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.
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
How do they get 3.5 km/s? I do see "relative to the inertial reference frame", but what does that mean?
$$Vo = √(GM/R)$$
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?