Past exam of the physics of the Earth as a Planet course of the University of Cambridge / 2015 / Paper 1 / 2 / a

Take upward deflection as positive, use $k>0$, and write $\Delta \rho ={\rho }_m-{\rho }_c>0$. A positive added load depresses the plate, so $w$ is negative beneath it. Because bending moves the original surface while the load adds thickness, the final topography is $e=s+w$. For the cosine mode the horizontal biharmonic operator multiplies $w$ by $k^4$, givingEliminate $s=e-w$ to obtainThe original loading equation contains ${\rho }_m$, whereas the relation after eliminating the added thickness contains $\Delta \rho$. These must not be interchanged prematurely.

At first order in the relief, represent a displaced density interface by a sinusoidal mass sheet. A sheet of surface-density amplitude $\Sigma$ at depth $z$ produces a downward gravity amplitude $2\pi G\Sigma e^{-kz}$ at the datum. This follows by solving the Fourier-transformed gravitational Poisson equation above and below the sheet: the decaying potential has amplitude $-2\pi G\Sigma e^{-k|z-z_0|}/k$, with the normal derivative jump fixed by the sheet mass.

The top interface contributes ${\rho }_{c}e$; the displaced crust–mantle interface contributes $\Delta \rho w$ at depth $t$. ThusThe second term is negative beneath a positive load: depression replaces denser mantle material by lighter crust. Substituting the deflection-to-topography ratio yieldsThis flexural gravity admittance combines mechanical compensation and attenuation of the buried interface's gravitational signal. It assumes small relief, a flat uniform thin plate, uniform density layers and the stated surface loading, without additional internal density anomalies. Here $t$ is the depth of the crust–mantle interface. It is not automatically the effective elastic thickness $T_e$; the latter enters through $D=E{T}_e^3/[12(1-{\nu }^2)]$.