Calculation of true thickness of the bed

 

 44) An infinite horizontal cylinder of radius 40 km is buried at a depth of 100 km and yields the same maximum gravity anomaly as that of an infinite horizontal cylinder of radius 1km buried at a depth of 1km having density contrast with the surrounding of 200kg/m³. The density contrast of the deeper cylinder with respect to the surrounding is ______kg/m³.

 (Thanks to Chandrasekhar, ANU)

Solution:

We know that the gravity anomaly of an infinite horizontal cylinder is

$\triangle g= 2\pi R^{2}\triangle\rho G(\frac{Z}{x^{2}+Z^{2}})$

 But the maximum gravity anomaly occur at X=0 then

 $\triangle g_{max}= 2\pi R^{2}\triangle\rho G(\frac{1}{Z})$------------------------------(1)

The given that $R_{1}$ =40km,                                         $R_{2}$ =1km

                        $Z_{1}$=100 km                                  $Z_{2}$=1 km

                       $\triangle\rho_{1}$=?                  $\triangle\rho_{2}$=200kg/m³      

By substituting the above values in the equation (1) and equate them because both the cylinders have same maximum gravity anomaly.

 

  $ 2\pi R^{2}_{1}\triangle\rho_{1} G(\frac{1}{Z_{1}})=2\pi R^{2}_{2}\triangle\rho_{2}G(\frac{1}{Z_{2}})$

 $ \Rightarrow R^{2}_{1}\triangle\rho_{1} (\frac{1}{Z_{1}})= R^{2}_{2}\triangle\rho_{2}(\frac{1}{Z_{2}})$

  $\Rightarrow \frac{(40)^{2}(\triangle\rho_{1})}{(100)}=\frac{(1)^{2}(200)}{(1)}$

  $\Rightarrow \triangle\rho_{1}=\frac{(200)(100)}{1600}$

   $\Rightarrow \triangle\rho_{1}=12.5 kg/m^{3}$