What should be the angular speed of the earth | vedclass

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(A) The effective acceleration due to gravity at a latitude $	heta$ is given by $g_{	heta} = g - \omega^2 R \cos^2 	heta$.At the equator,the latitu

Vedclass · Accepted Answer

$\frac{1}{800}\,rad/s$

Vedclass · Answer

(A) The effective acceleration due to gravity at a latitude $	heta$ is given by $g_{	heta} = g - \omega^2 R \cos^2 	heta$.At the equator,the latitude $	heta = 0^{\circ}$,so $\cos 0^{\circ} = 1$.For a body to appear weightless,the effective gravity must be zero,i.e.,$g_{equator} = 0$.Substituting these values into the equation: $0 = g - \omega^2 R(1)^2$.Rearranging for angular speed $\omega$: $\omega = \sqrt{\frac{g}{R}}$.Given $g = 10\,m/s^2$ and $R = 6400\,km = 6400 	imes 10^3\,m = 6.4 	imes 10^6\,m$.$\omega = \sqrt{\frac{10}{6.4 	imes 10^6}} = \sqrt{\frac{1}{6.4 	imes 10^5}} = \sqrt{\frac{1}{640000}} = \frac{1}{800}\,rad/s$.

What should be the angular speed of the earth,so that a body lying on the equator may appear weightless? $(g = 10\,m/s^2, R = 6400\,km)$

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