Question

The figure below shows Bouguer gravity anomaly values (in mGal) at a grid interval of 1 km, centred on point P (r1 and r2 are radial distances from P).

The second vertical derivative at P is _________ mGal/km\(^2\) (rounded off to two decimals).

Hint:

Use Laplace's equation (g_zz = -(g_xx+g_yy)) and estimate the horizontal Laplacian from both the near ring (r1) and far diagonal ring (r2) of grid values around P, then combine the two estimates to cancel truncation error.

Answer 1

Correct Answer: 0.16

Alternative — apply the standard 3×3 combined-ring second-vertical-derivative template directly in one step (this is algebraically the same result as the two-step Richardson extrapolation above, just packaged as a single operator, the way it is usually quoted in gravity-interpretation texts):

\[ (g_{xx}+g_{yy})=\dfrac{1}{s^2}\Big[2\,\Sigma(\text{Ring1})-\tfrac12\,\Sigma(\text{Ring2})-6P\Big] \]

Plug in the numbers (\(s=1\) km):
\(2\times30.99=61.98\)
\(0.5\times30.94=15.47\)
\(6\times7.78=46.68\)
\[ (g_{xx}+g_{yy})=61.98-15.47-46.68=-0.17 \]

By Laplace's equation, \(g_{zz}=-(g_{xx}+g_{yy})=0.17\ \text{mGal/km}^2\), reproducing the two-step result in a single line and confirming the value sits inside 0.16–0.18 mGal/km\(^2\).