The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to…

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, $\cos 0^{\circ}=1, \cos 90^{\circ}=0$ )
  1. $R \omega^2 \cos ^2 \theta$
  2. $R \omega^2$
  3. $\frac{R \omega^2}{g^2}$
  4. $\omega \cos ^2 \theta$

Solution

Acceleration due to gravity at a place of latitude $\theta$ due to rotation of earth is $g^{\prime}=g-R \omega^2 \cos ^2 \theta$ At equator, $\theta=0, \cos 0=1$ $\therefore g^{\prime}=g_e=g-R \omega^2$ At poles, $\theta=90, \cos 90=0$ $\therefore g^{\prime}=g_p=g$ $\therefore g_p-g_e=g-\left(g-R \omega^2\right)=R \omega^2$ :

Asked in: MHT CET 2022 (05 Aug Shift 2)

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