# Could a space elevator really be built from the top down?

I think most of us have this picture of a space elevator being built by lowering a cable from a platform in orbit until it reaches the ground. Leaving aside the technological hurdle of building a cable strong enough— how would this be possible given basic laws of orbital motion?

How is it possible to keep the cable pointed down towards earth when the end of the cable is at a lower altitude with a corresponding higher orbital velocity? It would seem that the end of the cable would start to move towards prograde the moment it begins to lower.

Does the end of the cable need a continuous source of thrust in the retrograde direction while it's being lowered?

• Yes I think it does. – Mike Dunlavey Mar 30 '17 at 19:36
• And the next question is 'could a space elevator really be build from the bottom up?'... – Jon Custer Mar 30 '17 at 20:06
• – Michael Seifert Mar 30 '17 at 21:00

If an object is in a circular orbit and gravity is the only force acting on it, then a lower elevations does indeed imply a faster orbit. If $\omega$ is the angular velocity of the orbit (radians/second around the circle), then the relationship between $r$ and $\omega$ can be expressed in the Physics-101-exercise way by equating centripetal force with gravitational force: $$F_R = F_G \quad \Rightarrow \quad m r \omega^2 = G \frac{m M_E}{r^2} \quad \Rightarrow \quad \omega = \sqrt{ \frac{G M_E}{r^3}}.$$
But the lower end of a descending space elevator doesn't only have gravity pulling it downwards; it also has an upward tension $T$ from the cable above it. We therefore have $$m r \omega^2 = G \frac{m M_E}{r^2} - T \quad \Rightarrow \quad T = G \frac{m M_E}{r^2} - m r \omega^2.$$ Thus, for a given radius $r$, we can get the object to orbit at any angular velocity $\omega$ by picking the tension correctly.