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(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.At the equator,the latitud

Vedclass · Accepted Answer

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

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(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.At the equator,the latitude $\lambda = 0^\circ$,so $\cos \lambda = 1$.Thus,the effective acceleration becomes $g' = g - \omega^2 R$.For the effective acceleration to be zero $(g' = 0)$,we have $0 = g - \omega^2 R$.This implies $\omega^2 = \frac{g}{R}$.Given $g = 10\,m/s^2$ and $R = 6400\,km = 6.4 	imes 10^6\,m$.$\omega = \sqrt{\frac{10}{6.4 	imes 10^6}} = \sqrt{\frac{1}{6.4 	imes 10^5}} = \sqrt{\frac{1}{64 	imes 10^4}} = \frac{1}{8 	imes 10^2} = \frac{1}{800}\,rad/s$.

In order to make the effective acceleration due to gravity equal to zero at the equator,the angular velocity of rotation of the earth about its axis should be $(g = 10\,m/s^2$ and radius of earth is $6400\,km)$.

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