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 - momega^(2) R" "implies" "omega = sqrt((2g)/(5R))`.