The angular velocity of the Earth at present | vedclass

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(C) The effective acceleration due to gravity at the equator is given by $g' = g - \omega'^2 R$, where $\omega'$ is the new angular velocity.For the w

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$17$

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(C) The effective acceleration due to gravity at the equator is given by $g' = g - \omega'^2 R$, where $\omega'$ is the new angular velocity.For the weight of a body at the equator to be zero, the effective gravity must be zero, so $g - \omega'^2 R = 0$.This implies $\omega' = \sqrt{g/R}$.Given the present angular velocity $\omega = \sqrt{g/R_e}$ is not the standard way to relate, we use the condition $g = \omega_0^2 R$ where $\omega_0$ is the required angular velocity.We know $g = 9.8 \ m/s^2$ and $R = 6.4 	imes 10^6 \ m$.Calculating $\omega_0 = \sqrt{9.8 / (6.4 	imes 10^6)} \approx 1.24 	imes 10^{-3} \ rad/s$.The present angular velocity of the Earth is $\omega = 2\pi / T$, where $T = 86400 \ s$.$\omega \approx 7.27 	imes 10^{-5} \ rad/s$.Ratio $\frac{\omega_0}{\omega} = \frac{1.24 	imes 10^{-3}}{7.27 	imes 10^{-5}} \approx 17$.Thus, the required angular velocity is $17\omega$.

The angular velocity of the Earth at present is $\omega$. With what angular velocity should it rotate so that the weight of a body at the equator appears to be zero (in $\omega$)? (Express the answer as a multiple of $\omega$)

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