Editing Space elevator (section)

===Apparent gravitational field===
An Earth space elevator cable rotates along with the rotation of the Earth. Therefore, the cable, and objects attached to it, would experience upward centrifugal force in the direction opposing the downward gravitational force. The higher up the cable the object is located, the less the gravitational pull of the Earth, and the stronger the upward centrifugal force due to the rotation, so that more centrifugal force opposes less gravity. The centrifugal force and the gravity are balanced at geosynchronous equatorial orbit (GEO). Above GEO, the centrifugal force is stronger than gravity, causing objects attached to the cable there to pull ''upward'' on it. Because the counterweight, above GEO, is rotating about the Earth faster than the natural orbital speed for that altitude, it exerts a centrifugal pull on the cable and thus holds the whole system aloft.

The net force for objects attached to the cable is called the ''apparent gravitational field''. The apparent gravitational field for attached objects is the (downward) gravity minus the (upward) centrifugal force. The apparent gravity experienced by an object on the cable is zero at GEO, downward below GEO, and upward above GEO.

The apparent gravitational field can be represented this way:<ref name="aravind"/>{{rp|Table 1}}

{{block indent|The downward force of actual [[Newton's law of universal gravitation|gravity]] ''decreases'' with height: [[Newton's law of universal gravitation|<math>g_r = -GM/r^2</math>]]}}
{{block indent|The upward [[centrifugal force]] due to the planet's rotation ''increases'' with height: [[Centrifugal force|<math>a = \omega^2 r</math>]]}}
{{block indent|Together, the apparent gravitational field is the sum of the two:
{{block indent|<math>g = -\frac{GM}{r^2} + \omega^2 r</math>}}}}

where
{{block indent|''g'' is the acceleration of ''apparent'' gravity, pointing down (negative) or up (positive) along the vertical cable (m s<sup>−2</sup>),}}
{{block indent|''g<sub>r</sub>'' is the gravitational acceleration due to Earth's pull, pointing down (negative)(m s<sup>−2</sup>),}}
{{block indent|''a'' is the centrifugal acceleration, pointing up (positive) along the vertical cable (m s<sup>−2</sup>),}}
{{block indent|''G'' is the [[gravitational constant]] (m<sup>3</sup> s<sup>−2</sup> kg<sup>−1</sup>)}}
{{block indent|''M'' is the mass of the Earth (kg)}}
{{block indent|''r'' is the distance from that point to Earth's center (m),}}
{{block indent|''ω'' is Earth's rotation speed (radian/s).}}

At some point up the cable, the two terms (downward gravity and upward centrifugal force) are equal and opposite. Objects fixed to the cable at that point put no weight on the cable. This altitude (r<sub>1</sub>) depends on the mass of the planet and its rotation rate. Setting actual gravity equal to centrifugal acceleration gives:<ref name="aravind"/>{{rp|p. 126}}
{{block indent|<math>r_1 = \left(\frac{GM}{\omega^2}\right)^\frac{1}{3}</math>}}

This is {{convert|35786|km|mi|0|abbr=on}} above Earth's surface, the altitude of geostationary orbit.<ref name="aravind"/>{{rp|Table 1}}

On the cable ''below'' geostationary orbit, downward gravity would be greater than the upward centrifugal force, so the apparent gravity would pull objects attached to the cable downward. Any object released from the cable below that level would initially accelerate downward along the cable. Then gradually it would deflect eastward from the cable. On the cable ''above'' the level of stationary orbit, upward centrifugal force would be greater than downward gravity, so the apparent gravity would pull objects attached to the cable ''upward''. Any object released from the cable ''above'' the geosynchronous level would initially accelerate ''upward'' along the cable. Then gradually it would deflect westward from the cable.