At the bottom, the acceleration due to gravity would be g. At the top, it would be something like .8g... But would the total acceleration effectively be 0 like on the ISS? How would the acceleration change, then, as it moves up? Or, because the top of the elevator would be well beyond the center of gravity of the system, would its acceleration have an upward component at the top?
2 Answers
$\begingroup$
$\endgroup$
Well if you are talking of orbit of 330 km like the International space station. the gravity difference will be least and the maths will show centripetal acceleration will point upwards
It is important to note that even thought the satellites are travelling in a circular orbit they do not experience the effect of the centripetal acceleration because the satellite is in free fall.
According to the newton's cannon experiment the object in orbit is in free fall and experiences zero gravity.
There will be a upward component of acceleration because the space lift is not in geostationary orbit. the space lift with respect to the earth at the altitude of 330-450 km has not attained the orbital velocity but is at the velocity of $R*\omega,R\to \text{radius of earth+lift length},\omega \to \text{angular velocity of earth}$ which is not sufficient to experience the effect of zero gravity.
In order for the space lift ,given that it is connected to the earth should be at the geostationary orbit altitude to attain the characteristics of a satellite or in this case the ISS to have no upward component of acceleration on its crew.
$\begingroup$
$\endgroup$
If you treat the entire elevator as one center of mass M1 then you would have one force pointing away from the center of gravity of Earth (pointing down) or any other big object M2, but you would have an equaly big force pointing in the other direction of M2 (Pointing up). These two forces counter act each other while the elevator is moving at constant velocity but when the elevator would slow down at the top there is a external force acting on M1 slowing it down so the force downwards toward the center of M2 would be greater than the force vector pointing in the other direction, and vice versa when the elevator begins to slow down at the bottom.
