The center of a 1.40 km diameter spherical pocket of oilis 1.50 km beneath the E

Question

The center of a 1.40 km diameter spherical pocket of oilis 1.50 km beneath the Earth's surface. Estimate by what percentage g directlyabove the pocket of oil would differ from the expected valueof g for a uniform Earth? Assume the density of oil is8.0*10^2kg/m^3. Express your answer using twosignificant figures. g/g =     ? % The center of a 1.40 km diameter spherical pocket of oilis 1.50 km beneath the Earth's surface. Estimate by what percentage g directlyabove the pocket of oil would differ from the expected valueof g for a uniform Earth? Assume the density of oil is8.0*10^2kg/m^3. Express your answer using twosignificant figures. g/g =     ? % Estimate by what percentage g directlyabove the pocket of oil would differ from the expected valueof g for a uniform Earth? Assume the density of oil is8.0*10^2kg/m^3. Express your answer using twosignificant figures. g/g =     ? % Estimate by what percentage g directlyabove the pocket of oil would differ from the expected valueof g for a uniform Earth? Assume the density of oil is8.0*10^2kg/m^3. Express your answer using twosignificant figures. g/g =     ? %

Explanation / Answer

The decrease in the value of g is given by .         g = G m /r2        where m is the difference between the mass of the sphere ofoil and the mass of an equivalent amount of rock. . So...       m = density of rock * volume - density of oil *volume = (4/3) R3 ( density rock -density oil) .                       = (4/3) 7003 * (2500 - 800) =    2.4425 x1012 kg . (note: I used 2500 for the density of rock; that valuecan vary a little, depending on the type of rock) . now...    g = 6.67 x10-11 * 2.4425 x 1012 /15002 = 7.24 x10-5     and so .      g / g = 7.24 x10-5 / 9.80 =    0.00074%

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.