geff at equator is given by g-ω2R
(draw the FBD at the equator to convince urself)
So geff is zero if g=ω'2R
So, ω' = √g/R
Now ω'/ω = x (given)
So, x = √g/R/ω
the speed of earth's rotation about its axis ω.its speed is increased to x times to make eff. acceleration zero at the equator then x=??????
geff at equator is given by g-ω2R
(draw the FBD at the equator to convince urself)
So geff is zero if g=ω'2R
So, ω' = √g/R
Now ω'/ω = x (given)
So, x = √g/R/ω