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Assuming an ion propulsion ship was leaving station in an orbit at the radius of the heliopause, and the most direct trajectory takes 18 months to get to Earth, would there be a launch window or could the ship leave at any time? I was figuring with that distance, and time to alter the flight path, a ship could leave anytime and adjust its acceleration for an optimal flight time to Earth.

My assumption about flight time is based on this chart. Propulsion method and flight times throughout the solar system enter image description here

I used the Ion B drive time to Pluto of 238 days as my starting point, an exaggeration, but not extreme.

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    $\begingroup$ What do you mean with a heliopause station? The heliopause is a sphere, do you want to put a station in an orbit at the radius of the heliopause? What are you even doing out there? Voyager 1 and 2 needed more than 30 years to get there $\endgroup$ – Whitecold May 21 at 14:25
  • $\begingroup$ I think you'll want to specify the flight path your spacecraft will be using, as well as the propulsion method and timescales (18 months seems a little short to travel that far, but hey, I'm not familiar with ion propulsion, so maybe ignore me on that one). We definitely can't say anything about the launch window for a trajectory if we don't know that trajectory. $\endgroup$ – HDE 226868 May 21 at 14:37
  • $\begingroup$ @HDE226868 Since ion engines are high exhaust velocity, low thrust and low reaction mass consumption, we should be able to treat OP's spacecraft as a low-thrust torchship. That said, I have a hard time coming to terms with low thrust and fast overall travel (the latter of which OP seem to posit). $\endgroup$ – a CVn May 21 at 15:00
  • $\begingroup$ It's a rendezvous, since Earth is moving. Without proper navigation (including a window), your ship will arrive at some point on at Earth's orbit...but the planet will be somewhere else on it's orbit. Or poor navigation may cause you to reach the Earth...but then flash past uselessly. If your propulsion method generates enough impulse, then you don't need windows - but ion propulsion is high efficiency but low impulse, so windows seem likely. $\endgroup$ – user535733 May 21 at 15:11
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    $\begingroup$ @Whitecold With all due respect, does what I am doing with space station at the heliopause have an impact on the answer to the question? $\endgroup$ – Bob516 May 21 at 15:43
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The answer is... yeeeeesss, ish?

tl;dr: there's always a window for the most fuel-efficient trajectory and another fore the quickest. They'll pop up once a year when earth is in the right spot, but your rocket is so powerful that even the worst case trajectory isn't much slower or less efficient that the best ones.


Lets take a quick stab at the performance of your rocket. Please note that these are very rough figures and don't fully describe all of the orbital trickery you require, but they do provide ballpark performance numbers which is all we really need here.

It looks like you're using a continuous thrust trajectory, or brachistochrone. The rough straight-line distance crossed is about $1.8*10^{10} km$, give or take an AU. You want to do it in a year and a half*. Using good old project rho to be my brain for me, this will give us an acceleration of $a = \frac{4 * d}{t^2}$, or about $0.03 m/s^2$. That's about 3 milligees. To run your engines at that thrust for that long, you'll have a $\Delta_v = 2\sqrt{da}$ or about $1521 km/s$. That's a lot. Modern day interplanetary missions would kill for a delta-V budget a hundredth of that.

With that much power behind you, you can in theory just point and shoot, because the sort of terrible trajectory planning that would be laughed out of NASA because it involved meeting earth in some kind of retrograde orbit and you literally have to burn up tens of km/s of delta-V in order to fix your ghastly mistake and match orbits with the earth... well, that would increase your delta-V requirements by under 5%.

In practice you can't be totally cavalier, because your engine only pulls a weedy 3 milligees and your orbital injection manoevers are going to have to be planned carefully and well in advance. You won't be just eyeballing this one.


I'd like to give you actual trajectory times and window periods, but lambert's theorem is hard and various helpful online and offline calculators only concern themselves with pitiful realistic rockets which aren't ever going to be able to end up in retrograde solar hyperbolic orbits, so that sort of thing will have to wait for another time, or a keener answerer.

* I'm mildly suspicious about your numbers, but in the absense of any better figures for your outrageously overpowered monster ion drive, I can't be more certain. If you can share where you got your numbers from, I might update my answer.

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  • $\begingroup$ This is the site where the table came from, astronomycafe.net/FAQs/BackTo343x.html. I don't necessarily think it is 100% reliable, but since I'm dealing with fiction I wasn't being too picky about it. $\endgroup$ – Bob516 May 22 at 0:27
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There would absolutely be a launch window for the most direct trajectory

This is a simplification, but it illustrates the point.

Assume the following:

  1. As in your post, the most direct trajectory gets you to earth in 18 months. We'll take this to mean that using the best maneuvers possible with your ion drive, it will take 18 months for your ship to intersect with earth's orbital path.
  2. It doesn't matter how fast you're going relative to the earth, so there is no need to slow down to match the earth's orbit. In other words, you can crash into the earth and it still counts as getting there. It would take additional time if this were not the case.

If you leave at the wrong time (really this is about the relative position of your station and the earth in relation to the sun), then you might encounter the earth's orbit when it is on the other side of the sun, you don't hit your target, you miss the earth completely. You are correct that you would be able to make course corrections regardless of the time you leave and still be able to get to the earth, however, the additional maneuvers that you would need would take more time. You would still get to the earth, and if you optimized you could even go in the most direct way possible for your departure time, but it would not be the most optimal for all possible departure windows.

This is mostly due to the fact that depending on where the earth and your station are to begin with can drastically change the length of the path you need to take. Note that you need to consider how long your path is from where you start (your station at a point in its orbit) to where you end (where the earth is, not when you leave, but where it will be after you spend that amount of time in flight). If you have to be on the opposite side of the sun for your encounter, then you're going to have to fly longer, if you don't have to go around the sun, then you can encounter the earth sooner.

There are other considerations of course to being able to get into orbit around the earth instead of just crashing into it, this could take considerably longer especially considering the low thrust available to your craft. But we're ignoring those for this discussion because it makes it even harder.

P.S. Play Kerbal Space Program for a while, and you'll be able to observe the issues involved first hand trough simulation.

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    $\begingroup$ I think that the window is not critical for a rocket of the sort of performance that the OP is suggesting. I think it might make a difference of maybe +10% on your delta-V budget, though there are a lot of unknowns I haven't quite worked out yet. $\endgroup$ – Starfish Prime May 21 at 21:08
  • $\begingroup$ @StarfishPrime Note that delta-V and most direct/shortest time are very different things and the original post specifies that time is the primary concern not delta-V $\endgroup$ – Mathaddict May 22 at 15:58
  • $\begingroup$ Over such a long distance and with such a high performing rocket, delta-V is about the only thing that changes significantly depending on your launch time. $\endgroup$ – Starfish Prime May 22 at 17:53
  • $\begingroup$ Except for the travel time due to the distance of your target. $\endgroup$ – Mathaddict May 22 at 21:37
  • $\begingroup$ Did I mention the long distance? You have to travel 119-121 AU. The extra distance is a very small fraction of the total, and because you'll be continuously boosting you'll reach a higher top speed and so journey time does not increase linearly with distance. The changes are not significant to the overall journey time. $\endgroup$ – Starfish Prime May 23 at 7:13

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