The calculations for the taper ratio of a space elevator don't seem to be all that difficult and it is easy to find some equations to use. You can even find an expression written out for the total mass (which involves more complicated mathematical functions).

But if we have an orbiting, tidally locked, partial space elevator what would the equations for those be? Searching around, I can't find any material on this. Presumably it would be slightly more complicated than a full geosynchronous space elevator because you have an additional independent variable. Does anyone have either an approach to this problem with some of the basic math hashed out, or just a straightforward equation for taper ratio?


1 Answer 1


I was over-complicating. It's the same equation as for a geostationary space elevator. In fact, the geostationary space elevator is just one specific case of a general equation for taper ratio. As a mental experiment, pretend that the space elevator equations were applied for a planet of the same mass but with a different rotational period. Clearly the geostationary point would lie at a different point, and you could pretend that such an imaginary planet exists for every non-geostationary orbit.

Borrowing from a previous answer, you could do this simply as follows. The one thing we're introducing is a new criteria for the neutral circular orbit based on the radius, r0, at the middle of the partial space elevator.

$$ U(r) = - \frac{ GM}{r } - \frac{1}{2} \omega^2 r^2 \\ \frac{ GM}{r_0^2} = \omega^2 r_0 \\ U(r) = - \frac{ GM}{r } - \frac{1}{2} \frac{ GM}{r_0^3 } r^2 \\ \frac{ \lambda(r_{0}) }{ \lambda(r) } = \exp{ \left( \frac{ U(r_{0}) - U(r) } { \left( \frac{ \sigma}{ \rho} \right) } \right) } $$

The two independent variables are r0 and r. The final expression is the taper ratio. You need to know the free-floating central point and you need to know how far it extends in either direction. Yes, it does need to be balanced on both ends, and that would be the more difficult requirement. In order to find the length on the "other side" that corresponded to a radius selection you would see to set the integral of the linear mass-thickness on both sides to be equal.

  • $\begingroup$ r is distance from planet center to tether foot? $\endgroup$
    – HopDavid
    Commented Dec 31, 2014 at 20:33
  • $\begingroup$ @HopDavid Yes, that was my intention. But it might also be for the tether "head"... provided that I got the signs right. $\endgroup$
    – AlanSE
    Commented Dec 31, 2014 at 20:36
  • $\begingroup$ The Aravind equations I use rely on r0 (geostationary radius), r (planet radius), Gm (Gravitational constant and mass of planet), density of tether material and tensile strength. So, for example, if I want the taper for a tether whose center is 1000 km altitude and whose lower end is 300 km altitude, I set r=6678 km (instead of 6378 km). I set r0=7378 km (instead of 42164 km). $\endgroup$
    – HopDavid
    Commented Dec 31, 2014 at 21:14
  • $\begingroup$ Of course ω becomes 2 pi radians each 1.75 hours instead of 2 pi radians per sidereal day. g is 8.93 m/s^2 instead of 9.8 m/s^2. I use the words "middle" or "center" hesitantly since the tether length above will be longer than than tether length below -- the acceleration gradient isn't symmetrical about the so called central orbit. $\endgroup$
    – HopDavid
    Commented Dec 31, 2014 at 21:15

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