Find the angular speed of rotation of the Ear | vedclass

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(C) The apparent acceleration due to gravity $g^{\prime}$ at the equator due to the rotation of the Earth is given by the formula:$g^{\prime} = g_0 -

Vedclass · Accepted Answer

$1.14 	imes 10^{-3} \ rad \ s^{-1}$

Vedclass · Answer

(C) The apparent acceleration due to gravity $g^{\prime}$ at the equator due to the rotation of the Earth is given by the formula:$g^{\prime} = g_0 - \omega^2 R$Given that the apparent $g^{\prime}$ is $(1/6)$th of its original value $g_0$,we have:$g^{\prime} = \frac{g_0}{6}$Substituting this into the equation:$\frac{g_0}{6} = g_0 - \omega^2 R$Rearranging the terms to solve for $\omega^2 R$:$\omega^2 R = g_0 - \frac{g_0}{6} = \frac{5}{6} g_0$$\omega = \sqrt{\frac{5 g_0}{6 R}}$Substituting the values $g_0 = 9.8 \ m/s^2$ and $R = 6.4 	imes 10^6 \ m$:$\omega = \sqrt{\frac{5 	imes 9.8}{6 	imes 6.4 	imes 10^6}}$$\omega = \sqrt{\frac{49}{38.4 	imes 10^6}} = \sqrt{1.276 	imes 10^{-6}}$$\omega \approx 1.13 	imes 10^{-3} \ rad \ s^{-1}$Thus,the angular speed is approximately $1.14 	imes 10^{-3} \ rad \ s^{-1}$.

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$

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