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I find time dilation confusing so I'm not sure if it is a relevant factor.

Lets say you have a spaceship. There's no such thing as faster than light travel, but you do have inertial dampener technology that allows you to accelerate at very high G's without obliterating yourself.

You want to get to a nearby star system as soon as possible. When I say "soon" I mean early on the time-line of the destination star system, not necessarily as quick as possible according to your own perspective of the voyage duration. In other words, you want to intervene in the events of the destination before they progress to far into the future.

Is there a maximum rate of acceleration or top speed you could reach before time dilation causes you to actually arrive later than you would have at a lower speed?

What about if you want to go there and then get back before your friends at home get any older?

Qmechanic
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  • For simplicity, suppose the origin and destination planets have no relative motion. We can work in just the frame where these two planets are at rest. It is clear in this frame that higher accelerations and top speeds (both as measured in this frame) will get the spaceship to its destination faster. Time dilation effects are only relevant if we wish to consider how much time the passengers of the spaceship have experienced during the trip, and you say you are not interested in this. There is additional analysis to be done if the acceleration is defined relative to the spaceship frame. – diracula Sep 02 '17 at 13:18
  • So, for my own curiosity, what if you were impatient, and didn't want to sit around on the ship for a long time? Is traveling faster going to make it seem to take longer in this case? – Lorry Laurence mcLarry Sep 02 '17 at 13:33
  • What the! How is this an exact duplicate? That question is pretty much unrelated. – Lorry Laurence mcLarry Sep 02 '17 at 14:14
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    I think you need to spend some time learning time dilation (well special relativity as whole, I suppose). I'd have closed your question as unclear, not as a duplicate, but John's answer there does provide the details for which you can use as a starting point to answer your question. – Kyle Kanos Sep 02 '17 at 16:56

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