= Solution
Work in the laboratory frame, with fluid at rest at infinity and the instantaneous <sphere> centre at the origin. In the <Unscaled Papkovich–Neuber representation>, take the harmonic potentials
$$
\mathbf A=-\frac{3a}{4r}\mathbf V,\qquad\phi=\frac{a^3}{4r^3}(\mathbf V\cdot\mathbf x),\qquad r=|\mathbf x|.
$$
Differentiating the two terms gives the <translating sphere in Stokes flow>:
$$
\boxed{\mathbf u=\frac{3a}{4r}(I+\mathbf n\mathbf n)\mathbf V+\frac{a^3}{4r^3}(I-3\mathbf n\mathbf n)\mathbf V,\qquad p-p_\infty=\frac{3\mu a}{2r^2}\mathbf V\cdot\mathbf n.}
$$
Here $\mathbf n=\mathbf x/r$. The potentials are harmonic outside the <sphere>, so they satisfy the <Stokes equations>. At $r=a$, the two coefficients of $\mathbf n\mathbf n$ cancel and the coefficient of $I$ is one, proving the <no-slip boundary condition> $\mathbf u=\mathbf V$; the field decays at infinity. Its leading part is a <Stokeslet> of <force> $6\pi\mu a\mathbf V$ exerted on the fluid.
Let $\widehat{\mathbf z}$ point upwards, $\delta=V_1-V_2$, and $D=a_2V_2-a_1V_1$. Each isolated <sphere> moves with <velocity> $-V_i\widehat{\mathbf z}$. To leading order its correction is the other's incident <Stokeslet> <velocity>:
$$
\dot{\mathbf x}_1=-V_1\widehat{\mathbf z}-\frac{3a_2V_2}{4r}(I+\mathbf n\mathbf n)\widehat{\mathbf z},\qquad\dot{\mathbf x}_2=-V_2\widehat{\mathbf z}-\frac{3a_1V_1}{4r}(I+\mathbf n\mathbf n)\widehat{\mathbf z}.
$$
The <forces> generating these fields are the fixed sedimenting loads, so their leading strengths are set by the isolated speeds. Subtraction yields the <leading interaction of two sedimenting spheres>:
$$
\boxed{\dot{\mathbf r}=-\delta\widehat{\mathbf z}-\frac{3D}{4r}(I+\mathbf n\mathbf n)\widehat{\mathbf z}.}
$$
Since $\widehat{\mathbf z}=\cos\theta\,\widehat{\mathbf r}-\sin\theta\,\widehat{\boldsymbol\theta}$, this reduces to
$$
\boxed{\dot r=-\left(\delta+\frac{3D}{2r}\right)\cos\theta,\qquad r\dot\theta=\left(\delta+\frac{3D}{4r}\right)\sin\theta.}
$$
The stipulated higher-order errors can be appended to both equations; they do not enter this leading calculation.
For the same-sign case, interchange labels if necessary to arrange $\delta>0,D>0$, and put $c=3D/4$. The vertical separation $z=r\cos\theta$ satisfies
$$
\dot z=-\delta-\frac c r(1+\cos^2\theta)\leq-\delta.
$$
If <sphere> one is initially above <sphere> two, its vertical order reverses; in every case $z\to-\infty$, implying $r\to\infty$. For a noncollinear trajectory there is no collision in the point-force equations. Indeed, divide the two polar equations and integrate:
$$
\frac{\delta r+c}{r(\delta r+2c)}dr=-\cot\theta\,d\theta,\qquad\boxed{r(\delta r+2c)\sin^2\theta=C>0.}
$$
At side-by-side passage, $\theta=\pi/2$, the closest separation is the positive root of $\delta r^2+2cr=C$. Before passage $\dot r<0$ and after passage $\dot r>0$, proving overtaking and unbounded subsequent separation. This <passing invariant for unequal point-force spheres> describes a physical far-field passage only when the closest separation also greatly exceeds the <sphere> radii. Exactly collinear approach instead reaches the near-contact regime, where this leading approximation cannot describe two finite <spheres> passing through one another.
For opposite signs, choose $\delta>0,D<0$ and consider the invariant configuration with <sphere> one vertically above <sphere> two, $\theta=0$. The radial equation is
$$
\dot r=-\delta+\frac{3|D|}{2r}=\delta\left(\frac{r_*}{r}-1\right),\qquad\boxed{r_*=-\frac{3D}{2\delta}.}
$$
If $r>r_*$, separation decreases; if $r<r_*$, it increases. Thus it approaches this <equilibrium> along the vertical line. <Linearization> gives $\dot{\delta r}=-(\delta/r_*)\delta r$. But a small tilt obeys $\dot\theta=(\delta/(2r_*))\theta$, so the <vertical bound pair in the point-force sedimentation model> is stable only against vertical perturbations and is a saddle in the full separation dynamics. The opposite vertical orientation has the reversed stability. Reversing gravity reverses both $\delta$ and $D$, leaves $r_*$ unchanged, and reverses every trajectory; radial attraction becomes repulsion. This is fully consistent with <kinematic reversibility of Stokes flow>.
There is an important restriction in the printed data. Its hierarchy $|D|\ll(a_1+a_2)|\delta|$ implies $r_*\ll a_1+a_2$, so the proposed <equilibrium> would overlap the physical <spheres> and lie outside the well-separated approximation. \b[With that hierarchy there is no justified physical far-field bound pair.] The intended far-field <equilibrium> instead requires the first inequality reversed, $|D|\gg(a_1+a_2)|\delta|$. The formal <equilibrium> and its vertical stability above show the intended mechanism, while also identifying the missing validity and transverse-stability qualifications.
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