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?

Question

anonymous · Answer

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.
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anonymous · Answer

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.