What does this equation mean?

An exact analytic evaluation, not a simulation, fixes the scale this actually requires. Take three clocks stacked vertically at heights -h , 0 , and +h near Earth’s surface, giving two edges $y_1$ = $\phi(0)$-$\phi(-h)$ and $y_2$ = $\phi(h)$-$\phi(0)$ . Their difference is an exact finite-difference identity up to a Taylor remainder of order $h^4$ : $y_2 - y_1 = \phi_{,zz}\,h^2 + O(h^4)$. For Earth modelled as a point mass, $\Phi(r)$ = -GM/r gives a radial second derivative of magnitude 2GM/$R^3$ and, since $\Phi$ is harmonic in vacuum, transverse second derivatives of magnitude GM/$R^3$ with the opposite sign — a trace-free tensor in the exact ratio 2:{-1}:{-1} . With the standard gravitational parameter G$M_\oplus$ = 3.986$\times$10^{14}\,$\mathrm{m^3\,s^{-2}}$ and mean radius $R_\oplus$ = 6.371$\times$10^{6}\,$\mathrm m$ , this gives G$M_\oplus$/$R_\oplus^3$ $\approx$ 1.541$\times$10^{-6}\,$\mathrm{s^{-2}}$ and a radial magnitude of 3.08$\times$10^{-6}\,$\mathrm{s^{-2}}$ , matching the standard geodetic free-air gravity gradient of about 3.086$\times$10^{-6}\,$\mathrm{s^{-2}}$ , or 0.3086 milligal per metre, to three figures [ 15 ] . So $\phi_{,zz}$ $\approx$ 3.08$\times$10^{-6}/$c^2$ $\approx$ 3.43$\times$10^{-23}\,$\mathrm{m^{-2}}$ .