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

Here, `omega = ? , mg' = (3)/(5) mg ,`
`R = 6400 km = 6.4 xx 10^(6)m`
Acceleration due to gravity at the equator is
`g' = g - R omega^(2)` or `mg' = mg - mR omega^(2)`
or `(3)/(5)mg = mg - mR omega^(2)`
or `mRomega^(2) = mg - (3)/(5) mg = (2)/(5) mg`
or `omega = sqrt((2g)/(5R)) = sqrt((2 xx 9.8)/(5 xx 6.4 xx 10^(6)))`
`= 7.8 xx 10^(-4) rad s^(-1)`