi }_{cd}$, with $y=M_c(x)$. Thus the order-only conclusion is not a universal necessary condition for every eccentric resonant chain.

Divide Newton's second law by the mass and resolve along the radial and transverse unit vectors in polar coordinates. A central force has no transverse component, soMultiplication of the second equation by $r$ shows that $d(r^2\dot{\theta })/dt=0$. Thuswhere $h$ is the signed specific angular momentum perpendicular to the orbital plane.

If $h\ne 0$ and $r>0$, the polar angle is locally a valid independent variable. For $u=1/r$, use primes for derivatives with respect to $\theta$. The chain rule givesSubstitution into the radial equation gives the Binet equationIf $h=0$, the orbit is radial and division by $h$ is unavailable; the first radial equation still applies with $\dot{\theta }=0$ away from the origin.

A geostationary orbit has the same angular velocity as the Earth. Let $M_E=4\pi \rho R^3/3$ and $r_g=\eta R$. Balancing the required centripetal acceleration with the exterior gravitational field givesThe orbital plane must be equatorial and the motion prograde for the satellite to remain over one point. The uniform density assumption supplies $M_E$ in terms of the specified $\rho$ and $R$.

Here $G$ is the gravitational constant, $r$ is the separation of the orbiting object from the central body's centre, and $J$ is the magnitude of the object's conserved orbital angular momentum, not angular momentum per unit mass. Approximate the much heavier central body as fixed, neglect other forces, and use the Newtonian gravitational potential energy $U=-GM{m}_0/r$, with zero at infinity.

A central force confines the motion to a fixed orbital plane. In plane polar coordinates its kinetic energy is $T=\frac{1}{2}{m}_0({\dot{r}}^2+r^2{\dot{\varphi }}^2)$. Conservation of angular momentum gives $J=m_0{r}^2\dot{\varphi }$; eliminating $\dot{\varphi }$ yieldsThe centrifugal term belongs to the radial effective potential; it is the eliminated transverse kinetic energy, not an additional physical interaction. If recoil of the central body mattered, the relative-coordinate reduction would use the reduced mass.

The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.

Differentiate the angular field equation to obtainConsequently $R_{00}=\Lambda A=\Lambda g_{00}$ and $R_{11}=-\Lambda /A=\Lambda g_{11}$. The other angular component follows from $R_{33}={\sin }^2\theta R_{22}$. Thus the single angular equation implies all the displayed vacuum equations wherever this coordinate chart is regular.

Integrating $(rA{)}^{\prime }=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq 1+2\Phi$ with the Newtonian gravitational potential $\Phi =-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the Schwarzschild-de Sitter spacetime hasNegative $\Lambda$ instead gives the Schwarzschild--anti-de Sitter black hole case. This matching fixes the normalization locally; a spacetime with nonzero $\Lambda$ is not asymptotically flat.

For timelike geodesics in the orbital plane, take proper time as parameter and use $L=(A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2)/2$. The cyclic coordinates give $A\dot{t}=k$ and $r^2\dot{\varphi }=h$; the angular canonical momentum is $-h$ in this signature. The four-velocity normalization $2L=1$ then givesHere $k$ is conserved Killing energy per unit rest mass in the chosen time normalization, and $h$ is conserved angular momentum per unit rest mass. There is no general interpretation of $k$ as energy measured at an asymptotically flat infinity when $\Lambda \ne 0$.

For radial motion $h=0$, differentiation gives $2\dot{r}\ddot{r}=-A^{\prime }\dot{r}$. The radial Euler-Lagrange equation extends the resulting expression through a turning point where division by $\dot{r}$ would fail. ThusThis is the proper-time second derivative of the areal radius, not a nonzero four-acceleration of the freely falling particle. Positive $\Lambda$ contributes an outward term; negative $\Lambda$ reinforces inward attraction. Restoring units in the slow-motion, weak-field Newtonian limit givesThe extra term describes the gravitational effect of a cosmological constant, not drag or a contact force. The approximation requires $GM/(r{c}^2)\ll 1$, $|\Lambda |r^2\ll 1$ and speed much less than $c$.

Put $f=1-2\mu /r$ and let dots now denote an affine parameter derivative. The null geodesic lagrangian in the orbital plane is $L=(f{c}^2{\dot{t}}^2-{\dot{r}}^2/f-r^2{\dot{\varphi }}^2)/2$. The time and azimuthal Euler-Lagrange equations integrate once; the null normalization givesFor completeness, the differential radial geodesic equation isthe remaining differential equations are $\ddot{t}+(f^{\prime }/f)\dot{t}\dot{r}=0$ and $\ddot{\varphi }+2\dot{r}\dot{\varphi }/r=0$. Spherical symmetry permits the equatorial choice.