What should be the angular velocity of the Ea | vedclass

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(C) The effective acceleration due to gravity at the equator is given by $g' = g - \omega^2 R$,where $\omega$ is the angular velocity and $R$ is the r

Vedclass · Accepted Answer

$7.8 	imes 10^{-4} \, rad/s$

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(C) The effective acceleration due to gravity at the equator is given by $g' = g - \omega^2 R$,where $\omega$ is the angular velocity and $R$ is the radius of the Earth.Given that the weight at the equator becomes $3/5$ of its initial weight,we have $g' = \frac{3}{5}g$.Substituting this into the equation: $\frac{3}{5}g = g - \omega^2 R$.Rearranging the terms: $\omega^2 R = g - \frac{3}{5}g = \frac{2}{5}g$.Solving for $\omega$: $\omega = \sqrt{\frac{2g}{5R}}$.Substituting the values $g = 10 \, m/s^2$ and $R = 6400 	imes 10^3 \, m$:$\omega = \sqrt{\frac{2 	imes 10}{5 	imes 6400 	imes 10^3}} = \sqrt{\frac{20}{32000 	imes 10^3}} = \sqrt{\frac{20}{3.2 	imes 10^7}} = \sqrt{6.25 	imes 10^{-7}} \approx 7.8 	imes 10^{-4} \, rad/s$.

What should be the angular velocity of the Earth due to rotation about its own axis so that the weight of an object at the equator becomes $3/5$ of its initial value? (Radius of Earth at the equator is $R = 6400 \, km$,take $g = 10 \, m/s^2$)

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