What should be the angular velocity of earth | vedclass

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

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$(2g/5R)^{1/2}$

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(A) The effective acceleration due to gravity at the equator is given by the formula $g' = g - \omega^2 R$, where $\omega$ is the angular velocity of the Earth and $R$ is the radius of the Earth.Given that the weight at the equator becomes $(3/5)^{th}$ of its initial value, we have $g' = (3/5)g$.Substituting this into the equation: $(3/5)g = g - \omega^2 R$.Rearranging the terms: $\omega^2 R = g - (3/5)g = (2/5)g$.Solving for $\omega$: $\omega^2 = (2g)/(5R)$.Therefore, $\omega = \sqrt{(2g)/(5R)}$ or $(2g/5R)^{1/2}$.

What should be the angular velocity of earth due to rotation about its own axis so that the weight at the equator becomes $(3/5)^{th}$ of its initial value? ($g$ = acceleration due to gravity, $R$ = radius of earth)

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