# Calculating momentum exchange of rotating skyhook

I am struggling to understand how momentum is exchanged during a rotating skyhook's interaction with a payload, during capture and after release. I am a layman in physics, please be forgiving if I've made any glaring errors.

Here is what I've got:

• In an ideal case, the suborbital vehicle matches speed with the hook and thereafter all the forces are directed along the tether's length. Because there is no net torque acting on the tether during and after the hooking (disregarding drag forces), angular momentum is unchanged. If angular momentum is the same while the mass of the tether system increases, then angular velocity must decrease. (I am unsure how to calculate this.) When the vehicle reaches the apex of the tether's swing, it simply lets go and again no net torque is applied. The tether system's mass decreases so angular velocity must increase, presumably back to what it was earlier.

• Initially, I thought the vehicle would depart at the tether's own tangential velocity, but now I'm not so sure. Suppose the tether has a tangential speed of $$v$$ km/s at the hook. The vehicle was caught when the hook was moving at -$$v$$ km/s and later was released when the hook was moving at +$$v$$ km/s, for a total of $$2v$$ km/s. If the vehicle leaves the tether with speed $$2v$$, then it has gained momentum $$2mv$$ from the tether. Because there is no net torque, that momentum should come from the linear/orbital momentum of the tether's center-of-mass, so the tether system gains -$$2mv$$.

Basically:

$$-mv + M(0) = mv + Mu;$$

where $$M(0)$$ is the initial momentum of the tether (in its own COM frame), $$-mv$$ is the initial momentum of the vehicle, and $$u$$ is the final velocity of the tether after release.

Is my understanding correct, or am I wildly off track? If the latter, then how should I go about calculating the exchange between the rotating tether satellite and payload?

The net effect of the vehicle gaining $$2mv$$ and the tether gaining $$-2mv$$ is correct. This is consistent with conservation of linear momentum.

I do however have an issue with this part:

Because there is no net torque acting on the tether during and after the hooking (disregarding drag forces), angular momentum is unchanged. If angular momentum is the same while the mass of the tether system increases, then angular velocity must decrease.

The angular momentum of the tether system does not stay the same when hooking a big moving mass at the end! You're adding a big chunk of angular momentum. ("matching speeds" can then be viewed as exactly matching the added mass and added angular momentum so that the angular velocity stays the same)

The description would make sense for picking up a stationary mass (or in this case, orbiting at the same speed as the centre of mass), because then no additional angular momentum is added, but the mass is increased, so the angular velocity goes down.

The assumption made here appears to be that "angular momentum can not change without torque", but that is only true when mass is conserved, which is not the case here.

The exact same thing applies to the release. A big chunk of angular momentum is lost, but the mass also goes down so the angular velocity stays the same.

• Oh, you're right! The payload adds a big chunk of angular momentum, which is now pretty obvious in hindsight. Thanks for clarifying that. Also, if you don't mind my asking, at the apex of the tether's swing the payload will fly off at a 45° angle relative to the tangent, right? So, the -$2mv$ momentum applied to the tether will be along that direction?
– BMF
Jan 30 at 18:07
• The payload will fly off exactly parallel to the tangent. When the tether no longer restrains it, the mass continues in the direction it is going. Which is tangential. Jan 30 at 18:33