Calculate the angular speed of rotation of the Earth so that the apparent g at the equator becomes half of its value at the surface. Also calculate the length of the day in this situation.

Explanation

The apparent acceleration due to gravity,
`g=g_(0)-omega^(2)R=g_(0)//2`
`rArromega=sqrt((g_(0))/(2R))=sqrt((9.8 "ms"^(-2))/(6.4 xx10^(6)"m" xx 2))=8.75 xx10^(-4) "rads"^(-1)`
The length of the day = Time period of rotation of the Earth
`=(2pi)/(omega)=2pi sqrt((2R)/(g_(0)))=2pisqrt((2xx6.4 xx10^(6))/(9.8))~~2h`.