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The key difference with a plain old-fashioned orbital rendezvous would seem to be that there'd be limited time to only briefly match velocities and trajectories:

Let's say a vehicle launched from the ground on a suborbital trajectory wants to meet a space tug that had slowed down from LEO for payload handover.

Assuming the tug had slowed to 5 km/s, and the ascender accelerated to exactly that velocity, just under 2/3rds of minimum orbital speed. If either the tug, or the ground-launched vehicle were just 1 second off, the vehicles would be 5 km apart, and you'd have to do the measuring, calculation and navigating to bridge that gap within a few minutes before their trajectories had to diverge again, as the tug would soon have to fire its thrusters to get back into orbit before entering the atmosphere together with the suborbital ascender.

If you were just a millisecond off, it'd be 5 metres, and considering that the relative trajectories were divergent, again there'd be little time to fix it.

I can't find anything on this online. It would seem to be a useful technique as it would lower the delta-v for the vehicle that lifts the payload from the ground to the rendezvous point relative to fully making orbit. (NOTE: Yes, you'd have to get extra fuel to the tug to perform this manoeuver but that's potentially a smaller problem.)

Is it just too hard so no one's ever contemplated trying?

EDIT I: The first answer (now deleted) made before this amendment of the explanation and of the title to make it clearer that the question isn't about planes meeting in the atmosphere was technically correct, aerial re-fueling in the atmosphere is a suborbital rendezvous manoeuver. However, I assume the upper vehicle here is on a trajectory to re-enter orbit either because it's the end of a tether that will spin up by virtue of its momentum or a space tug which would have to fire its engines before entering the atmosphere.

EDIT II: As @bitchaser pointed out below in the comments to the first answer, the slowing down might happen via aerobraking to save fuel, and engines on the tug could be optimized for the purpose of re-injection into orbit.

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  • $\begingroup$ I've removed two irrelevant tags. $\endgroup$ – Magic Octopus Urn Apr 5 at 15:11
  • $\begingroup$ One such "suborbital rendezvous" $\endgroup$ – Chris Apr 5 at 15:15
  • $\begingroup$ I am adding the tags again as I don't see what else you could meet that was going back into orbit. $\endgroup$ – Prototypist Apr 5 at 15:25
  • $\begingroup$ Neither skyhook nor space-tug is applicable to this topic, you can click on each individual tag to see what they actually relate to. Skyhooks are a theoretical "crane from orbit" sky hook and space tugs are orbital transfer vehicles. But I won't edit it again. Also, for reference, the fastest Soyuz rendezvous with the ISS was 3 hours and 46 minutes MET (time from launch). Meaning they likely did 2 or more orbits before docking. $\endgroup$ – Magic Octopus Urn Apr 5 at 15:44
  • $\begingroup$ The definition for skyhook attached to this tag here is "A proposed method for orbital launch and transfers. Utilizes a long wire with fixed rotation to manage momentum." which is exactly what I had in mind, transfer momentum to augment a sub-orbital trajectory so that it becomes orbital. Also, I mean space tug in the sense of sub-orbital to orbital transfer which I think is covered by the meaning. $\endgroup$ – Prototypist Apr 5 at 16:24
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I can't find anything on this online. It would seem to be a useful technique as it would lower the delta-v for the vehicle that lifts the payload from the ground to the rendezvous point relative to fully making orbit. (NOTE: Yes, you'd have to get extra fuel to the tug to perform this manoeuver but that's potentially a smaller problem.)

This is the opposite of a useful technique.

The speeds have to be matched, or you have a collision instead of a rendezvous. An ascending ship on a 5km/s suborbital trajectory meeting an orbiting station at 7.7km/s yields a massive cloud of debris. You can't just toss the payload from one airlock to the other as you go by; it's still got a relative velocity of 2.7km/s.

So in saving X amount of delta-v on the ascending ship, the destination station would have to spend X delta-v to slow down to match, and then spend X delta-v again to get back to orbital speed, plus a little more to recover the altitude lost during the slowdown. The destination station is probably substantially larger than the ascender, so that 2X delta-v probably means much more fuel expenditure as well.

On top of that, if there's any delay in achieving rendezvous, or any problem restarting the circularization thruster on the station, you've lost the whole thing.

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    $\begingroup$ I picture a space barge / space tugboat leaving ISS, slowing, picking up load, regaining speed. I don't think that would change the answer much, but conceivably the barge could use aerobraking to slow, and/or be extra efficient using fuel. $\endgroup$ – b and d restore Monica Apr 5 at 20:19
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    $\begingroup$ @bitchaser Any feature you give to that barge, it's more efficient to give to the ascender. $\endgroup$ – Russell Borogove Apr 5 at 20:24
  • $\begingroup$ I agree that's true, at least in any configuration I can conceive. $\endgroup$ – b and d restore Monica Apr 5 at 20:26
  • $\begingroup$ I wrote space tug as I meant barge, as in a specialized vehicle for this purpose. And yes you'd have to match velocities or you get debris, same with any kind of rendezvous. You'd have less time to match velocities as I wrote, that's why I theorized that it would be more difficult than doing an orbital rendezvous. The question actually was if there is existing research/experiments. Possibly not then, perhaps as the usefulness itself needs to be established first. $\endgroup$ – Prototypist Apr 6 at 4:45
  • $\begingroup$ @RussellBorogove replying to your reply to bitchaser's comment about the ascender being more efficient at what the barge would do: Assuming the same fuel and engines as with an upper stage that does the orbital injection, you might be right in that the barge would require extra air braking and docking gear, making it heavier. However the barge wouldn't need any kind of air drag optimizing fuselage, making it potentially lighter again, and could be re-used as many times as the engines last for. $\endgroup$ – Prototypist Apr 6 at 5:50

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