Given two spacecraft (let's say, the ISS and an unmanned resupply capsule), how do the two spacecraft rendezvous with each other in space? If they're moving at the same speed in space, they'll never catch up; if one tries to move faster, it will be boosted into a higher orbit as a consequence of its speedup. Do they simply create a very slight speed differential (so the difference in altitude will be negligible) and then wait for the two to sync up?
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1$\begingroup$ Related: Why did the rendezvous attempt fail on Gemini 4? $\endgroup$– userJul 17, 2014 at 21:44
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1$\begingroup$ Here's a video showing how to rendezvous in orbit. It's from Kerbal Space Program, so the altitudes won't match Earth orbit, but it answers the core of your question quite well. $\endgroup$– JoeJan 26, 2015 at 19:00
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2$\begingroup$ Somebody has to say it :-) - "East takes you out, out takes you west, west takes you in, in takes you east, port and starboard bring you back." - Larry Niven, Integral Trees. And see larryniven.net/physics/img34.shtml" $\endgroup$– Russell McMahonJan 28, 2015 at 22:44
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$\begingroup$ very carefully! $\endgroup$– uhohDec 10, 2021 at 10:47
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3 Answers
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They use orbital mechanics, never fight them. First of all let's be clear about directions in Space:
The X-vector of the RIC (radial, in-track, cross-track) frame is called radial as it points along the radial position of the object w.r.t. central body. The Y-vector is called in-track and points, for circular orbits, along the orbital velocity direction. The Z-vector is orbit normal.
First approach method is V-bar that relies on changing the relative altitude. Essentially one of the S/C goes lower/higher than the other and hence it has a slightly different period than the other one. This causes relative motion between the two. They often make "hops" using this technique. So change the altitude a little, drift closer together, return to the same altitude, hold for a while to make sure everything is fine, and then keep hopping. Read about the ATV approach technique to learn more about this.
Sometimes they also do a R-bar rendezvous. In this case the secondary spacecraft approaches the primary along the local radial direction (X-direction in the RIC frame), but this requires many engine firings in the radial direction rather than only several altitude (and orbital velocity of the secondary) change manoeuvres.
But both techniques have been used many times and are generally well-understood.
answered Jul 13, 2014 at 15:55
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$\begingroup$ If hypothetical craft one approaches hypothetical spacecraft two along the X axis, isn't one still drifting with respect to the other? In which case, the rendezvous abrupt: sc one fires once it's exactly aligned for docking on sc two. That sounds risky, so I think I misunderstand how the R-bar rendezvous work. Could you clarify? $\endgroup$– ChrisRJul 19, 2014 at 0:02
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1$\begingroup$ Yes, when the secondary approaches along R-bar they still drift, so S/C has to compensate for this with thrust. And, in fact, it's pretty safe, because as soon as something goes wrong the secondary can just stp firing and passively drift away. $\endgroup$ Jul 19, 2014 at 8:51
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$\begingroup$ Okay, I think I understand now. Thanks. That also means the docking maneuver will fire along +X and -Y then (with respect to the chasing vehicle), right? $\endgroup$– ChrisRJul 20, 2014 at 0:38
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1$\begingroup$ The drift due to different altitudes will be in the Y-direction (+ve Y if the chaser is below the primary), so that's where most of the burns will be made during the approach. But controlling the relative velocity and distance will require burns in the X-direction (could be in +ve and -ve directions depending on what the situation requires), yes. When I get to work tomorrow I may show you how this whole drift thing works (I need full version of STK that I don't have on my personal laptop). $\endgroup$ Jul 20, 2014 at 10:01
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Here's a fairly abstract way to see how it can be done: consider two ships fairly close together, on similar but not identical orbits.
There's one vector running from the current position of ship A to ship B. If you thrust along that vector, the ships will tend to get closer.
There's another vector that represents the difference in velocities between the two ships. If you thrust along that vector, you can reduce the difference in velocities, i.e. make the two ships more stationary relative to one another.
It should be clear that if those two vectors aren't directly opposed, there's some vector in between those two, which if you thrust along, will both close the range and reduce the relative speed. If you repeatedly compute and use those vectors, eventually you can get both range and relative speed arbitrarily close to zero, at which point you're docked.
answered Aug 26, 2014 at 20:34
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$\begingroup$ So, should I just to go to needed orbit at random point and then slowly decrease a distance between them? $\endgroup$– RobotexOct 18, 2021 at 13:30
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Rendezvous can be divided into 4 main phases:
1) Phasing: to approach the target orbit
2) Initial approach: to acquire a stable orbit with respect to the target (distance in the order of magnitude of $10^3$)
3) Final approach: to get closer to the target (distance in the order of magnitude of $10^2$)
4) Final translation: to finalize the contact (distance in the order of magnitude of $10^1$)
Usually every phase starts with a $\Delta v$: therefore the propulsion system is turned on (impulsive manoeuvres) to change the velocity (in magnitude and/or direction) of the spacecraft.
