Variation of g with depth and terrain?

Answer 1

Variation of "g" with height:

Consider the body which is at height h from the surface of earth whose radius is considered to be R .

Let the mass of earth be M .

so,

acceleration due to gravity acted on a body at surface of earth = GM / R 2

or, g = GM / R 2 ............ ... ... ... (i)

again,

acceleration due to gravity acted on a body at the height h above the surface of earth = GM / (R+h)2

or, g ' = GM / (R+h) 2 ... ... ... ... ... (ii)

Now, dividing (i) by (ii), we get,

g/g ' = (GM / R 2 ) / (GM / (R + h) 2 )

or, g/g ' = (GM x (R+h) 2 ) / (GM x R 2 )

or, g/g ' = (R+h) 2 / R 2

or, g/g ' = (R/R + h/R) 2

or, g/g ' = (1+h/R) 2

or g ' = g(1+h/R) -2

Now, expanding binomially, we get,

or, g ' = g(1- 2h/R)

Variation of g with depth:

Consider the body at the depth of x from the surface of earth whose mass is considered to be M and radius to be R .

But at this situation body will experience acceleration due to gravity only by that mass in which it is standing. Let that mass is considered to be M ' . Similarly radius will be (R-x).

Now, acceleration due to gravity at the surface of earth = GM / R 2

or, g = GM / R 2 ... ... ... ... ... ... ... ... (i)

Similarly, acceleration due to gravity below the depth of x from surface of earth = GM ' / (R-x) 2

or, g ' = GM ' / (R-x) 2 ... ... ... ... ... ... (ii)

Dividing (ii) by (i), we get,

g '/ g =( GM ' / (R-x) 2 ) / (GM / R 2 )

or, g '/ g = M ' R 2 / (M (R-x) 2 )... ... ... ... (iii)

Now, Density of earth ( ρ ) = M/V

ρ = M/ (4/3 π R 3 )

M = ρ x (4/3 π R 3 ) ... ... ... ... (iv)

Density of earth below x meter from the surface of earth (ρ) = M '/V '

ρ = M ' / (4/3 π (R - x) 3 )

M ' = ρ x (4/3 π (R - x) 3 ) ... ... ... (v)

Putting (iv) and (v) in (iii), we get,

g '/ g = (ρ x (4/3 π (R - x) 3 ) x R 2 )/ ( ρ x (4/3 π R 3 ) (R-x) 2 )

or, g ' = g(1- x/R)