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

GATE-2016

Q.33: A spherical cavity of radius 8m has its center 15m below the surface. If the cavity is full of sediments of density 1.5*10^3 kg/m^3 and is in a rock body of density 2.4*10^3 kg/m^3, the maximum value of its gravity anomaly is.........m Gal

Solution;

Gravity anomaly of the Spherical body (g_Z) = ( (G*M_S*Z) / (x^²+ z^²)^^3/2 )------------------(1)

Excess mass = M_S

Center is at a depth =z

At maximum gravity anomaly z= 0

substitute values in eq(1)

g_Z =( G*M_S*Z) / (z^³) ----------------(2)

Density = mass/ volume

mass = density * volume

Spherical body volume = (4/3) *П*R^³

mass = (4/3) *П*R^{^3} * Δρ

g_Z =( ((4/3)* ∏*R^³* Δρ*G) / (z^{^2}) )

Universal Gravitational constant (G) = 6.67* 10^{^-11} N m^{^2} / kg^{^2}

Density contrast (Δρ) = 2.4 *10^³ – 1.5*10^³

                       

                                 = 900 kg / m^³

Radius of the sphere (R) = 8m

Depth to the center (z) = 15 m

Maximum Gravity anomaly (g_m) = ( ((4/3)*3.14*8^{^3}* 900*6.67*10^{^-11}) / (15^²) )

              = (38626762.752 * 10^-11)/ 675

               = 57224.833 *10^-11 N/ kg

              = 57224.833 * 10^-11 m/s^{^2}

( 1 Gal = 1 cm/ s^{^2}

= 10^^-2 m/s^²

1 m Gal = 10^^-5 m/s^² )

           g_m = 57224.833 * 10^{^-11} *10^{^5} mGal

           g_m = 57224.833 * 10^^-6 mGal

                            

           g_m = 0.057224 mGal

Maximum gravity anomaly (g_m) = 0.06 mGal