6
$\begingroup$

Let's assume that

  • a spaceship (capsule) is about to dock to the ISS
  • the distance between spaceship (capsule) and ISS is 20 meters
  • the relative velocity between the spaceship (capsule) and ISS is nearly 0
  • the orbits' focii (center of earth) and spaceship and the docking adapter are sitting one and same (radial) line

What kind of burns would the spaceship then perform in order to reach the docking adapter? In other words, how would the spaceship (capsule) then change its height? Would it perform a mini-version of a Hohmann Transfer? Or just burn radially?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ FWIW, this is yet another question where "play Kerbal Space Program and find out for yourself" would be a reasonable suggestion. The game even comes with a built-in rendezvous and docking tutorial. (To be fair, the "magnetic" docking ports in KSP are totally unrealistic in their simplicity and robustness, but that's just the last fraction of a meter. And of course KSP players tend to take much bigger risks and fly faster approaches than real astronauts ever would, since they can just reload a save if things go wrong. But other than that, the orbital mechanics are quite realistic.) $\endgroup$
    – Ilmari Karonen
    Jun 6, 2020 at 22:29
  • $\begingroup$ There's a guy who was named Edwin who wrote a very related thesis. His nickname, unitl he changed it to his first name is "Buzz" and his last name is Aldrin. He wrote his MIT PhD these on orbital rendezvous, then got accepted to be an astronaut, where he applied some of his theory. airandspace.si.edu/stories/editorial/buzz-aldrins-phd-thesis $\endgroup$
    – Adam
    Jun 7, 2020 at 16:30

2 Answers 2

Reset to default
15
$\begingroup$

Circular orbits at different altitudes require different speeds, so if you start with a radial separation, the spacecraft and station will tend to drift further apart unless they accelerate themselves radially to close the distance. The effect is small at small distances, larger at long distances. To a first approximation, the separation is the difference between the force of gravity at the altitudes of the two. (Despite the frequent use of "zero gravity" and "microgravity", there's plenty of gravity in orbit -- about 88.5% of Earth's surface gravity, at the ISS' 400km altitude.)

At 20m of radial separation, this gravitational gradient causes the spacecraft and station to drift apart by about 50 micrometers per second squared - 5 millionths of a g. This is small enough that it can be largely ignored -- it is "lost in the noise" of thruster variability and inaccuracy of measurement of speed and distance. It can be counteracted with very small pulses of thrust, and the docking spacecraft can maneuver directly for its destination, i.e. with a radial burn.

At greater distances, the gradient is more significant. At 40km separation, the effect is a relative acceleration of about 0.1 m/s2, which is about the maximum that a loaded Crew Dragon could achieve firing 4 small Draco thrusters continuously -- it could just hold the distance and couldn't approach closer. So at distances like that, approaches are done by firing prograde and retrograde, Hohmann-style; you fire retrograde to lower your perigee by 40km, wait half an orbit, then fire prograde to circularize at the lower altitude.

Somewhere in the middle is a crossover point, where your pilot (human or computer) can begin treating the space between the spacecraft and station as "flat" and ignore the gravitational gradient. I believe the approach and docking process with the real ISS, which is more complicated than I'm going into here, defines a number of "hold points" starting at around 250m separation where the gradient is about 0.6 mm/s2; for a hold point to make sense, the gradient has to be small enough that you don't spend significant fuel fighting it.

Improve this answer
$\endgroup$
9
  • 1
    $\begingroup$ thanks for the answer @Russell! but why is is possible to use radial burns for small orbit "height" changes (20m diff) but not for big "height changes" where a Hohmann is needed? $\endgroup$
    – engineerelectic553
    Jun 6, 2020 at 14:36
  • 1
    $\begingroup$ Hohmann is the most efficient option when going between circular orbits at different altitudes. You can go between circular orbits with radial burns, but it wastes fuel. $\endgroup$
    – Russell Borogove
    Jun 6, 2020 at 14:41
  • $\begingroup$ So for the last 20 meters in radial separation of a spacecraft and ISS, how would you decide whether to make use of a - Hohmann Transfer or - Radial Burn Purely fuel economy? $\endgroup$
    – engineerelectic553
    Jun 6, 2020 at 14:49
  • 3
    $\begingroup$ @engineerelectic553 At 20m away, you're not going to have sufficient precision on the burn you can do to close the distance in LEO with a Hohmann-like trajectory. The atmosphere may be extremely rarified, but it's still there, the Earth's gravitational field is lumpy, and there's a lot of breakable stuff on the outside of the Space Station. It's hard to believe anyone would trust a pure Hohmann to close twenty meters.. $\endgroup$
    – notovny
    Jun 6, 2020 at 15:07
  • 2
    $\begingroup$ @engineerelectic553 No, for a big maneuver you care about fuel economy so you use the Hohmann. For the last 20m, you're going to do only tiny maneuvers with tiny fuel expenditure. It's easier on the pilot (human or computer) to treat that 20m gap as flat space between the spacecraft and the station -- that 50µm/s^2 is going to be lost in the noise of measurement error and thruster variability -- so you just fly direct radial. $\endgroup$
    – Russell Borogove
    Jun 6, 2020 at 15:40
3
$\begingroup$

The basic idea is that if you can get there within a small fraction of the orbit, the orbital dynamics do not matter at all. You are close enough that the two spacecraft will essentially be in the same orbit. As a low orbit is 90 minutes or so, if you can get there in 5 minutes you don't have to worry about what the gravity does. If you can get much closer in both space and velocity within 5 minutes the gravity won't matter in the next 5 minutes. You will use more fuel than the ideal transfer orbit, but it isn't much fuel anyway. You get there much faster as you don't have to wait a half orbit, 45 minutes or so, for the gravitational differences to take effect.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.