Solution (source code)

= Solution

At each fixed <semimajor axis>, let $B=b_{3/2}^{2}(\alpha)/b_{3/2}^{1}(\alpha)$ and $Z_f=B e_{\rm pl}$. The <planet>'s <longitude of pericentre> is the zero of longitude. The <secular perturbation> equation becomes $\dot z=iA(z-Z_f)$. Integrating it with the initial condition $z(0)=0$ gives
$$
\boxed{z(t)=Z_f[1-e^{iAt}]}.
$$
The <forced eccentricity> vector is the constant real $Z_f$; the <free eccentricity> vector is $-Z_f e^{iAt}$. Consequently the <complex eccentricity> follows a circle in the <Argand plane> centered at $Z_f$, with radius $Z_f$, starting at the origin and moving counterclockwise. Its initial velocity is downward, $\dot z(0)=-iAZ_f$. The secular period is $2\pi/A$, and
$$
e(t)=|z(t)|=2Z_f|\sin(At/2)|,\quad e_{\max}=2Z_f,\quad
\overline e=\frac{4Z_f}{\pi},\quad \sqrt{\overline{e^2}}=\sqrt2Z_f.
$$
The <longitude of pericentre> is undefined at the origin, but the <complex eccentricity> passes smoothly through it. At $At=\pi$, $z=2Z_f$ and the orbit is maximally eccentric and aligned with the <planet>.

\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-64-eccentricity-circle.png]
{title=Forced and free eccentricity vectors tracing a secular circle in the Argand plane}

Using the small-$\alpha$ <Laplace coefficient> expansion gives
$$
b_{3/2}^{1}=3\alpha+O(\alpha^3),\qquad
b_{3/2}^{2}=\frac{15}{4}\alpha^2+O(\alpha^4),\qquad
Z_f=\frac54\alpha e_{\rm pl}[1+O(\alpha^2)].
$$
Also $n=n_{\rm pl}\alpha^{-3/2}$ to leading order in the <planet>-to-<star> <mass> ratio. The <secular forcing by a distant outer planet> is therefore
$$
A=\frac34n_{\rm pl}\frac{M_{\rm pl}}{M_\star}\alpha^{3/2}[1+O(\alpha^2)],
\qquad
\boxed{z(t)\simeq\frac54\alpha e_{\rm pl}\left[1-\exp\left(i\frac34n_{\rm pl}\frac{M_{\rm pl}}{M_\star}\alpha^{3/2}t\right)\right]}.
$$
A relative $O(\alpha^2)$ correction to $A$ can accumulate a phase error at very long times; the leading-frequency formula is not a uniform-in-time expansion.

If $a_2=a_1(1+\delta)$ with $\delta=\delta a/a_1\ll1$, then $Z_{f,2}/Z_{f,1}=1+\delta+O(\delta^2)$ and $A_2/A_1=1+3\delta/2+O(\delta^2)$. The outer <planetesimal> therefore has a slightly larger <forced eccentricity circle> and a shorter secular period. Its accumulated secular phase lead is $\Delta(At)\simeq3A_1t\delta/2$. Initially both circles start at the origin, but their <eccentricity vectors> progressively lose alignment. Linearizing the exponential in $\delta$ additionally requires $|A_1t\delta|\ll1$.

The <orbit crossing> is caused by this accumulating differential secular phase, even though individual <orbital eccentricities> remain small. To first order in <orbital eccentricity>, the radius of an orbit at fixed inertial longitude $\theta$ is
$$
r(a,\theta,t)=a\left[1-\operatorname{Re}(z(a,t)e^{-i\theta})\right]+O(ae^2).
$$
For an infinitesimally neighboring orbit,
$$
\frac{\delta r}{\delta a}\simeq1-\operatorname{Re}\left[(z+a\partial_a z)e^{-i\theta}\right].
$$
This derivative is evaluated within the first-order orbital-shape model; derivatives of the neglected quadratic terms give relative $O(Z_f)$ corrections near crossing. The least separation over longitude is consequently $\delta a[1-|z+a\partial_a z|]$. The <differential secular orbit-crossing criterion> is $|z+a\partial_a z|=1$. Since $Z_f\propto a$ and $A\propto a^{3/2}$,
$$
z+a\partial_a z=2Z_f(1-e^{iAt})-\frac32iZ_fAt\,e^{iAt}.
$$
The first term stays bounded by $4Z_f\ll1$, whereas the second grows in proportion to time. To leading order in $Z_f$, crossing begins when $(3/2)Z_fAt=1$. Thus
$$
\boxed{t_{\rm cross}\simeq\frac{2}{3Z_fA}
=\frac{32}{45}n_{\rm pl}^{-1}\frac{M_\star}{M_{\rm pl}}e_{\rm pl}^{-1}\left(\frac{a_1}{a_{\rm pl}}\right)^{-5/2}}.
$$
More precisely, in this first-order orbital-shape approximation the exact first crossing lies within a fractional $O(Z_f)$ window of this value, since the magnitude differs from $(3/2)Z_fAt$ by at most $4Z_f$. Higher-order <Laplace coefficients> give additional corrections. The local derivation takes the neighboring-orbit limit before the long-time limit; for a finite separation it requires $\delta a/a_1\ll Z_f$ near crossing. A broad continuous disk contains such neighboring orbits. The crossings first appear in directions selected by the phase of $z+a\partial_a z$, not simultaneously at every longitude.

For the <collision> energy estimate adopt the phase-averaged mean <orbital eccentricity> $\overline e=4Z_f/\pi\simeq5\alpha e_{\rm pl}/\pi$. The specified velocity estimate gives
$$
v_{\rm col}\sim\overline e\sqrt{\frac{GM_\star}{a_1}},\qquad
v_{\rm col}^2\sim\frac{25}{\pi^2}\alpha e_{\rm pl}^2\frac{GM_\star}{a_{\rm pl}}.
$$
Using the projectile <kinetic energy> per target <mass> convention for <specific impact energy>, equal-size and equal-density bodies have equal <masses> and $Q=\tfrac12v_{\rm col}^2$. A <catastrophic planetesimal collision> requires $Q\gtrsim Q_D^*$, so
$$
\boxed{\frac{a_1}{a_{\rm pl}}\gtrsim\frac{2\pi^2}{25}\frac{Q_D^*a_{\rm pl}}{GM_\star e_{\rm pl}^2}}.
$$
The numerical coefficient is only illustrative because the <collision> speed was specified only to order of magnitude. If the disruption threshold is instead defined using the <centre of mass> <kinetic energy> per combined <mass>, equal <masses> give $Q_R=v_{\rm col}^2/8$ and the coefficient becomes $8\pi^2/25$ for a threshold expressed in that convention. The robust result is a lower bound of order $Q_D^*a_{\rm pl}/(GM_\star e_{\rm pl}^2)$.

\b[Within the inner-disk approximation, catastrophic <collisions> require sufficiently large $a_1/a_{\rm pl}$, not arbitrarily small radii.] Although the Keplerian speed rises inward, the secularly induced <orbital eccentricity> falls faster: $v_{\rm col}^2\propto\alpha$. The lower bound must be much smaller than one to leave a domain compatible with $\alpha\ll1$. A small <planet> <orbital eccentricity>, a large disruption threshold, or a distant <planet> can eliminate that domain. The <planet> <mass> controls the time to crossing but cancels from the eventual <forced eccentricity> amplitude and this <collision>-energy criterion. The threshold alone does not ensure that crossing has occurred within the disk's available lifetime.