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}})$
$\Rightarrow
\frac{(40)^{2}(\triangle\rho_{1})}{(100)}=\frac{(1)^{2}(200)}{(1)}$
$\Rightarrow
\triangle\rho_{1}=\frac{(200)(100)}{1600}$
$\Rightarrow
\triangle\rho_{1}=12.5 kg/m^{3}$
Refer material
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