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The methoxy substituent donates electron density into the coordinated arene by chemical resonance. This increases the electron density available at chromium for pi backbonding to CO, weakening C–O and lowering its stretching frequencies. The ester substituent withdraws electron density from the arene, reducing pi backbonding and raising the carbonyl stretching frequencies. For both observed bands, the methoxy-substituted compound has the lowest carbonyl frequency, the unsubstituted compound is intermediate, and the ester-substituted compound has the highest.

In the harmonic oscillator approximation,so the trend reflects an increasing effective C–O bond force constant rather than a change in the CO reduced mass. For example, comparison of the higher-frequency bands gives $k_{\mathrm{ester}}/k_{\mathrm{methoxy}}\approx (1992/1983{)}^2=1.0091$. This is only an approximate local-bond interpretation because the observed normal modes couple the three CO oscillators. In the idealized local $C_{3v}$ geometry they span $A_1+E$, giving two infrared-active frequencies. A substituted arene can lower the exact molecular symmetry and split the $E$ pair, even if that splitting is unresolved experimentally.

$\mathrm{OP}(\mathrm{OMe}{)}_3$ also has approximately tetrahedral molecular geometry at phosphorus, now with three P–O–C sigma linkages and a terminal phosphoryl O. Only the latter is being compared with the terminal O of Me${}_3$PO; the three P–O–C bonds are chemically distinct.

The electronegative alkoxy substituents withdraw electron density from P and improve its capacity to accept donation from terminal O into the surrounding antibonding framework. The elementary substituent argument therefore predicts a stronger, stiffer terminal P–O bond in $\mathrm{OP}(\mathrm{OMe}{)}_3$ than in $\mathrm{M}{\mathrm{e}}_3\mathrm{PO}$. A description invoking polar bonding and negative hyperconjugation avoids assigning a literal localized d-orbital double bond solely because the formula is written P=O.

Measure the terminal P–O stretching vibration by infrared or Raman spectroscopy under comparable, non-hydrogen-bonding conditions. In an approximate local-mode comparison the reduced mass is the same, so a larger terminal stretching wavenumber indicates a larger force constant, $k\propto {\tilde{\nu }}^2$. Mode assignment can be checked by ${}^{18}$O substitution at the terminal oxygen or by a normal-mode calculation; otherwise an alkoxy P–O–C vibration could be compared by mistake. X-ray diffraction can supply a supporting bond-length comparison, but neither bond length nor a single coupled vibrational frequency directly measures an isolated bond-dissociation enthalpy.

For the same pair of atoms, the harmonic approximation gives $\tilde{\nu }=(2\pi c{)}^{-1}\sqrt{k/\mu }$, where $k$ is the bond force constant and $\mu$ the reduced mass. The large shifts mainly reflect changes in $k$ from CO pi backbonding; the complexes are not distinguished simply by their total number of valence electrons.

$\mathrm{W}(\mathrm{CO}{)}_6$ has stronger net backbonding than $\mathrm{Cr}(\mathrm{CO}{)}_6$. Both are group-6 M(0), $d^6$, 18-electron complexes. The more spatially extended tungsten $5d$ orbitals provide effective overlap with CO ${\pi }^{\ast }$, consistent with the lower stretching frequency. As a simple same-mode force-constant comparison, $(1977/2000{)}^2=0.977$, about a 2.3% reduction.

$\mathrm{Ni}(\mathrm{CO}{)}_3(\mathrm{PM}{\mathrm{e}}_3)$ has appreciable but less C–O weakening than those hexacarbonyls. Ni(0) is $d^{10}$ and the complex is also 18-electron. The more contracted, lower-energy late-3d occupied orbitals give less effective donation into CO ${\pi }^{\ast }$ than the group-6 carbonyls in this comparison. Trimethylphosphine donates electron density and increases backbonding relative to an otherwise comparable less-donating nickel complex; that effect does not force its frequency below chromium's. Comparing different metals requires both orbital energies and overlap.

$[\mathrm{Cu}(\mathrm{CO}{)}_4{]}^{+}$ has very weak backbonding and a frequency above free CO. Cu(I) is $d^{10}$ and 18-electron here, but positive charge lowers its occupied orbital energies and suppresses backdonation. Sigma donation and the cation's electrostatic polarization can strengthen the internal C–O bond. Consequently the usual free-CO reference, about $2143\mathrm{c}{\mathrm{m}}^{-1}$, lies below this cation's band. A blue shift is compatible with carbon coordination; it does not imply absence of a metal–CO bond. Coupled normal modes and measurement conditions limit literal extraction of a local bond force constant from one band.

For a hydrogen-like ion, the Coulomb potential is stronger by the nuclear atomic number $Z$, and the energy levels scale as $Z^2$:Here $Z$ is the number of protons in the atomic nucleus; the ion has exactly one electron, so its net electric charge is $(Z-1)e$. The zero of energy is a separated stationary electron and atomic nucleus at infinite distance. This is the nonrelativistic hydrogenic approximation using the supplied $R_H$; a more precise isotope-specific result also corrects the reduced mass.

Express the rotational constant as a spectroscopic wavenumber $\tilde{B}=h/(8{\pi }^{2}cI)$. The rotational absorption selection rule $\Delta J=+1$ givesSuccessive lines therefore have spacing $2\tilde{B}$, regardless of the starting $J$. Both observed spacings are $20.88\mathrm{c}{\mathrm{m}}^{-1}$, giving $\tilde{B}=10.44\mathrm{c}{\mathrm{m}}^{-1}$. The three lines are consequently the $5\to 6$, $6\to 7$ and $7\to 8$ transitions within the stated precision; the first observed line is not the $0\to 1$ line.

Using the given isotopic masses, the reduced mass isWith $\tilde{B}=1044{\mathrm{m}}^{-1}$ and $c$ in $\mathrm{m}{\mathrm{s}}^{-1}$, the bond length follows fromThe small common $0.01\mathrm{c}{\mathrm{m}}^{-1}$ offset relative to integer multiples of the spacing is immaterial at the precision of this model. A real ground-vibrational-state rotational constant determines an effective vibrationally averaged bond length; it need not equal the exact equilibrium separation.

The sharp solid curve is the $4.2$ K spectrum; the broadened dashed curve is the $77$ K spectrum. At low temperature, a strong discrete exciton transition appears just below the continuum edge, with weaker higher members converging toward the band gap. Heating broadens those resonances through scattering, reduces their spectral resolution, and shifts the band edge to lower photon energy. The lower-energy broad dashed maximum is therefore not evidence for a second unrelated low-temperature band gap.

Read approximate energies from the native figure: the main low-temperature peak is near $1.610$ eV and a weaker feature is near $1.617$–$1.618$ eV, with the continuum edge around $1.620$ eV. A three-dimensional hydrogenic series $E_n=E_g-R^{\ast }/n^2$ then givesFor example, $E_2-E_1=3R^{\ast }/4$ predicts about $7.5$ meV, consistent with those feature spacings. These are graphical estimates at roughly meV precision, not extra digits extracted from the line widths. The strong direct interband onset and subgap series support a direct-gap Wannier-Mott interpretation. The continuum absorption scale is of order $10^6$ m${}^{-1}$, corresponding to an optical absorption length of order a micrometre. At $77$ K, $k_{B}T\simeq 6.6$ meV is already comparable with the inferred binding scale, explaining why a sharp Rydberg series is poorly resolved; at $4.2$ K, $k_{B}T\simeq 0.36$ meV.

With $m_e^{\ast }=m_h^{\ast }=0.2m_e$, the reduced mass is ${\mu }^{\ast }=0.1m_e$. In the screened hydrogen model,It follows thatA radius of several nanometres is consistent with an extended exciton in an ordinary semiconductor lattice. The relative permittivity inferred here is the effective dielectric screening parameter of that model; identifying the material chemically or determining a full dielectric spectrum is not possible from these data alone.

Put $M=m_1+m_2$, define the center of mass $\mathbf{R}=(m_1{\mathbf{x}}_1+m_2{\mathbf{x}}_2)/M$, and let $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$ be relative position. Adding the two Newton's second law equations givesThe center of mass has constant velocity, by conservation of total momentum.

Subtracting the particle accelerations instead givesThus relative motion is a one-particle problem with the reduced mass $\mu$, while center-of-mass motion is uniform. The same reduced mass appears in the kinetic energy decomposition $T=\frac{1}{2}M|\dot{\mathbf{R}}{|}^2+\frac{1}{2}\mu |\dot{\mathbf{r}}{|}^2$.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$, with attractive Coulomb potential $V(r)=-\kappa /r$, where $\kappa =e^2/(4\pi {\epsilon }_0)$. Separating $\psi =R(r)Y_{\ell }^m(\theta ,\phi )$ uses the spherical harmonic angular eigenvalue $-\ell (\ell +1)$ of the unit-sphere Laplacian. The radial kinetic contribution is $-{\hslash }^2(R^{\prime \prime }+2R^{\prime }/r)/(2m)$, the angular kinetic contribution is the centrifugal term ${\hslash }^2\ell (\ell +1)R/(2m{r}^2)$, and the remaining term is the attractive potential energy. The right side is the energy eigenvalue times the radial wavefunction. Here the stated $m$ neglects nuclear recoil; including it replaces the electron mass by the reduced mass.

For $\ell =0$, substitution of the ground-state exponential givesIts $1/r$ coefficient must vanish and its constant coefficient determines the energy, soHere $a_0={\hslash }^2/(m\kappa )$ is the Bohr radius.

For $R_2=N_2(r+b)e^{-\beta r}$, direct differentiation yieldsCancel the nonzero exponential in the radial Schrödinger equation and compare coefficients of $r$, the constant term, and $1/r$. They give, respectively,ConsequentlyThe positive decay constants ensure normalizability; the second radial wavefunction has one positive-radius node at $2a_0$, as appropriate to the first radial excitation.

If the entire energy gap is carried by one photon, neglecting atomic recoil, conservation of energy gives the requested frequencyThere is a physical qualification: these are the $2s$ and $1s$ states, so the usual single-photon electric-dipole transition is forbidden by the electric-dipole selection rules for hydrogen. Explicitly, both angular factors are $Y_0^0$, and $\int |Y_0^0{|}^2\hat{\mathbf{r}}d\Omega =0$, so every component of $⟨1s|\mathbf{r}|2s⟩$ vanishes. The leading isolated-atom decay is instead two-photon emission, for which ${\nu }_1+{\nu }_2={\nu }_{\mathrm{gap}}$ and individual frequencies are not fixed. This metastability and the two-photon channel are documented in NIST's hydrogen spectral-data compilation. Thus the boxed answer is the energy-gap frequency under the printed single-photon assumption, rather than an assertion of an allowed electric-dipole line.

Thermal energy versus electrostatic repulsion. The mean translational kinetic energy of one nonrelativistic thermal proton isUsing the center-to-center touching separation $2r_p$, the Coulomb barrier estimate isThus $E_C/⟨E⟩\simeq 274$. Using $r_p$ rather than $2r_p$ as the rough interaction distance doubles the estimate but leaves the conclusion unchanged. Typical solar thermal energies are far below the classical barrier. The two essential ideas are the high-energy tail of the thermal distribution and quantum tunnelling through the Coulomb barrier: neither a mean-energy estimate nor a sharp classical cutoff describes the reacting population.

For two equal-mass protons, let $v$ be their relative speed. In the center-of-mass frame each has speed $v/2$, so their total collision energy isHere ${\mu }_r$ is the reduced mass, not the stellar mean molecular weight.

In the low-energy, nonresonant approximation, the geometrical quantum-wave factor in a collision cross-section is proportional to the squared de Broglie wavelength, hence to $1/E$. The slowly varying short-range reaction probability is absorbed into $S_0$; for proton–proton fusion it also includes the small weak-reaction probability. The rapidly varying exponential is the quantum tunnelling probability. For a Coulomb potential $V(r)=e^2/r$, its leading semiclassical exponent isThe integral follows from $r=(e^2/E){\sin }^2\theta$; its dimensionless integral is $\pi /2$. A finite nuclear radius modifies the short-distance factor absorbed into $S_0$. In this convention $E_B={\pi }^2{\mu }_r{e}^4/(2{\hslash }^2)$: it is a penetration-energy scale, distinct from the classical contact energy $E_C$.

Thermal averaging and pair counting. Each proton samples collision partners with number density $N$ and collision flux $v$, so its collision rate is $N⟨\sigma v⟩$. Multiplying by the number of protons counts each identical pair twice. The thermonuclear reaction rate is thereforeThe relative speed of two independent thermal protons has a Maxwell-Boltzmann velocity distribution with reduced mass $m_p/2$. Changing variables with $E=m_p{v}^2/4$ gives the normalized energy densitySubstituting $v=2\sqrt{E/m_p}$ and the collision cross-section causes the energy powers to cancel:

The Gamow peak and its Gaussian width. Set $\Phi (E)=E/(kT)+2\sqrt{E_B/E}$. It diverges at either endpoint and has its unique minimum whereThus the energy called $E_G$ in this question is the Gamow peak energy,In the stellar tunnelling regime $E_B\gg kT$,This establishes the requested hierarchy in its physical regime; it is not an inequality valid at arbitrary temperature.

At the minimum, $\Phi (E_G)=3E_G/(kT)$ and ${\Phi }^{\prime \prime }(E_G)=3/(2kT{E}_G)$. HenceThe Gaussian standard deviation is $\sqrt{2kT{E}_G/3}$, small compared with $E_G$ when $kT\ll E_G$. Extending the lower endpoint to $-\infty$ relative to the peak then givesCombining this with the thermal prefactor yields the Gaussian approximation to a nonresonant thermonuclear rate,At fixed reactant density and fixed $S_0,E_B$, comparison with the requested temperature form givesThe constant $\beta$ here has dimensions of temperature; it is not inverse temperature. Resonances, plasma screening or variation of $S(E)$ would require corrections to this particular approximation.

Put $M=m_1+m_2$ and let $\mathbf{R}=(m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2)/M$ be the center of mass. Applying Newton's second law separately to the two point masses givesThus the center of mass has constant velocity, $\mathbf{R}(t)=\mathbf{R}(0)+t\dot{\mathbf{R}}(0)$. For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, subtraction givesHere $\mu$ is the reduced mass; it converts the relative motion into a one-body equation of motion.

For the given linear force, the relative position obeys simple harmonic motion, $\ddot{\mathbf{r}}+{\Omega }^2\mathbf{r}=0$, where ${\Omega }^2=k/\mu$. The initial rest condition givesFor $d>0$, collision first occurs when the relative position vanishes, namely when $\Omega t=\pi /2$. ThereforeThe separation $d$ cancels because the frequency of this simple harmonic motion is independent of its amplitude.

By Newton's third law, the force on the second particle is $-F_{12}$. For $M=m_1+m_2$ and centre of mass $R=(m_1{r}_1+m_2{r}_2)/M$, Newton's second law gives $M\ddot{R}=F_{12}-F_{12}=0$. ThusThe centre moves on a straight line with constant velocity; zero velocity is allowed. For the relative position $r=r_1-r_2$, subtraction of the two equations givesHence the two-body problem reduces to a single particle of reduced mass $\mu$, together with the elementary centre motion. Reconstruct the positions as $r_1=R+(m_2/M)r$, $r_2=R-(m_1/M)r$.

Yes: the two equal masses can follow the same fixed circle in diametrically opposite positions. Put the stationary centre of mass at the circle's center and choose tangential velocities with the same sense of rotation. If the circle has radius $a$, the separation is $2a$ and the mutual gravitational force is directed toward that center, with magnitude $G{m}^2/(4a^2)$. The circular orbit condition is $ma{\omega }^2=G{m}^2/(4a^2)$, so an equal-mass circular binary hasThese initial data maintain the opposite positions and give the same constant angular speed for both particles.

Let $M=M_1+M_2$, and let $\mu =M_1{M}_2/M$ be the reduced mass. The stellar distances from the center of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. For a circular Kepler orbit, ${\Omega }^2=GM/a^3$, so summing the two orbital contributions gives the circular-binary orbital angular momentumThe separation $a$ is inside the square root. This also has the required dimensions of angular momentum.

For the homologous rotating collapse, label a fluid element by $s=r/R$ and its polar angle $\theta$. Its enclosed mass $m(s)$ is constant during stellar homology, as are the dimensionless density profile and the inertia coefficient $\alpha$. Conservation of angular momentum and solid-body rotation giveThe magnitude of the centrifugal acceleration is ${\Omega }^{2}r\sin \theta$, while the Newtonian gravitational field has magnitude $Gm(s)/r^2$. ThereforeUsing only the radial component of centrifugal force replaces $\sin \theta$ by ${\sin }^2\theta$ and leaves the scaling unchanged. The centrifugal force vanishes on the rotation axis; the central point is understood through the limiting field rather than a ratio of two zero forces.

At the equator of the cloud, critical rotation of a spherical cloud means ${\Omega }_c^2{R}_c=GM/R_c^2$, where $R_c=R_{\mathrm{crit}}$. ConsequentlyImmediately after the first fission, each daughter has mass $M/2$ and radius $a/2$ because the spheres touch. Their orbital moment of inertia is $M{a}^2/4$, while their combined spin moment of inertia is $\alpha M{a}^2/4$. With a common spin and orbital frequency ${\Omega }_f=\sqrt{GM/a^3}$, the synchronous fission model for binary formation givesThe common frequency can change during fission; corotation requires equal instantaneous frequencies, not an unchanged frequency from before the rearrangement. Equating the two values of $J$ yieldsFor a nonzero positive inertia coefficient, the necessary and sufficient condition in this algebraic model for $a<R_c$ isFor example, a moment of inertia of a uniform solid sphere has $\alpha =2/5$ and gives $a/R_c=64/49>1$; it fails the required compact-fission condition. A sufficiently centrally concentrated cloud can have smaller $\alpha$.

For the next collapse, neglect exchange of spin with the unchanged outer orbit, so each daughter's spin angular momentum is separately conserved. Immediately after the first fission that spin isFor a daughter of mass $m=M/2$ to reach critical rotation of a spherical cloud at radius $R_{c,d}$, the same relation gives $S_d=\alpha m\sqrt{Gm{R}_{c,d}}$. HenceEquivalently, its centrifugal-to-gravity ratio starts at $1/4$ when its radius is $a/2$ and reaches one after contraction by a factor four. Applying the same fission rule to this daughter produces an inner pair with separationCorotation at this second fission is local to each newly formed inner pair; it is not a single common frequency for the entire four-star hierarchy.

This spin assumption matters. If tidal locking instead kept each shrinking daughter synchronized to the fixed outer orbital period throughout the intervening collapse, its angular velocity would remain fixed. Its centrifugal-to-gravity ratio would then scale as $R_d^3$ and decrease, so it would never reach the proposed second fission. This is a counterexample to extending the corotation assumption through the whole contraction. The printed ratio describes the separately spin-conserving interpretation of the repeated fission hierarchy.

The model can generate a compact hierarchy algebraically, but is not a general physical account of multiple-star formation. In the allowed range $\alpha <1/3$, one has $a^{\prime }/a<1/8$, suggesting well-separated inner and outer scales. However, rapidly rotating gas deforms, touching daughters are tidally distorted, and pressure, gas flows, dissipation and spin-orbit torques cannot generally be ignored. The assumed identical profiles, equal mass splits, instantaneous corotation and later torque-free contractions are restrictive. Star formation can involve gravitational fragmentation and redistribution of angular momentum; the toy budget neither proves that fission occurs nor predicts the population of real multiple systems.

Take $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $r=|\mathbf{r}|$ and $M=M_1+M_2$. Newton's law of universal gravitation gives ${\ddot{\mathbf{x}}}_1=-G{M}_2\mathbf{r}/r^3$ and ${\ddot{\mathbf{x}}}_2=G{M}_1\mathbf{r}/r^3$. Subtracting yields the relative two-body equationThe reduced mass is $\mu =M_1{M}_2/M$. The orbital energy and angular momentum in the center of mass frame areDifferentiation givesThus both are conserved. The eccentricity vector isFor $\mathbf{h}=\mathbf{J}/\mu$, one has $\ddot{\mathbf{r}}\times \mathbf{h}/(GM)=\dot{\mathbf{r}}/r-\mathbf{r}(\mathbf{r}\cdot \dot{\mathbf{r}})/r^3=d(\mathbf{r}/r)/dt$, so $\dot{\mathbf{e}}=0$. In particular the orbital eccentricity is constant, withFor a bound Kepler orbit, $E=-GM\mu /(2a)$ and $J^2=GM{\mu }^{2}a(1-e^2)$.

For the far-field multipole expansion, set the center of mass to zero. Then ${\mathbf{x}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{x}}_2=-(M_1/M)\mathbf{r}$. At $x=|\mathbf{x}|\gg r$, expand each point-mass Newtonian gravitational potential:Repeated indices are summed. The dipole vanishes since ${\sum }_A{M}_A{\mathbf{x}}_A=0$, and ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$. The second-order gravitational quadrupole potential of a point-mass binary is thereforeBoth tensors specified in the question are trace-free. Direct contraction givesso to the requested orderFor equal masses the cubic multipole also vanishes, but it need not vanish for unequal masses.

To evaluate the quadrupole formula, use $k=GM$, $\mathbf{v}=\dot{\mathbf{r}}$, $\mathbf{n}=\mathbf{r}/r$, $u=\dot{r}=\mathbf{n}\cdot \mathbf{v}$, and $\mathbf{w}=\mathbf{v}-u\mathbf{n}$. The acceleration and its derivative areThe third derivative of $r_i{r}_j$ is ${\dot{b}}_i{r}_j+r_i{\dot{b}}_j+3b_i{v}_j+3v_i{b}_j$. Its trace is $d^3(r^2)/d{t}^3=-2ku/r^2$. Substituting in $q_{ij}=\mu (3r_i{r}_j-r^2{\delta }_{ij})/2$ givesBecause $\mathbf{n}\cdot \mathbf{w}=0$, the cross tensor contraction is zero. The squared norms of the two bracketed pieces are $6u^2$ and $72w^2$, respectively. Thusand the instantaneous energy loss isThe parenthesis is $w^2+u^2/12\ge 0$, as required for positive radiated power. This is the instantaneous quadrupole luminosity of a Kepler binary with loss sign for the orbital energy.

A slowly evolving initially circular binary is expected to follow a quasi-circular inspiral at leading radiation order. One needs angular-momentum loss as well as energy loss for this conclusion: the rotating quadrupole has frequency $2\Omega$ and azimuthal harmonic number two, so its fluxes satisfy the gravitational-wave energy and angular-momentum balance $\dot{E}=\Omega \dot{J}$. Along the circular family,The flux relation is precisely the tangent condition for remaining on that family; equivalently it keeps $e^2=1+2E{J}^2/(G^2{M}^2{\mu }^3)$ equal to zero at leading secular order. The shrinking orbit is not an exact fixed-radius circle: its small radial drift is neglected at this adiabatic order.

For a circular orbit of instantaneous separation $a$, $u=0$ and $v^2=GM/a$. ThereforeDifferentiating $E=-GM\mu /(2a)$ at fixed masses now givesWriting $C=64G^3{M}^2\mu /(5c^5)$, the circular gravitational-wave inspiral obeys $a^4=a_0^4-4Ct$. The formal point-mass coalescence time is $a_0^4/(4C)$, and $\dot{P}/P=(3/2)\dot{a}/a$ gives a decreasing orbital period and increasing gravitational-wave frequency.

Gravitational radiation tightens a detached close binary and becomes more effective at small separation. It can bring a stellar donor into Roche-lobe overflow or drive compact objects toward merger. Finite radii, tides and mass transfer can then alter the evolution; the fixed-mass inspiral law is not the entire evolution law once these matter. Near merger the weak-field, slow-motion quadrupole formula also ceases to be sufficient.

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.

Write $M=M_1+M_2$, $a=a_1+a_2$, and $\mu =M_1{M}_2/M$, the reduced mass. In the center of mass frame, $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Both components of the circular orbit have the same angular velocity $\Omega$. Thus their total orbital angular momentum isUsing Kepler's third law, ${\Omega }^2=GM/a^3$, and the orbital period $P=2\pi /\Omega$, gives the circular-binary orbital angular momentum