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I am exploring some of the properties of a vertical orbiting tether (or non-rotating skyhook). This particular tether has a quite large mass at the location with a circular orbital velocity, with the mass of the tether itself being much smaller and ignored.

orbiting tether

If something, such as a shuttle docks at the tether foot, the mass of the docking object is going to cause a periapsis drop.

periapsis of tether has dropped

Given the central mass, the mass of the shuttle, and of course all desirable parameters such as foot altitude etc. how can I calculate the drop in periapsis?

My initial strategy was that both energy and angular momentum in the new orbit is always going to be the same, making it easy to set up an equation with one solution in apopasis and periapsis. But those equations quickly grew out of hand.

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  • $\begingroup$ Are you taking into consideration the flexibility of the tether? In addition, is the length of the tether non-negligible compared to the size of the overall system? My initial hunch here is that if the tether is used mostly for gravity torque control, then it's relatively small, and you can approximate the overall system to a point mass which would lie at the barycenter of both the shuttle and the mass at the end of the tether. $\endgroup$
    – ChrisR
    Mar 29, 2017 at 20:37
  • $\begingroup$ @ChrisR Having a flexible tether is not a requirement, but using "realistic" properties would be most useful. The length of the tether is not negligible (potentially several times larger than the radius of the central planet). Also, as both having the shuttle outmass the tether by several orders of magnitude and or the opposite makes the situation trivial, that is not the case. $\endgroup$
    – SE - stop firing the good guys
    Mar 29, 2017 at 20:43

1 Answer 1

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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

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  • $\begingroup$ I think it would be a mistake to "assume tether mass is negligible" - however the tether/orbital base can be considered a system with a single mass/CoM, so the calculations are reasonable. While the calculation should account for the stretch in the tether, the hundred meters or so is negligible. $\endgroup$
    – user20636
    Oct 23, 2017 at 8:08

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