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

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Question

$g^{\prime}=g-R \omega^2 \cos (2 \lambda)$

$g^{\prime}=g-R \omega 2 \cos 2$ where $\omega \omega$ is the angular velocity of rotation of earth abou

Vedclass · Accepted Answer

Decreases

Vedclass · Answer

$g^{\prime}=g-R \omega^2 \cos (2 \lambda)$

$g^{\prime}=g-R \omega 2 \cos 2$ where $\omega \omega$ is the angular velocity of rotation of earth about its polar axis, $R R$ is the radius of the earth and $\lambda \lambda$ is the latitude of a plac

At poles, $\lambda=90$

$\lambda=90$

$	herefore g_p 0 l e=g-R \omega^2 \cos (2 	imes 90)$

$	herefore$ gpole $=g-R \omega 2 \cos 2900=g$

at equator $\lambda=0$

gequator $= g - R \omega 2 \cos 200= g - R \omega 2< g$ gequator $= g - R \omega 2 \cos 200= g - R \omega 2$

Thus, the acceleration due to gravity decreases from poles to equator.

Hence, when a body is taken from poles to equator on the earth, its weight decreases.

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

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