anonymous
  • anonymous
Why does the gravitational acceleration (g) decrease at the distance further from 4500 m below the sea level? and what is the value of g is at the center of the earth?
MIT 18.01 Single Variable Calculus (OCW)
schrodinger
  • schrodinger
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anonymous
  • anonymous
The value of g at the center of the earth is approximately 0. If we take a hollow sphere as in the diagram, all the particles of the sphere exert a force \[G m_{1}m_{2} \div r ^{2}\] These forces can be cancelled up resulting in no force acting at the center. When we consider infinite number of spheres from r=0 to r=R(R=Radius of Earth), we get a resultant force of 0. |dw:1404915773985:dw|
anonymous
  • anonymous
May I suggest another perspective? Newton developed the concept that gravitational FORCES are related not only to the square of the radius of a planetary body, but also related to it's mass. This in turn affects the acceleration of a free-falling object. The average radius of the Earth is 6367 kilometers. 4.5 kilometers (4500 meters) below sea level means that the radius of the Earth at that spot is (6367 - 4.5) = 6362.5 km, so the radius has decreased and the gravitational force should therefore be reduced by a small fraction. 6367^2 = 40538689 6362.5^2 = 40481406.25 If the radius zero, then gravitational forces should be zero. Of course, no one has been there so this remains our understanding at this time. I hope that I have helped and not hindered your understanding of this problem.

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