Solution:
The effective acceleration due to gravity at latitude \(\lambda\) is \(g' = g - \omega^2 R \cos^2 \lambda\). At the poles, \(\lambda = 90^\circ\), so \(g' = g\). Thus, rotation has no effect at the poles.
The effective acceleration due to gravity at latitude \(\lambda\) is \(g' = g - \omega^2 R \cos^2 \lambda\). At the poles, \(\lambda = 90^\circ\), so \(g' = g\). Thus, rotation has no effect at the poles.
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