If the angular speed of earth is increased so much that the objects start flying from the equator, then the length of the day will be nearly : 

1.5 hrs

At equator, \(g_{\lambda} = g_e - R_e \omega^2\)

... where \(R_e = 6400 km \text{ ; } \omega = \frac{2 \pi}{T} \text{ ; } g_e = 10 m/s^2\)

Now, for the objects to fly at equator, \(g_{\lambda} = 0 \Rightarrow R_e \omega^2 = g_e\)

Time period for the given condition :

\(T = 2\pi \sqrt{\frac{R}{g}}\)

Putting the value : T = 5024 sec ≈ 1.4 hrs.