Why does gravity decrease as we go down into the Earth? We all know that gravity decreases as the distance between the two increases. Hence F = G (Mm)/(r^2).Hence the acceleration due to gravity g =F/m= G M/r^2 increases as r decreases. Then why does it decrease as we go deep into the earth?

e3r2a1cakCh7

e3r2a1cakCh7

Answered question

2022-11-23

Why does gravity decrease as we go down into the Earth?
We all know that gravity decreases as the distance between the two increases. Hence
F = G M m r 2 .
Hence the acceleration due to gravity
g = F m = G M r 2
increases as r decreases. Then why does it decrease as we go deep into the earth?

Answer & Explanation

Russell Knox

Russell Knox

Beginner2022-11-24Added 15 answers

It's actually not entirely true that the strength of the Earth's gravitational field decreases as a function of depth. It is true for certain regions in the Earth, but it's untrue for others because of the non-trivial dependence of the Earth's density on depth.
To see what's going on, assume that the Earth is a sphere whose density is spherically symmetric.
Now consider a mass m at some radius r from the center of the Earth. Using Newton's Law of Gravitation, one can show that that given the spherical symmetry, the gravitational attraction on m of all mass with radii greater than r exert no net force on it. It follows that only the mass with radii less than or equal to r contribute to the gravitational force on m, which, by the Law of Gravitation is
F ( r ) = G M ( r ) m r 2
where M ( r ) is the mass of stuff at radii less than or equal to r. Notice, then, that F ( r ) will be an increasing function of r (and will decrease as r 0), provided M ( r ) / r 2 is an increasing function of r
Now, If the Earth were uniformly dense with density ρ 0 , then the mass within a radius r would be
M ( r ) = 4 3 π r 3 ρ 0
namely just the density times the volume of a sphere of radius r, and in this case the strength of the gravitational field as a function of radius would be
g ( r ) = F ( r ) m = G 1 r 2 4 3 π r 3 ρ 0 = ( 4 3 π g ρ 0 ) r
However, the Earth's density is not constant, and instead has some non-trivial dependence on r.

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