Regarding deep space missions, like the Rosetta mission, how much of the journey to the final destination is pre-programmed vs. recalculated on the fly? For the Rosetta example, was the whole sequence of gravity assists (and burns to align the trajectory in order to do the assists) pre-programmed, or was the trajectory recalculated after every maneuver? Could the whole journey be executed in a "hands off" fashion (not counting the other various stuff that needs to be done during the journey like doing science, telemetry checks etc., just the trajectory planning and maneuver execution)?

  • $\begingroup$ The "on the fly" usually means something between a couple days and a couple months before given maneuvers are executed. For example, in case of Rosetta, not knowing the shape of the comet they were unable to calculate the orbits ahead of time, but they created a plan of months worth of orbital maneuvers once the data became available. $\endgroup$
    – SF.
    Mar 10 '17 at 10:43
  • $\begingroup$ @SF Well, of course the operations at the target are "on the fly" since we often don't know what the target is like (that's the reason we launch such missions). I was more concerned about the journey from the Earth to the target and the planning of it. $\endgroup$
    – zegkljan
    Mar 10 '17 at 10:45

The journey to the destination is about always completely pre-planned. All the gravity assists, close fly-bys, and so on, are planned before launch - and often long before the probe design is completed, as often the requirements of the trajectory influence design considerations: ability to hibernate, delta-V of the engines, tolerance to heat if the probe is to get assists from Venus, equipment to pick science specific from given targets-of-opportunity (asteroid flybys), and all equipment lifetime, as the trajectory means trip time.

This doesn't mean the probe can do everything fully autonomously.

In the mission plan, there are events of "course adjustment". Each maneuver, like a gravity assist, departure burn, or plane change is performed with a certain error. At these distances, a couple milliseconds of error at departure burn will mean hundreds of kilometers of error at the destination. After a maneuver was performed, the deviation from planned trajectory is determined. Parameters of the corrective burn are established, and the probe performs the maneuver at the pre-planned location and time, but according to parameters that were obtained only after the maneuver that introduced the error.

  • 3
    $\begingroup$ Thanks for the answer, but the course adjustment part is what I'm interested in. Who (i.e. the vessel or the mission control) measures the trajectory deviation and who calculates the adequate corrective burn parameters? Could the vessel itself do all of this (i.e. is there any kind of auto-correction) or does it rely on the mission control to do this stuff for it? $\endgroup$
    – zegkljan
    Mar 10 '17 at 11:06
  • 5
    $\begingroup$ @zegkljan: Mission control. Probe position determination involves signal to/from Earth and is performed on Earth, plus the probes don't have the supercomputers the control center has to calculate the adjustment. Technically, if probes were equipped with doppler interferometry equipment and Earth ground stations would just bounce their signals without processing them, that would be doable, but rather pointless. More on the subject tl;dr: probe doesn't know where it is, control center does. $\endgroup$
    – SF.
    Mar 10 '17 at 11:15
  • 2
    $\begingroup$ @Coxy : ​ I know very little about this, but would imagine SF. is referring to the probe being significantly closer to the sun at that point. ​ ​ ​ ​ $\endgroup$
    – user2822
    Mar 10 '17 at 13:26
  • 3
    $\begingroup$ @RickyDemer: This is correct. Generally, dipping below Venus orbit distance from Sun requires extra considerations regarding heat management. $\endgroup$
    – SF.
    Mar 10 '17 at 13:50
  • 1
    $\begingroup$ The Sun is hot enough to melt me at Earth's orbit: I would certainly need an air conditioned spaceship to get as close as Venus. $\endgroup$
    – user5932
    Mar 10 '17 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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