The angular speed of rotation of the Earth ab | vedclass

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

Vedclass · Accepted Answer

$8.75 	imes 10^{-4} \ rad \ s^{-1}$

Vedclass · Answer

(B) The effective acceleration due to gravity at the equator $(g_E)$ is given by $g_E = g_P - \omega^2 R$,where $g_P$ is the acceleration due to gravity at the poles,$\omega$ is the angular speed of the Earth,and $R$ is the radius of the Earth.Given that the weight at the equator is half the weight at the poles,we have $g_E = \frac{g_P}{2}$.Substituting this into the equation: $\frac{g_P}{2} = g_P - \omega^2 R$.Rearranging for $\omega$: $\omega^2 R = g_P - \frac{g_P}{2} = \frac{g_P}{2}$.Thus,$\omega = \sqrt{\frac{g_P}{2R}}$.Using $g_P = 9.8 \ m/s^2$ and $R = 6400 \ km = 6.4 	imes 10^6 \ m$:$\omega = \sqrt{\frac{9.8}{2 	imes 6.4 	imes 10^6}} = \sqrt{\frac{9.8}{12.8 	imes 10^6}} = \sqrt{0.7656 	imes 10^{-6}} \approx 8.75 	imes 10^{-4} \ rad/s$.

The angular speed of rotation of the Earth about its axis,at which the weight of a man standing on the equator becomes half of his weight at the poles,is given by:

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