Gravitation Near Earth's Surface — Question 7

Question 7

The width of a gravity anomaly can reveal a buried depth

After subtracting the background field, a compact excess mass $\Delta M$ is modeled as a point at depth $d$ below a flat surface. Let $x$ be horizontal distance from the point directly above it.

1. Derive the downward gravity anomaly $\Delta g_z(x)$.

2. If the anomaly falls to half its central value at $x=w$, infer $d$ in terms of $w$.

Question 7 - Solution

The width of a gravity anomaly can reveal a buried depth

(a) Derivation and calculation

Let $s=\sqrt{x^2+d^2}$. The downward projection of the point-source field is $$\Delta g_z=\frac{G\Delta M}{s^2}\frac ds =\boxed{\frac{G\Delta M d}{(x^2+d^2)^{3/2}}}.$$ At the central point, $$\Delta g_z(0)=\frac{G\Delta M}{d^2},\quad \frac{\Delta g_z(x)}{\Delta g_z(0)}=[1+(x/d)^2]^{-3/2}.$$

(b) Result and interpretation

The half-height condition gives $$[1+(w/d)^2]^{-3/2}=\frac12 \quad\Longrightarrow\quad 1+(w/d)^2=2^{2/3}.$$ $$\boxed{d=\frac{w}{\sqrt{2^{2/3}-1}}\simeq1.30477w},\qquad \Delta M=\frac{d^2\Delta g_z(0)}G.$$ Width determines depth and amplitude then determines excess mass within the stated model.

Physical check

The anomaly is symmetric in $x$ and decreases to zero far from the buried source.