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After one halted attempt at docking with the ISS, the CRS-10 Dragon capsule successfully completed the maneuver. In the time between the two attempts, Dragon must have maintained some kind of orbit with some relationship to ISS's orbit, to make a time for the next attempt available and yet to remain a safe distance away from the ISS without need of active control.

What orbit did CRS-10 maintain during the wait for the 2nd attempt, and how did it move relative to the ISS during that time?

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    $\begingroup$ Theoretically (putting aside pesky realities like drag) if the capsule has the same orbital period as the space station it will either stay the same distance all the time (if just ahead or behind the station on the same orbit), or at worst will re-unite with the station after each orbit. It'd be logical to match orbital period with the station, but I have no idea if SpaceX actually did that, they might have just let it drift away then re-rendezvoused. $\endgroup$ – Blake Walsh Feb 24 '17 at 12:26
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    $\begingroup$ @OrganicMarble I saw that too. I'd guess it's the same period, plane an phasing, so from the ISS it would look like its orbiting the station, but that's just my guess. If there's no answer I'll check the public TLEs and see if they're updated frequently enough to show it during this period. $\endgroup$ – uhoh Feb 24 '17 at 16:13
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    $\begingroup$ You can find a list of the TLE here: satellite-calculations.com/11parameter/ephemeris/… Maybe someone can plot a nice 3D chart that demonstrates this orbit... $\endgroup$ – Polygnome Feb 24 '17 at 17:05
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    $\begingroup$ Think of an oval NASCAR racetrack, where the drivers only turn left. If the Dragon is in the same orbital plane as the ISS (same ascending node, same inclination), the same semi-major axis, almost the same argument of perigee, and almost the same eccentricity, the Dragon relative to the ISS will follow an NASCAR-like oval track about the ISS. It's not quite as simple as that due to the differential drag suffered by the Dragon versus the ISS, and the need for 24 hour safety should the Dragon go completely brain dead. $\endgroup$ – David Hammen Feb 25 '17 at 5:57
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    $\begingroup$ The exact details of these racetracks are hard (impossible?) to find on the 'net, which might mean that this might be ITAR, which in turn might mean that writing anything more could get me in trouble. $\endgroup$ – David Hammen Feb 25 '17 at 6:04
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In the reference frame of the ISS, the course looks approximately like this:

Racetrack mission profile

(Note: distances not to scale.)

In an earth-bound reference frame, this is simply a pair of Hohmann transfers between two orbits below and above the ISS:

As the spacecraft was below the ISS, its orbital period was slightly shorter than ISS' orbit, so it eventually came out in a safe distance ahead (and still below) the ISS.

A slight prograde burn raised the apogee of its orbit above the ISS. In the apogee, a slight prograde burn circularized the orbit.

Now being in an orbit above the ISS with an orbital period was slightly longer than ISS' orbit, it eventually came out behind (and still above) the ISS.

Now another pair of burns, this times retrograde, lowered the orbit back below the ISS, still behind but chasing up, ready for another rendezvous attempt.

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  • $\begingroup$ Hmm, any idea why a method that requires regular burns would be chosen over the orbit described in the other answer which does not need any navigational control at all? Why so many transitions - 1.6 to 6.2 mi and 4.3 to 6.2 mi? Can you provide the source/credit for the image, and something I can read further? Thanks! $\endgroup$ – uhoh Feb 28 '17 at 0:19
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    $\begingroup$ I got the image from spaceflight101.com/… they credit NASA for it, but don't provide a link :-( $\endgroup$ – oefe Feb 28 '17 at 19:25
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    $\begingroup$ OK, found the source nasa.gov/exploration/commercial/cargo/graphics_041612.html page loads rather slowly, and images don't load at all, however this archive works: web.archive.org/web/20160330023856/http://www.nasa.gov/… unfortunately there isn't much of an explanation there either $\endgroup$ – oefe Feb 28 '17 at 19:58
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Note well: I'm using the terms "chaser" and "target" in this answer. These are technical terms with a long heritage (relatively to the space era). The chaser vehicle acts to change its orbit. The target vehicle keeps its orbit.

Suppose the chaser and target are orbiting an object with no atmosphere, and that the chaser is in the same orbital plane as is the target, and that the chaser has the same semi-major axis as the target. Tweak the chaser's other orbital parameters just right and the chaser will follow a more or less oval shape about the target. Think NASCAR, where drivers only turn left. (I prefer non-oval races. Then again, F1 is boring.)

Tweak it just right and, ignoring atmospheric drag, the chaser will "orbit" the target in an oval centered on but always missing the target. Atmospheric drag is not something to ignore. The vehicles that bring crew and resupplies to the ISS are bullets. In comparison, the ISS is a huge butterfly. Fortunately, the ISS is a cooperative target. (Other targets: Not so much.)

Given that the Internet is chock full of mentions about racetracks but never mentions the details, I gather that those details are not to be released. But the summary description as an oval where the vehicle only turns in one direction thanks to gravitation: that's just Kepler's laws.


Note well: If the DoD had it's druthers, $F=ma$ and $F=GMm/r^2$ would be classified as TS NOFORN, as would be Kepler's laws. That's one of those details.

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  • $\begingroup$ Thanks very much for the explanation and subtleties! Do I understand correctly that the chaser is seen to orbit the target in a target-centric rotating or "synodic" frame, but in a target-centric inertial frame it orbits off to one side of the target? $\endgroup$ – uhoh Feb 25 '17 at 23:20
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The answer is "essentially the same orbit". When you're that close to another orbiting object, slight differences in distance from the body being orbited have only a tiny impact on orbital velocity and thus rate of drift away from the rendezvous target.

You enter a lower or higher orbit to catch the object you're trying to rendezvous with, then "approximately the same orbit" when you're preparing for docking.

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  • $\begingroup$ From here - ditto here. "essentially the same" is not a useful term. Once again, slight, tiny, approximately, essentially... don't have any utility without numerical values. In this case however, your statements are just plain wrong. $\endgroup$ – uhoh Feb 24 '17 at 10:14

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