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

Question

The difference in the acceleration due to gravity at the pole and equator is (g= acceleration due to gravity, R= radius of earth, = latitude, = angular velocity, cos ⁡0^ =1 , cos ⁡90^ =0 )

Accepted Answer

Correct answer: C. (c) The variation of the acceleration due to gravity as per the latitude is given by, g=g_0-R ^2 cos^2 ⁡ where, g_0= accleration due to gravity at the equator R= Radius of earth = Angular velocity of earth = latitude of the place At poles: =90^ ⇒ cos^2 ⁡ =0 So from equation (i), g_p=g_0 At equator: =0^ ⇒ cos^2 ⁡ =1 ;g_e=g_0-R ^2 The difference, g_p-g_e=g_0-g_0+R ^2g_p-g_e=R ^2

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

Solution

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