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The useful high-density core bound from TOV compactness follows without choosing the unknown high-density equation of state. Let $a$ be the radius at which the decreasing energy density first reaches ${\rho }_0$, and put $\mu =m(a)$, $p_0=p({\rho }_0)$. The Tolman–Oppenheimer–Volkoff equation gives $p^{\prime }<0$ away from a nonempty centre; since $dp/d\rho >0$, energy density decreases outward. HenceFor the local stellar compactness, if $z\ge 0$, $1-z+\sqrt{1+z}\le 2$, since its derivative is $-1+(2\sqrt{1+z}{)}^{-1}<0$. Applying the supplied compactness inequality at the core boundary therefore gives $\mu /a<4/9$. Combining the two bounds yieldsThe pressure-dependent form further restricts the allowed core-boundary data:In particular, positivity of its right side gives $6\pi a^2{p}_0<3$, so $a^2<1/(2\pi p_0)$ when $p_0>0$. These restrictions depend only on known quantities, not on the equation of state above ${\rho }_0$.

The intended stellar-mass argument now attaches the known low-density envelope to each admissible pair $(a,\mu )$ with pressure $p_0$. Its Tolman–Oppenheimer–Volkoff equation involves only the known equation of state. For a physical low-density envelope with a uniformly bounded mass on this bounded set of core data, taking the supremum of the envelope-plus-core masses gives an upper mass limit independent of any high-density continuation. Stars with ${\rho }_c\le {\rho }_0$ are also determined entirely by the known equation of state. Unknown dense-core physics cannot evade the displayed core bounds.

A qualification is necessary for a claim about the entire star: the printed monotonicity assumptions alone do not guarantee a finite maximum total mass for an arbitrary known low-density equation of state. An explicit low-density polytropic mass divergence occurs for $p=K{\rho }^{5/4}$ with fixed $K>0$. In the low-density Newtonian limit, this is a stellar polytrope of index $n=4$, whose Lane-Emden equation has a finite first zero. With $\rho ={\rho }_c{\vartheta }^4$ and $r=\ell \xi$, hydrostatic balance fixeswhere ${\mu }_4>0$ is the dimensionless Lane-Emden surface mass constant. Thus $M\to \infty$ as ${\rho }_c\to 0$, although $M/R\propto {\rho }_c^{1/4}\to 0$. The first Lane-Emden zero is simple. The continuous dependence of an ODE solution on parameters therefore preserves this finite surface for sufficiently small relativistic parameter. The relativistic correction is controlled by $p_c/{\rho }_c=K{\rho }_c^{1/4}\to 0$, so the corresponding regular finite-radius Tolman–Oppenheimer–Volkoff equation solutions have the same limiting scalings. They never sample $\rho >{\rho }_0$, and satisfy the local positivity and monotonicity assumptions in their nonvacuum interiors. This shows exactly why a suitable bounded-envelope hypothesis, or an appropriate restriction to physically stable stellar models, must accompany the intended total-mass conclusion. The compactness inequality supplies a bound on the unknown core, not a bound on the mass of every conceivable dilute envelope.

A regular central expansion of the TOV equations requires $m(0)=0$. Once ${\rho }_c$ is chosen, the barotropic stellar equation of state fixes $p_c=p({\rho }_c)$. Smoothness at the centre forces the radial first derivatives of $p$ and $\rho$ to vanish. Expanding the Tolman–Oppenheimer–Volkoff equation givesSince $dp/d\rho >0$, the equation of state is locally invertible, and these coefficients determine the corresponding energy density expansion as well. Integrating the regular solution outward gives the surface at the first zero of the pressure, when such a finite surface exists. The ordinary differential equation uniqueness theorem then fixes the interior and its exterior mass from ${\rho }_c$.

The additive constant ${\Phi }_c$ is a time-coordinate normalization, not a second physical parameter. Matching to the Schwarzschild metric with time normalized at infinity fixes it. Therefore the regular stellar solutions, modulo time rescaling, have one physical parameter: ${\rho }_c$. This statement uses a suitably smooth fixed equation of state and refers to central densities for which the solution actually forms a regular star; not every central energy density need give a finite-radius configuration.

In the vacuum exterior, the Tolman–Oppenheimer–Volkoff equation gives $m^{\prime }=0$, so $m(r)=M$. The remaining metric function obeysThus $e^{2\Phi }=C(1-2M/r)$. A constant rescaling of the time coordinate removes $C$; equivalently, asymptotic flatness fixes $C=1$ when time is normalized at infinity. Consequently the exterior is the Schwarzschild metric:The mass $M=m(R)$ includes the full perfect fluid in general relativity interior. No equation of state beyond the surface enters this vacuum result.

For the Komar integral of a five-dimensional Reissner--Nordstrom metric, use the one-form metrically dual to the stationary Killing vector field: $k^{♭}=-f(r)dt$. Its exterior derivative is $d{k}^{♭}=f^{\prime }(r)dt\wedge dr$. The stated orientation gives volume formBecause the squared norm of $dt\wedge dr$ is $g^{tt}{g}^{rr}=-1$, the defining wedge identity for the Hodge star operator givesThus the Komar integral over the positively oriented angular three-sphere issince the unit three-sphere has volume $2{\pi }^2$. Write $s=r_{+}^2+r_{-}^2$, $q=r_{+}^2{r}_{-}^2$. Then $f=1-s/r^2+q/r^4$ and $r^3{f}^{\prime }=2s-4q/r^2$. With the normalization specified for the Komar mass,The orientation makes this positive. The problem's coefficient $1/(16\pi )$ must be retained: it is not the usual dimension-adjusted five-dimensional ADM-normalized Komar coefficient. With that conventional normalization the ADM mass would instead be $3\pi s/8$.

Let $\mu =M_p/M_{\star }$, $q=\sqrt{1-\beta }$, and restrict first to $0\le \beta <1$. The radiation-pressure coefficient reduces the grain's central gravitational parameter to $G{M}_{\star }{q}^2$, whereas the planet's is approximately $G{M}_{\star }$. Their mean motions obey $n/n_p=q(1+ϵ{)}^{-3/2}$. Setting this ratio to $j/(j+1)$ gives the radiation-pressure shift of a mean-motion resonance:It lies interior to the planet exactly when $q(j+1)<j$, or $j>q/(1-q)$. Equality puts the frequency commensurability at the planet's radius. For $\beta =0$ every such commensurability is exterior.

There is a frame qualification in the printed encounter formula. The circular inertial speed of the grain at $a_p(1+{ϵ}_1)$ is $v_1=q{v}_p(1+{ϵ}_1{)}^{-1/2}$. Subtracting the planet's inertial velocity at conjunction gives the local planet-relative inertial speedThis is the velocity used in the question's two-body gravitational assist calculation. In a frame actually rotating about the star at $n_p$, the grain's tangential velocity is instead $v_1-n_p{a}_p(1+{ϵ}_1)$, with magnitude $v_p[1+{ϵ}_1-q(1-{ϵ}_1/2)]$. In particular its nonradiating shear is $3v_p{ϵ}_1/2$, rather than $v_p{ϵ}_1/2$. Thus the requested expression uses a local translating, inertial encounter convention despite the source's rotating-frame wording. The following derivation consistently uses that convention, and its coefficients are an encounter estimate rather than an exact rotating-frame result.

Take impact parameter $b\simeq a_p{ϵ}_1$ and a small gravitational scattering angle $\theta$. Expanding the supplied hyperbolic scattering relation gives $\theta \simeq 2G{M}_p/(b{v}_{\infty }^2)$. Initially the relative tangential velocity is $-v_{\infty }$; after the deflection it has tangential component $-v_{\infty }\cos \theta$. The relative speed is unchanged in the local hyperbolic Kepler orbit, but the stellar-frame kinetic energy changes. Taking the dot product with the planet's velocity givesThe specific orbital energy of the grain is $-G{M}_{\star }(1-\beta )/(2a)$. Comparing it immediately before and after the encounter at the same position givesConsequentlyThe encounter increases the grain's semi-major axis within this geometry. These steps assume ${ϵ}_1\ll 1$, $\theta \ll 1$, and an orbital kick small enough for the subsequent linearization.

Choose the initial conjunction longitude as zero. The synodic period is $2\pi /(n_p-n)$, and the planet's longitude accumulated by the next conjunction isThese are unwrapped longitudes, counting complete revolutions. The second uses the new mean orbital period while neglecting the encounter's immediate phase offset and the difference between true and mean conjunction. A comparison reduced modulo $2\pi$ could not implement the stated one-turn sensitivity test.

For $\beta =0$, $v_{\infty }\simeq v_p{ϵ}_1/2$, so $\delta ϵ\simeq 32{\mu }^2{ϵ}_1^{-5}$ and ${\lambda }_c\simeq 4\pi /(3ϵ)$. Conjunction-longitude sensitivity to an encounter givesRequiring this to exceed $2\pi$ gives ${ϵ}_1^7<64{\mu }^2/3$. At a high-index first-order mean-motion resonance, ${ϵ}_j=(1+1/j{)}^{2/3}-1\simeq 2/(3j)$, henceThe interpretation of the resonance-overlap encounter estimate is loss of coherent encounter phases: a small kick changes the next synodic encounter time by more than one planetary revolution. Resonant protection by repeatedly meeting at controlled orbital phases can then fail, allowing irregular kicks and diffusion. This sensitivity criterion motivates an unstable region; it is not a proof that every orbit or resonant phase in it must escape.

For a quantitative sketch with radiation, put $h=1-q(1-ϵ/2)$ and $d=1-q(1+ϵ{)}^{-3/2}$. Differentiating the conjunction formula and substituting the kick givesEvaluate this at the positive ${ϵ}_j$ and take $\mathcal{S}=1$ as the critical-index curve. At small $\beta$, increased encounter speed weakens planetary scattering, moving the threshold to larger $j$. As $\beta$ rises further, the radiation-pressure shift of a mean-motion resonance brings each commensurability closer to the planet, strengthening the close encounters. The curve turns over and approaches the exterior-limit boundary $j=q/(1-q)$. Larger planetary mass produces stronger kicks and a lower critical index. These competing effects are shown below; the curves retain the factors in $\mathcal{S}$, so their $\beta =0$ intercepts differ slightly from the leading high-$j$ formula. The boundary for disappearing exterior commensurabilities is geometric; divergence of the encounter approximation near $ϵ=0$ should not be read as a precise physical scattering rate.

Continue with downward coordinate $y=H-z$ and $W=W_s(a)>0$. By Stokes drag law, the small-particle isolated settling velocity is $W/4$. Normalize concentrations by ${\varphi }_{\max }$: write $q_i={\varphi }_i/{\varphi }_{\max }$. In suspension the specified independent hindered settling laws give fluxes $j_1/{\varphi }_{\max }=W{q}_1(1-q_1)$ and $j_2/{\varphi }_{\max }=(W/4)q_2(1-q_2)$. The initial state is $(q_1,q_2)=(1/8,1/8)$.

Two upper sedimentation shocks clear the species separately. The faster front removes large particles, leaving $(0,1/8)$ behind it; the small concentration is unchanged across this front because the species fluxes are independent. The slower front separates that small-only suspension from clear fluid. Their downward velocities areThere is also a growing mixed deposit at the bottom. Let its normalized solid concentrations be $(p_1,p_2)$, with $p_1+p_2=1$, and let its upward speed magnitude be $s=-{\hat{U}}_3>0$. Applying the Rankine-Hugoniot condition to each species givesAdding these relations provesThe deposited flux is zero even though each species occupies only part of the packing: the settling law applies to the suspension, while the deposit is a separate stationary state. These jump relations determine the deposit composition from sedimentation jump conditions.

The large-particle clearing front has height $H-7Wt/8$ and meets the mixed-deposit front at height $35Wt/192$. ThusAt this merger all large particles have entered the deposit. Only the small particles remain suspended, at normalized concentration $1/8$. The new material deposited above the mixed layer is pure small-particle solid, at total concentration ${\varphi }_{\max }$. Its sedimentation shock therefore has downward speedFor $t>t_1$, the deposit top and the remaining clearing front have heightsTheir meeting, or the total solid-volume balance, givesThis is two-stage bidisperse batch sedimentation: the fast species finishes first, followed by deposition of the remaining slow species.

The final solid composition by volume, and also by mass because the particle densities are identical, isThe upper pure-small layer has thickness $9H/116$. As checks, $(29/40)h_1=H/8$ accounts for all large-particle solid volume, and $(11/40)h_1+(h_2-h_1)=H/8$ accounts for all small-particle solid volume, in units of ${\varphi }_{\max }$. If “percentage of particles” means number rather than volume, a large sphere occupies eight times the volume of a small sphere: the mixed bottom layer contains $29/117\simeq 24.8\%$ large particles by number and $88/117\simeq 75.2\%$ small. The upper layer remains entirely small under either convention.

At a fixed height, first determine the stored quantities per unit vertical height and per unit length along the line source. With $\Delta \rho =\rho -{\rho }_0$, the triangular-profile line plume has the useful profile integralThe geometric volume is $\mathcal{A}=2b$; the mass is $\mathcal{D}={\rho }_0{I}_0+\Delta \rho I_1=b(\rho +{\rho }_0)$. The upward momentum is $\mathcal{P}=W({\rho }_0{I}_1+\Delta \rho I_2)$, while the buoyancy force is $\mathcal{B}=-g\Delta \rho I_1$. ThusHere positive buoyancy corresponds to a rising light plume. If instead buoyancy is defined kinematically, divide $\mathcal{B}$ by ${\rho }_0$; the density-weighted convention here is the one matching the printed equations.

Multiply each local density by the vertical velocity to obtain the fluxes through a horizontal section, still per unit source length:In particular, the stored momentum equals the mass flux $Q$, and the buoyancy flux is $F=g({\rho }_{0}V-Q)$. The different triangular-profile factors in storage and flux must not be replaced by top-hat coefficients.

Apply volume conservation, mass conservation and vertical momentum flux balance to a slice of the triangular-profile line plume. Ambient ingestion contributes $2e$ of volume and $2{\rho }_{0}e$ of mass per unit height and source length. The ambient is quiescent, so it supplies no leading vertical momentum. The integrated driving force is the stored buoyancy $\mathcal{B}$. Using the quantities from part (a), the non-Boussinesq triangular-profile line-plume balances areTogether with Batchelor entrainment, these are three equations for $W,\rho ,b$. They neglect viscous boundary stresses and streamwise pressure-force corrections within the integral-plume approximation.

Multiplying volume conservation by $g{\rho }_0$ and subtracting $g$ times mass conservation cancels the ambient sources exactly, givingThis is buoyancy conservation in the homogeneous ambient; it is a consequence of the first two balances, not an additional independent equation.

Now take the Boussinesq approximation in inertia and entrainment, retaining the small density deficit in buoyancy. Then $Q={\rho }_{0}bW$, $M=(2/3){\rho }_{0}b{W}^2$ and $F=(2/3)gbW({\rho }_0-\rho )$. ConsequentlyThe mass source is $2\alpha {\rho }_{0}W=3\alpha {\rho }_{0}M/Q$. Substitution into the mass conservation, momentum and buoyancy balances gives the unsteady Boussinesq triangular-profile line-plume balancesThese use density-weighted mass and momentum fluxes. Removing ${\rho }_0$ from some flux definitions while retaining it in the source would mix incompatible conventions.

For a smooth test function $\phi$, integration by parts gives the moment identityThe assumed decay removes all boundary terms. Apply it to $1$, $v_i$ and $|v{|}^2/2$. Writing $m_i=\int v_{i}f=M{u}_i$ gives the Ornstein-Uhlenbeck moment equationsAssume $M(0)\ne 0$ when using the normalized mean $u$; otherwise $u$ is undefined, while the unnormalized momentum equation still holds. Since $M$ is constant, $u_i^{\prime }=-u_i$. Their solutions areThus mass is conserved, the mean velocity tends to zero, and $E(t)\to d{M}_0/2$. This includes the mean velocity conclusion omitted from the converted TeX.

For positive smooth $f$ with the stated decay, integration by parts givesand, using unit mass,For zero values of $f$, these calculations can be made with the positive unit-mass approximation $(f+\epsilon \gamma )/(1+\epsilon )$ and then passed to the limit whenever the quantities are finite. Positive-time solutions also have the usual Gaussian smoothing.

Since $-\log \gamma =v^2/2+\frac{1}{2}\log (2\pi )$, mass conservation givesThe derivative of the first term is $\int {\partial }_t{f}_t\log f_t$, because $\int {\partial }_t{f}_t=0$. The two identities above and the energy equation in (b), with $d=M=1$, yieldNow expand the relative Fisher information:where $\int v{f}^{\prime }=-1$. Thus the entropy dissipation identity for Ornstein-Uhlenbeck flow is

For unit mass, the inverse-square force has potential energy $-k/r$, with zero potential at infinity. Conservation of angular momentum gives $r^2\dot{\theta }=h$. Differentiate the polar equation of the Kepler orbit:Consequently the total mechanical energy iswhere $k=h^2/\ell$. This also explains the sign distinction between an elliptic orbit, a parabolic Kepler orbit, and a hyperbolic Kepler orbit.

At the original periapsis, the parabolic Kepler orbit has $e=1$, $r=d=\ell /2$, and purely tangential speed $V$. ThusThe outward radial impulse has zero moment about the Sun, so it preserves angular momentum and hence $\ell$. Its additional velocity is perpendicular to the original velocity, so the new mechanical energy is $E^{\prime }={\alpha }^2{V}^2/2$. Applying the energy formula givesFor $\alpha >0$, the new orbit is a hyperbolic Kepler orbit; $\alpha =0$ leaves the parabola unchanged.

To fix the orientation of the sketch, put the Sun at the origin, the impulse point at $(d,0)$, and take the original motion there upwards. The new velocity is $(\alpha V,V)$. The eccentricity vector isIt points towards the new periapsis, which is rotated clockwise through $\beta =\arctan (2\alpha )$ from the original periapsis direction. Thus, measured using the original polar angle $\theta$, the new orbit isIn particular, the impulse point is already on the outgoing part of the new orbit and is not its new periapsis. The old parabola satisfies $x=d-y^2/(4d)$. Both orbits pass through $(d,0)$ and have the same focus, but the hyperbola's new periapsis axis is tilted.

Use Newton's second law, $F=ma$. Acceleration has physical dimension $L{T}^{-2}$, so multiplying by mass gives the dimension of forceThe bracket notation records a physical dimension, rather than the numerical value or the choice of units.

From Newton's law of universal gravitation, $F=G{m}_1{m}_2/r^2$, the gravitational constant has physical dimensionHere the two factors of mass and the inverse square of the separation determine the powers of $M$ and $L$.

Work in one fixed inertial frame, and take positive velocity in the rocket's direction. During $d\tau$, the rocket ejects the positive amount of mass $-m^{\prime }(\tau )d\tau$. Its exhaust has inertial velocity $v(\tau )-u$, by the Galilean transformation of velocities. Once expelled, that material retains its velocity because it experiences no external force. The system's momentum is therefore the rocket momentum plus the sum of all these exhaust momenta:The minus sign comes from the positive expelled mass being $-m^{\prime }(\tau )d\tau$.

Differentiate, applying the product rule to the rocket momentum and the fundamental theorem of calculus to the integral:The momentum conservation law thus gives the nonrelativistic rocket equation

The same bookkeeping for kinetic energy givesIts derivative iswhere the rocket equation cancels the first two terms. Thus kinetic energy increases during mass ejection, with $\boxed{K^{\prime }=-u^2{m}^{\prime }/2>0}$ when $u>0$ and $m^{\prime }<0$. This does not violate conservation of energy: the increase comes from stored internal energy.