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If release imparts no velocity kick, a grain's initial Kepler orbit under reduced stellar attraction has $q=r_b$, $Q=r_b/(1-2\beta )$, and $e=\beta /(1-\beta )$ for $0\le \beta <1/2$. Its pericentre distance remains at the dust birth ring, while its apocentre distance grows with the radiation-pressure coefficient.

Put $\mathcal{G}=G(M_{\star }+M_{\mathrm{pl}})$, $h=\sqrt{\mathcal{G}a(1-e^2)}$, $\theta =\omega +f$ and $L=n_{\mathrm{pl}}t$. Here $h$ is the specific angular momentum, $f$ is the true anomaly and $\omega$ is the planar longitude of periapsis. Differentiating the Kepler orbit equation with $\dot{f}=h/r^2$ givesResolve the velocity along the radial and tangential unit vectors, $\hat{\mathbf{r}}=(\cos \theta ,\sin \theta )$ and $\hat{\mathbf{\theta }}=(-\sin \theta ,\cos \theta )$. The identities $\sin f\cos \theta -\cos f\sin \theta =-\sin \omega$ and $\sin f\sin \theta +\cos f\cos \theta =\cos \omega$ yieldThis is the Kepler velocity hodograph: the velocity runs around a circle whose centre is displaced by the orbital eccentricity.

The centre of mass condition gives the star's barycentric position ${\mathbf{r}}_{\star }=-\mu a_{\mathrm{pl}}(\cos L,\sin L)$. Translating to the star without rotating the axes gives ${\mathbf{r}}^{\prime }=\mathbf{r}-{\mathbf{r}}_{\star }$ and ${\mathbf{v}}^{\prime }=\mathbf{v}-{\dot{\mathbf{r}}}_{\star }$. From Kepler's third law, $n_{\mathrm{pl}}^2{a}_{\mathrm{pl}}^3=\mathcal{G}$, soConsequentlyThe small displacement and the larger reflex velocity have different orders: $|{\mathbf{r}}^{\prime }-\mathbf{r}|/a=\mu \alpha$, but the reflex speed divided by the circular speed is $\mu {\alpha }^{-1/2}$. These are the source of the astrocentric osculating-element oscillations.

For the coordinate ratio, the exact result isFor fixed $e<1$, $r/a\ge 1-e$, so this is $O(\mu \alpha )$ on any angular region with $|\cos \theta |$ bounded away from zero. The printed coordinate claim is not uniform over a full orbit: at $x=0$ its ratio is undefined. The nonsingular statement is $|{\mathbf{r}}^{\prime }-\mathbf{r}|/r\le \mu \alpha /(1-e)$, assuming the pericentre distance remains much larger than the binary separation.

Squaring the two components of the velocity givesAt $\mu =0$, $\kappa =0$ and $\mathcal{G}=G{M}_{\star }$. Sincethis reduces to the vis-viva equation, $v^{\prime 2}=G{M}_{\star }(2/r-1/a)$.

Now let the barycentric Kepler orbit be circular, write $\Delta =L-\theta$ and $\epsilon =\mu {\alpha }^{-1/2}\ll 1$, and put $v_0=\sqrt{\mathcal{G}/a}$. The osculating orbital elements about the star use the gravitational parameter $G{M}_{\star }=(1-\mu )\mathcal{G}$. To the required order, $r^{\prime }=a[1+O(\mu \alpha )]$ and $v^{\prime 2}=v_0^2[1+2\epsilon \cos \Delta +O({\epsilon }^2)]$. The specific orbital energy therefore impliesHere $\mu =o(\epsilon )$ and ${\epsilon }^2=o(\epsilon )$ under the stated hierarchy. Even at a phase where the leading cosine vanishes, the absolute error bound remains valid.

Finally, use the eccentricity vector ${\mathbf{e}}^{\prime }=({\mathbf{v}}^{\prime }\times {\mathbf{h}}^{\prime })/(G{M}_{\star })-{\hat{\mathbf{r}}}^{\prime }$. In radial and tangential components it isThe translation changes the radial basis only by $O(\mu \alpha )$, whereas $v_r^{\prime }=-v_0\epsilon \sin \Delta +O(v_0\mu \alpha )$ and $v_{\theta }^{\prime }=v_0(1+\epsilon \cos \Delta )+O(v_0\mu \alpha )$. ThusFor the original circular Kepler orbit, $\omega$ and $f$ are separately arbitrary; their sum $\theta$ is well defined. Using the eccentricity vector avoids introducing an undefined pericentre for that circular Kepler orbit.

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.

Interpret projection from very far away as the asymptotic incoming state at $r\to \infty$. For $v>0$ and impact parameter $b>0$, conservation of energy and conservation of angular momentum give $E=v^2/2$ and $|h|=bv$. At the closest approach the radial velocity vanishes. With $k=-\kappa <0$, the central-force radial turning point equation isOnly the positive root is physical. Put $q=\kappa /v^2$. The pericentre distance and the purely tangential speed there areThus $p<b$ and $u>v$: attraction bends the path inward and increases the speed. With positive energy, the Kepler orbit is a hyperbola. Measured from the pericentre direction, its polar coordinates satisfy $r=h^2/[\kappa (1+e\cos \varphi )]$ with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$.

The hyperbola shown has a nonzero impact parameter. For $b=0$, the attractive radial orbit instead reaches the singular origin; there is no regular turning point with a finite closest-approach speed.

Work in the planet's rest frame, neglecting its recoil and tidal forces during the close encounter. A hyperbolic Kepler orbit has positive specific orbital energy $\epsilon =v_{\infty }^2/2$ and specific angular momentum $h=b{v}_{\infty }$, where $b$ is the impact parameter. Its orbital eccentricity isAt the pericentre distance, $r_{\min }=h^2/[\mu (1+e_h)]$. Substituting $h^2={\mu }^2(e_h^2-1)/v_{\infty }^2$ yieldsThe two asymptotes have true anomalies $\pm f_{\infty }$, where $\cos f_{\infty }=-1/e_h$. Their directions of motion give the gravitational scattering angle $\theta =2f_{\infty }-\pi$, so $\sin (\theta /2)=-\cos f_{\infty }=1/e_h$. ThereforeAs a useful independent form, $b^2=r_{\min }^2+2\mu r_{\min }/v_{\infty }^2$, displaying gravitational focusing. Here $\mu =G{M}_p$ to the stated small-comet accuracy; the exact relative-coordinate formula replaces it by $G(M_p+M)$.

For an unperturbed Kepler orbit, successive revolutions advance the planet-relative orientation by $-n_p{T}_c$ modulo $2\pi$, where $T_c$ is the comet's orbital period. An irrational period ratio makes these orientations dense; sufficiently distant mean-motion resonance and adequate observation time justify the corresponding phase mixing approximation.

At each radius between its pericentre distance and apocentre distance, the comet eventually visits every azimuth in the rotating reference frame. The spatial projection is thereforeIt is an annulus, with a nonuniform radial residence probability, rather than a uniformly filled area. Strictly, the full position-velocity phase space is not this annulus: at each radius the energy and angular momentum constrain the velocity, with inward and outward branches. The annulus is its position-space projection. This description assumes the orbital elements have not yet been substantially changed by planetary scattering.

Parameterize the trajectory by the decreasing orbital eccentricity. The Poynting–Robertson drag invariant givesFor $Q_0\gg q_0$, $C\simeq 2q_0$. Initially $e\simeq 1$, and the orbit moves almost vertically down a plot of $Q/q_0$ against $q/q_0$: the apocentre distance shrinks rapidly while the pericentre distance changes little. Integrating the high-eccentricity slope givesOnce the orbit has moderate orbital eccentricity, both distances change appreciably. For example $e=1/2$ gives $q\simeq 0.766q_0$ and $Q\simeq 2.30q_0$ in the $Q_0/q_0\to \infty$ limit. Eventually $e\ll 1$, $q\simeq Q\simeq C{e}^{4/5}$, and the trajectory approaches the diagonal $Q=q$ before reaching the origin. Thus the two approximate phases are apocentre contraction at nearly fixed pericentre, followed by nearly circular inward migration. They are a smooth crossover, not two separate exact solutions.

Using $q=a(1-e)$ and $Q=a(1+e)$, differentiate each apsidal distance. The apsidal evolution under Poynting–Robertson drag isBoth the pericentre distance and apocentre distance decrease. Since $a=(Q+q)/2$ and $e=(Q-q)/(Q+q)$, the same rates areTheir ratio isIn particular $Q\gg q$ gives $dQ/dq\simeq (5/3)(Q/q{)}^2$, whereas the circular limit gives slope $1$.

Retain the unchanged-angular momentum assumption of the preceding part. The initial orbit has radius $r_o=6400+500=6900\mathrm{km}$, and the desired pericentre distance is $r_p=6400+100=6500\mathrm{km}$. Using $r_p=r_o/(1+e)$ givesThe required fractional increase in initial kinetic energy is thereforeHere $\Delta E=\Delta K>0$ because the instantaneous kick does not change potential energy. Since the initial total orbital energy is $E_0=-K_0$, an alternative convention for the requested fractional energy change givesThe total orbital energy becomes less negative. Explicitly specifying its denominator avoids interpreting a positive energy input as a decrease of energy.

For a Kepler orbit with pericentre distance $q$, apocentre distance $Q$, and semi-major axis $a=(q+Q)/2$, a uniform mean anomaly givesThe two radial passages per orbital period are included, and ${\int }_q^Q{p}_r(r)dr=1$.