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What sorts of unique problems are there with planning a trajectory to Proxima Centauri, besides it being out of the plane of the ecliptic? (I'm guessing if you've got the juice to make it there, inclination's the least of your worries) So...cumulative errors? Unknown perturbations in gravity (rogue planets, unseen dark matter)? Not to mention velocity - I'm assuming an Orion-drive or Magneto-Inertial Fusion of similar power that could eventually (hopefully) accelerate to around 0.10c after an Oberth maneuver either at the Sun or Jupiter.

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    $\begingroup$ Cumulative errors can be dealt with by cumulative correction, just as they are in lunar and interplanetary missions. $\endgroup$ Apr 23, 2017 at 0:04
  • $\begingroup$ @RussellBorogove were does the navigation data come from? How do you know you are off of the ideal trajectory? You could be on the right path but not the right velocity, or you could be off the "glide path" a bit. Where specifically (my new replacement word for "actually") would the navigational information come from? $\endgroup$
    – uhoh
    Apr 23, 2017 at 2:22
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    $\begingroup$ If someone has time to read through these, an excellent answer can be generated. I can't do it myself until middle of next week, so if someone else would like to have a go at it... arxiv.org/abs/1704.03871 and arxiv.org/abs/1701.08803 $\endgroup$
    – uhoh
    Apr 23, 2017 at 2:31
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    $\begingroup$ @RussellBorogove This is my quest, to follow that star! No matter how hopeless, no matter how far... youtu.be/RfHnzYEHAow?t=55 So that gives a direction, but this goes back to another question I had - it does not tell you if you are moving in the right direction to high precision, or how far you are or how fast you are going. It only really tells you if your telescope is pointed in the right direction. Considering the distance, this difference can add up. One can simply say it's easy, but over a few light years, errors can add up. $\endgroup$
    – uhoh
    Apr 23, 2017 at 15:19
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    $\begingroup$ Track the angles between multiple nearby stars to fix your position. Differentiate to get velocity. Make sure Proxima isn't moving the right amount relative to the distant stars and that it keeps getting brighter. Repeat until finished. $\endgroup$ Apr 23, 2017 at 15:27

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Only pure $\Delta v$ matters, none of your optimizations are significant:

  • Inclination: The initial velocity due to our location is about $30,000m/s$. Compare that with $0.1c$, about $300,000,000m/s$, only $0.01\%$.
  • Oberth manoeuvre: The best we can possibly do is burning close to the surface of the Sun (there are some problems with that, but we can ignore them for now). The $\Delta v$ saved, ignoring relativistic effects, is:

$$0.1c - \sqrt{(0.1c)^2 - v_e^2} \approx 0.00002c \text{ or about 0.02%}$$

I would say the greatest complication is to stop at Proxima Centauri, as you will have a velocity of $0.1c$ when you arrive. You need a similar amount of $\Delta v$ to stop.

Perturbation in gravity should be easy to compensate for as you have already shown you have enough boost to escape the gravitational field of an entire star by many orders of magnitude.

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  • $\begingroup$ 0.0002c savings is 6km/sec if I've done my math right -- much less than the delta-v you'd need to get close to the sun in the first place. I suppose you could do a cheaper maneuver at Jupiter flyby that could put you on a close approach to the sun? $\endgroup$ Apr 23, 2017 at 0:07
  • $\begingroup$ So only 2/100 of a percent dV saved...yeah, so not worth it except maybe the free inclination change. And yes, a Jupiter flyby is in the cards as well. $\endgroup$
    – PHChilly
    Apr 23, 2017 at 0:09
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    $\begingroup$ @RussellBorogove You have to reach escape velocity anyway, and from there a close encounter with the Sun is almost free. You can combine that with the inclination change, and congrats! You have saved another 0.0001%! $\endgroup$ Apr 23, 2017 at 0:17
  • $\begingroup$ @PHChilly Why do you think a free inclination change is worth it? The savings from that are ridiculously small. A Jupiter flyby gives you even less. $\endgroup$ Apr 23, 2017 at 0:19

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