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Under spherical symmetry, the continuity equation and Darcy's law giveHence $q_r=C/r^2$, and integrating the pressure gradient from infinity givesThe boundary condition at $r=a$ fixes $C={\Pi }_{1}a\Delta p/\eta$. Therefore spherical Darcy injection hasHere $q_r$ is the superficial Darcy velocity, not the pore velocity $q_r/\phi$.

For the comparison, assume geometrically similar pore networks, equal porosity, the same source radius and far boundary geometry, and equal pump efficiency. Dimensional analysis then gives ${\Pi }_1=C_{\mathrm{geom}}{d}^2$ with the same dimensionless factor. ConsequentlyAt a fixed desired volumetric flow rate, the required excess pressure and pumping power areAquifer A requires half the pumping power at equal throughput, or half the energy to inject the same volume over the same time. This compares the excess pressure-volume work dissipated by the Darcy flow; any different baseline compression or other plant costs need separate accounting.

The imposed comparison matters. At fixed $\Delta p$, A takes twice the volumetric flow rate and twice the instantaneous power, but the same excess energy per injected volume. The pore spacing alone does not determine permeability of a porous medium if network geometry or porosity changes; without the geometric assumptions the preference cannot be inferred from $d$ alone.

For the common constant mass density, the background hydrostatic pressure satisfies $d{p}_0/dz=-\rho g$. Define the excess pressure $P=p-p_0(z)$. In Darcy's law, the background pressure gradient cancels gravity: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: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 thenand thereforeThis 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 $Q=4\pi {\Pi }_{1}a\Delta p/\eta$ still holds. Equal injected and resident mass densities are essential to the background cancellation.