Determine the speed with which the earth would have to rotate on its axis , so that a person on the equator would weigh `(3)/(5)` th as much as the person. Take `R = 6400 km`.

Explanation

Weight on the equator `W=(3)/(5)W`
`implies(3)/(5)"mg"="mg"-m omega^(2)Rimpliesomega=sqrt((2g)/(5R))`.