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Question
Explain the variation of g with latitude.
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Solution
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’.
OPz, 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
aPQ = 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.
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