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 the body to be weight

Vedclass · Accepted Answer

$2.5 	imes 10^{-3} \, rad/sec$

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 the body to be weightless,the effective gravity must be zero,so $g' = 0$. Substituting the values: $0 = g - \omega^2 R \cos^2(60^{\circ})$. Since $\cos(60^{\circ}) = 1/2$,we have $\cos^2(60^{\circ}) = 1/4$. Thus,$g = \omega^2 R (1/4) \implies \omega^2 = 4g/R$. Taking the square root,$\omega = 2 \sqrt{g/R}$. Using $g = 9.8 \, m/s^2$ and $R = 6.4 	imes 10^6 \, m$: $\omega = 2 \sqrt{9.8 / (6.4 	imes 10^6)} \approx 2 \sqrt{1.53 	imes 10^{-6}} \approx 2 	imes 1.237 	imes 10^{-3} \approx 2.47 	imes 10^{-3} \, rad/s$. Rounding to the nearest option,$\omega = 2.5 	imes 10^{-3} \, rad/s$.

What should be the angular speed of the Earth so that a body at $60^{\circ}$ latitude becomes weightless? (Radius of Earth $R = 6400 \, km$)

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