Homework Help: Physics

Posted by Mandy on Friday, March 18, 2011 at 1:54pm.

The center of a 1.50 diameter spherical pocket of oil is 1.20 beneath the Earth's surface. Estimate by what percentage directly above the pocket of oil would differ from the expected value of for a uniform Earth? Assume the density of oil is 800 kg/m^3. Solve for delta g/ g = %

Physics - bobpursley, Friday, March 18, 2011 at 4:19pm
do it this way.

you know g for a solid earth.

Now find g for a solid sphere of Earth the size of the bubble, 1.20km away.

Then, find g for a sphere of oil the size of the bubble, 1.20km away.

you have three numbers:

Take the first, subtract the second, and add the third. That is the resultant g value.

Remember in working these fictional bubbles, the mass is volume*density.
g'=GM'/distance^2 where M' is the mass of the bubble (Earth, then oil).

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