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)]