What does this equation mean?

Earth’s own centrifugal contribution is the concrete case worth sizing, precisely because it is not negligible enough to ignore by assumption. With $\Omega$ $\approx$ 7.292$\times$10^{-5}\,$\mathrm{rad\,s^{-1}}$ fixed by astrometry to far more figures than any clock network could hope to determine on its own [ 10 ] , the centrifugal Hessian has magnitude $\Omega^2$ $\approx$ 5.32$\times$10^{-9}\,$\mathrm{s^{-2}}$ — smaller than the radial tidal magnitude computed above by a factor of nearly six hundred, but not so much smaller that a network aiming to certify the 2:{-1}:{-1} ratio to good precision can skip subtracting it. Left in, it would masquerade as roughly a fifth of one percent of the genuine tidal signal, concentrated entirely in the horizontal plane and constant across the whole network — exactly the flat, source-independent signature the falloff-and-orientation test above exists to catch, and exactly why $\Omega$ has to be subtracted using its own independently known value rather than absorbed into whatever the least-squares fit finds convenient. A measured tensor that fails any one of those three tests — wrong falloff, wrong orientation relative to the vertical, or an eigenvalue ratio left over after subtraction that does not converge to 2:{-1}:{-1} — has not detected spacetime curvature, whatever its magnitude.