Determine the speed with which the earth have to roate on its axis so that a person on the equator would weigh `2//5th` as much as at present. Take the equatorial radius as `6400 km`.

Explanation

True weight at equator, `W = mg`, Observed wieght at equator, `W' = mg' = (2)/(5) mg`
At equator latitude `lambda = 0`
As `mg' = mg - m R omega^(2) cos^(2) lambda`
`(2)/(5) mg = mg - m R omega^(2) cos^(2) 0 = mg - mR omega^(2)`
or `m R omega^(2)` or `omega = ((3)/(5) (g)/(R ))^(1//2) = ((3xx9.8)/(5xx 6.4 xx 10^(6)))^(1//2) = 9.6 xx 10^(4) rad//s`.