GATE-2018 (24) :Wyllie – Time average equation

Gate_2020

28) The Gravity anomaly over a spherical ore body is shown in the figure below. The calculated excess mass due to the ore body will be _______ * 10¹⁰ kg.

(Thanks to Chandrasekhar, ANU)

Solution:

The excess mass of a spherical ore body is 

 

$M_{e} =\frac{g_{max}Z^{2}}{G} kg$ 

 

(0r)

$M_{e} = 253.26\times g_{max}\times (x_{1/2})^{2} tones$

Given that 

 

$g_{max} = 4 m gals =4 \times 10^{-5} m/sec^2$

$x_{1/2} = 150 m$

$M_{e}=?$

By using 

 

$M_{e} =\frac{g_{max}Z^{2}}{G} $ 

 

$M_{e} =\frac{4 \times 10^{-5}\times (1.3\times x_{1/2})^{2}}{G} $

$M_{e} =\frac{4 \times 10^{-5}\times (1.3\times 150)^{2}}{6.673\times 10^{-11}} $

$M_{e} =\frac{4 \times 195 \times 195 \times 10^{-5} }{6.673\times 10^{-11}} $

$M_{e} =\frac{152100 \times 10^{-5} }{6.673\times 10^{-11}} $

$M_{e} =22793.346 \times10^{6} kg $

$M_{e} =2.2793 \times10^{10} kg $