= Solution
{c}
For the common constant <mass density>, the background <hydrostatic pressure> satisfies $dp_0/dz=-\rho g$. Define the excess <pressure> $P=p-p_0(z)$. In <Darcy's law>, the background <pressure gradient> cancels <gravity>:
$$
\mathbf q=-\frac{\Pi_1}{\eta}(\nabla p+\rho g\hat{\mathbf z})
=-\frac{\Pi_1}{\eta}\nabla P,\qquad \nabla^2P=0.
$$
If the source imposes the same excess <pressure> relative to the local <hydrostatic pressure> everywhere on its spherical surface, $P(a)=\Delta p$, the <spherical Darcy injection> result survives exactly:
$$
\boxed{p(\mathbf x)=p_0(z)+\frac{a\Delta p}{r},\qquad
\mathbf q=\frac{\Pi_1a\Delta p}{\eta r^2}\hat{\mathbf r}.}
$$
It is important to specify what is maintained on the source. A literally uniform absolute <pressure> $p_s=p_0(0)+\Delta p$ on a finite sphere centered at $z=0$ instead gives $P(a,\theta)=\Delta p+\rho ga\cos\theta$, where $\theta$ is measured from the upward vertical. The decaying <harmonic function> outside the sphere is then
$$
P=\frac{a\Delta p}{r}+\frac{\rho ga^3\cos\theta}{r^2},
$$
and therefore
$$
q_r=\frac{\Pi_1}{\eta}\left(\frac{a\Delta p}{r^2}+\frac{2\rho ga^3\cos\theta}{r^3}\right),\qquad
q_\theta=\frac{\Pi_1\rho ga^3\sin\theta}{\eta r^3}.
$$
This supplies a concrete exception to an unqualified claim that <gravity> leaves the radial flow unchanged. The additional dipolar contribution has zero net <volumetric flow rate>, so $Q=4\pi\Pi_1a\Delta p/\eta$ still holds. Equal injected and resident <mass densities> are essential to the background cancellation.
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