The difference in the acceleration due to gra | vedclass

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Question

(B) The effective acceleration due to gravity at a latitude $	heta$ due to the rotation of the Earth is given by $g^{\prime} = g - R \omega^2 \cos^2

Vedclass · Accepted Answer

$R \omega^2$

Vedclass · Answer

(B) The effective acceleration due to gravity at a latitude $	heta$ due to the rotation of the Earth is given by $g^{\prime} = g - R \omega^2 \cos^2 	heta$.At the equator,the latitude $	heta = 0^{\circ}$,so $\cos 0^{\circ} = 1$.Thus,the acceleration due to gravity at the equator is $g_e = g - R \omega^2$.At the poles,the latitude $	heta = 90^{\circ}$,so $\cos 90^{\circ} = 0$.Thus,the acceleration due to gravity at the poles is $g_p = g$.The difference in the acceleration due to gravity between the pole and the equator is $g_p - g_e = g - (g - R \omega^2) = R \omega^2$.

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

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