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Why value of 'g' is zero at center of Earth?
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When we move towards centre of earth, the mass is equally distributed in all directions.
The mass beneath you = Mass in front of you = Mass behind you
Thus, all the gravitational forces applied cancel each other and acceleration due to gravity (g) at centre of earth on centre of earth becomes zero (0).
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The general equation of acceleration due to gravity is;-
g = ![\frac{GM}{r^2}](https://tex.z-dn.net/?f=%5Cfrac%7BGM%7D%7Br%5E2%7D%20)
Assuming earth to be a perfect sphere and considering its uniform density.
We make the following adjustments in the following equation.
g = ![\frac{GM}{r^2}](https://tex.z-dn.net/?f=%5Cfrac%7BGM%7D%7Br%5E2%7D%20)
Multiplying and dividing RHS by volume 'V'.
g = ![\frac{GM}{V} × V × \frac{1}{r^2}](https://tex.z-dn.net/?f=%5Cfrac%7BGM%7D%7BV%7D%20%C3%97%20V%20%C3%97%20%5Cfrac%7B1%7D%7Br%5E2%7D%20)
We know that;-
= Density, ρ
Therefore,
g = ![\frac{ρGV}{r^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%CF%81GV%7D%7Br%5E2%7D)
For sphere;-
V =
π![r^{3}](https://tex.z-dn.net/?f=r%5E%7B3%7D)
This makes
g =
π
× ![\frac{ρG}{r^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%CF%81G%7D%7Br%5E2%7D)
g = ![\frac{4πρG}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%CF%80%CF%81G%7D%7B3%7D)
So, at center of Earth
since, r = 0
so, g = 0.
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