Question

The figure shows a homogeneous rock layer of thickness \(100\ \text{m}\). A vertical borehole is drilled through the layer and gravity is measured at the top point \(A\) and at the base point \(B\). If the difference in measurements between \(A\) and \(B\) is \(5\ \text{mGal}\), find the density \(\rho\) of the rock layer in g/cc (ignore terrain corrections). 

Hint:

Between two levels separated by a uniform layer, the gravity change equals the free-air term minus \(4\pi G\rho d\). Using \(4\pi G \approx 0.08386\ \text{mGal}/(\text{m}\cdot\text{g/cc})\) makes quick estimates easy.

Answer 1

Step 1: Account for free-air change between \(A\) and \(B\).
Going down by \(d=100\ \text{m}\) increases \(g\) by the free-air gradient \[ \Delta g_{\text{FA}} \approx 0.3086\ \frac{\text{mGal}}{\text{m}} \times 100 = 30.86\ \text{mGal}. \] 

Step 2: Account for the rock layer mass between \(A\) and \(B\).
The slab attraction of a layer of thickness \(d\) at a point above it is \(\ +2\pi G \rho d\), and at a point below it is \(-2\pi G \rho d\).
Therefore, moving from \(A\) to \(B\) changes the slab contribution by \[ \Delta g_{\text{slab}} = (-2\pi G\rho d) - (+2\pi G\rho d) = -4\pi G\rho d. \] In convenient units, \[ 4\pi G \approx 0.08386\ \frac{\text{mGal}}{\text{m}\cdot\text{(g/cc)}}. \] 

Step 3: Match the observed difference and solve for \(\rho\).
Observed \( \Delta g_{B-A} = 5\ \text{mGal} \) (downward increase): \[ 5 = \Delta g_{\text{FA}} + \Delta g_{\text{slab}} = 30.86 - (0.08386)\,\rho \times 100. \] \[ 0.08386\times 100\,\rho = 30.86 - 5 = 25.86 \;\Rightarrow\; \rho = \frac{25.86}{8.386} \approx 3.0859. \] \[ \boxed{3.09\ \text{g/cc}} \]