Variation of ‘g’ with Depth

by • 04/07/2012 • GeneralComments (0)571

Variation of ‘g’ with Depth

Suppose earth is perfectly spherical in shape with uniform density ρ.

Let

Re = Radius of earth

Me = Mass of earth

d = Depth (between P and Q)

Me = Mass of earth at a depth ‘d’

At the surface of earth,

g = G Me / Re2 ……………………………….. (1)

At a depth ‘d’, acceleration due to gravity is

g = G Me / (Re – d)2 ……………………… (2)

Me = ρ x Ve = ρ x (4/3) π Re3 = 4/3 π Re3 ρ

Me = ρ x Ve = ρ x (4/3) π (Re – d)3 = 4/3 π (Re – d)3 ρ

Ve = Volume of earth

Substitute the value of Me in (1),

(1) => g = (G / Re2) x (4/3) π Re3 ρ

g = 4/3 π Re ρ G …………………………… (3)

Substitute the value of Me in (2)

g = [G / Re – d)2] x (4/3) π (Re – d)3 ρ

g = 4/3 π (Re – d) ρ G

Dividing (4) by (3)

g / g = [4/3 π (Re – d) ρ G] / [4/3 π Re ρ G]

g / g = (Re – d) / Re

g / g = 1 – d/Re

g / g = g (1 – d / Re) ……………………… (5)

Equation (5) gives the value of acceleration due to gravity at a depth ‘d’ below the surface of earth

From (5), we can conclude that as the value of ‘d’ increases, value of ‘g’ decreases.

At the centre of the earth.

d = Re => Re / d = 1

Therefore,

(5) => g = g (1-1)

g = 0

Thus at the centre of the earth, the value of gravitational acceleration is zero.

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