Question

A spherical ore body produces a maximum gravity anomaly of 18 mGal when its centre is at a depth of 2 km from the surface. Assuming that the density contrast and the radius of the body remain unchanged, the ore body will produce a maximum gravity anomaly of 2 mGal if the depth to its centre in km is _______ (in integer).

Hint:

For spherical ore bodies, the gravity anomaly at the surface decreases rapidly with increasing depth, following an inverse square law: \(\Delta g \propto 1/z^2\). Doubling the depth reduces the anomaly by a factor of 4.

Answer 1

Step 1: Recall gravity anomaly relation for a buried sphere
The maximum gravity anomaly due to a buried sphere is inversely proportional to the square of the depth of burial (\(z\)): \[ \Delta g \propto \frac{1}{z^2} \] Step 2: Set up ratio of anomalies at two depths
Let \(\Delta g_1 = 18 \, \text{mGal}\) at depth \(z_1 = 2 \, \text{km}\).
Let \(\Delta g_2 = 2 \, \text{mGal}\) at depth \(z_2 = ? \, \text{km}\).
Using proportionality: \[ \frac{\Delta g_1}{\Delta g_2} = \frac{z_2^2}{z_1^2} \] Step 3: Substitute values
\[ \frac{18}{2} = \frac{z_2^2}{(2)^2} \] \[ 9 = \frac{z_2^2}{4} \] \[ z_2^2 = 36 \] \[ z_2 = 6 \, \text{km} \] Step 4: Final Answer
Thus, the required depth is: \[ \boxed{6 \, \text{km}} \]