Show that the object lying at the equator will fly off the surface of earth, if the speed of rotation of the earth increase seventeen times its present speed.

Explanation

Due to rotation of earth, the value of acceleration due to gravity `(g')` at a place of latitude `lambda` is given
by `g' = g [1 - (R omega^(2) xx cos^(2) lambda)/(g)]`
At equator, `lambda = 0^(@), cos lambda = 1`
So, `g' = g [1 - (R omega^(2) xx 1^(2))/(g)] = g (1- (R omega^(2))/(g))`
Here, `(R omega^(2))/(g) = ((6.4 xx 10^(6)) xx [2 pi//(24 xx 60 xx 60)]^(2))/(9.8) = (1)/(289)`
When the speed of rotating of earth becomes seventeen times its present speed, then
`g' = g [1- (R(17 omega)^(2))/(g)] = g [1- 289 (R omega^(2))/(g)]`
`= g [1- (289)/(289)] = g [1 - 1] = 0`
It means a body lying at the equator will just fly off because it experiences no resultant force towards the centre of the earth.