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The probability that an isotropically oriented circular orbit produces an exoplanet transit, assuming $a>R_{\star }+R_p$. Uniformity of $|\cos i|$ gives the displayed result. Orbital eccentricity, a restriction to full rather than grazing crossings, or a non-isotropic orientation distribution changes the expression.

Spread mass $M$ uniformly along $-a<z<a$ on an axis, with linear mass density $\lambda =M/(2a)$. In its perpendicular bisector plane, axial field components cancel, and the radial component isThis is an exact line mass model and a thin-rod approximation for a physical cylinder. A centred circular orbit in this symmetry plane has $v^2=GM/\sqrt{R^2+a^2}$. Thus $v\to \sqrt{GM/a}=\sqrt{2G\lambda }$ for $R/a\to 0$, and $v\sim \sqrt{GM/R}$ for $R/a\to \infty$. For a finite-width rod, the first limit is useful only while its transverse size remains much smaller than $R$.

A circular infinite cylinder of radius $b$ and uniform mass density $\rho$ has linear mass density $\lambda =\pi b^2\rho$. A coaxial Gaussian surface of length $L$ has inward radial gravitational field and no end-cap flux. Gauss's law for gravity gives $2\pi rL{g}_r=-4\pi G\lambda L$ outside, hence the displayed result. Inside, enclosed linear mass density is $\pi r^2\rho$, so $g_r=-2\pi G\rho r$. An exterior circular orbit has speed $\sqrt{2G\lambda }$ independent of radius. The exterior formula presupposes circular axial symmetry; area alone would not specify the near field of a noncircular rod.

An orbit around an isolated source constrains its standard gravitational parameter $\mu =GM$ rather than independently measuring the source's mass and the gravitational constant. In the test particle approximation, a circular orbit of radius $a$ and orbital period $T$ satisfiesThe left equation equates the required centripetal acceleration to the gravitational acceleration. More generally, a Kepler orbit depends on the same $\mu$. In the exact relative two-body problem, replace $GM$ by $G(M+m)$.

For any positive dimensionless number $q$, replacing $G$ by $qG$ and $M$ by $M/q$ leaves $GM$ and every predicted test particle trajectory unchanged. This is the gravitational mass degeneracy: more observations of the same orbit can refine $GM$ without separating its factors. A laboratory source whose mass is calibrated independently supplies the missing information.

Consequently an astronomical mass expressed in kilograms as $M=\mu /G$ inherits measurement uncertainty from $G$, even if $\mu$ is known much more precisely. To first order,When the relative standard uncertainty of $\mu$ is negligible, that of $M$ is approximately $u(G)/G$. Conversely, the ratio of two source masses inferred using the same $G$ is $M_1/M_2={\mu }_1/{\mu }_2$, so this common factor cancels. That distinction explains how excellent orbit determination can coexist with less precise absolute astronomical masses.

For a binary star viewed far from its centre of mass, let $\mathbf{r}$ be the separation and $\mu$ the reduced mass, $M=M_1+M_2$ the total mass, $r=|\mathbf{r}|$ and $x=|\mathbf{x}|$. The Taylor expansion of the point-mass Newtonian gravitational potential has zero dipole because ${\sum }_A{M}_A{\mathbf{x}}_A=0$. Its second moment is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$, soThis holds for an instantaneous separation in any orbit, not only a circular orbit. The mass quadrupole moment convention $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ gives ${\Phi }_2=-3G{Q}_{ij}{x}_i{x}_j/(2x^5)$.

Facing a companion and facing the source of light are different relations. The Moon keeps its near side of the Moon approximately toward Earth, while the direction to the distant Sun is nearly fixed during the idealized orbit shown below. A fixed surface patch therefore turns toward, across and away from sunlight. The far side of the Moon has day and night too; its name does not mean permanent darkness. ^[ref].

If the Sun direction has fixed angle $\alpha$ and the marked surface patch has outward normal vector at angle $\theta$, its direct illumination is proportional to $\max (0,\cos (\theta -\alpha ))$. With synchronous rotation, $\theta =f+\pi$, so the patch goes through day and night even though it always faces Earth. In reality the Earth-Moon system advances around the Sun: the lunar spin and orbital cycle relative to distant stars is about $27.3$ Earth days, while the lunar phase cycle has a synodic period of about $29.5$ Earth days. These numbers describe the actual system, not the fixed-sunlight cartoon. ^[ref].

For a synchronously rotating exoplanet whose orbital companion is itself the illuminating star, the direction to the star turns with the orbit. On an ideal circular orbit, with perpendicular spin axis and no additional oscillations, this direction is fixed relative to the surface, giving persistent day and night hemispheres. Day-night heat redistribution then helps determine the nightside temperature; permanent lack of direct starlight does not imply zero temperature. Orbital eccentricity and departures from this ideal spin geometry modify that simple picture.

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.

For an isolated two-body problem in Newtonian gravity, the relative elliptic orbit's orbital period $T$ obeysHere $a$ is the relative semi-major axis, and $M+m$ is the sum of the two masses. The separation $r$ varies during an elliptic orbit, so replacing $a$ by $r$ works only for a circular orbit, where $r=a$. Likewise, either body's individual semi-major axis about the centre of mass must first be converted to the relative one. When $m$ is negligible compared with $M$, the denominator can be approximated by $GM$.

For fixed gravitational parameter, the orbital period depends on the semi-major axis and is independent of orbital eccentricity. Our $a=1$, $\mu =1$ example gives $T=2\pi$ both for $e=0.6$ and for a circular orbit with $e=0$. The eccentric body's rapidly changing speed compensates for its different path. A compact way to see the cancellation uses Kepler's second law: divide the ellipse's area $\pi ab$ by its areal velocity $h/2$. With $b=a\sqrt{1-e^2}$ and specific angular momentum $h=\sqrt{\mu a(1-e^2)}$, the result is $2\pi \sqrt{a^3/\mu }$.

Compare two separate Kepler orbits with the same gravitational parameter, $a_1=1$ and $a_2=4$:With $\mu =1$, their orbital periods are $2\pi$ and $16\pi$; measured in units of $T_1$, these are $1$ and $8$. Quadrupling the size therefore takes eight times as long to complete a circuit. This power law makes comparisons possible without first choosing physical units.

The same relation can be inverted to weigh a system: $M+m=4{\pi }^2{a}^3/(G{T}^2)$. An observed relative semi-major axis and orbital period determine the total mass, while additional information is needed to separate the two masses. This is useful for binary stars, as discussed in NASA, Orbits and Kepler's Laws: measuring masses. Geometry and a clock together become a weighing instrument.

As a heuristic approximation, inserting $r=6GM/c^2$ into the circular orbit form of Kepler's third law, with $M$ here denoting the total mass, givesThe dominant quadrupole gravitational-wave frequency is twice the orbital frequency, $f_{\mathrm{GW}}=2/T$.

For a spherical body of radius $R$, mass $M$ and mean density $\bar{\rho }=3M/(4\pi R^3)$, a low circular orbit has period $P\simeq 2\pi \sqrt{R^3/(GM)}$, by equating the gravitational acceleration from Newton's law of universal gravitation to the centripetal acceleration. Substitution proves the displayed relation. Thus similar low-orbit periods indicate similar mean densities, even for bodies with very different masses and radii. At altitude $h$, the right side is multiplied by $(1+h/R{)}^3$; the simple density inference requires $h\ll R$ and neglects rotation, nonsphericity and other gravitating bodies.

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 Newtonian potential of a point mass $M$ at the origin isIt is the free-space Green function of the Laplacian with the normalization of the Poisson equation for Newtonian gravity. Differentiating gives inward gravitational acceleration $-GM/r^2$. With $z=r/r_0$, the normalized Newtonian gravitational potential is $\Phi /(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.

For a radially outward launch from $r_0$, conservation of mechanical energy in the fixed-source test particle approximation gives the minimum escape velocity:At this threshold the speed tends to zero only at infinite separation. The model neglects air resistance, external tidal forces, and general relativity. ^[ref]

A circular orbit at $r_0$ instead requires tangential speed $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt{2}\simeq 1.41421$. A radial launch at speed $1$ reaches $r=2$ and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

Orbital speed is the magnitude of the velocity of an orbiting body in a specified reference frame. For a Newtonian circular orbit around a point source of mass $M$, equating $v^2/r$ to $GM/r^2$ gives $v=\sqrt{GM/r}$. In a curved spacetime, a speed inferred from coordinate radius and coordinate time generally differs from a speed measured by a local observer. For a circular orbit in the Schwarzschild metric, a static local observer measures $v_{\mathrm{local}}=r|d\varphi /dt|/\sqrt{1-2GM/(c^{2}r)}$; $r|d\varphi /dt|$ uses the clock normalization at infinity instead.

Let $r_S=2GM/c^2$. In the Paczyński-Wiita potential, circular orbit balance and the radial epicyclic frequency formula giveThus the innermost stable circular orbit lies at $r=3r_S$. Its specific orbital energy is $-c^2/16$, giving a zero-torque radiative efficiency of black-hole accretion of $1/16$. The model reproduces the Schwarzschild inner orbit exactly but approximates its relativistic energy and frequency. Abramowicz's derivation explains the potential's relation to the Schwarzschild effective potential.

Expand a unit-mass particle's Lagrangian about an axisymmetric circular orbit of radius $r_0$ and frequency ${\Omega }_0$, using $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Circular-orbit balance removes the linear radial term; total-time-derivative invariance of a Lagrangian removes $r_0{\Omega }_0\dot{y}$. To second order,Here $q$ is the orbital shear parameter and ${\Omega }_z$ is the vertical epicyclic frequency. The Euler-Lagrange equations retain the local Coriolis acceleration and tidal gravity.

For a Newtonian circular orbit, $v_{\varphi }^2=GM/r$, so the specific orbital energy is $e=v_{\varphi }^2/2-GM/r=-GM/(2r)$. Half of the released potential energy remains in circular orbital motion. The binding-energy estimate between the specified edges is thereforeUsing $M_{\odot }=1.98847\times 10^{30}\mathrm{kg}$ and one Julian year $=3.15576\times 10^7\mathrm{s}$ gives $\dot{m}=6.30\times 10^{22}\mathrm{kg}{\mathrm{s}}^{-1}$. Thus the radiative luminosity scale is $L\sim 1.9\times 10^{38}\mathrm{W}$, approximately $1.9\times 10^{45}\mathrm{erg}{\mathrm{s}}^{-1}$ or $4.9\times 10^{11}{L}_{\odot }$. The black-hole mass cancels because both edge radii scale with $M$. The specified inner edge is $6R_s$; it must not be silently replaced with a different relativistic orbit radius.

For a finite viscous disc, the precise radiated power also depends on boundary work: torque transport of energy in a steady accretion disk means that orbital binding-energy release need not all be radiated inside the two stated edges. If the intended model is a radiatively efficient steady Keplerian accretion disk with a zero-torque inner boundary condition, integrate the one-face flux $F=3GM\dot{m}[1-\sqrt{r_{\mathrm{in}}/r}]/(8\pi r^3)$ over both faces. It givesThe difference from $L_{\mathrm{bind}}$ is the outward torque power at the outer edge. The question supplies neither boundary torque nor the external material-supply energy budget, so the two calculations state their assumptions explicitly: about $2\times 10^{38}$ W is the binding-energy estimate, while $1.37\times 10^{38}$ W is the finite-disc zero-inner-torque dissipation result. Both neglect radiation advection, winds and relativistic corrections as required for this model.^[ref]

A stable sinusoidal radial-velocity semi-amplitude can be stellar reflex motion caused by a companion on a nearly circular orbit. The amplitude is compatible with a giant planet; a low-mass star or brown dwarf viewed nearly face-on can mimic the same projected motion, but requires an increasingly unlikely orientation. Starspots and magnetically suppressed convection can produce a rotational stellar activity signal on a ten-day timescale; contemporaneous photometry, activity indicators and stellar spectral-line bisector and other profile changes distinguish these from a clean common Doppler shift. Solar-like stellar oscillations ordinarily have much shorter periods, so a coherent ten-day oscillation of an otherwise ordinary solar-type dwarf is less plausible. Instrumental drift, sampling aliases and blended spectra also need checking. A sinusoid alone does not establish a planetary origin, and the data do not assign numerical probabilities to these alternatives.

For the minimum companion mass from radial velocity, take a circular two-body orbit and initially $M_{\star }=M_{\odot }$. The exact relation and its low-mass limit areWith $\sin i=1$,Here $a=[G{M}_{\star }{P}^2/(4{\pi }^2){]}^{1/3}$ is the relative orbital semi-major axis. Nonzero orbital eccentricity multiplies the inferred $M_p\sin i$ by $\sqrt{1-e^2}$ at fixed measured $K,P$, but the sinusoidal model motivates $e=0$ here.

For the radial-velocity sensitivity limit at fixed planet mass, use $K=M_p\sin i\sqrt{G/(M_{\star }a)}$ and interpret the stated detection scale as a semi-amplitude threshold $K_{\min }$. The largest distance occurs for $\sin i=1$:For an inclination $i$, this boundary is smaller by ${\sin }^{2}i$. The result assumes adequate time coverage and neglects the stipulated stellar noise. If the detection scale instead denotes a full peak-to-peak excursion, set $K_{\min }$ to half that scale; $a_{\max }$ then increases fourfold.

The stellar-mass scaling of radial-velocity planet estimates changes these inferences because Kepler's third law and the stellar reflex response depend on the stellar mass. At fixed $K,P$, $M_p\sin i\propto M_{\star }^{2/3}$ and $a\propto M_{\star }^{1/3}$, whereas the fixed-mass detectability boundary scales as $M_{\star }^{-1}$. ThereforeThe observed period and amplitude are unchanged. The stellar radius does not enter these leading dynamical estimates, but is crucial for the exoplanet transit inference.

For the transiting companion, assume an opaque spherical planet, negligible transit dilution and negligible limb darkening. Its radius ratio is $k=R_p/R_{\star }=\sqrt{\delta }=0.03873$, henceUsing the revised stellar mass and a circular orbit gives $a=0.01850\mathrm{au}$ and $A=a/R_{\star }=7.957$. Interpret the duration as the total first-to-fourth-contact duration $T_{14}$. The impact parameter from total transit duration follows from the total duration of a circular planetary transit:This is noncentral but not grazing: $b<1-k$. The small-planet straight-chord approximation gives $b\simeq 0.845$; retaining finite $R_p$ and orbital curvature gives the stated value. A duration measured between other contacts needs the corresponding contact formula. Unknown orbital eccentricity and argument of periapsis change the transverse speed and prevent a unique unconstrained $b$; the numerical estimate is conditional on the stated circular model.^[ref]

Because the transit constrains $\sin i\simeq 0.994$, the RV nondetection now bounds the actual planetary mass, rather than only its projected mass. The bulk-density upper limit from a transit and an RV nondetection isThese bounds assume the instrumental threshold can be applied to a fitted signal at the transit period; no statistical confidence level is specified. Such a large radius at this low mass is inconsistent with an ordinary compact rock-and-iron body. A low-density volatile-rich planet, possibly with an extended H/He envelope, is indicated under the planetary interpretation. The observations do not uniquely determine interior composition, atmospheric abundance or formation history; blended eclipses and activity still require validation. There is no luminosity or effective temperature supplied for the host, so a numerical planetary equilibrium temperature cannot be inferred from these data alone.

The numerical conversions above use externally documented $(GM{)}_{\odot }=1.3271244\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, $R_{\odot }=6.957\times 10^8\mathrm{m}$, $M_{\oplus }=5.9722\times 10^{24}\mathrm{kg}$, $R_{\oplus }=6.371\times 10^6\mathrm{m}$, $M_J=1.89813\times 10^{27}\mathrm{kg}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $1\mathrm{au}=149597870700\mathrm{m}$. They are conversion constants, not additional measured system parameters.^[ref][ref]

Adopt circular orbits, negligible planetary masses, spherical bodies, uniform stellar brightness and no transit dilution. Interpret the measured duration as the total duration of a circular planetary transit, from first to fourth contact, $T_{14}$. These conventions matter especially for the off-centre crossing: finite planet size cannot automatically be ignored there. Unknown orbital eccentricity or limb darkening would require further information to infer unique stellar radii.

First convert the apparent magnitude difference into a fractional exoplanet transit depth, rather than treating a milli-magnitude as a part in a thousand:Both crossings are non-grazing because their transit impact parameters satisfy $b<1-k$, so the full projected planetary disc can block the star. Equal depths determine equal radius ratios, not equal absolute planet radii.

For orbital inclination $i$ measured from the sky plane, $b=(a/R_{\star })\cos i$. Choosing mid-transit as orbital phase zero, the projected separation obeysAt first and fourth contact, $z/R_{\star }=1+k$. The full crossing duration therefore isFor the numerical scale use solar radius $R_{\odot }=6.957\times 10^8\mathrm{m}$ and nominal solar mass parameter $(GM{)}_{\odot }=1.3271244\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, with one day equal to $86400\mathrm{s}$. These are stated conversion choices, not additional measurements of the two systems.^[ref]

Kepler's third law gives $a=[G{M}_{\star }{P}^2/(4{\pi }^2){]}^{1/3}$. For a solar-mass, solar-radius host, this gives $a/R_{\odot }=12.3051$ at five days and $6.68023$ at two days. Substitution in the contact formula yieldsThe transit duration of a small planet instead takes both $k\to 0$ and $R_{\star }/a\ll 1$, giving $T\simeq P{R}_{\star }\sqrt{1-b^2}/(\pi a)$. It predicts $3.10\mathrm{h}$ and $1.37\mathrm{h}$, respectively. The difference is appreciable for B because the fractional change in its crossing chord is larger. Durations defined by mid-ingress to mid-egress or by second to third contact should not be silently equated with $T_{14}$.

The original diagram uses a straight projected track to display the contact geometry; the duration formula above retains orbital curvature.

To infer the actual stellar radii, invert the contact formula before applying the supplied mass-radius scaling. Let $A=a/R_{\star }$ and $s=\sin (\pi T_{14}/P)$; thenThe measured durations imply $A_A=8.34661$ and $A_B=7.55743$. Normalize the stipulated constant mass-radius ratio to the Sun, so $M_{\star }/M_{\odot }=R_{\star }/R_{\odot }\equiv r_{\star }$. Kepler's third law then givesThus the stellar radius from transit duration and a linear mass-radius relation isThe same assumed scaling gives $M_A=1.79M_{\odot }$ and $M_B=0.831M_{\odot }$. A's longer crossing requires a lower mean stellar mass density, while B's shorter full-contact crossing requires a higher one. They are not both solar replicas under this model. Their planets have $R_{p,A}=0.1213R_{\odot }$ and $R_{p,B}=0.05633R_{\odot }$.

For a consistently applied point-planet approximation, the duration scales as $T\propto P^{1/3}{r}_{\star }^{2/3}\sqrt{1-b^2}$ and the inferred radii would instead be $R_A\simeq 1.98R_{\odot }$, $R_B\simeq 1.07R_{\odot }$. The latter is close to solar. Consequently the assertion of a substantial non-solar radius for B depends on which duration convention and finite-size approximation are intended. The contact-based values above use the measured depth rather than discarding it. If eccentric orbits are allowed, the transit speed introduces an additional factor depending on eccentricity and orientation, so the two durations alone cannot establish those radii uniquely.

For an outer coplanar planet around B, the common orbital inclination gives $b_2/b_B=a_2/a_B=(P_2/P_B{)}^{2/3}$. A point-planet centre must cross the stellar disc, requiring $b_2<1$. The usual coplanar transit limit is thereforeThis is a limiting value approached from below; a tangent path has no duration. For an outer planet of finite, unspecified radius ratio $k_2$, any overlap instead requires $b_2<1+k_2$, giving $P_2<P_B[(1+k_2)/b_B{]}^{3/2}$. If it had the same ratio $k_2=k$, that geometric limit would be $3.08$ days; requiring full overlap would give $2.52$ days. Because neither its radius nor a photometric detection threshold is supplied, no unique finite-size practical maximum period can be inferred beyond this conditional expression. The stellar radius cancels from the point-planet period-ratio bound.

For B's inner planet, a reflected-light planetary phase curve haswhere $A_g$ is geometric albedo and $\alpha$ is phase angle. Near exoplanet secondary eclipse the visible hemisphere is almost fully illuminated. Using the standard full-phase approximation and the contact-based $A_B$,The exact lost fraction of combined stellar and planetary light is $F_p/(F_{\star }+F_p)$, which changes the result negligibly at this contrast. The corresponding magnitude depth is about $0.0437$ milli-magnitudes. The circular orbit has the same impact distance at eclipse as at transit, and $b_B<1-k$, so the star can hide the entire planetary disc.

The inclination inferred for B is $83.92^{\circ }$, putting closest approach to full phase at $\alpha \simeq 6.08^{\circ }$. A fully precise prediction would therefore also require the reflected-light phase function, which the question does not specify; the boxed value uses $\Phi \simeq 1$. The consistent point-planet duration approximation yields about $56.7\mathrm{ppm}$ instead, through its different inferred $a/R_B$. It is important to keep the geometry used for the radius inference and the reflected-light calculation consistent.

Finally, the stellar radial-velocity semi-amplitude for a low-mass companion on a circular orbit isThe two stars' equal apparent magnitudes make a comparison at similar photon statistics plausible, though spectral-line widths, stellar variability and observing cadence can still differ. With the contact-based stellar estimates,If the planets have comparable mean densities, A is easier: their equal depth then gives $M_{p,A}/M_{p,B}\simeq (R_A/R_B{)}^3\simeq 9.99$, so $K_A/K_B\simeq 4.44$. The larger planetary mass outweighs A's longer orbital period and larger host mass. For equal planetary masses, B is easier by a factor $2.25$ in amplitude. The supplied transit data give planet radii, not masses, so the ranking is conditional on a planetary mass-radius assumption; a unique unconditional ranking would invent information.

Sub-Keplerian rotation means that the gas rotates more slowly than a pressure-free particle on a circular orbit at the same cylindrical radius. Let $v_K=(G{M}_{\star }/r{)}^{1/2}$. Neglecting radial acceleration, radial force balance in the gas isAn outwardly decreasing pressure, $d{P}_g/dr<0$, supplies part of the inward-gravity balance through an outward pressure force. Less centrifugal support is needed, so $v_{g\varphi }<v_K$. Using $c_s^2=P_g/{\rho }_g$, define the pressure-gradient parameter of a protoplanetary disk byThenFor a thin disk and a smooth radial pressure profile, $\eta$ is small, of order $(H/r{)}^2$. The factor of two is part of this explicit convention; changing the convention for $\eta$ changes the drift formula correspondingly.

A solid particle feels essentially no continuum gas-pressure support. Left alone, it would orbit at $v_K$, so in the gas it feels a headwind. Drag removes angular momentum, causing radial drift of solids in a gas disk. Define Stokes number $\mathrm{St}={\Omega }_K{t}_s$ using the aerodynamic stopping time $t_s$. In a dilute test-particle limit with linear drag, negligible radial gas flow and small departures from a circular Keplerian orbit, let $u_{\varphi }=v_{p\varphi }-v_K$. The steady local radial and azimuthal balances areSolving givesThe sign is inward for the usual negative pressure gradient. Very small particles, $\mathrm{St}\ll 1$, closely follow the gas and have a small relative drift $v_{pr}\simeq -2\eta v_K\mathrm{St}$. Very weakly coupled large bodies, $\mathrm{St}\gg 1$, also drift slowly, with $v_{pr}\simeq -2\eta v_K/\mathrm{St}$. The fastest pressure-driven inward drift occurs at $\mathrm{St}=1$, where $|v_{pr}|=\eta v_K$. The corresponding physical grain or aggregate size depends on gas density, orbital radius and the drag regime; “metre-size barrier” is a conventional description of some disk models, not a universal size.

The drift timescale in this model isIt can be much shorter than the disk gas lifetime, threatening loss of the material needed to make planetesimals. Growth through the vulnerable coupling range must compete with this loss, while differential drift generates collisions that may stick, bounce or fragment. Radial redistribution also changes solid-to-gas ratios and moves material across condensation fronts. These are consequences for planet formation, rather than merely a small correction to an orbital frequency.

The same mechanism can help concentrate solids. At a local radial pressure maximum, $d{P}_g/dr=0$, so this component of drift vanishes. On the inside, a positive pressure gradient gives outward drift; on the outside, a negative gradient gives inward drift. Particles therefore converge toward a dust trap. Turbulent diffusion and inward gas advection can limit retention. With a nonzero radial gas speed $v_{gr}$, the test-particle result generalizes to $v_{pr}=[v_{gr}-2\eta v_K\mathrm{St}]/(1+{\mathrm{St}}^2)$, so gas advection need not vanish even at a pressure maximum.

Finally, if local solid loading becomes large, backreaction on the gas can no longer be neglected. Coupled gas-particle drift can support streaming instability and strong solid concentrations, potentially enabling gravitational collapse into planetesimals. The simple test-particle drift formula is a useful starting point, but not a description of that nonlinear high-loading regime. Pressure support creates both a drift obstacle to growth and opportunities for concentration where the pressure structure changes.

Radial epicyclic motion. At fixed specific angular momentum, use ${\Phi }_{\mathrm{eff}}=\Phi +{\ell }^2/(2r^2)$ and $\ddot{r}=-{\Phi }_{\mathrm{eff}}^{\prime }$. A circular orbit at $r_c$ obeys ${\Phi }^{\prime }(r_c)={\ell }^2/r_c^3=r_c{\Omega }^2$. Expanding $r=r_c+\xi$ gives $\ddot{\xi }+{\kappa }^2\xi =0$, whereFor ${\kappa }^2>0$ the radial epicyclic frequency describes simple harmonic motion. The angular correction follows from conservation of angular momentum, $\dot{\varphi }\simeq \Omega (1-2\xi /r_c)$, and describes an epicyclic motion around the guiding circular orbit.

Stabilizing terms. In the given local fluid dispersion relation, ${\kappa }^2$ is rotational support, important at long wavelength; $a_0^2{k}^2$ is random-motion or pressure support, important at short wavelength. The term $-2\pi G\Sigma |k|$ is destabilizing self-gravity. Although the wording refers to stellar dispersion, this algebra uses the supplied fluid approximation; a collisionless stellar treatment has a reduction factor and a different numerical normalization.^[ref]

Axisymmetric threshold. For $n=0$, set $q=|k|\ge 0$. Completing the square givesThus $Q>1$ gives local axisymmetric oscillatory stability; $Q<1$ gives a growing band, withAt $Q=1$, the minimizing mode has zero frequency and is marginal, not a strictly restoring oscillation. Local axisymmetric stability does not establish stability to every global nonaxisymmetric mode.

long-wave and short-wave branches at marginal Toomre stability. For $Q=1$ and $n\ge 1$,The long-wavelength branch, using the minus sign, requiresThese endpoints are the inner and outer Lindblad resonance conditions. The stated dispersion relation does not restrict every wave solution to that interval. Its plus-sign short-wavelength branch has $q>0$ even outside it: for example, ${\Omega }_P=\Omega +2\kappa /n$ gives $q=3\kappa /a_0$ and satisfies the equation. Additional branch selection or resonant absorption physics would be needed for an absolute confinement claim. No such condition is supplied.

Toomre stability radius of a flat-rotation disk. With $\Omega =v_0/r$, $\kappa =\sqrt{2}{v}_0/r$. Using $G=4.3009\times 10^{-3}\mathrm{pc}(\mathrm{km}{\mathrm{s}}^{-1}{)}^2{M}_{\odot }^{-1}$,The supplied local model therefore predicts stable radii below about $4.19\mathrm{kpc}$, marginality at that radius, and axisymmetric instability farther out while its assumed constant surface density and thin-disk approximation remain applicable. An infinite constant-density, flat-rotation disk is an idealization, not a global self-consistent galaxy model.