Homework Help: Physics Problem 6.20

Posted by Michael Moskvich on Sunday, April 4, 2010 at 1:31pm.

The center of a 1.40 km diameter spherical pocket of oil is 1.50 km beneath the Earth's surface.

Estimate by what percentage g directly above the pocket of oil would differ from the expected value of g for a uniform Earth? Assume the density of oil is 8.0*10^2kg/m^3.

Express your answer using two significant figures.

Physics Problem 6.20 - bobpursley, Sunday, April 4, 2010 at 2:04pm
Gravitational field= GMe/re^2-M!@#$%^&/700^2 + G Massoil/700^2

= 9.8N/kg+ G/700^2 (volumemhole)(densityoil-densityEarth)
= 9.8+G/490000 * 4/3*PI*700^3 (800-6000)
(I estimated Earth density at 6times that of water)
= 9.8+6.67E-11/4900 (1.44E9)(-5200)
=9.8-.12 or 9.7N/kg

check all this.

Answer this Question

Related Questions

Physics - The center of a 1.50 diameter spherical pocket of oil is 1.20 beneath ...
Physics - The center of a 1.50 diameter spherical pocket of oil is 1.20 beneath ...
Physics Problem 6.20 - The center of a 1.10 diameter spherical pocket of oil is ...
Physics - The center of a 1.10km diameter spherical pocket of oil is 1.10km ...
Phsyics - The center of a 1.10km diameter spherical pocket of oil is 1.10km ...
physics - Careful measurements of local variations in the acceleration due to ...
Physics - The Near Earth Asteroid Rendezvous (NEAR), after traveling 2.1 billion...
physics - The Near Earth Asteroid Rendezvous (NEAR), after traveling 2.1 billion...
physics - The Near Earth Asteroid Rendezvous (NEAR), after traveling 2.1 billion...
Physics please help - How far from the center of the Earth (in km) is the center...

For Further Reading

First Name:
School Subject:
Answer: