I am working out the math behind the concept of a rotating skyhook and have quite a few questions which warrant multiple posts, but for this question I am trying to understand how the tip of a rotating skyhook can travel slower than when at the top (in a higher orbit). All resources I am reading mention that every time the tip completes a rotation and dips closer to the atmosphere, ~200km, it'll be traveling at ~3.5 km/s and at its highest point, ~610 km, it'll be traveling at ~7.6 km/s.

Here is a snippet I am trying to breakdown mathematically:

The baseline Rotovator™ space tether facility shown in Figure 7 is composed of a tether control station (containing power supplies, tether reel, command and control, and ballast mass) , a 600 km long tapered tether, and a grapple assembly at the end of the tether. The tether facility is placed in a slightly elliptical orbit (e=0.0062) with a CM apogee altitude of 700 km, a perigee altitude of 610 km, and a perigee velocity of a little over 7.6 km/s. The orbit was chosen to be elliptical and payload capture was performed at perigee in order to reduce the amount of total facility mass needed to keep the facility and tether above the atmosphere after the facility captures a payload. The tether is set into rotation with a tip velocity of a little over 3.5 km/s. The center of mass of the tether facility is located about 90 km from the tether control station, so when the facility is at perigee altitude of 610 km, the tether control station is at an altitude of 700 km and the tether tip is at an altitude of 100 km, moving at a velocity of approximately 7.6 km/s - 3.5 km/s = 4.1 km/s relative to the inertial reference frame, thus matching the speed of the hypersonic airplane. The atmospheric drag on the tether at 100 km altitude was calculated and found to be negligible.

  1. How do you get "little over 7.6 km/s" - when using the below formula I am getting ~7.55 km/s, just below 7.6 km/s

  2. How do they get 3.5 km/s? I do see "relative to the inertial reference frame", but what does that mean?

Given that

$$Vo = √(GM/R)$$

R and V have an inverse relationship. Would angular momentum and moment of inertia play a part in calculating specifically the velocity at apogee and perigee since the tether is rotating?


1 Answer 1


Roll a wheel down the road. If it’s really rolling, the place where the wheel touches the road doesn’t move relative to the road: it’s just sets down as the wheel gets to a spot, then lifts up after. On the other hand, the top of the wheel is moving twice as fast as the axle.

The skyhook is like that. It’s engineered to be rotating and orbiting at just the rates to come to a stop as it gets close to Earth.

  • $\begingroup$ If I understand this correctly, the bottom of the hook isn't moving relative to the Earth? Am I mistaken in saying that it is moving relative to the center of rotation and the center of mass is moving relative to the Earth? $\endgroup$
    – joethemow
    Jan 20, 2020 at 1:45
  • $\begingroup$ Not sure what you’re asking. Both CoR and CoM are moving. The ends of the skyhook move relative to them. Above, those add; below the subtract to zero. $\endgroup$ Jan 20, 2020 at 16:37
  • $\begingroup$ I added a snippet above (basically what I am trying to mathematically prove). I am unsure how they got 3.5 km/s given the apogee and perigee of the "tether facility". I read up on inertial reference frame which makes sense (can't be accelerating), but it should be measured relative to some point. In the snippet, what is it relative to? $\endgroup$
    – joethemow
    Jan 21, 2020 at 1:02

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