Explain the variation of 'g' with altitude.

Explanation

Variation of g with altitude:
Consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is
`g'= (GM)/((R_(E)+h)^(2))`
`R_(E) to ` Radius of Earth
`g'=(GM)/(R_(E)^(2)(1+(h)/(R_(E)))^(2))`
`=(GM)/(R_(E)^(2))(1+(h)/(R_(E)))^(-2)`
If `h lt lt R_(E)`
We can use Binomial expansion. Taking the terms upto first order
`g'=(GM)/(R_(E)^(2))(1-2(h)/(R_(E)))`
`=g(1-2(h)/(R_(E)))`
We find that `g' lt g`. This means that as altitude h increases, the acceleration due to gravity g decreases.
Variation of g with depth:
Consider a particle of mass m which in in a deep mine on the Earth. (Example: coal mines -in Neyveli).Assume the depth of the mine as d. To calculate g' at a depth d, consider the following points.
The part of the Earth which is above the radius `(R_(E)-d)` do not contribute to the acceleration. The result is proved earlier and is given as

`g'=(GM')/((R_(E)-d)^(2))`
Here M' is the mass of the Earth of radius `(R_(E)-d)`
Assuming the density of Earth `rho` to be constant, `rho=(M)/(V)`
where M is the mass of the Earth and V its volume,
Thus,
`rho=(M')/(V'), (M')/(V')=(M)/(V)` and
`M'=(M)/(V)V'`
`M'=((M)/((4)/(3)pi R_(E)^(3)))((4)/(3)pi (R_(E)-d)^(3))`
`M'=(M)/(R_(E)^(3))(R_(E)-d)^(3)`
`g'= G(M)/(R_(E)^(3))(R_(E)-d)^(3). (1)/((R_(E)-d)^(2))`
`g'=GM (R_(E)(1-(d)/(R_(E))))/(R_(E)^(3))`
`=GM(1-(d)/(R_(E)))/(R_(E)^(2))`
Thus
`g'=g(1-(d)/(R_(E)))`
Here also `g' lt g`. As depth increases, g' decreases.