document.write( "Question 1090222: Given the Earth's radius of 4000 miles, and a mass of 200 pounds:
\n" ); document.write( "------
\n" ); document.write( "The weight on the surface is 200 pounds.
\n" ); document.write( "What is the weight in an elevator to the Earth's center at various depths?
\n" ); document.write( "Depth of 500 miles, 1000 miles, 2000 miles, etc.
\n" ); document.write( "-\Ignore any effects of pressure and temperature.
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #704664 by Alan3354(69443)\"\" \"About 
You can put this solution on YOUR website!
Given the Earth's radius of 4000 miles, and a mass of 200 pounds:
\n" ); document.write( "------
\n" ); document.write( "The weight on the surface is 200 pounds.
\n" ); document.write( "What is the weight in an elevator to the Earth's center at various depths?
\n" ); document.write( "Depth of 500 miles, 1000 miles, 2000 miles, etc.
\n" ); document.write( "-\Ignore any effects of pressure and temperature.
\n" ); document.write( "==================================\r
\n" ); document.write( "\n" ); document.write( "Inside a solid sphere of constant density, the gravitational force varies linearly with distance from the center, becoming zero by symmetry at the center of mass. This can be seen as follows: take a point within such a sphere, at a distance r from the center of the sphere. Then you can ignore all the shells of greater radius, according to the shell theorem. So, the remaining mass m is proportional to r^3, and the gravitational force exerted on it is proportional to m/r^2, so to r^3/r^2 = r, so it is linear in r.
\n" ); document.write( "-------------------------------------------
\n" ); document.write( "-->
\n" ); document.write( "Weight = 200*(4000 - d)/4000 where d = the depth in miles.
\n" ); document.write( "
\n" ); document.write( "
\n" );