What should be the angular speed of the Earth | vedclass

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(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - \omega^2 R_e \cos^2 \lambda$.At the equa

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$\sqrt{\frac{g}{R_e}}$

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(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - \omega^2 R_e \cos^2 \lambda$.At the equator,the latitude $\lambda = 0^\circ$,so $\cos 0^\circ = 1$.Substituting this into the equation,we get $g' = g - \omega^2 R_e$.For the acceleration due to gravity to be zero at the equator,we set $g' = 0$.Therefore,$0 = g - \omega^2 R_e$.Rearranging the terms,we get $\omega^2 R_e = g$.Solving for $\omega$,we find $\omega = \sqrt{\frac{g}{R_e}}$.

What should be the angular speed of the Earth about its axis so that the acceleration due to gravity at the equator becomes zero?

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