Solution:
The effective gravity is \(g' = g - \omega^2 R \cos^2 \theta\). At the poles, \(\theta = 90^\circ\), so \(g' = g\), which is independent of the earth's angular velocity.
The effective gravity is \(g' = g - \omega^2 R \cos^2 \theta\). At the poles, \(\theta = 90^\circ\), so \(g' = g\), which is independent of the earth's angular velocity.
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