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A disrupted binary star with positive post-impulse specific orbital energy ${\epsilon }^{\prime }$ has relative speed $V=\sqrt{2{\epsilon }^{\prime }}$ at infinite separation, because the Newtonian gravitational potential energy tends to zero. With the notation of the kick-direction binary survival criterion, $V=v\sqrt{1+2\alpha \cos \psi +{\alpha }^2-2\beta }$. This is not a systemic recoil velocity. A marginal parabolic Kepler orbit has $V=0$ at infinity.

An external pressure changes the balance even if the system remains confined. For a material boundary acted on by spatially uniform inward pressure $P_s$, the surface force contributes $-P_s{\int }_{\partial \mathcal{V}}\mathbf{r}\cdot \mathbf{n}dS=-3P_s\mathcal{V}$, by the divergence theorem. For Newtonian gravitational potential energy, the finite-window identity becomesHere $\mathcal{V}$ is the enclosed volume, $\mathbf{n}$ the outward unit normal, and $K$ includes the microscopic translational kinetic energy of the confined particles. Do not also add the same random motions as a separate internal pressure term. A nonuniform external stress needs its surface integral; a fixed observational boundary crossed by matter also needs flux terms.^[ref]

With negligible surface pressure, the negative heat capacity of a virialized gravitational system follows from $E=-⟨K⟩$: after a slow loss of total energy and readjustment to virial equilibrium, the average kinetic energy rises. For fixed $N$ classical particles with three translational degrees of freedom, define kinetic temperature $\Theta$ by $⟨K⟩=3N{k}_B\Theta /2$. Along this isolated equilibrium sequence, $dE/d\Theta =-3N{k}_B/2$. The system contracts and its binding potential energy becomes more negative, supplying both the outgoing energy and the increased kinetic energy. This is an endpoint comparison, not a claim that every transient contraction obeys the average relation. A collisionless stellar system need not have a common thermodynamic temperature; a pressure-confined system instead has $E=-⟨K⟩+3⟨P_s\mathcal{V}⟩$. A stable canonical ensemble in contact with a large thermal reservoir does not support this negative heat capacity branch.^[ref]

The virial theorem balances a relaxed galaxy cluster's orbital kinetic energy, of order $M{\sigma }_v^2$, against its Newtonian gravitational potential energy, of order $-G{M}^2/R$. Their ratio gives the displayed mass scale. The numerical coefficient depends on the density profile, radius definition and conversion from line-of-sight velocity dispersion to three-dimensional orbital motion. Membership contamination, anisotropy and nonequilibrium dynamics can bias the inference.

The escape velocity is the least initial speed that permits a freely coasting body to move arbitrarily far from an isolated attracting body. Despite its conventional name, it is a speed, not a unique velocity vector. For a fixed point mass $M$, or outside a body with spherical symmetry, set Newtonian gravitational potential energy zero at infinity. An exterior test particle of constant mass $m\ll M$ then hasHere $\mu =GM$ is the standard gravitational parameter, and $r$ is distance from the centre, not height above the surface. Set $r=R$ for surface escape velocity. Evaluate velocity in the body's rest inertial frame. Neglect drag, later rocket thrust, other bodies' gravity, and relativistic effects.^[ref]

Conservation of energy gives the threshold by requiring $E\ge 0$: at infinity the potential energy tends to zero and the kinetic energy cannot be negative. If $E=0$, the outward motion loses speed continuously toward zero. If $E>0$, the asymptotic speed is $v_{\infty }=\sqrt{2E/m}=\sqrt{v_0^2-v_{\mathrm{esc}}(r_0{)}^2}$. If $E<0$, infinity is inaccessible. For a radial outward launch, its outer turning point is $r_{\max }=-GMm/E$. For nonradial motion use angular momentum and the escape energy threshold to distinguish zero radial velocity from zero total speed.

Direction still determines the orbit and whether the body hits an obstacle. An initially inward velocity can cause collision even when $E\ge 0$. For an isolated spherical body, an exterior outward launch, or a tangential launch at or above escape velocity, has an outward escape branch without intersecting the body. A parabolic trajectory is the nonradial threshold path; a hyperbolic Kepler orbit has positive energy. The formula specifies their required local speed, not their orientation.^[ref]

When two masses occupy spatial superpositions, their branch-dependent Newtonian gravitational potential energy produces relative phases. If those phases cannot be separated into one phase for each mass, the interaction turns an initial product state into an entangled state.

For a self-gravitating system of mass $M$ and Newtonian gravitational potential energy $W<0$, its gravitational radius is defined byIt converts the total binding energy into a characteristic size without assuming a particular density profile.

A mass-preserving stellar homology with scale $x$ has mass density proportional to $x^{-3}$. For an adiabatic process with constant adiabatic index $\gamma$, its internal energy scales as $x^{-3(\gamma -1)}$ and its Newtonian gravitational potential energy as $x^{-1}$. Writing $a=3(\gamma -1)$, stationarity at $x=1$ requires $W_0=-a{U}_0$, and the second scale derivative is $a(a-1)U_0$. Thus this particular displacement is restoring for $\gamma >4/3$, destabilizing for $1<\gamma <4/3$, and marginal at $\gamma =4/3$. This is a test of homologous radial stability; it does not by itself test every stellar displacement.

Take an initially isolated virialized self-gravitating stellar population with kinetic energy $T=|W|/2$ and Newtonian gravitational potential energy $W<0$. Instantaneous spatially uniform removal of mass, without kicks to surviving material, leaves fraction $f$ with $T^{\prime }=fT$ and $W^{\prime }=f^{2}W$. Hence $E^{\prime }=|W|f(1/2-f)$: the homologous whole-system energy is negative only for $f>1/2$. Positive whole-system energy does not forbid a bound core after selective escape, and slow mass loss follows the adiabatic mass-loss expansion law instead.

Here inverse-square potential means the potential energy associated with an attractive inverse-square law for a central force: the potential energy itself varies as inverse distance. For $k>0$ and position $\mathbf{r}$ with $r=|\mathbf{r}|>0$,The gradient makes the distinction explicit: doubling $r$ halves the magnitude of potential energy but quarters the magnitude of force. The neighboring repulsive inverse-square potential instead has potential energy proportional to $r^{-2}$ and an inverse-cube force; the two naming conventions must be distinguished.

For Newton's law of universal gravitation, $k=GMm$, where $G$ is the gravitational constant. Treat the attracting mass $M$ as an isolated fixed point mass, or as a body with spherical symmetry and exterior test particle of constant mass $m\ll M$. The spherical shell theorem gives the same exterior Newtonian gravitational potential energy $U=-GMm/r$ without requiring uniform mass density. Here $r$ is measured from the centre, and the formula applies outside the body; it does not describe its interior. The Newtonian gravitational potential per unit mass is $\Phi =U/m=-GM/r$.^[ref]

The negative sign records binding relative to infinite separation, rather than a negative kinetic energy. Moving from $r_1$ to $r_2>r_1$ increases potential energy byBecause the force is a conservative force, this work depends on the endpoints, including for a curved path. Integrating $dV/dr=k/r^2$ with zero at infinity gives $V=-k/r$; adding a constant leaves the force unchanged but shifts the numerical escape energy threshold by that same constant. This is why the reference at infinity must accompany an escape velocity calculation.

For an exact synthetic comparison, stipulate the standard gravitational parameter $\mu =GM=8{\mathrm{m}}^3/{\mathrm{s}}^2$ and a $1\mathrm{kg}$ test particle. At $r=2\mathrm{m}$, $U=-4\mathrm{J}$ and the inward force has magnitude $2\mathrm{N}$; at $r=4\mathrm{m}$, $U=-2\mathrm{J}$ and the inward force has magnitude $0.5\mathrm{N}$. The increase is $2\mathrm{J}$ of potential energy. These stipulated parameters are an arithmetic example, not observations of a particular body. The escape-energy balance diagram uses the same parameters.

A magnetic buoyancy instability is a gravitational instability of a magnetized stratified fluid in which displacements of the gas and magnetic field can release Newtonian gravitational potential energy. Magnetic pressure supports part of the weight. Both interchange displacements and undular displacements can enter this broad category; Parker instability is the undular mechanism involving drainage along bent magnetic field lines.

The Newtonian gravitational potential $\Phi$ is potential energy per unit mass of a test particle. Its gradient determines the Newtonian gravitational field:In SI, $U$ has units $\mathrm{J}$, while $\Phi$ has units $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$; its gradient has units of acceleration, not force. Adding a constant to $\Phi$ leaves the Newtonian gravitational field unchanged. For an isolated point mass, the convenient reference is $\Phi (\infty )=0$. ^[ref]

Increasing the radius from $r_0$ to $r_1>r_0$ raises the Newtonian gravitational potential energy:An external agent supplies this positive work if the initial and final kinetic energies agree; gravity does work $-\Delta U$. For $r_1=2r_0$, the required work is $GMm/(2r_0)$. ^[ref]

The Newtonian potential of a point mass $M$ at the origin isIt is the free-space Green function of the Laplacian with the normalization of the Poisson equation for Newtonian gravity. Differentiating gives inward gravitational acceleration $-GM/r^2$. With $z=r/r_0$, the normalized Newtonian gravitational potential is $\Phi /(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.

For a radially outward launch from $r_0$, conservation of mechanical energy in the fixed-source test particle approximation gives the minimum escape velocity:At this threshold the speed tends to zero only at infinite separation. The model neglects air resistance, external tidal forces, and general relativity. ^[ref]

A circular orbit at $r_0$ instead requires tangential speed $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt{2}\simeq 1.41421$. A radial launch at speed $1$ reaches $r=2$ and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

Use conservation of energy for an isolated spherical perturbation of fixed mass $M$, with negligible initial random motion and no significant surface-pressure term. For a homologous density profile its Newtonian gravitational potential energy is $U=-\alpha G{M}^2/R$, with the same positive structure coefficient $\alpha$ at both stages; a uniform sphere has $\alpha =3/5$.

At spherical-collapse turnaround, coherent radial motion momentarily vanishes, so $E=U_{\max }=-\alpha G{M}^2/R_{\max }$. In the final virialized state, the virial theorem gives $2T_{\mathrm{vir}}+U_{\mathrm{vir}}=0$, hence $E=T_{\mathrm{vir}}+U_{\mathrm{vir}}=U_{\mathrm{vir}}/2$. ThereforeThis is the virial radius from turnaround energy in the spherical-collapse model. It is an idealized virial-radius estimate rather than the radius of the formal pressureless point collapse. If the profile coefficient changes, the same argument instead gives $R_{\mathrm{vir}}/R_{\max }={\alpha }_{\mathrm{vir}}/(2{\alpha }_{\max })$. Significant energy loss, mass loss or boundary work also changes the result.

For an ideal gas with constant adiabatic index $\gamma >1$, let $c_v$ and $c_p$ be specific heat capacities and $\mathcal{R}=c_p-c_v$. The ideal gas law and the internal energy of an ideal gas give $P=\rho \mathcal{R}T$ and specific internal energy $e=c_{v}T$. Since $\gamma =c_p/c_v$, the internal energy per unit volume is $\rho e=P/(\gamma -1)$. Consequently a star with spherical symmetry hasThe radial integral requires the volume measure $dV=4\pi r^{2}dr$: the first printed expression omits this factor if its differential is read literally as $dr$. The mass-coordinate expression confirms the required volume interpretation.

Define the enclosed mass $M(r)$, total mass $M_{\mathrm{tot}}$ and dimensionless shell label $m=M(r)/M_{\mathrm{tot}}$. Mass conservation gives $M_{\mathrm{tot}}dm=dM=\rho dV$, so the change of variables givesHere $m$ labels a fixed material shell rather than its radial position. In a mass-preserving stellar homology, $r(m)=x{r}_0(m)$ with $x=R/R_0$. Both a shell's radius and thickness gain a factor $x$, so its volume gains $x^3$ and its mass density becomes $\rho (m)=x^{-3}{\rho }_0(m)$. An adiabatic process keeps $P/{\rho }^{\gamma }$ fixed separately in each shell, even if that constant differs between shells. Thus $P=x^{-3\gamma }{P}_0$ and $P/\rho =x^{-3(\gamma -1)}{P}_0/{\rho }_0$. Integration at fixed mass labels provesThe spherical shell theorem gives the Newtonian gravitational potential energy by assembling shells against the mass already inside them:Every radius gains the same factor $x$, while $M$ and $dM$ remain fixed. HenceThis is the homologous adiabatic stellar stability test. The derivative with respect to the physical stellar radius is ${\partial }_R=R_0^{-1}{\partial }_x$. Stationarity at $x=1$ therefore requires $-a{U}_0-W_0=0$, orAfter imposing this equilibrium relation, $E/U_0=x^{-a}-a/x$ andSince $a>0$, a strict energy minimum under this homologous displacement requiresFor $a>1$, contraction sends $E$ to $+\infty$, expansion sends it to $0$ from below, and $x=1$ is the unique minimum, with $E(1)=(1-a)U_0<0$. For $0<a<1$, contraction sends $E$ to $-\infty$, expansion sends it to $0$ from above, and $x=1$ is the unique maximum. When $a=1$, the equilibrium relation makes $E(x)=0$ for every $x$: the scale direction is marginal, not strictly stable. The plotted curves display these three possibilities. This checks homologous radial stability; stability against every stellar displacement needs a fuller perturbation analysis.

The regional mass density is $3H_0^2{\Omega }_m/(8\pi G{a}^3)$, while the Einstein-de Sitter universe has ${\rho }_{\mathrm{bg}}=1/(6\pi G{t}^2)$. Substituting the parametric spherical collapse from part (a) yieldsAt spherical-collapse turnaround, $\theta =\pi$, and henceFor an isolated uniform sphere, the Newtonian gravitational potential energy is $U=-3G{M}^2/(5R)$. At spherical-collapse turnaround the coherent kinetic energy is zero, so conservation of energy fixes $E=U_m$. At virial equilibrium, the virial theorem gives $2T_v+U_v=0$, hence $E=U_v/2$. Equating these energies proves the virial radius from turnaround energy:This uses the same mass and gravitational structure coefficient before and after collapse and neglects boundary pressure, mass loss and external energy exchange.

If virial equilibrium is assigned to $2t_m$, the sphere's density is eight times its turnaround density, whereas the background mass density has decreased by four. ConsequentlyThe often quoted value $18{\pi }^2$ is a density ratio; the density contrast defined as $(\rho -{\rho }_{\mathrm{bg}})/{\rho }_{\mathrm{bg}}$ is smaller by one. This finite virialized estimate and the infinite cold-collapse contrast in part (b) describe different physical endpoints.

First specify the dynamical estimate: assume an initially isolated virialized system, stars of different masses well mixed with the same velocity distribution, and ejecta escaping rapidly compared with a stellar crossing time, without imparting kicks to surviving stars. If $f$ is the retained stellar mass fraction, the original virial theorem gives $2T+W=0$, where $T$ is kinetic energy and $W<0$ is Newtonian gravitational potential energy. Immediately after loss, positions and velocities are unchanged, soThus impulsive disruption by stellar mass loss makes the energy of the whole remaining system nonnegative when at least half the initial mass is removed.

For the power-law initial mass function, the fraction removed is a mass fraction, not a number fraction:For integer slopes, $F(0)\simeq 0.990$, $F(1)\simeq 0.901$, $F(2)=0.333$, and $F(3)\simeq 0.009$. Interpolating between $\alpha =1$ and $2$ puts the threshold around $1.7$--$1.8$. Solving $F=1/2$ more accurately givesThe rearranged equation has a spurious root at $\alpha =2$ introduced when the vanishing denominator was cleared; the logarithmic limit $F(2)=1/3$ excludes it. Increasing $\alpha$ shifts the mass distribution towards lower masses, so the top-heavy side, $\alpha <{\alpha }_{\mathrm{crit}}$, has nonnegative remnant energy in the impulsive model; equality is marginal. One can see strict monotonicity from the covariance obtained by differentiating $F$: $F^{\prime }=-\mathrm{Cov}(1_{x>10},\ln x)<0$ under the normalized mass-weighted distribution.

The initial mass function alone does not guarantee complete disruption. If losses occur slowly, the adiabatic mass-loss expansion law gives expansion rather than the instantaneous half-mass threshold, and positive total energy can still leave a bound core after selective escape. The stated value is therefore the conventional global impulsive estimate, with no external confining Newtonian gravitational potential.

In the spherical-collapse model, a bound overdense region initially expands with the background, but its additional Newtonian gravity slows that expansion. Its outer shell reaches a maximum turnaround radius, where its radial velocity is zero, and then falls inward. A perfectly spherical pressureless solution formally collapses to zero radius. In a realistic collisionless system, deviations from that idealization, shell crossing and violent relaxation convert coherent infall into random motions, producing a roughly stationary bound object.

Let the Newtonian gravitational potential energy of the homologous sphere be $U=-aG{M}^2/R$, where $a=3/5$ for a uniform sphere. At spherical-collapse turnaround, the kinetic energy of the bulk motion vanishes, so $E=U_{\mathrm{ta}}$. For the settled object, the virial theorem gives $2K_{\mathrm{vir}}+U_{\mathrm{vir}}=0$ and hence $E=U_{\mathrm{vir}}/2$. If energy and mass are conserved and the structure coefficient is unchanged,This is the virial radius from turnaround energy. It assumes negligible external pressure, mass loss and energy exchange. If the final structure coefficient differs, $R_{\mathrm{vir}}=(a_{\mathrm{vir}}/2a_{\mathrm{ta}})R_{\mathrm{ta}}$ instead. Half the turnaround radius is an idealized equilibrium size, not the radius at which each collisionless particle stops moving. The plotted settling time is schematic; the usual collapse-time convention is $t_{\mathrm{coll}}=2t_{\mathrm{ta}}$.

Let $X$ be the hydrogen mass fraction, $f_g={\rho }_g/{\rho }_{\mathrm{tot}}$ the gravitating gas fraction, and $\chi =n_{\mathrm{tot}}/n_H$ the total-particle-to-hydrogen ratio. The optically thin gas cooling time and free-fall time of a uniform sphere arewhere ${\rho }_{\mathrm{tot}}=m_p{n}_H/(X{f}_g)$. Equating them gives the cooling-to-free-fall equality curve:At fixed composition, its shape is $n_{\mathrm{crit}}\propto T^2/{\Lambda }^2$. Gas above this curve cools faster than it falls; gas below it cools more slowly. Strong line cooling produces a deep low-density trough. In the high-temperature free-free regime, $\Lambda \propto T^{1/2}$ gives $n_{\mathrm{crit}}\propto T$.

The displayed uniform self-gravitating-cloud example uses $f_g=1$, $X=0.76$, $\mu \simeq 0.59$ and $\chi \simeq 2.24$. Holding these ionized-gas factors fixed makes the diagram transparent, but near the atomic threshold their actual temperature dependence changes numerical normalizations. With these conventions the supplied cooling values imply $n_{\mathrm{crit}}(1.5\times 10^4\mathrm{K})\simeq 9.6\times 10^{-10}\mathrm{c}{\mathrm{m}}^{-3}$ and $n_{\mathrm{crit}}(10^7\mathrm{K})\simeq 0.107\mathrm{c}{\mathrm{m}}^{-3}$. For gas supported in a dark-matter-dominated dark matter halo, take $f_g<1$: at fixed $n_H$ the free-fall time of a uniform sphere is shorter and the critical curve shifts upward by $1/f_g$.

To add virial mass contours in a cooling diagram, use the uniform self-gravitating sphere consistently. Its Newtonian gravitational potential energy is $-3G{M}^2/(5R)$ and its thermal kinetic energy is $(3/2)M{k}_{B}T/(\mu m_p)$. The virial theorem gives $k_B{T}_{\mathrm{vir}}=\mu m_{p}GM/(5R)$, while $M=(4\pi /3)R^3{\rho }_{\mathrm{tot}}$. Eliminating $R$ givesThus constant total gravitating mass gives slope three in the logarithmic density-temperature plane, with larger mass contours lying at smaller density. A different density profile changes the structural coefficient; the customary halo temperature scale $k_B{T}_{\mathrm{vir}}=\mu m_{p}GM/(2R)$ shifts the mass contours but preserves their slopes. The horizontal illustrative formation-density lines are $n_{H,\mathrm{vir}}\simeq {\Delta }_v(2\times 10^{-7})(1+z{)}^3\mathrm{c}{\mathrm{m}}^{-3}$, using ${\Delta }_v=18{\pi }^2$. They show why increasing formation cosmological redshift helps gas reach the rapid-cooling region.

At the low-mass end, a dark matter halo's virial temperature may not reach the atomic threshold. The atomic and molecular cooling thresholds for galaxy formation therefore favor an atomic-cooling mass of order $10^8{M}_{\odot }$ at early formation epochs, with its precise value varying as approximately $(1+z{)}^{-3/2}$. Smaller primordial systems require molecules; photoheating and stellar feedback further suppress their luminous output.

At the high-mass end, the virial temperature at a fixed assembly density scales as $M^{2/3}$. Beyond the line-cooling interval the gas becomes increasingly hot, while thermal bremsstrahlung improves only as $T^{1/2}$. Hence at fixed assembly density $t_{\mathrm{cool}}/t_{\mathrm{ff}}\propto T^{1/2}\propto M^{1/3}$ in that regime. Large systems can remain pressure-supported hot atmospheres for many dynamical times, instead of promptly fragmenting into one stellar object. This is the upper galaxy mass from gas cooling: it associates efficient condensation with roughly galactic, rather than group or cluster, masses and motivates the broad $10^8$–$10^{12}{M}_{\odot }$ interval.

The diagram gives characteristic cooling scales, not exact universal mass bounds. Density profiles, gas fraction, enrichment and formation epoch move the boundaries; the two supplied values alone do not determine a complete numerical cutoff. Some gas can cool in dense cores of larger dark-matter haloes, early smaller progenitors can form stars before merging, and subsequent galaxy mergers can assemble more massive galaxies. Long-lived hot atmospheres also require heating to prevent excessive late-time cooling. These qualifications explain why the cooling criterion for galaxy formation is useful without equating each dark dark matter halo to one galaxy.

Use the paper's positive binding-energy convention $U>0$; the physical Newtonian gravitational potential energy is $-U$. Assume positive masses, an isolated system, and no collision during the time interval under consideration. Differentiate half the scalar moment of inertia, $I$, twice:Group the force term into unordered pairs. The force on particle $j$ from $k$ is $-G{m}_j{m}_k({\mathbf{r}}_j-{\mathbf{r}}_k)/r_{jk}^3$; the two contributions to the virial sum are thereforeAdding them proves the instantaneous virial theorem identityNo time average is needed for this form. A vanishing average of $\ddot{I}$ would require additional boundedness assumptions.

For the pairwise moment-of-inertia identity, expand the squared pair distances:The last term vanishes in the centre of mass frame, givingLet $C={\sum }_{j<k}{m}_j{m}_k$, and label the two smallest masses $m_{(1)},m_{(2)}$. If $r=\min r_{jk}$, every term in $U$ is at most $G{m}_j{m}_k/r$, so $U\le GC/r$. One pair attains $r$, and its mass product is at least $m_{(1)}{m}_{(2)}$, so $U\ge G{m}_{(1)}{m}_{(2)}/r$. HenceThese constants show the minimum-separation binding-energy bounds: inverse binding energy is comparable to the nearest-pair distance, independently of configuration.

Similarly, every separation is at most $R=\max r_{jk}$ and at least one pair attains $R$. The pairwise identity gives the maximum-separation inertia boundsBoth lower bounds rely on actual extremal separations, rather than an arbitrary numerical lower or upper estimate for all the pair distances.

Negative energy does not imply a positive lower bound on the minimum separation. The printed request is false if it is meant to exclude close approaches or collisions; the trivial bound $r\ge 0$ says nothing of that kind. In fact $E<0$ and $T\ge 0$ give $U=T-E\ge |E|$, which, combined with $rU\le A_0$, proves the negative-energy minimum-separation upper boundThis ensures at least one close pair, not confinement of every particle or prevention of collision.

An explicit negative-energy gravitational collision is a pair initially at rest with separation $a$. Its total energy is $E=-G{m}_1{m}_2/a<0$. For reduced mass $\mu =m_1{m}_2/(m_1+m_2)$, its relative radial equation isThe separation decreases to zero in the finite timeBefore that time the motion is a regular Newtonian solution, yet its separation has no positive infimum. Even at fixed negative energy, bound Kepler orbits with eccentricity approaching one have arbitrarily small pericentre distance. Thus a missing angular momentum or collision-exclusion hypothesis cannot be supplied by the energy sign alone.

For $E>0$, the instantaneous virial theorem gives $\ddot{I}=T+E\ge E$. Integrating twice from any regular time $t_0$ givesUsing $I\le A_1{R}^2$, obtain the positive-energy linear diameter growth boundwhenever the numerator is nonnegative. Consequently, for a solution existing for arbitrarily large future times,This is the precise at-least-linear expansion statement. It does not assert that $R$ is monotone at every instant, nor that all individual particles escape. The large-time conclusion presupposes continued existence of the trajectory; the inequality itself holds on every nonsingular time interval.

Assemble the star from concentric shells. A shell of mass $dm$ at radius $r$ has interaction energy $-Gm(r)dm/r$ with the already assembled interior. This counts each gravitational pair once. Hence the Newtonian gravitational potential energy isUsing $M=4\pi \rho R^3/3$ givesThe negative sign expresses gravitational binding. A factor of one half should not be inserted again: the shell assembly already avoids double counting.