Chapter 6 Problem 67

Problem

A geologist searching for oil finds that the gravity at a certain location is 2 parts in $10^{7}$ smaller than average. Assume that a deposit of oil is located 2000m directly below. Estimate the size of the deposit, assumed spherical. Take the density (mass per unit volume) of rock to be 3000 kg/m$^{3}$ and that of oil to be 800 kg/m$^{3}$.

Solution

Since all of the masses (or mass holes) are spherical, and g is being measured outside of their boundaries, we can use the simple Newtonian gravitation expression. In the diagram, the distance $r=2000$ m. The radius of the deposit is unknown.

$g_{\text{actual}}=g_{\text{full}}-g_{\text{dirt in cavity}}+g_{\text{oil}}$ $=g_{\text{full}}-{\frac {GM_{\text{dirt in cavity}}}{r^{2}}}+{\frac {GM_{\text{oil}}}{r^{2}}}$ $=g_{\text{full}}-{\frac {G\left(M_{\text{dirt in cavity}}-M_{\text{oil}}\right)}{r^{2}}}$

$\Delta g=g_{\text{full}}-g_{\text{actual}}={\frac {G\left(M_{\text{dirt in cavity}}-M_{\text{oil}}\right)}{r^{2}}}$ $={\frac {G}{r^{2}}}\left(V_{\text{dirt in cavity}}\rho _{\text{dirt in cavity}}-V_{\text{oil}}\rho _{\text{oil}}\right)$ $={\frac {GV}{r^{2}}}\left(\rho _{\text{dirt in cavity}}-\rho _{\text{oil}}\right)$ $={\frac {2}{10^{7}}}g$

$V_{\text{oil}}=V={\frac {2}{10^{7}}}g{\frac {r^{2}}{G}}{\frac {1}{\left(\rho _{\text{dirt in cavity}}-\rho _{\text{oil}}\right)}}$ $={\frac {2}{10^{7}}}\left(9.80{\text{m/s}}^{2}\right){\frac {\left(2000{\text{m}}\right)^{2}}{\left(6.67\times 10^{-11}{\text{Nm}}^{2}/{\text{kg}}^{2}\right)}}{\frac {1}{\left(3000-800\right){\text{kg/m}}^{3}}}$ $=5.34\times 10^{7}{\text{m}}^{3}\approx 5\times 10^{7}{\text{m}}^{3}$

$r_{\text{oil}}=\left({\frac {3V_{\text{oil}}}{4\pi }}\right)^{1/3}=234{\text{m}}\approx 200{\text{m}}$

$m_{\text{oil}}=V_{\text{oil}}\rho _{\text{oil}}=4.27\times 10^{10}{\text{kg}}=4\times 10^{10}{\text{kg}}$