At what rate should the earth rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?

Explanation

At equator the apparent weight will become zero if `g' = 0`. Let it be for amgular speed of rotation of earth be `omega'`. Then `g' = g - R omega'^(2) = 0`
or `omega' = sqrt((g)/(R )) = sqrt((9.8)/(6.37 xx 10^(6)))`
Present speed of rotation of earth,
`omega = (2pi)/(T) = (2 xx 22//7)/(24 xx 60 xx 60)`
`:. (omega')/(omega) = sqrt((9.8)/(6.37 xx 10^(6))) xx (24 xx 60 xx 60 xx 7)/(2 xx 22)`
`= 17.047 ~~17`
or `omega' = 17 omega` or `(2pi)/(T') = 17 xx (2pi)/(T) =(17 xx 2pi)/(24)`
or `T' = (24)/(17) = 1.41h`