text: Acceleration towards the Earth is equal to the Earth's geocentric gravitational constant* (3.986 x 10^14 meters^3/sec^2) divided by the square of distance from the center of Earth (6378000 meters)^2.  \n\nIn other words:  (3.986 x 10^14)/(6.378 x 10^6)^2\n\nThat comes to about 9.8 meters/sec^2.\n\nAcceleration away from Earth due to the rotation of the Earth is equal to the square of the velocity divided by the radius of the Earth (6.378 x 10^6 meters). \n\nIn other words, v^2/r.\n\nThe Earth rotates 2(pi) radians per day, or about 7.292115 x 10^-5 rad/sec.  You convert that to a linear velocity by multiplying the radius (6.378 x 10^6 meters) by the angular velocity (7.292115 x10^-5 rad/sec).  At the equator, that comes to 465 meters/sec.\n\nSquaring the linear velocity (465 meters/sec) and dividing by the radius (6.378 x 10^6) gives you an acceleration of only 3.4 cm/sec^2 away from the Earth.\n\nIn other words, the acceleration towards the Earth due to gravity is about 288 times the centripetal acceleration away from the Earth.  It's there, but the force of gravity is much stronger.\n\n* - the geocentric gravitational constant is just the universal gravitational constant multiplied by the mass of the Earth.