Explain the variation of g with altitude from the Earth's surface.

Explanation

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)) , g. = (GM)/(R_(e)^(2) (1+h/(R_(e)))^(2))`
`g. = (GM)/(R_(e)^(2)) (1+h/(R_(e)))^(-2)`
If `h lt lt R_(e)` : We can use Binomial expension. Taking the terms up to first order
`(1+x)^(n)=1+nx `
`[1+h/(R_(e))]^(-2) = [ 1-2 (h/(R_(e)))] `
Replacing this value and we get
`g. = (GM)/(R_(e)^(2)) [ 1- (2h)/(R_(e))] `
` g . = g (1-2 h/(R_(e)))`
We find that `g. < g`. This means that as altitude h increase the acceleration due to gravity g decreases.