On the right are the equations.
On the left is an example based on LEO around Terra.
5.97E24 kg Mc = Central mass
900000 kg Mb = mass of tether base
100000 kg Ms = mass of shuttle
Assume tether mass is negligible
6.67E-11 m^3/(s^2*kg) G = Newton’s constant
3.986E14 m^3/(s^2) mu = G*Mc
7000000 m Rb = distance from center of Mc to center of Mb
100000 m Lt = length of tether
With Mb in circular orbit it has speed
7546 m/s Vb = sqrt[mu/Rb] and flight angle zero
If tether hangs straight down, speed at foot is
7438 m/s Vf = Vb *((Rb-Lt)/Rb)
The term “dock” indicates that the shuttle matches velocity with the foot
of the tether, so that it is moving horizontally at Vf, and it stays on
the line between Mc and Mb.
The center of mass of the combination sits at
6990000 m Ca = Rb – Lt*(Ms/(Ms+Mb)) and has speed less than circular
7535 m/s Va = Vb * Ca/R , moving horizontally.
Find the eccentricity of the center of mass from
0.004280 e = sqrt[1–(Va^2*Ca/mu)*(2-Va^2*Ca/mu)*(cos(0))^2],
and the argument of the cosine is zero since the
velocity is horizontal.
Then the periapsis of the center of mass is
6930426 m Cp = Ca*(1-e)/(1+e) and the shuttle hangs below at
6840426 m Ds = Cp - Lt*(Mb/(Ms+Mb)) hwile the base is above the
center of mass at
6940426 m Db = Cp + Lt*(Ms/(Ms+Mb))
The speed of the center of mass at periapsis is
7600 m/s Vp = sqrt[(mu/Cp)*(1+e)]
Caveat, this assumes that Mc, Mb, and Ms always remain in a line. I think there is an oscillation about the center of mass, as the angular motion of the center of mass around Mc is not a constant. - MBM