What should be the angular speed of the earth | vedclass

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Question

$\mathrm{g}_{	heta}=\mathrm{g}-\omega^{2} \mathrm{R} \cos ^{2} 	heta$

at equator $	heta=0^{\circ} $ and for weight lessness

${g_{equator}}$ $

Vedclass · Accepted Answer

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

Vedclass · Answer

$\mathrm{g}_{	heta}=\mathrm{g}-\omega^{2} \mathrm{R} \cos ^{2} 	heta$

at equator $	heta=0^{\circ} $ and for weight lessness

${g_{equator}}$ $=0$

$\Rightarrow 0=\mathrm{g}-\omega^{2} \mathrm{R} \cos ^{2} 0$

$\Rightarrow \omega=\sqrt{\frac{g}{R}}=\sqrt{\frac{10}{6400 	imes 10^{3}}}=\frac{1}{800} \mathrm{\,rad} / \mathrm{sec}$

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

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