Question

A geologist searching for oil finds that the gravity at a certain location is 2 parts in 10 smaller than average. Assume that the location contains oil and the oil is located 3100 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 1000 kg/m3 kg

Answer 1

Solution)

Given,

density of rock = 3200 kg/m3

density of the oil = 1000 kg/m3

value of g due to oil deposit

g=Gm_oil/D^2

Let,

g=g_oil-g_rock

So, g=GV(g_oil-g_rock)/D^2

V=gD^2/G(g_oil-g_rock)

So, V=9.8*(-2*10^-7)(2500m)^2/(6.67*10^-11(1000-3200 ))

V=8.34*10^7 m^3

Here, mass of the oil m_oil=g_oil*v

=8.34*10^7 m^3*1000

=8.34 *10^10 kg (Ans)

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