# 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.

• Makes sense. But in order to have tension in the first place, doesn't one end of the cable need to have a continual thrust applied to it? In other words— we can't get around the fact that some part of the system will need to have thrust until it reaches the ground (or atmospheric drag is high enough)? – Jim Heising Mar 30 '17 at 19:51
• @JimHeising: I believe this is why you have to simultaneously build a tether out from geosynchronous orbit. If you play the inward and outward tethers out at the appropriate speed, the inward force needed to keep the outer tether moving fast enough is exactly balanced by the outward force needed to keep the inner tether from moving too fast. (I think.) – Michael Seifert Mar 30 '17 at 20:04
• Ahhh. That makes sense now. Reminds me of one of those retractable charger/audio cables :D – Jim Heising Mar 30 '17 at 20:13
• Actually, now that I think about this more— I'm not sure this will work. Think of it like a geostationary satellite. A geostationary satellite must spin 360 degrees over a 24 hour period in order to stay pointed at earth. So if you imagine this elevator cable system is just like a geostationary satellite— its cables are just like REALLY long antennas— it too has to rotate once every 24 hours to keep pointed in the right direction. You could never get a system that large to rotate that quickly without immense amounts of force. Am I missing something? – Jim Heising Mar 30 '17 at 21:18
• You can spin a geostationary satellite at the beginning and it will keep going for the most part with minimal ongoing correction. But this cable system will act like a figure skater spreading his/her arms. There's no way to keep it spinning without continual increasing force. – Jim Heising Mar 30 '17 at 21:24