The angular speed with which the earth would | vedclass

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(B) The effective acceleration due to gravity at the equator is given by $g' = g - \omega^2 R$, where $g$ is the acceleration due to gravity at the po

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$\sqrt{2g/5R}$

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(B) The effective acceleration due to gravity at the equator is given by $g' = g - \omega^2 R$, where $g$ is the acceleration due to gravity at the poles (or if the earth were stationary), $\omega$ is the angular velocity, and $R$ is the radius of the earth.Given that the weight at the equator becomes $3/5$ of its present weight, 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 $\omega$: $\omega^2 = 2g / (5R)$, which gives $\omega = \sqrt{2g/5R}$.

The angular speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $3/5^{th}$ as much as at present is ($g$ = gravitational acceleration, $R$ = equatorial radius of the earth.)

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