Solution (source code)

= Solution

Write $\ell_g$ for the dimensional offset of the <centre of mass>, reserving $h$ below for the dimensionless swimming parameter. The weight $-mg\mathbf k$ acts at $-\ell_g\mathbf p$, so its <torque> about the geometric centre is $mg\ell_g\mathbf p\times\mathbf k$. Buoyancy acts through the geometric centre and has no such <torque>. For a spherical cell, the ambient angular <velocity> is $\boldsymbol\omega/2$ and the rotational resistance is $\zeta_r=\rho\nu\alpha_\perp v$. Neglecting angular inertia, <torque> balance gives
$$
\boldsymbol\Omega_{\mathrm{cell}}=\frac12\boldsymbol\omega+\frac1B\mathbf p\times\mathbf k,\qquad B=\frac{\rho\nu\alpha_\perp v}{mg\ell_g}.
$$
Since $\dot{\mathbf p}=\boldsymbol\Omega_{\mathrm{cell}}\times\mathbf p$, the <bottom-heavy spherical-cell orientation dynamics> is
$$
\boxed{\dot{\mathbf p}=\frac1B[\mathbf k-(\mathbf k\cdot\mathbf p)\mathbf p]+\frac12\boldsymbol\omega\times\mathbf p.}
$$
Thus a steady orientation satisfies
$$
\boxed{\frac1B[\mathbf k-(\mathbf k\cdot\mathbf p)\mathbf p]=-\frac12\boldsymbol\omega\times\mathbf p.}
$$
\b[The printed algebraic equation has the opposite sign on its right side], if <vorticity> means the usual $\nabla\times\mathbf u$. Its later concentration equation is consistent with the corrected sign. In particular, linearizing the corrected equation about $\mathbf p=\mathbf k$ gives $\mathbf p_\perp=(B/2)(\boldsymbol\omega\times\mathbf k)_\perp$, whereas the printed sign would reverse that response. Merely changing the definition of <vorticity> would also require changing the ensuing fluid and orientation calculations.

The <gyrotactic reorientation time> $B$ measures gravitational restoration against rotational viscous resistance. In still fluid a tilt angle obeys $\dot\theta=-\sin\theta/B$, so
$$
\tan\frac{\theta(t)}2=\tan\frac{\theta(0)}2e^{-t/B}.
$$
For small tilt it is precisely the exponential relaxation time. Applying the algebraic orientation balance to a time-dependent disturbance additionally assumes that this relaxation is fast enough to neglect orientation lag. Keeping the full dynamic orientation equation would introduce an additional relaxation factor into the perturbation problem.