What should be the angular speed of the Earth | vedclass

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(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.For a body to appear weigh

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.For a body to appear weightless at the equator,the effective gravity $g'$ must be zero.At the equator,the latitude $\lambda = 0^\circ$,so $\cos \lambda = 1$.Substituting these values into the equation: $0 = g - \omega^2 R$.Rearranging for $\omega$,we get $\omega = \sqrt{\frac{g}{R}}$.Given $g = 10\,m/s^2$ and $R = 6400\,km = 6.4 	imes 10^6\,m$.$\omega = \sqrt{\frac{10}{6.4 	imes 10^6}} = \sqrt{\frac{1}{6.4 	imes 10^5}} = \sqrt{\frac{1}{64 	imes 10^4}} = \frac{1}{8 	imes 10^2} = \frac{1}{800}\,rad/s$.

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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