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I have been working on a sci-fi story for a little while now, and have recently discovered that I made a very big error in assuming the position of Eris. The main issue is that I had forgotten to take into account the fact that Eris's orbit is radically askew from the solar system's plane, 44 degree inclination askew. I had figured it would be slightly off like Pluto, but calculating its position in the year 2428 puts it at roughly 84 AU from the sun and about as far away from the rest of the solar system as it gets. It would be almost perfectly at the bottom of the image below.

enter image description here

With that understanding, I have to get a group of ships from Mars at 1.5 AU to Eris at 86.5 AU (in the year 2428). This puts me at an average speed of roughly 1 AU per day. This is kind of ridiculous, considering I am trying to work within a certain level of reasonableness, and had previously expected my ships to meet an average speed of 1 million kph, not 150 million kph. What I really need is a constant acceleration to the midpoint, and then deceleration to destination.

What would the constant acceleration be given those perameters?

Also, might it be more efficient and/or economical to slingshot off a gas giant after leaving Mars instead? If so, which might be the most reasonable?

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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    $\begingroup$ It would really depend on relative positions of planets. Can your travellers wait a few years for the best alignment? $\endgroup$ – Bald Bear Dec 20 '18 at 17:14
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    $\begingroup$ 3 months from whose standpoint? 3 months of travel time to the traveler or a bystander? At that speed, relativistic time dilation effects might be seen. $\endgroup$ – Anoplexian Dec 20 '18 at 23:58
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    $\begingroup$ @Anoplexian with a constant boost, peak velocity would be less than 1% of light speed. They might need to reset their atomic clocks when they get home, but they would't miss their grandchildren's funerals... $\endgroup$ – Zeiss Ikon Dec 21 '18 at 13:03
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If you have to travel 84 AUs in three months your average speed will be about 28 AUs per month. With an average of about 30.5 days per month or 91.5 days total the average speed will be about 0.918 AUs per day, 0.03825 AUs per hour, 0.0006375 AUs per minute, or 0.0000106 AUs per second.

To achieve such speeds with the least possible acceleration, the ship would have to accelerate constantly until it reached twice the average speed midway through the journey and then decelerate constantly for the rest of the voyage until it reached Eris.

Twice an average speed of about 0.0000106 AUs per second would be about 0.0000212 AUs per second, or 3,171,474.85884 meters per second. The middle point of a voyage of 91.5 days would be at about 45.75 days, or 1,098 hours, or 65,880 minutes, or 3,952,800 seconds.

Thus your space ship would have to accelerate at a rate of 0.8023362 meters per second per second, which is about 0.0818155 of one gravity.

One gravity of acceleration is 9.80665 meters per second per second, so accelerating at one gravity for 161,700.21 seconds would reach a speed of 1,585,737.42942 meters per second necessary to reach Eris in three months. 161,700.21 seconds would be 2,695.0035 minutes, 44.916725 hours, or 1.8715302 days.

Thus if your space ship can accelerate at one gravity, it could accelerate to the needed average speed in about 1.87 days, coast for a little less than 91.5 days, and then decelerate for about 1.87 days, to reach Eris.

Humans can and have survived accelerating at more than one gravity for short periods, but I wouldn't want to accelerate at even as little as 1.25 g for even as short a time as one day until and unless tests prove that would be safe, so one g can be considered the maximum possible acceleration for the mission to Eris.

Thus, according to my calculations, your space ship needs to accelerate constantly at 0.0818155 to 1.00 g, or 0.8023362 to 9.80665 meters per second per second, for days or months at a time in order for the mission to Eris to take three months.

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    $\begingroup$ Looks like a correct calculation, but as my engineering professors would have said (back around 1980): "You need to trim your significant digits." $\endgroup$ – Zeiss Ikon Dec 20 '18 at 18:26
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    $\begingroup$ Surviving acceleration of 1g is something I do every day when I get out of bed. It really is no big deal, just take a nap for a couple days on the wall. $\endgroup$ – Trevor D Dec 20 '18 at 18:39
  • $\begingroup$ TLDR, but the end result (about 0.08g) does look right to me. $\endgroup$ – Alexander Dec 20 '18 at 22:15
  • $\begingroup$ @Alexander that's constant acceleration, it adds up to moving pretty fast. After a day of constant acceleration at 0.8 m/s (~0.08g) you'd be going almost 250000 km/h or 155000 mph... $\endgroup$ – Spoki0 Dec 21 '18 at 9:52
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    $\begingroup$ This answer is worthy of a bounty. $\endgroup$ – Renan Dec 21 '18 at 13:21
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If you have a constant-boost drive, you can get quite a long way in three months, even at just 1 G. Eliminate turnover, and you'd be close to 1/4 light speed in that time; assuming a turnover and stop at the end,you'd peak at around 12% c -- or average 6% c over the whole trip (roughly 14 light hours, you'd need about 224 hours). That means that Eris is less than two weeks at 1 G constant boost, from anywhere in the Solar system. Getting there in twelve weeks would require significantly less acceleration: average of .006 c, so peak of .012 c, 45 days to turnover and the same to stop, gives 1.8e6 m/s over 3.88e6 s, about .46 m/s^2, or just under .05 G.

At this kind of boost, there's no point in trying to tack on a mass like Jupiter -- you'll gain virtually nothing by it at best.

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    $\begingroup$ This. People underestimate just how terrifyingly fast constant acceleration gets you going if there’s nothing opposing you. $\endgroup$ – Joe Bloggs Dec 20 '18 at 20:38
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    $\begingroup$ Sorry, your math does not look right. What formula do you use to get 0.7g? $\endgroup$ – Alexander Dec 20 '18 at 22:14
  • $\begingroup$ @Alexander Probably a typo for .07g if I were to hazard a guess. $\endgroup$ – Logan Clark Dec 21 '18 at 4:49
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    $\begingroup$ @JoeBloggs : what people also underestimate just how mindbogglingly huge amounts of fuel would be needed for such a thing. And, considering the tyranny of the rocket equation, good luck taking all that mass of fuel out of the Earth's gravity well. $\endgroup$ – vsz Dec 21 '18 at 5:23
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    $\begingroup$ @vsz: Well, yes. You kinda need a reactionless drive or some stupidly efficient nuclear propulsion method to even think of such a thing!! $\endgroup$ – Joe Bloggs Dec 21 '18 at 8:37

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