Explain the variation of 'g' with latitude.

Explanation

(i) Latitute : When and object is on the surface of the Earth , it experiences a centrifugal force that depends - on the latitude of the object on Earth . If the Earth .If the Earth were not spining . The force on the object experiences an additional centrifungal force due to spining of the Earth .
This centrifungal force latitute . The conponent of centrifungal acceleration experiend by the object in the direction opposite to g is
` a_(PQ) = omega^(2) R cos lamda = omega^2 R cos ^2 lamda `
since 1 R. = `R cos lamda `
therefore ` g. = g - omega^2 R cos ^2 lamda`
from the above expression , we can infer that at equator `lamda =0 , g. = g- omega^2 R ` the equator g. is minimum .
(ii ) alitiude : consider an object of mass m at a height h from the surface of the earth accelartion experiened by the object due to Earth is
` g. ( GM )/( (R_e + h )^2) : g . = (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 ) [ 1 + h/(R_e) ]^(-2)`
if ` h lt lt R_e ` we can use Binomial expansion . 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.=(GM )/(R_(e )^(2))(1-(2h )/(R_e ))`
` g. = 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.