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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.

Let $f=0.46$. At Roche-lobe overflow contact, the supplied equilibrium radius relation givesKepler's third law then impliesThus the Roche-lobe period-mass relation for a linear donor radius law isHere $R_{\odot }$ and $M_{\odot }$ denote the solar radius and solar mass. Cancellation of the total mass is the special Roche-lobe-filling period-density relation combined with $R_2\propto M_2$.

Between eruptions the conservative binary mass transfer has $M$ constant and ${\dot{M}}_1=-{\dot{M}}_2$. Neglecting spin, writeIts logarithmic derivative givesDifferentiating the Roche lobe relation consequently givesFor $\dot{J}=0$, the Roche-lobe radius response exponent is ${\zeta }_L=2q-5/3$. If the donor follows its specified equilibrium radius sequence, ${\dot{R}}_2/R_2={\dot{M}}_2/M_2$. Its logarithmic overflow therefore changes bySince ${\dot{M}}_2<0$, $q<4/3$ gives negative feedback in this equilibrium-response model, whereas $q>4/3$ gives increasing overflow. For $q>4/3$, loss of donor mass shrinks its Roche lobe faster than its modeled stellar radius, increasing transfer and hence amplifying the original disturbance. The $q>4/3$ comparison is a formal extension; the initial system is specified to have $q<1$.

The word dynamically needs a qualification. Thermal equilibrium specifies ${\zeta }_{\mathrm{eq}}=1$, but actual dynamical stability of binary mass transfer depends on the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, which need not equal one. Stability after a rapid small mass loss requires ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, because $d\log M_2<0$. The equilibrium relation alone does not supply this adiabatic response. As a formal counterexample to deriving a universal dynamical threshold from it, a radius model with a dimensionless entropy label $S$, $R(M_2,S)=C{M}_2^2{e}^S$, can have equilibrium $S_{\mathrm{eq}}=S_0-\log (M_2/M_{\ast })$, giving $R_{\mathrm{eq}}\propto M_2$ but ${\zeta }_{\mathrm{ad}}=2$ at fixed $S$. At $q=3/2$, ${\zeta }_L=4/3<2$, so that model is dynamically stable despite the extrapolated $q>4/3$ condition. This shows the missing response hypothesis, not a claim that this toy radius law describes the given low-mass star.

A more relevant illustration is an ideal fully convective adiabatic stellar polytrope of index $3/2$. At fixed entropy its polytropic mass-radius relation is $R\propto M_2^{-1/3}$, giving ${\zeta }_{\mathrm{ad}}=-1/3$ and a dynamical critical ratio $q=2/3$ under the same Roche lobe approximation. For example, $q=0.9<4/3$ can be unstable dynamically even though the equilibrium comparison has negative feedback. The requested $4/3$ threshold is the intended result with instantaneous response $R_2\propto M_2$; it is not a universal dynamical criterion. The subsequent cycle calculation uses the supplied equilibrium sequence and presumes a physically stable, sufficiently slow transfer regime.

Two important angular-momentum-loss mechanisms are gravitational-wave emission from a binary system and magnetic braking of a binary star. Gravitational waves directly remove orbital energy and angular momentum. A magnetized stellar wind removes donor spin angular momentum with a large magnetic lever arm; tidal synchronization replenishes that spin from the orbit, so the net effect is orbital angular-momentum loss. Treating the wind mass loss as small compared with the Roche-transfer flow is consistent with the conservative approximation used here; otherwise the mass and angular-momentum equations must both be modified. Neglecting the stored spin does not eliminate the torque transmitted through it.

Such losses shrink the Roche lobe and offset the donor's tendency to move inside it. Maintaining contact on the equilibrium sequence requires ${\dot{R}}_L/R_L={\dot{M}}_2/M_2$, hence the binary mass-transfer contact equation becomesFor $q<4/3$ and $\dot{J}<0$, this drives ${\dot{M}}_2<0$ and maintains the modeled semi-detached binary.

Now take $\delta m>0$ for the mass expelled from the accretor in one classical nova. The white dwarf loses $\delta M_1=-\delta m$, while the donor has $\delta M_2=0$ and the total mass change is $\delta M=-\delta m$. Isotropic ejection in the accretor rest frame supplies no mean kick and carries the accretor's specific orbital angular momentum. Its orbital radius is $a_1=a{M}_2/M$, soThis is the angular-momentum budget for isotropic re-emission from a binary star, without an additional spin or frictional torque on the ejecta. The general logarithmic change of $J=M_1{M}_2\sqrt{Ga/M}$ isUsing $M/M_1=1+q$ gives the first-order nova changesThe donor mass and its assumed equilibrium radius are unchanged during the eruption. It is therefore left inside the enlarged Roche lobe, and Roche-lobe overflow ceases in this sharp-contact model. An extended atmosphere, irradiation-induced donor expansion or additional ejecta torques would require a different treatment.

The stipulated circular orbit is an important idealization. Isotropic loss that is slow compared with the orbital period can preserve circularity. If it is truly impulsive compared with that period, an initially circular point-mass orbit instead acquires eccentricity. With unchanged velocity and position, its new specific orbital energy is ${\epsilon }^{\prime }=GM/(2a)-G(M-\delta m)/a=-G(M-2\delta m)/(2a)$ and its squared specific angular momentum stays $h^2=GMa$. The Kepler orbit energy and eccentricity relations then give $a^{\prime }=a(M-\delta m)/(M-2\delta m)$ and $e^{\prime }=\delta m/(M-\delta m)$. The latter still has the same first-order separation increase. The question assumes circularity; its first-order circular budget is consistent with adiabatic orbital mass loss or with subsequent circularization that changes the separation only at second order. Work symbolically with $\delta m$, since the printed numerical layer mass has no explicit unit.

Finally put ${\mathcal{L}}_J=-\dot{J}>0$, constant over both intervals as specified, and let $J,M,q$ denote their leading-order values over one small cycle. While detached, masses and donor radius are fixed, so ${\dot{R}}_L/R_L=2\dot{J}/J=-2{\mathcal{L}}_J/J$. The fractional gap $4\delta m/(3M)$ closes afterOnce contact resumes, the contact equation gives $-{\dot{M}}_2=3M_2{\mathcal{L}}_J/[J(4-3q)]$. Accumulation of the next layer takesDividing yields the detached-to-semidetached time ratioBoth times are first-order in $\delta m/M$. Corrections to either time are second-order in that small mass fraction, so the displayed ratio has relative corrections of order $\delta m/M$. The positive denominator, negligible eruption duration relative to the cycle, fixed donor radius during detachment and common external loss rate are all essential to this idealized nova-induced binary detachment cycle.

For Roche-lobe overflow in the stipulated stellar thermal equilibrium, $R_2=R_L$. Using the mass-radius relation and Kepler's third law,The total mass cancels, givingHere $R_{\odot }$ is the solar radius. This is the Roche-lobe-filling period-density relation specialized to $R_2\propto M_2$.

Neglecting spin, the angular momentum of a circular orbit is $J=M_1{M}_2\sqrt{Ga/M}$. Between classical novae the transfer conserves total mass, so $\dot{M}=0$ and ${\dot{M}}_1=-{\dot{M}}_2$. Its logarithmic derivative givesIn particular, for conservation of angular momentum,The source PDF has $\dot{J}=0$ here; the TeX's $J=0$ is a transcription defect. For $q>4/3$, the Roche-lobe radius response exponent exceeds the equilibrium stellar radius response exponent $1$. Since ${\dot{M}}_2<0$, the Roche lobe then shrinks faster than the donor star; the overfill increases and drives more transfer, giving positive feedback. Equality at $q=4/3$ is marginal in this linear response test.

Two qualifications matter. First, $q>4/3$ is a formal extrapolation outside the stated $q<1$ range, and the supplied Roche lobe approximation need not remain accurate there. Second, true dynamical stability of binary mass transfer compares ${\zeta }_L$ with the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, not ${\zeta }_{\mathrm{eq}}=1$. The stellar thermal equilibrium argument supplies the displayed feedback threshold under the imposed radius law, rather than a necessary and sufficient physical dynamical threshold. For example, the polytropic mass-radius relation for a fully convective adiabatic stellar polytrope with index $3/2$ gives ${\zeta }_{\mathrm{ad}}=-1/3$; within this same Roche lobe approximation its dynamical threshold is $q=2/3$. Thus $q<4/3$ by itself does not guarantee dynamical stability.

Gravitational-wave emission from a binary system removes orbital energy and angular momentum. Magnetic braking of a binary star provides another important loss: a magnetized stellar wind carries away donor spin, while tidal locking couples that spin to the orbit. Tides alone redistribute angular momentum and are not an external sink. Matter expelled from the system can also carry orbital angular momentum, although appreciable continuous mass loss would require nonconservative modifications of the preceding equations. Setting ${\dot{R}}_L/R_L={\dot{R}}_2/R_2$ gives the binary mass-transfer contact equationFor $q<4/3$, a negative external $\dot{J}$ therefore sustains negative ${\dot{M}}_2$ along the stipulated stellar thermal equilibrium contact sequence, provided the system is otherwise stable and can remain thermally relaxed.

During the classical nova, the ejecta carry the white dwarf's specific angular momentum, not zero angular momentum. With ${\Omega }^2=GM/a^3$ and the accretor's distance from the centre of mass $a_1=a{M}_2/M$, isotropic escape givesThe slow ejection relative to the orbital period permits the adiabatic circular orbit approximation; the donor star does not transfer appreciable additional mass during it. Take $\delta M_1=-\delta m$, $\delta M_2=0$ and $\delta M=-\delta m$. Differentiating $J=M_1{M}_2\sqrt{Ga/M}$ givesSince $M/M_1=1+q$, this yieldsThese are first-order formulae with errors of order $(\delta m/M{)}^2$. The orbital widening agrees with Jeans-mode mass loss. The donor star's radius is unchanged to this order by the assumed ejection, so its Roche lobe expands away from it, causing nova-induced binary detachment. This idealization neglects changes in the donor from irradiation or interaction with ejecta; those are not supplied in the model.

Let $\Gamma =-\dot{J}>0$ be the fixed external loss rate. During the detached binary phase both component masses are fixed, and ${\dot{R}}_L/R_L=2\dot{J}/J=-2\Gamma /J$. Closing the fractional gap $4\delta m/(3M)$ takesIn the subsequent semidetached binary, the binary mass-transfer contact equation gives $-{\dot{M}}_2=3M_2\Gamma /[J(4-3q)]$. Accumulating a fresh layer of mass $\delta m$ therefore takesThroughout these first-order estimates $J,M,q$ can be evaluated at the start of the cycle: their fractional changes during it are of order $\delta m/M$. The limit $\Gamma =0$ is excluded, since no finite reconnection time follows without an external shrinkage mechanism.

A star with broadly Sun-like properties; the range depends on the stated classification criterion. In elementary planetary estimates, a solar mass, solar radius, and solar effective temperature are common approximate inputs, rather than consequences of the name alone.