Minimum angular speed of Earth for a body to escape its gravity At what angular speed must the Earth rotate for a body to just fly off the surface at the equator?
 A: $$g=\omega^2r$$
$$\omega=\sqrt{9.81/6.4\times10^6}= 1.24\times 10^{-3}$$
The Earth rotates $7.27\times 10^{−5}\:\mathrm{rads^{−1}}$ roughly 17 times slower than the minimum speed to fly off the Earth.
A: To escape the gravity of earth from its surface, a body has to have at a minimum the known escape speed ve=11.2km/s in any direction away from the surface. The escape speed derives from the kinetic energy that is equal to the (negative) gravitational potential energy of the body at the surface of the earth according to Newton's gravitational law. Therefore, the angular velocity  of the rotation of earth at the equator to achieve this is given by =ve/r, where r= 6378 km is the earth radius at the equator. This yields an angular escape velocity at the equator of  =11.2kms^-1/6378km=1.80·10^-3s^-1. For comparison, the angular velocity of the earth's natural rotation is 7.27·10^-5s-1. Thus the earth's rotation would have to speed up by a factor of 24.8 to let bodies at the equator escape the gravity of earth. This corresponds to reducing the day to 58.2 minutes.  
