Calculation of true thickness of the bed

27) A micro gravity survey with appropriate station spacing is performed to detect a subsurface spherical cavity in a bedrock of density 2500 kg/m^3. The depth to the center of the cavity is 4 m from the surface and the elevation measurement accuracy of the surveying instrument is 0.1m. The smallest cavity that can be detected by the survey must have a radius greater than ______m.

 

 (Thanks to Chandrasekhar, ANU)

(Thanks to Srirama Madhusudhan)

 

Solution:

 Given that Density= 2500 kg/m³

 

               Depth (Z) = 4 m

 

               Radius(R) =?

 

 We know that gravity for a buried sphere at a depth is(X=0)

 

$\triangledown g_{z}=\frac{4}{3}\pi\rho G\frac{R^{3}}{Z^{2}}$-----------------(1)

 

 Here $\triangledown g_{z}$ is the lowest gravity value that we can measure with the elevation

accuracy 0.1 m (From Free-air correction we know that=0.3086 h mGal/m)

 

     =$0.3086\times 0.1$

 

    =$0.03086 mgals$

 

    =$0.03086\times 10^{-5} m/sec^{2}$

 

 Now substitute this value and the given values in the equation (1) then

 

$\triangledown g_{z}=\frac{4}{3}\pi\rho G\frac{R^{3}}{Z^{2}}$

 

$0.03086\times 10^{-5}=\frac{4}{3}(2500)(3.14)(6.673\times 10^{-11}) \frac{R^{3}}{4^{2}}$

 

$0.03086\times 10^{-5}=4365.2541\times 10^{-11}\times R^{3}$

 

$R^{3}=7.06\times 10^{-6}\times 10^{-5} \times 10^{11}$

 

$R^{3}=7.06$

 

$R=\sqrt[3]{7.06}$

 

$R=1.918 m$