If the Earth has no rotational motion, the weight of a person on the equator is \(W\). Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weight \(\frac{3}{4}\,W\) . Radius of the Earth is \(6400\, km\) and \(g = 10\, m/s^2\)

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Answer & Solution

Correct Answer

(C) \(0.63 \times {10^{ - 3}}\,rad/s\)

Step-by-step Solution

Detailed explanation

We know, \(g' = g - {\omega ^2}R{\cos ^2}\theta \)

\(\frac{{3g}}{4} = g - {\omega ^2}R\)

\(Given,\,g' = \frac{3}{4}g\)

\({\omega ^2}R = \frac{g}{4}\)

\(\omega = \sqrt {\frac{g}{{4R}}} = \sqrt {\frac{{10}}{{4 \times 6400 \times {{10}^{ - 3}}}}} \)

\( = \frac{1}{{2 \times 8 \times 1000}} = 0.6 \times {10^{ - 3}}\,rad/s\)

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