explain why the value of g is zero at the centre of the earth using mathematical concept derive it
Answers
g = GM/r^2
Assuming Earth to be a perfect sphere, and considering it's uniform density
we make the following adjustments in the above equation.
g= GM/r^2
multiplying and dividing the RHS value by v
g= GM/v ×v ×1/r^2
we know that M/V is density p
therefore , g= pGv//r^2
For a sphere , v= 4/3πr^3. this makes
g= 4/3πr^3 ×pG/r^2
g=4πprG/3
at the centre of the earth, r=0
Thus, g= 0
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Why value of 'g' is zero at center of Earth explain and derive it mathematically?
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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 =
Assuming earth to be a perfect sphere and considering its uniform density.
We make the following adjustments in the following equation.
g =
Multiplying and dividing RHS by volume 'V'.
g =
We know that;-
= Density, ρ
Therefore,
g =
For sphere;-
V = π
This makes
g = π
×
g =
So, at center of Earth
since, r = 0
so, g = 0.
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