Monday

December 22, 2014

December 22, 2014

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

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 2:04pmGravitational 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 Problem 6.20 - The center of a 1.10 diameter spherical pocket of oil is ...

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 - 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 ...

College Physics - When searching for gold, measurements of g can be used find ...

physics - Careful measurements of local variations in the acceleration due to ...

math - in his left pocket ralph has 4 quarters and 5 nickels. in his right ...

Physics - Calculate the percentage of Earth’s gravitational force on the ISS ...

spanish - Could someone please tell me what the word of 'pocket' as in 'pants ...