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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.

Let $d$ be an impactor diameter and assume it is much smaller than target diameter $D$, with equal material mass density. Neglect gravitational focusing and take the encounter speed to be the stated target-size law $v_{\mathrm{rel}}=v_p{D}^p$. The specific impact energy isA catastrophic planetesimal collision requires $Q\ge Q_D^{\ast }=Q_b{D}^b$. Therefore the catastrophic projectile diameter isUse $\max (D_{\mathrm{cc}},D_{\min })$ as the actual lower size cutoff when integrating a rate. If this exceeds $D_{\max }$, the specified population contains no destructive impactors. The small-impactor approximation itself requires $D_{\mathrm{cc}}\ll D$.

Assume spherical bodies of common mass density $\rho$, a well-mixed belt, and that $n(D)dD$ counts bodies in the entire belt rather than per unit volume. A body of diameter $D$ has mass $\pi \rho D^3/6$. The mass normalization of a power-law size distribution isConsequentlyThe thin-torus volume is circumference times radial width times full vertical height,The local number-density distribution needed for a planetesimal collision rate is $n(D)/V$. Thus belt geometry does not enter this total-number $K$ but does enter collision rates. This approximation treats the stated height as uniform, independent of size.

Let $\omega =-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$ be the Lane-Emden surface mass constant. The Lane-Emden mass formula givesThus $R\propto {\rho }_c^{-1/6}$ while $M\propto {\rho }_c^{1/2}$ at fixed composition. Eliminating central mass density gives the polytropic mass-radius relationMore explicitly,Increasing the mass of a nonrelativistic white dwarf decreases its radius. This scaling ceases to apply when relativistic electron degeneracy pressure becomes important.

For $P=K{\rho }^2$, hydrostatic equilibrium gives $2K\nabla \rho =-\nabla \Phi$. The Poisson equation for Newtonian gravity therefore gives the Helmholtz equationFor spherical symmetry, the regular polytrope of index one solution isThe singular $\cos (kr)/r$ solution is excluded by finite central mass density. Taking the first zero as the stellar surface yieldsThe mass integral isDividing by the volume $4\pi R^3/3$ gives

Let a representative fixed volume have mixture mass density $\bar{\rho }={\rho }_s\phi +{\rho }_l(1-\phi )$. Since the crystal framework is rigid and stationary, the liquid mass flux is ${\rho }_l\mathbf{u}$, where $\mathbf{u}$ is Darcy velocity. Mass conservation givesThe constant phase densities therefore imply the phase-change volume source in a rigid mushFor ice less dense than brine, increasing solid fraction creates an expansion flow; when the phase densities agree, this source vanishes.

Use a common factor $e^{i(kx-\omega t)}$, with $k=k_0>0$ and $\omega ={\omega }_0$. For the upward propagating branches with $0<{\hat{\omega }}_2=\omega -k{U}_2<N_2$, defineThese are the inclinations of upward-sloping constant-phase lines of an internal gravity wave to the horizontal. Write the incident, reflected and transmitted complex velocity amplitudes as $A_0,A_r,A_t$. Incompressibility gives $(u_i,w_i)=A_0(\cos {\theta }_1,\sin {\theta }_1)$, $(u_r,w_r)=A_r(\cos {\theta }_1,-\sin {\theta }_1)$, and $(u_t,w_t)=A_t(\cos {\theta }_2,\sin {\theta }_2)$.

The jump conditions for stratified inviscid shear flow require a common interface displacement, not a common vertical velocity: $w=-i\hat{\omega }\eta$ on each side. Since the background mass density is continuous, pressure is continuous without a hydrostatic jump. The horizontal momentum equation gives $p/{\rho }_0=(\hat{\omega }/k)u$. HenceThese two matching equations determine internal-wave transmission across a velocity jump:The amplitude formula printed in the PDF is inconsistent with these material-interface matching conditions. In particular, identical layers have $A_t/A_0=1$ and $A_r/A_0=0$, whereas its expression labelled “reflected” equals one. The displayed results above distinguish reflection from transmission and retain the intrinsic-frequency factors required by the kinematic boundary condition and pressure balance.

For equal buoyancy frequencies and the specified opposing current, ${\theta }_1=\pi /6$, ${\theta }_2=\arcsin (3/4)$, andThe incident phase lines have slope $1/\sqrt{3}$, the reflected lines slope $-1/\sqrt{3}$, and the transmitted lines slope $3/\sqrt{7}$: they steepen above the interface. For the other specified current, ${\hat{\omega }}_2=0$. The transmitted vertical wavenumber then diverges and there is no regular propagating upper-layer wave of that frequency: this is the critical level of an internal gravity wave limit. Approaching it from ${\hat{\omega }}_2>0$ gives $A_r/A_0\to -1$ and vanishing transmitted vertical energy flux; setting the intrinsic frequency to zero directly is outside the regular matching calculation.

For a dilute particle suspension, take $W_s>0$ as the downward speed relative to the carrier fluid. The particle deposition flux through a horizontal absorbing bed is $W_s{\varphi }_b$, where ${\varphi }_b$ is the near-bed particle volume fraction. Without stirring, settling can create a particle-free upper layer and a descending concentration front, so uniform concentration throughout the remaining fluid is not automatic. Weak turbulence can maintain a nearly uniform suspended concentration while leaving the mean settling velocity approximately equal to the isolated-particle value; this justifies using $W_s\varphi$ in a mixed-layer balance. Strong stirring can produce resuspension and requires a separate erosion model.

For a bed inclined at angle $\vartheta$ to the horizontal, the normal settling speed is $W_s\cos \vartheta$, giving deposition flux $W_s{\varphi }_b\cos \vartheta$ per unit actual bed area. There is also downslope particle motion, and near-wall effects or resuspension can alter this simplest absorbing-boundary model. Integrating the normal projection over a sloping bed is equivalent to integrating $W_s{\varphi }_b$ over its horizontal projection.

The mixture mass density is $\rho ={\rho }_0+({\rho }_p-{\rho }_0)\varphi$, so its reduced gravity isThe Boussinesq approximation requires $({\rho }_p-{\rho }_0)\varphi /{\rho }_0\ll 1$, in addition to the dilution assumption.

Put $z=x_3$. The body is a solid of revolution with disk radiusIts volume and uniform mass density are

By the continuous center of mass formula, the distance of the center of mass from the pointed origin is

For a disk of radius $R$, $\int r^{2}dA=\pi R^4/2$. Therefore the axial moment isFor the transverse $x_1$-axis, $\int x_2^{2}dA=\pi R^4/4$, while every slice also contributes $z^2$ times its mass. HenceBy axial symmetry $I_2=I_1$. These values agree with the moment of inertia of a pointed cubic-profile solid of revolution.

Writing $x_i=(x_i,y_i,z_i)$, rotation about the $z$-axis gives particle $i$ the speed $\omega \sqrt{x_i^2+y_i^2}$. Its kinetic energy is therefore $m_i{\omega }^2(x_i^2+y_i^2)/2$, soThis is the defining mass sum for the moment of inertia about the $z$-axis.

For the uniform cuboid, the mass density is $\rho =M/(8abc)$ and

The Schechter function has logarithmic slope $\alpha \simeq -1$ for $M\ll M_{\ast }$ and an exponential cutoff for $M\gg M_{\ast }$. Thus a log-log plot is approximately a straight line of slope $-1$ at low mass and bends sharply downward near $M_{\ast }$.

For a collisionless self-gravitating system, define the inertia tensorTwo time derivatives, the collisionless Boltzmann equation, and integration by parts give the tensor virial theoremwhereFor an isolated steady state, or a long-time average of a bounded system, ${\ddot{I}}_{ij}=0$ and surface terms vanish, soTaking the trace gives $2K+W=0$. Therefore

Introduce the gravitational radius by $W=-G{M}^2/R$. Virial equilibrium givesAssume dissipationless accretion, negligible escaping mass, no external work, and final re-virialization. The accreted systems bringWith $\eta =M_A/M_I$ and $ϵ=⟨v_A^2⟩/⟨v_I^2⟩$,Since $M_F=M_I(1+\eta )$ and $E_F=-M_F⟨v_F^2⟩/2$,Using $E=-G{M}^2/(2R)$ also gives

Many minor galaxy mergers have dynamically cold accreted material, $ϵ\ll 1$. If they double the mass, $\eta =1$, then $R_F/R_I\simeq 4$, andThe steep low-mass side of the Schechter function supplies many small satellites. Repeated dry galaxy mergers can therefore add stars mainly at large radius, increasing size much faster than mass and lowering mean density. This explains how compact, dense redshift-two galaxies can evolve into larger present-day elliptical galaxies without requiring comparable in-situ star formation.

For an inviscid fluid of constant mass density $\rho$ with no body force, the Euler equations for an inviscid fluid areThe vector identityand irrotational flow give $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$. Writing $u=\nabla \varphi$ in terms of a velocity potential, we obtainThus the bracket depends only on time:Since adding a function of time to $\varphi$ leaves $u=\nabla \varphi$ unchanged, that function may be chosen to absorb $C(t)$. This is the Unsteady Bernoulli equation.

The physical filament ends when the mass density first reaches zero. If $x_1\simeq 2.4$ is the first positive zero of the Bessel function $J_0$, thenThe first zero must be used because continuing to the next lobe would make the density negative.

The Navier-Stokes equation for constant mass density $\rho$, dynamic viscosity $\mu$, and body force per unit mass $\mathbf{f}$ iswhere the Newtonian fluid stress tensor isTaking the dot product with $\mathbf{u}$ givesIncompressibility impliesand converts the left side into a local time derivative plus the divergence of kinetic-energy flux. The divergence theorem therefore gives the kinetic-energy balance for an incompressible Newtonian fluidThe four terms are respectively outward advective transport of kinetic energy, mechanical power supplied by surface traction, power supplied by the body force, and irreversible viscous dissipation into heat.

Let the belt contain $n(D)dD=C{D}^{-\alpha }dD$ bodies. Its total mass fixesFor equal mass density, an impactor of diameter $D$ supplies specific impact energyThus catastrophic disruption requires $D\ge D_{\mathrm{cc}}=X_c{D}_t$, whereThe impactor number density per diameter is $C{D}^{-\alpha }/V$. Neglecting gravitational focusing, the geometric collision cross-section is $\pi (D_t+D{)}^2/4$, so the exact catastrophic planetesimal collision rate in this model isEquivalently, with $A=\pi C{v}_{\mathrm{rel}}/(4V)$,If $D_{\min }\ll X_c{D}_t\ll D_t\ll D_{\max }$, the steep distribution makes impactors just above $X_c{D}_t$ dominate and their collision cross-section is approximately $\pi D_t^2/4$. HenceThis approximation also assumes a common $v_{\mathrm{rel}}$ and size-independent catastrophic disruption threshold $Q_D^{\ast }$.

The density anomaly of water makes mass density increase with temperature up to $T_m$ and decrease above $T_m$. Since temperature increases with depth in the stated layer, the density profile first increases to its maximum ${\rho }_m$ at the $T=T_m$ isotherm and then decreases toward the warm bottom.

The upper region has denser water below lighter water and hence stable density stratification. Below the density maximum, density decreases downward, so lighter water lies beneath denser water and gives unstable density stratification. The lower region therefore overturns by thermal convection, while the cold upper region remains a stagnant cap. This coexistence is penetrative convection.

The unperturbed temperature is linear in each material. Continuity of the conductive heat flux at $x=h$ givesOn definingthis becomes $T_m-T_h=ϵ(T_h-T_a)$. Therefore

Let ${\rho }_i$ be the ice mass density and $L_f$ its latent heat of fusion per unit mass. The ice is isothermal in this model, so the Stefan condition equates latent-heat production to the heat flux conducted through the water:Consequently the unperturbed lateral solidification speed isThe material parameters are the thermal conductivities $k_w,k_a$, ice density ${\rho }_i$, and specific latent heat $L_f$; the geometric thermal length is $\delta$.

For equal mass density, the specific impact energy isThus the smallest catastrophic impactor has $D_{\mathrm{im}}=X_{c}D$, whereIntegrating the collision rate givesFor $X_c\ll 1$, the last term is the dominant lower-limit contribution, so the catastrophic planetesimal collision rate is

For constant phase mass densities ${\rho }_s,{\rho }_l$, a fixed solid framework has mixture density $\bar{\rho }={\rho }_s\phi +{\rho }_l(1-\phi )$. Mass conservation giveswhere $\mathbf{u}$ is Darcy flux. Equal phase densities give a divergence-free flux. Unequal densities require expansion or contraction flow during freezing or melting.

For a planar adiabatic explosion with energy per unit area $E$ in a uniform medium of mass density ${\rho }_0$, the conserved energy has form $E=I{\rho }_{0}Z{\dot{Z}}^2$, with a finite positive integral $I$ of the similarity profiles. Hence $Z=C(E/{\rho }_0{)}^{1/3}{t}^{2/3}$. If $E$ is shared by two fronts, its fraction assigned to either side is absorbed into $C$.