Calculate that imaginary angular velocity of the Earth for which effective acceleration due to gravity at the equator becomes zero. In this condition, find the length (in hours) of a day? Radius of Earth `= 6400 km. g = 10 ms^(-2)`.

Explanation

At equator, latitude `lambda = 0^(@)`. Let `g` be the acceleration due to gravity if the Earth were at rest. The acceleration due to gravity `(g')` at the equator, when Earth is roating with angular velocity `omega` be zero.
`:. g' = g - R omega^(2) cos^(2) lambda = 0`
or `g = R omega^(2) cos^(2) 0^(@) = R omega^(2) ( :' lambda = 0^(@))`
or `omega = sqrt((g)/(R )) = sqrt((10)/(6400xx 10^(3)))`
`:.` New time period rotation of Earth
`T = (2 pi)/(omega) = (2xx 3.14)/(1.25 xx 10^(3)) = 5024 s = 1.4 h`