Question

A geologist searching for oil finds that the gravity at a certain location is 2 parts in 107 smaller than average. Assume that the location contains oil and the oil is located 3300 m below the Earth's crust. Estimate the size of the deposit. Take the density of rock to be 3200 kg/m3 and that of oil to be 1060 kg/m3. kg

Answer 1

g� = 2/10^7 g d = 3300 m rho_1 = 3200 Kg/m^3 rho_2 = 1060 kg/m^3 rho - M/V rightdoublearrow M = (3200 - 1060) V = 2140 V F = GMm/d^2 Mg� = GMm/d^2 rightdoublearrow g� = GM/d^2 9.8 times 2/10^7 = (6.67 times 10^-11) (2140 V)/(3300)^2 V = 14953512.61 m^3 1m^3 = 2406.53 kg 149535512.61 m^3 = 359861697161.34 rightdoublearrow 359861697161.34 kg