Why the Space Elevator's Center of Mass is not at GEO

I have often encountered people who think that the Space Elevator's center of mass must be located at the altitude corresponding to geosynchronous orbit, because it rotates synchronously with the Earth. This assumption that people make is actually incorrect for two reasons as we shall see.

The Space Elevator is Connected to the Earth

One reason that people easily accept for the center of mass not being at GEO, is that the Space Elevator is in fact not a free body. It is attached to the Earth, and there is a significant tension on it at its base. This is equivalent to having an elevator that is not attached to the Earth but that has an anchor mass attached at its base, which is sufficient to provide the appropriate tension in the base of the cable. This mass effectively lowers the center of gravity of the elevator. At this point, some people are ready to consider that the center of mass of the cable-counterweight-anchor mass system has to be at GEO, rather than the one of the cable-counterweight system. These people are still wrong.

The Space Elevator is not a Point Mass

Kepler's laws of motion, which many people are familiar with, only apply to masses that are small compared with the variations of the gravity field they are moving in. In the case of the Space Elevator, this is far from being the case; gravity goes down by more than a factor of 300 over its height. For non point masses, the orbital characteristics deviate from those of point masses (in fact the term of orbit isn't really well suited anymore). Here is a simple example that proves my point, it comes from one of my messages on the Yahoo Groups space-elevator mailing list.

Take two identical masses m attached by a length 2L of massless cable. Place one mass at a distance R+L from the center of the Earth, and the other at a distance R-L. Put them in a circular orbit with angular velocity w in such a way that the axis of the cable always goes through the center of the Earth.

• The centrifugal acceleration on these masses is:
  • (R-L)w^2 for the low one
  • (R+L)w^2 for the high one
• The gravity acceleration on these masses is:
  • GM/(R-L)^2 for the low one
  • GM/(R+L)^2 for the high one

In order for your system to be in equilibrium at angular velocity w you need these accelerations to compensate each other:

(R-L)w^2+(R+L)w^2=GM/(R-L)^2+GM/(R+L)^2

This leads to w^2=GM/2R * (1/(R-L)^2 + 1/(R+L)^2)

If you set L=0, you get w^2=GM/R^3 which is Kepler's third law.

If L is small then 1/(R+L)^2 is approximately 1/R^2-2L/R^3, so we once again get Kepler's third law. However, for large values of L you find that w is greater than Kepler's third law would predict for an object at altitude R where the center of gravity of the system is located. This is because the increase in gravity for the low mass is greater than the decrease for the high mass.

So in general, for an orbit that extends over a great range of altitudes, the orbit will be faster than the orbit at the center of mass altitude. In conclusion, even for an elevator not attached to the Earth, the center of mass should be above GEO.