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A natural interpretation of the short-wavelength peak is reflected starlight, while the longer-wavelength peak is planetary thermal radiation. A reflected spectrum approximately follows the stellar spectrum multiplied by the wavelength-dependent geometric albedo; thermal radiative flux approximately follows the planet's Planck function, modulated by molecular opacity. Thus two peaks need not represent two planetary surface temperatures.

For a quantitative estimate assume both are broad $F_{\lambda }$ peaks, reflection has a slowly varying geometric albedo, and star and planet have approximately blackbody spectral envelopes. Wien's displacement law then givesThe stellar estimate is compatible at order of magnitude with an old, relatively small main-sequence star; equal age does not mean equal temperature to the Sun. Assume the far-infrared signal is thermal and sufficiently long-wavelength that the Rayleigh-Jeans law applies to both bodies. The planet-star radius estimate in the Rayleigh-Jeans limit givesand thereforeThis is a small volatile-rich-planet-sized estimate, not a Jupiter-sized one. It is conditional: peaks caused by molecular windows, strongly chromatic reflection, or a spectrum expressed as $F_{\nu }$ rather than $F_{\lambda }$ do not support those two Wien estimates. Without $T_p/T_{\ast }$, the far-infrared ratio only fixes $R_p^2{T}_p/T_{\ast }$.

For a close-in hot planet around a small star, transit-based atmospheric observations are the natural route if the orbital geometry allows them. Exoplanet transmission spectra gain from the small stellar radius, with a limb signal scaling as $2R_{p}H/R_{\ast }^2$; they probe composition at the day-night terminator. The stated $10^{-3}$ thermal contrast also makes exoplanet secondary eclipse measurements a particularly useful route to the dayside exoplanet emission spectrum. Resolving such a close-in small planet by exoplanet direct imaging is much harder. Transit and secondary-eclipse spectroscopy are favoured for a transiting close-in interpretation; the supplied spectrum alone does not determine the orbital geometry or a unique best technique.

In the collisionless exosphere, an atom escapes if its outward trajectory has positive total mechanical energy. Neglect tides and stellar forces and use Newtonian gravity at exobase radius $r_e$:With thermal speed $v_{\mathrm{th}}=\sqrt{2k_B{T}_e/m}$, the Jeans escape parameter isA thermal distribution always has an escaping tail; Jeans escape is exponentially suppressed for ${\lambda }_e\gg 1$, with Jeans escape flux proportional to $(1+{\lambda }_e)e^{-{\lambda }_e}$. Efficient escape requires ${\lambda }_e$ of order a few or smaller, with an order-unity energetic estimate $k_B{T}_e\sim G{M}_{p}m/r_e$.

Assume a Neptune-like mass and radius, $r_e\simeq R_p$, and atomic hydrogen. Using the supplied rounded constants givesHenceUsing the mean kinetic energy $3k_{B}T/2$ instead gives an order-unity coefficient and $T\simeq 2.3\times 10^4\mathrm{K}$. An expanded exobase has weaker binding and lowers the estimate by $R_p/r_e$. A comet-like tail can also be shaped by radiation pressure and stellar-wind interactions; it does not by itself measure $T_e$ or prove that a hydrostatic Jeans model is valid. At ${\lambda }_e\sim 1$, $H/r_e\sim 1$ and hydrostatic equilibrium fails as a global description: substantial mass loss must usually be treated as hydrodynamic atmospheric escape.

The exobase is defined by mean free path $\ell \sim H$, not by a universal pressure. For a neutral hydrostatic gas with collision cross-section ${\sigma }_c$,For example, explicitly assuming ${\sigma }_c\sim 10^{-19}\text{–}10^{-20}{\mathrm{m}}^2$ gives $P_e\sim 2\times 10^{-7}\text{–}2\times 10^{-6}\mathrm{Pa}$, or $2\times 10^{-12}\text{–}2\times 10^{-11}\mathrm{bar}$. These are representative extremely dilute neutral-exobase pressures, with orders of magnitude varying with composition, cross-sections and expansion. The supplied constants contain no collision information, so they cannot uniquely determine an exobase pressure; ionization or a non-hydrostatic density profile changes this estimate.

Assume the present epoch has the same age as present-day Jupiter, the host now has solar luminosity, and moving inward did not change the stipulated intrinsic cooling law or its normalization. This neglects persistent tidal heating and irradiation-induced suppression of cooling. The intrinsic planetary luminosity is then approximately $L_{\mathrm{int},\mathrm{J}}$, not its value at the migration epoch.

Define incident irradiation as intercepted power before reflection,Comparing with present-day Jupiter at $a_J\simeq 5\mathrm{AU}$ givesFor equal radii,Retaining unequal radii multiplies the right side by $(R_J/R_p{)}^2$. If irradiation instead denotes incident radiative flux per area, the orbital factor is still $10^4$, but it must be compared with an intrinsic flux consistently. Absorbed power also includes $1-A_B$, so comparisons of absorbed irradiation require the two Bond albedos.

Close-in gas giants motivate planetary migration when compared with formation models. Two useful observational diagnostics are then orbital eccentricities and spin-orbit geometry. An eccentric population of wider potential progenitors together with circular short-period orbits supports eccentricity excitation followed by tidal dissipation. The second diagnostic is stellar obliquity, including misaligned or retrograde orbits measured through the Rossiter-McLaughlin effect; these can favour scattering or secular pathways over smooth coplanar migration. Conversely, aligned resonant architectures are compatible with disc-driven migration. None is unique: primordial disc tilt or alternative formation can mimic some signatures. The expected present-day ratio is suppressed by the inverse-square orbital factor, while eccentricities and spin-orbit geometry test migration pathways.

Assume a Newtonian, spherically symmetric, nonrotating body in hydrostatic equilibrium, supported by a polytropic equation of state $P=K{\rho }^{1+1/n}$ with constant $K>0$ and $n>0$. The enclosed mass and hydrostatic pressure support equation areEliminate $m$ by first writing $r^2{\rho }^{-1}dP/dr=-Gm$ and then differentiating:Let ${\rho }_c$ be central mass density, and introduce the Lane-Emden variables for a stellar polytropewhereThen $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Substitution cancels the dimensional factors and yields the Lane-Emden equationThe central conditions enforce the chosen central mass density and regular spherical symmetry. The local regular expansion is $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. If a finite first zero ${\xi }_1$ exists and surface pressure is negligible, the physical radius is $R=a{\xi }_1$, and the Lane-Emden mass formula is $M=4\pi a^3{\rho }_c[-{\xi }_1^2{\theta }^{\prime }({\xi }_1)]$. For $0<n<5$ the standard isolated solution has a finite surface; $n=5$ has infinite extent, so a finite surface must not be assumed for all indices. Irradiation, composition stratification and non-polytropic equations of state require more general structure equations.

Hot-Jupiter radius inflation is the excess radius of many irradiated gas giants relative to standard cooling models of the same mass, composition and age. A young gas giant starts with high entropy and a large radius, but ordinarily contracts through Kelvin-Helmholtz contraction. An old inflated object must either retain that thermal reservoir unusually well or receive power that affects the deep interior.

The first class is delayed cooling of an inflated giant planet. Two specific mechanisms are increased atmospheric opacity, which makes internal radiation escape less readily, and inhibited interior convection caused by composition gradients, potentially with layered transport. Both slow entropy loss. Stellar irradiation can also maintain an outer radiative blanket, but an appropriate irradiated boundary is already included in many baseline models and does not alone explain every extreme radius.

The second class is heating of an inflated giant planet. Two mechanisms are tidal heating from eccentricity, obliquity or other time-dependent tidal forcing, and Joule heating from wind-driven electric currents in an ionized atmosphere coupled to the magnetic field. Mechanical energy carried inward from atmospheric circulation is another candidate. Heating must be deposited sufficiently deeply and with sufficient power; superficial absorption that is promptly reradiated is not equivalent to deep interior heating. A circular synchronized isolated orbit does not automatically provide a persistent tidal source.

The classes can coexist and are tested through radius-age-irradiation trends, orbital properties and the required energy budget. Merely changing an observed transit altitude by a few atmospheric scale heights is generally distinct from inflating the bulk giant-planet interior.

At fixed mass, three broad controls of a super-Earth-sized body's observed radius are:

Bulk composition, envelope fraction/composition, and thermal/irradiation history are three independent controls. If mass is not fixed, the mass itself is an additional major variable. The label super-Earth does not ensure a rocky composition or an Earth-like atmosphere, and a radius alone does not determine which of these effects dominates.

Three research directions natural in the 2017 setting are:

Small-planet atmospheres, formation through composition, and atmospheric dynamics/variability provide three concrete emerging directions. The examples describe research aims as of the paper's date, rather than treating later discoveries or later operating missions as established in 2017.

The endpoint radii are $5$ and $13$, but substituting those two numbers in the finite conical frustum formula is invalid. A rotation about the $z$-axis preserves the horizontal angle between the endpoint position vectors; here $(3,4)$ and $(12,5)$ are not parallel. The radius along the generating segment is not linear in height.

Parametrize the actual segment byIts rotated circle at height $1+6t$ has radiusThus the rotated image is the surface of revolution $\mathbf{R}(t,\theta )=(\rho (t)\cos \theta ,\rho (t)\sin \theta ,1+6t)$. Applying the conical frustum formula to infinitesimal pieces of this radial profile, or directly using the cross product as in ii, gives the surface area of a surface of revolutionTo evaluate it explicitly, put $u=82t+31$. Then $9676t^2+7316t+1861=(59u^2+19602)/41$. DefineThe chain rule gives $G^{\prime }(u)=2\sqrt{59u^2+19602}$, so the answer isThe positive logarithm arguments make this a real-valued expression. Rotating the segment's endpoints independently into one vertical plane before joining them would instead create a different surface, with surface area $180\pi \simeq 565.4867$.

Interpolate the radius and height byA parametrized surface for the conical frustum isThe strict positivity of $a,c$ gives $\rho (t)>0$. Since $d>b$, the height identifies $t$ uniquely; the angle then identifies a point of each circle uniquely modulo $2\pi$. Hence this covers the lateral surface of revolution once. If $a=c$, it is the lateral surface of a circular cylinder.

On the surface of revolution, a normal vector is proportional to $-H^{\prime }(r){\mathbf{e}}_r+{\mathbf{e}}_z$. Thus the normal force and gravity have no ${\mathbf{e}}_{\theta }$ component. The ${\mathbf{e}}_{\theta }$ component of Newton's second law, using the given linear drag force, isMultiplication by $r$ therefore givesThe printed target equation omits the factor $1/m$. The matching original PDF, page 5, gives both ${\mathbf{F}}_D=-\alpha \dot{\mathbf{r}}$ and $\dot{h}=-\alpha h$, so this is a source inconsistency. If its $\alpha$ in the target equation is intended to mean the drag coefficient per unit mass, put $\lambda =\alpha /m$ and write $\dot{h}=-\lambda h$. Equivalently, if the intended force was ${\mathbf{F}}_D=-m\alpha \dot{\mathbf{r}}$, the requested convention givesThese formulas determine $\dot{\theta }$ once $r(t)$ is known; they do not assume a constant radius.

Initially $\dot{r}=0$ also gives $\dot{z}=H^{\prime }(b)\dot{r}=0$. Thus the initial velocity is azimuthal, with $|h|=bv$. By conservation of energy, the supplied mechanical energy divided by $m/2$ givesThe first term is nonnegative. ConsequentlyFor a positive launch speed $v>0$, introduce twice the radial effective potential on a surface of revolution per unit mass,It satisfiesso it is a strictly convex function. Also $S(r)\to \infty$ as $r\to \infty$. To justify the other endpoint carefully, convexity gives $H(r)\ge H(b)+H^{\prime }(b)(r-b)$ for $0<r<b$. This supplies a finite lower bound near $0$, and the positive $r^{-2}$ term consequently forces $S(r)\to \infty$ there.

Thus $S$ has a unique minimum at some $r_m>0$. Its derivative is strictly increasing and zero there, so $S$ decreases to $r_m$ and increases afterwards. Its sublevel set $S(r)\le B$ is a single interval $[b_1,b_2]$, with $0<b_1\le b\le b_2<\infty$. If $B>S(r_m)$, the two boundary equations $S(b_i)=B$ have distinct solutions; if $B=S(r_m)$, both boundaries are $r_m$. The conservation of energy identity therefore provesThe two radii coincide exactly when $b=r_m$, equivalently $S^{\prime }(b)=0$. This givesThe equation in b then has the constant solution $r=b$, with $\dot{\theta }=\pm v/b$: this is a circular orbit.

A positive inner bound requires $v>0$. If zero launch speed is allowed, $h=0$ and the displayed inequality only gives $r\le b$ because $H$ is strictly increasing; one may take the limiting inner bound $b_1=0$. For example, $H(r)=r^2$ satisfies all the stated hypotheses but permits inward motion from rest all the way towards the excluded axis. The two-positive-turning-radii argument therefore uses the natural interpretation of “launched with speed $v$” as $v>0$.

The quadric surface $x^2+y^2+z^2=1$ is the unit sphere centered at the origin. Each section $z=c$, for $|c|<1$, is a circle of radius $\sqrt{1-c^2}$.

The quadric surface $z=x^2+y^2$ is a circular elliptic paraboloid, opening in the positive $z$ direction with vertex at the origin. Its horizontal sections $z=c>0$ are circles of radius $\sqrt{c}$, and its vertical sections through the $z$ axis are parabolas.

For $a{x}^2+y^2=0$, $z$ is unrestricted. If $a>0$, both nonnegative terms must vanish, so the locus is the $z$ axis, a straight line. If $a=0$, it is the plane $y=0$, represented by the repeated factor $y^2$. If $a<0$, factorization givesso the locus is two distinct planes, $y=\pm \sqrt{-a}x$, meeting along the $z$ axis. These are degenerate quadrics, not ordinary curved cylinders.

With $L=1/\sqrt{n^2-1}$, one has $\nu =n$. Since $n\ge 3>\sqrt{3}$, ii proves boundedness. Write the positive rational number $n=p/q$ in lowest terms. The radius and its angular derivative repeat after $\Delta \theta =2\pi /n$. After $p$ such radial oscillations,so both the position and the radial and transverse velocities return to their initial values. The elapsed time iswhich is finite and positive because $r$ is bounded above and below by positive constants. Thus the motion is periodic. The orbit is a closed, generally self-intersecting rosette orbit, with $p$ radial oscillations during $q$ complete revolutions; for integral $n$ it has $n$ evenly spaced lobes around the origin. It never reaches the origin and is not a circular orbit.

The oscillatory numerator $\cos (\nu \theta )+\nu \sin (\nu \theta )$ has amplitude $\sqrt{1+{\nu }^2}$. Therefore the reciprocal radius stays bounded away from zero exactly whenIn this case the radial bounds areFor completeness, the potential energy per unit mass is $\Phi =-1/(ar)+1/(2r^2)$, and conservation of mechanical energy givesThis effective potential diverges positively at zero, has one minimum at $r=a(1+L^2)$, and tends to zero from below at infinity. Since the initial radial velocity is negative, the particle first reaches an inner turning point. If $L^2\ge 1/2$, the subsequent outward motion has no finite outer turning point and escapes. At equality the escape speed tends to zero, but the radius is still unbounded; infinity is approached only as time tends to infinity. HenceThe minimum of $U$ permits a circular orbit for other initial data, but the nonzero initial radial velocity here excludes that exceptional solution.

The initial transverse velocity gives $h=a(L/a)=L$. From $\dot{r}=-h{u}^{\prime }$, the initial conditions are $u(0)=1/a$ and $u^{\prime }(0)=1/a$. DefineThe Binet equation becomes the constant-coefficient linear ordinary differential equationIts particular solution is $({\nu }^2-1)/(a{\nu }^2)$. Applying the two initial conditions to the complementary solution givesThe physical branch is the connected range containing $\theta =0$ on which the right-hand side is positive. A zero corresponds to escape to infinite radius, rather than continuation to negative $r$.

The source calls $F$ an inward force magnitude, but here $F(r)=(r-a)/(a{r}^3)$ changes sign at $r=a$. The stated initial condition has inward radial velocity, so the particle immediately enters $r<a$, where the force is outward. The consistent interpretation is that $F$ is the signed inward radial component of the force per unit mass.

The rod and disc contribute restoring torques $-mg(\ell /2)\sin \theta$ and $-Mgd\sin \theta$, respectively. Using i in the rotational equation of motion givesOnly the disc's center of mass enters its gravitational torque; its finite radius enters its moment of inertia. The formula is also regular at $M=m$, although the source happens to exclude that value.

The disc is rigidly attached, so its rotation axis is perpendicular to its face. Its moment of inertia about its center is $M{a}^2/2$: with surface mass density $M/(\pi a^2)$, an annulus of radius $s$ contributes $s^2[M/(\pi a^2)]2\pi sds$, whose integral is $M{a}^2/2$. The parallel axis theorem adds $M{d}^2$ at the pivot. Adding the rod's moment of inertia gives

Completing the square puts the equation of the circle in the formHence its centre is $(0,-3)$ and its radius is $2$.

Call the centre of the circle $O$, the left endpoint on the diameter $A$, the point of tangency $T$, and the external intersection $E$. Since $OA=OT$ are radii, the isosceles triangle $AOT$ has equal base angles: $\angle OAT=\angle ATO=30^{\circ }$. Therefore $\angle AOT=120^{\circ }$, and, because $OA$ and $OE$ point in opposite directions, $\angle TOE=60^{\circ }$. The tangent to a circle $TE$ is perpendicular to the radius $OT$, so in the right triangle $OTE$,

The distance from the origin is $r=4\pi -t$. The direction angle, measured counterclockwise from the positive horizontal axis, is $\theta =\pi /2-t$ modulo $2\pi$. Thus the path is an Archimedean spiral, turning clockwise through two complete revolutions as its radius decreases linearly from $4\pi$ to zero.

It starts at $(0,4\pi )$, reaches $(7\pi /2,0)$ at $t=\pi /2$, crosses $(0,-3\pi )$ at $t=\pi$, returns to the positive vertical axis at $(0,2\pi )$ when $t=2\pi$, and ends at the origin when $t=4\pi$. These crossings fix both the scale and direction of the sketch.