94. Gravity anomalies at two points, distant 2units and 5 units respectively from the position of the maximum gravity anomaly along a profile across a spherical ore deposit are in the ratio 8:1. Then, the depth(units) to the center of the ore body is:
Solution:
Above problem clearly mention that the ratio of the gravity anomalies are 8:1
Therefore,
$$\frac{Δg_{1}}{Δg_{2}} =\frac{8}{1}$$
Gravity anomaly of the spherical body is given by
$$Δg=\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(x_{1}^{2}+z^{2})}^{3/2}}$$
$$\frac{\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(x_{1}^{2}+z^{2})}^{3/2}}}{\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(x_{2}^{2}+z^{2})}^{3/2}}} =\frac{8}{1}$$
$$\frac{\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(4+z^{2})}^{3/2}}}{\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(25+z^{2})}^{3/2}}} =\frac{8}{1}$$
$$\frac{{(25+z^{2})}^{3/2}}{{(4+z^{2})}^{3/2}} =\frac{8}{1}
$$
$$\frac{{(25+z^{2})}}{{(4+z^{2})}} =\frac{8}{1}^{2/3}$$
$$\frac{{(25+z^{2})}}{{(4+z^{2})}} =4$$
$$25+z^{2} =4z^{2}+16
$$
$$3z^{2} =9
$$
$$z =\sqrt{3}units$$
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