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Question

(A) 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 pol

Vedclass · Accepted Answer

$\sqrt{\frac{5}{6} \frac{g}{R}}$

Vedclass · Answer

(A) 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 without rotation),$\omega$ is the angular velocity,and $R$ is the radius of the earth.Given that the weight at the equator becomes $\frac{1}{6}$ of its present value,we assume the present weight is approximately $mg$ (since $\omega^2 R$ is very small compared to $g$).Thus,$g' = \frac{1}{6} g$.Substituting this into the equation: $\frac{1}{6} g = g - \omega^2 R$.Rearranging for $\omega^2 R$: $\omega^2 R = g - \frac{1}{6} g = \frac{5}{6} g$.Solving for $\omega$: $\omega = \sqrt{\frac{5g}{6R}}$.Therefore,the required angular speed is $\sqrt{\frac{5}{6} \frac{g}{R}}$.

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{1}{6}$ th as much as at present is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

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