Question

Explain the variation of g with latitude.

Answer 1

When an 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 were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to the spinning of the Earth.

This centrifugal force is given by mω²R’.

OP_z, cos λ = `"PZ"/"OP" = "R’"/"R"`

R’ = R cos λ

       Variation of g with latitude

where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is

a_PQ = ω²R’ cos λ = ω²R cos² λ

Since R’ = R cos λ

Therefore, g’ = g – ω² R cos² λ

From the above expression, we can infer that at the equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g’ = g, it is maximum. At the equator, g’ is minimum.