 Solution

Ridge push is a horizontal pressure force integrated with depth, equivalently a difference in gravitational potential energy per horizontal area. It is not the weight of a slab multiplied by the surface slope at one chosen point.

For an explicit compensated-column derivation, let $y$ measure depth below the ridge seafloor; the old seafloor is at $y=h$. Set $b={\rho }_m\alpha \Delta T$. The hot column has greater pressure. Its pressure excess over the old column isThe second line follows by integrating $\delta \rho (s)=b(1-s/a)$ upwards from the compensation level. The two expressions agree at $y=h$ because $({\rho }_m-{\rho }_w)h=ba/2$. Integrating this positive pressure excess gives the ridge-push force in a plate model, directed away from the ridge:Consistently retaining only first order in thermal expansion givesIf the topographic term is retained within the same linear-density column model, it adds $1.52\times 10^{11}\mathrm{N}{\mathrm{m}}^{-1}$, giving $F\simeq 2.75\times 10^{12}\mathrm{N}{\mathrm{m}}^{-1}$. This is a useful small correction, not a claim that all other second-order material effects have been included. Both values use the intended hot initial plate, not the inconsistent printed zero-age profile.