The angular speed of the Earth in rad/s,so th | vedclass

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(A) For a body on the equator to appear weightless,the effective gravity must be zero.The condition for weightlessness due to the rotation of the Eart

Vedclass · Accepted Answer

$1.25 	imes 10^{-3}$

Vedclass · Answer

(A) For a body on the equator to appear weightless,the effective gravity must be zero.The condition for weightlessness due to the rotation of the Earth is given by $g - \omega^2 R = 0$.Therefore,$\omega^2 R = g$,which implies $\omega = \sqrt{g / R}$.Given $g = 10\, m/s^2$ and $R = 6.4 	imes 10^3\, km = 6.4 	imes 10^6\, m$.Substituting the values: $\omega = \sqrt{\frac{10}{6.4 	imes 10^6}}$.$\omega = \sqrt{\frac{10}{6.4} 	imes 10^{-6}} = \sqrt{1.5625 	imes 10^{-6}}$.$\omega = 1.25 	imes 10^{-3}\, rad/s$.

The angular speed of the Earth in $rad/s$,so that bodies on the equator may appear weightless is: [Use $g = 10\, m/s^2$ and the radius of the Earth $R = 6.4 \times 10^3\, km$]

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