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First solve the hinted fixed-gap problem. Let $y$ measure distance across the gap and $H(x)=h+x^2/(2a)$ be its leading local height. The lubrication approximation gives $\mu u_{yy}=p_x$, with $u=0$ at $y=0,H$, soThe constant flux per unit transverse length is $q$. Integrating its pressure gradient gives the pressure-driven flux through a parabolic lubrication gap:With $x=\sqrt{2ah}t$, the integral is $\sqrt{2a}{h}^{-5/2}{\int }_{-\infty }^{\infty }(1+t^2{)}^{-3}dt$. The substitution $t=\tan \chi$ gives ${\int }_{-\pi /2}^{\pi /2}{\cos }^4\chi d\chi =3\pi /8$, and therefore

Now take the sphere speed to be $U>0$ downwards and assume zero net liquid through-flow, so the displaced liquid must pass through the annular gap. Near the equator, the sphere's surface radius is $R-z^2/(2R)+\cdots$, giving $H(z)=d+z^2/(2R)$. In the sphere frame the relative volume flux is $U\pi (R+d{)}^2$; dividing by circumference $2\pi R$ gives $q\simeq UR/2$. The Couette contribution from the moving tube wall is only order $Ud$, smaller by $d/R$, so the fixed-gap pressure result givesThe pressure changes principally within axial distance $\sqrt{Rd}$ around the equator. It is nearly constant on each hemisphere, so its leading force is the pressure difference times projected area, $F_p\simeq \pi R^2\Delta p$. Direct viscous shear contributes a relative order $d/R$ and is smaller. Balance with the buoyant weight:Thus the settling speed of a nearly occluding sphere without through-flow isThe printed speed lacks the factor $1/\pi$. The parabolic-gap integral above fixes the normalization: its $3\pi /8$ cannot be dropped. With standard dynamic viscosity and Stokes drag law conventions, the requested coefficient must be corrected by dividing the printed expression by $\pi$. Also, the bypass result assumes no net through-flow; allowing a piston-like displacement of the entire fluid column would require specifying tube end conditions and a different resistance balance.

A sphere of radius $d$ in an unbounded fluid has sedimentation speed $U_d=2\Delta \rho g{d}^2/(9\mu )$. ThereforeThe large confined sphere falls much more slowly even than this much smaller unconfined sphere.

For radial offset $e<d$, let $ϵ=e/d$. To leading order the narrowest gap varies as $d(\varphi )=d(1+ϵ\cos \varphi )$. All circumferential strips share the same pressure drop and act as parallel channels. Inverting the gap-resistance formula shows that each carries flux proportional to $d(\varphi {)}^{5/2}\Delta p$. At fixed buoyant load, the leading pressure drop is fixed, so the speed ratio isThe eccentricity increases narrow-gap bypass mobility result holds because the gain through the wider side outweighs the loss through the narrower side. Strict convexity of $x^{5/2}$ and the zero mean of cosine prove the inequality. Pairing the positive- and negative-cosine contributions in the derivative proves monotonicity with offset. The broader hint's assertion for every exponent $\alpha >0$ is not correct for $0<\alpha <1$, when concavity reverses the inequality; the required exponent $5/2$ is in the valid convex range.