Exoplanet transit photometry discovers a giant planet through a periodic flux decrement and measures $R_p/R_{\star }\simeq \sqrt{\delta }$. Stellar-radius estimates turn this into a planetary radius. The radial-velocity method discovers the stellar orbital reflex motion and constrains $M_p\sin i$. Follow-up exoplanet transit photometry is required to measure a geometric radius when such a planet also transits; a radial-velocity detection alone gives no direct size. Alternatively, exoplanet direct imaging finds young luminous giants, whose sizes are inferred less directly from luminosity, temperature, distance, and a planetary mass-radius relation or atmosphere model.

The two broad explanations for hot-Jupiter radius inflation are retention of primordial heat and addition of new interior power.

Delayed cooling of an inflated giant planet can arise from enhanced atmospheric opacity, which slows radiative escape; an irradiation-maintained radiative blanket, which insulates the convective interior; or compositional stratification and inefficient layered convection, which inhibit the outward transport of heat. These alter the rate of Kelvin-Helmholtz contraction.

Heating of an inflated giant planet can arise from tidal heating maintained by eccentricity or obliquity; Joule heating of currents driven by atmospheric winds through a magnetic field; or downward transport and dissipation of atmospheric mechanical energy generated by irradiation. To affect radius, the energy must be deposited at a depth and rate that changes the interior cooling balance. Simply absorbing starlight high in the atmosphere does not automatically supply deep heating. These are proposed mechanisms with different efficiencies, not six universally established contributions in every inflated planet.

At $10\mathrm{au}$ from a four-solar-luminosity star, the incident flux is proportional to $4/10^2=1/25$, approximately the same as for Jupiter at $5\mathrm{au}$ around the Sun. Thus Jupiter supplies a possible old comparison at roughly the same mass and irradiation.

Assume the radius law holds from the young epoch to an old age of $4.5\mathrm{Gyr}$, that the irradiation history can be represented by the same fixed value, and that the present old radius is $R_J$. If $t$ is approximated by stellar age, thenIf $t$ instead starts at completion of formation, the model's formation delay of up to $3\mathrm{Myr}$ leaves the young planet with thermal age between $2$ and $5\mathrm{Myr}$. Taking the old thermal age as approximately $4500\mathrm{Myr}$ then gives $\alpha \simeq 0.090$–$0.102$.

Thus an exponent of order 0.1 is a reasonable conditional estimate. A single radius at one age cannot determine both the exponent and normalization of a power law; the old Jupiter comparison and the assumptions about formation time and irradiation are essential.

For a spherical planet, combine the hydrostatic pressure support equation with $dm/dr=4\pi r^2\rho$:Neglect surface pressure. Since $r(m)\le R$, the hydrostatic lower bound on planetary central pressure isFor a physically usual mass density decreasing outward, the mean interior density exceeds the global mean. Hence $r(m)\le R(m/M{)}^{1/3}$, giving the sharper minimum within this class,Uniform mass density attains the sharper bound; direct integration with $m(r)=M{r}^3/R^3$ gives $P_c=3G{M}^2/(8\pi R^4)$. Real planets are centrally concentrated through compression and dense cores, so their central pressure is higher.

Using $M_{\oplus }=5.97\times 10^{24}\mathrm{kg}$, $R_{\oplus }=6.37\times 10^6\mathrm{m}$, $M_J=1.90\times 10^{27}\mathrm{kg}$, and $R_J=7.15\times 10^7\mathrm{m}$, the universal bounds are approximately $5.75\times 10^{10}\mathrm{Pa}$ for Earth and $3.66\times 10^{11}\mathrm{Pa}$ for Jupiter. The uniform-density estimates are respectively $1.72\times 10^{11}\mathrm{Pa}$ and $1.10\times 10^{12}\mathrm{Pa}$, or $1.72$ and $11.0$ megabars. Either comparison gives a Jupiter minimum about 6.4 times the Earth minimum.

Assume all three planets retain approximately their original hydrogen-helium bulk composition. Stellar encounters change orbital energy and irradiation, not automatically the planet's elemental abundances. Their atmospheric structures can become approximately stationary long before their interiors finish cooling.

For the unperturbed Jupiter-like planet at $a\simeq 5.2\mathrm{au}$, the zero-albedo globally averaged planetary equilibrium temperature is about $122\mathrm{K}$. Stellar light heats the outer atmosphere, while internal cooling supports a deeper temperature gradient and convection. At cooler pressures, chemical equilibrium favors methane and ammonia; condensate clouds can include ammonia at high levels and water deeper down. Disequilibrium chemistry in an exoplanet atmosphere can preserve carbon monoxide or other species from deeper layers.

For the inward-migrated planet at $a\simeq 0.052\mathrm{au}$, the irradiation-only planetary equilibrium temperature is ten times higher, about $1220\mathrm{K}$, because $T_{\mathrm{eq}}\propto a^{-1/2}$. Its irradiated planetary atmosphere has a heated radiative exterior, potentially strong day-night differences, and a deep radiative-convective boundary. At suitable pressures, chemical equilibrium increasingly favors carbon monoxide over methane; water remains important, while alkali absorption and high-temperature condensates can matter. An atmospheric thermal inversion depends on absorbers, clouds, and irradiation and is not guaranteed simply by migration.

The ejected object is a rogue planet. Its irradiation-based planetary equilibrium temperature becomes very small, but its actual emitting temperature is set primarily by internal cooling and Kelvin-Helmholtz contraction. After only several million years it can remain warm and self-luminous, with deep convection and an outward-cooling radiative atmosphere. Its photospheric chemistry and clouds depend on that cooling temperature; they cannot be inferred from the absence of a host star alone. None of these cases fixes an exact pressure-temperature profile without a cooling model, opacities, and atmospheric composition.

For $a\gg a_{AB}$, both stars illuminate the circumbinary planet at approximately distance $a$. Treat them as blackbodies, neglect internal heating, and assume Bond albedo $A_B$ and uniform global reradiation. The absorbed and emitted powers areThus the equilibrium temperature of a circumbinary planet isFor an example, take $R_A=R_{\odot }$, $R_B=0.3R_{\odot }$, $T_A=6000\mathrm{K}$, $T_B=3000\mathrm{K}$, $a=1\mathrm{au}$, $a_{AB}=0.05\mathrm{au}$, circular coplanar orbits, and $A_B=0$. ThenThe cooler star contributes only $0.3^2(3000/6000{)}^4\simeq 0.0056$ of the hotter star's luminosity in this example.

An opaque isothermal atmosphere emits $F_{\lambda }=\pi B_{\lambda }(T_p)$, with no absorption or emission bands. The Wien displacement law puts the maximum of this wavelength spectrum near ${\lambda }_{\max }\simeq 10\mu \mathrm{m}$. At observer distance $d$, the planetary spectral flux is $\pi B_{\lambda }(T_p)(R_p/d{)}^2$.

Assume local thermodynamic equilibrium, negligible scattering of thermal radiation, and a plane-parallel atmosphere. With inward optical depth ${\tau }_{\lambda }$ and outward ray cosine $\mu$, the radiative transfer equation isFor a deep atmosphere, its formal solution of the radiative transfer equation isThe weighting samples ${\tau }_{\lambda }$ of order unity. The Eddington-Barbier relation makes this precise when the source function is nearly linear: $I_{\lambda }(0,\mu )\simeq B_{\lambda }[T({\tau }_{\lambda }=\mu )]$.

A molecular band with greater opacity reaches unit optical depth higher than the nearby continuum. If temperature decreases upward, that band samples cooler gas and appears in absorption. If an atmospheric thermal inversion makes the upper gas hotter, the band appears in emission. If both depths have the same temperature, an opaque isothermal atmosphere hasand no opacity-dependent features. This last conclusion assumes all wavelengths are opaque or the lower boundary emits the same Planck function; an optically thin isothermal layer over a different background need not be featureless.

In the annulus model for transmission spectroscopy, an added aerosol optical depth changes the signal to $A[1-e^{-({\tau }_{\mathrm{gas}}+{\tau }_{\mathrm{aerosol}})}]$. Large particles can supply nearly wavelength-independent extinction. An opaque high exoplanet cloud deck then masks deeper gas, flattens the optical exoplanet transmission spectrum, and weakens atomic or molecular features.

Small particles can instead produce a rising transit radius toward short wavelengths. In an isothermal atmosphere, the slant optical depth of an isothermal atmosphere isTaking the effective radius near ${\tau }_{\lambda }\sim 1$ gives $z(\lambda )=H\log {\sigma }_{\lambda }+\mathrm{constant}$. For Rayleigh scattering, ${\sigma }_{\lambda }\propto {\lambda }^{-4}$, so the scattering slope of a transmission spectrum isA steeper-than-expected optical slope or suppressed gas features can therefore indicate atmospheric haze or exoplanet clouds. These signatures are not unique: high mean molecular mass reduces $H$, gas itself can produce Rayleigh scattering, and stellar surface heterogeneity can mimic slopes. Consistent optical and infrared features help distinguish these explanations.

At 10 parsecs, $0.01$ arcseconds corresponds to $a=0.10$ astronomical units. Take a solar-radius star with $T_{\star }=5772\mathrm{K}$, zero Bond albedo, full day-night heat redistribution, negligible internal heating, and a hydrogen-helium atmosphere with mean particle mass $2.3m_H$. The planetary equilibrium temperature isFor Jupiter mass and radius, $g\simeq 24.8\mathrm{m}{\mathrm{s}}^{-2}$. Its atmospheric scale height isAssume a strong band spans $N_H=5$ atmospheric scale heights and saturates in the annulus model for transmission spectroscopy. The atmospheric spectral-feature amplitude isFor a five-standard-deviation detection, the uncertainty of the measured differential contrast must satisfyThis estimate scales linearly with the assumed feature height; one atmospheric scale height would require about $7.5\mathrm{ppm}$. The question gives no opacity or abundance from which to fix $N_H$, so an atmospheric detection threshold is necessarily assumption-dependent.

For thermal emission at $10\mu \mathrm{m}$, assume the planet and star emit as blackbodies. The thermal eclipse depth from the Planck law isThus the uncertainty required for a five-standard-deviation exoplanet secondary eclipse detection isThese are uncertainties of the final transit or eclipse contrasts, including the uncertainty of their reference levels. The distance affects photon counts and observing time but cancels from the flux ratios. An opaque exactly isothermal atmosphere emits a featureless blackbody spectrum: an eclipse detects its thermal light, while identifying atmospheric composition requires spectral features and a nonisothermal structure or other diagnostics.

Assume a uniform stellar disc of radius $R_{\star }$, an opaque planetary radius $R_p$, and a thin atmospheric annulus of thickness $z\ll R_p$. Neglect planetary light during transit and scattering back into the beam. The attenuation part of the radiative transfer equation gives $I_{\lambda }=I_{\lambda ,0}{e}^{-{\tau }_{\lambda }}$, where ${\tau }_{\lambda }$ is the slant optical depth along a stellar ray through the atmosphere. The fraction of light removed from that annulus is therefore $1-e^{-{\tau }_{\lambda }}$.

If the annulus is represented by one effective optical depth, its area divided by the stellar area isThe annulus model for transmission spectroscopy consequently gives the extra transit depthHere ${\delta }_{\lambda }$ is normalized to the unobscured stellar flux and excludes the opaque-disc depth $(R_p/R_{\star }{)}^2$. For a spectral feature measured relative to a continuum with slant optical depth ${\tau }_c$, the corresponding contrast is $A(e^{-{\tau }_c}-e^{-{\tau }_{\lambda }})$; the displayed formula takes the annular continuum to be transparent.

A real exoplanet transmission spectrum has an impact-parameter-dependent optical depth. Its more accurate expression isStellar limb darkening and horizontally varying clouds further modify the weighting. The single-annulus approximation is useful for estimating an atmospheric spectral-feature amplitude, rather than predicting every spectral line.

The dipole magnetic-field line geometry gives the angular radius at the sonic point as ${\theta }_s\simeq {\theta }_{\star }\sqrt{r_s/R_{\star }}$. The sonic region must remain a narrow flux tube for the one-dimensional transonic branch equations to apply. Thus a necessary small-angle constraint isThis also makes the transverse sound-crossing time $r_s{\theta }_s/c_s$ small compared with the radial flow time $r_s/c_s$. The restriction must hold at any larger matching radius as well. To match to a reservoir with negligible gravitational potential within the same narrow dipolar approximation, one needs an overlap regionHence the stronger useful sufficient scaling is ${\theta }_{\star }^2\ll R_{\star }{c}_0^2/(G{M}_{\star })$. The small-angle dipolar model should not be extrapolated to literal infinity, where its boundary magnetic field line turns toward the equatorial plane.

The transonic accretion in a dipolar flux tube values areThe mass density follows from $c^2/c_0^2=(\rho /{\rho }_0{)}^{\gamma -1}$ for the same polytropic equation of state. If $A_{\star }$ is the total loaded area at the surface, mass conservation at the sonic point givesFor two equal polar caps with surface angular radius ${\theta }_{\star }$, $A_{\star }\simeq 2\pi R_{\star }^2{\theta }_{\star }^2$, so the total mass accretion rate isFor only one loaded polar cap, divide this expression by two. Expressing the result first in $A_{\star }$ also covers a bundle whose loading angle is defined at another radius. The smooth isothermal limit $\gamma \to 1$ has $r_s=G{M}_{\star }/(3c_0^2)$, $c_s=c_0$, and ${\rho }_s=e^{5/2}{\rho }_0$.

A regular sonic point must make both sides of the flow equation vanish:Substituting into the Bernoulli equation yieldsFor $\gamma >1$, positive finite sound speed therefore requires the critical adiabatic index for dipolar accretionTo check that this gives real regular crossings, differentiate the flow equation at the sonic point and put $x=r_s{v}_s^{\prime }/c_s$. The transonic accretion in a power-law tube calculation givesThe minus sign gives the transonic branch whose Mach number increases inward. At $\gamma =7/5$ the positive-energy reservoir cannot match a finite sonic point; for larger $\gamma$ the required $c_s^2$ is negative. A physical surface-crossing solution also requires $r_s>R_{\star }$ and validity of the narrow flux tube approximation up to the sonic region.

In the narrow polar flux tube, the magnetic field lines are almost radial and the transverse width is much smaller than the radial scale. The Lorentz force density has no component along the magnetic field, so the longitudinal magnetically channelled accretion is hydrodynamic. Let $v(r)>0$ denote inward speed. Mass conservation, the Euler equations for an inviscid fluid, and the polytropic equation of state giveUsing the adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$ and $A\propto r^3$,The Bernoulli equation iswhere the right-hand side comes from matching to a nearly stationary reservoir with negligible gravitational potential. The fixed magnetic field provides transverse confinement; neglecting magnetic forces in this one-dimensional equation concerns their longitudinal component.

Align the polar axis with the magnetic dipole moment. The magnetic dipole field hasThe magnetic-field-line equation gives $dr/(rd\theta )=2\cot \theta$, hence the dipole magnetic-field line is $r=C{\sin }^2\theta$. Normalize the loaded bundle by its small surface angular radius ${\theta }_{\star }$ at $r=R_{\star }$. Its boundary obeysThe dipolar flux-tube area for one polar cap is consequentlyEquivalently, $BA$ is constant by conservation of magnetic flux and $B\propto r^{-3}$ near the axis. Two equal loaded caps double the total area. The $r^3$ scaling is independent of the normalization; if the loading angle is specified at another radius $r_L$, use $A(r)=\pi {\theta }_L^2{r}^3/r_L$ instead.

Setting $I=0$ in cylindrical magnetostatic pressure balance gives $p+B_z^2/(8\pi )=\mathrm{constant}$. The axial boundary value sets this constant to $p_0$. Thus the magnetic pressure support with vanishing axial field solution isThe azimuthal component of the magnetostatic Ampère-Maxwell equation givesSince $1-e^{-R^2/a^2}=R^2/a^2+O(R^4)$, the radial limit isThere is a regularity subtlety: $B_z\sim \sigma \sqrt{8\pi p_0}R/a$ is not a smooth function of Cartesian position at the axis, and the nonzero limiting $j_{\varphi }$ multiplies an azimuthal unit vector with no unique direction there. These formulas solve the radial problem for $R>0$ and give the requested scalar limit, but the data in this part do not admit the fully smooth vector-field regularity assumed in part (a).

For constant $B_z$, cylindrical magnetostatic pressure balance givesRegularity gives $I(0)=0$. With $x=R^2/a^2$, integration yields the power-law pressure-supported axial currentThe sign of $I$ may be chosen either way and fixes the sense of the toroidal magnetic field. For $p_0>0$ and nontrivial decreasing pressure ($k>0$), the integral converges at infinity precisely whenAt $k=1$ it diverges logarithmically; for $0<k<1$ it diverges as $R^{2(1-k)}$. The degenerate case $k=0$ has constant pressure and zero electric current.

The remaining magnetic field has $\mathbf{B}=B_{\varphi }(R){\mathbf{e}}_{\varphi }+B_z(R){\mathbf{e}}_z$. The radial Lorentz force density separates into a magnetic pressure gradient and inward magnetic tension:Here $(\mathbf{B}\cdot \nabla )\mathbf{B}$ has radial component $-B_{\varphi }^2/R$ because ${\partial }_{\varphi }{\mathbf{e}}_{\varphi }=-{\mathbf{e}}_R$. Thus magnetostatic equilibrium givesIntegrating the axial component of the magnetostatic Ampère-Maxwell equation gives $I(R)=cR{B}_{\varphi }/2$, since regularity removes the integration constant. Substituting $B_{\varphi }=2I/(cR)$ combines the toroidal terms intoThis is the cylindrical magnetostatic pressure balance equation in terms of enclosed electric current.

With no dependence on the azimuthal or axial coordinates, Gauss's law for magnetism in cylindrical coordinates givesRegularity at the axis forces $C=0$, so $B_R=0$. The radial component of the magnetostatic Ampère-Maxwell equation isConsequentlyThis is also the starting point of cylindrical magnetostatic pressure balance.

For $N$ equal weak shocks, $(1+\delta {)}^N=P_s$, henceAdding the individual increases in specific entropy gives the entropy production in successive weak shocksBy contrast, a single normal shock wave with the same total pressure ratio givesThus a single strong shock produces more entropy than the chain of sufficiently small weak shocks. At fixed $P_s$, the chain approaches zero entropy production as $\delta \to 0$, while the single-shock result stays positive. The gradual compression approaches reversible isentropic flow.

Put $S(P)=[s]/c_v$. Differentiating the expression in part (c) givesFor $P=1+\delta$, the denominator is $4{\gamma }^2+O(\delta )$, soIntegrating from the unshocked state, where $S(1)=0$, proves the cubic entropy production in a perfect-gas shock:The first-order and second-order terms vanish: a weak shock agrees with reversible compression through second order.