When a body is taken from the pole to the equ | vedclass

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(C) The acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - R \omega^2 \cos^2 \lambda$,where $\omega$ is the a

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Decreases

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(C) The acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - R \omega^2 \cos^2 \lambda$,where $\omega$ is the angular velocity of the Earth's rotation and $R$ is the radius of the Earth.At the poles,the latitude $\lambda = 90^\circ$. Substituting this into the formula: $g_{	ext{pole}} = g - R \omega^2 \cos^2(90^\circ) = g - 0 = g$.At the equator,the latitude $\lambda = 0^\circ$. Substituting this into the formula: $g_{	ext{equator}} = g - R \omega^2 \cos^2(0^\circ) = g - R \omega^2$.Since $g - R \omega^2 Weight is defined as $W = mg$. Since $g$ decreases when moving from the pole to the equator,the weight of the body decreases.

When a body is taken from the pole to the equator,its weight:

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