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Explain the variation of g with lattitude.

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प्रश्न

Explain the variation of g with latitude.

संक्षेप में उत्तर
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उत्तर

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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अध्याय 6: Gravitation - Evaluation [पृष्ठ ४४]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 6 Gravitation
Evaluation | Q III. 8. | पृष्ठ ४४

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