Calculation of true thickness of the bed

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

$$Δg=\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(x_{1}^{2}+z^{2})}^{3/2}}$$


$$\frac{\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(x_{1}^{2}+z^{2})}^{3/2}}}{\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(x_{2}^{2}+z^{2})}^{3/2}}} =\frac{8}{1}$$  

$$\frac{\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(4+z^{2})}^{3/2}}}{\frac{4}{3}\pi R^{3} G\rho \frac{z}{{(25+z^{2})}^{3/2}}} =\frac{8}{1}$$

$$\frac{{(25+z^{2})}^{3/2}}{{(4+z^{2})}^{3/2}} =\frac{8}{1}$$


$$\frac{{(25+z^{2})}}{{(4+z^{2})}} =\frac{8}{1}^{2/3}$$


$$\frac{{(25+z^{2})}}{{(4+z^{2})}} =4$$


$$25+z^{2} =4z^{2}+16$$


$$3z^{2} =9$$



$$z =\sqrt{3}units$$