The angular speed of the Earth,such that an o | vedclass

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(A) For an object on the equator to appear weightless,the effective acceleration due to gravity must be zero.The effective gravity $g'$ at the equator

Vedclass · Accepted Answer

$1.25 	imes 10^{-3}\,rad/s$

Vedclass · Answer

(A) For an object on the equator to appear weightless,the effective acceleration due to gravity must be zero.The effective gravity $g'$ at the equator is given by $g' = g - R\omega^2$.For weightlessness,$g' = 0$,which implies $g = R\omega^2$.Therefore,the angular speed $\omega$ is given by $\omega = \sqrt{\frac{g}{R}}$.Given $g = 10\,m/s^2$ and $R = 6400\,km = 6.4 	imes 10^6\,m$.$\omega = \sqrt{\frac{10}{6.4 	imes 10^6}} = \sqrt{\frac{1}{0.64 	imes 10^6}} = \sqrt{\frac{1}{64 	imes 10^4}} = \frac{1}{8 	imes 10^2} = \frac{1}{800} = 0.00125\,rad/s$.Thus,$\omega = 1.25 	imes 10^{-3}\,rad/s$.

The angular speed of the Earth,such that an object on the equator may appear weightless,is ($g = 10\,m/s^2$,radius of the Earth $R = 6400\,km$).

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