Explain the variation of 'g' with latitude.

Explanation

When an object is on the surface for the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth. This centrifugal force is given by `m omega^2 R `
`OP_2 cos lamda = (PZ)/(OP) = (R.)/(R ) `
` R. = R cos lamda `
where `lamda ` is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g. is
`a_(PQ) = omega^2 R cos lamda = omega^2 R cos^2 lamda`
Since ` 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 acceleration due to gravity is minimum. At poles `lamda = 90 , g. = g ` , it is maximum. At the equator, g. is minimum.