# At 1.5 g of constant acceleration, how long does it take to get to 0.93c?

I just finished reading Andy Weir’s “Project Hail Mary”. He crafted a situation where a ship with was able to travel to Tau Ceti, some 13 light years distant. Having crew, he limited the acceleration to 1.5 g, with the maximum velocity of 0.93c.

How long would it take to get up to speed? How long would they coast? What would the overall trip time be?

• Searching for "relativistic acceleration calculator" or "space travel calculator" turns up plenty of tools to do the math for you, e.g. omnicalculator.com/physics/space-travel Nov 27 '21 at 19:46
• Thanks, but I tried that one and couldn’t make it work to come up with the answer I wanted. Same with another calculator I tried. It might be my lack of understanding of the subject matter. At any rate, @L.Dutch found an appropriate calculator (answer below) and plugged in numbers for me (extra credit). Nov 27 '21 at 19:55
• This is just simple math, not a reasonable worldbuilding question. Nov 28 '21 at 17:43
• I note that in Habitable Planets for Man, Stephen H. Dole estimated that few people would want to settle a planet where the surface c gravity was more than aobut 1.25 or 1.5 g. In this case the crew would spend a year or ship time at 1.5 g, which seems to be about the limit of human tolerance. Nov 28 '21 at 18:43

When I input your parameters into the one I linked, I get

• acceleration time: 1.07 years in ship time, 1.6 years in earth time
• Thanks. I looked at several calculators, and none of them provided useful information given what I knew. Looks like if I’d found this one, it would have done the trick. Nov 27 '21 at 19:52

L.Dutch's answer is good for using off-hand, but I did want to show the equations behind it. There are three steps to the calculation:

1. Calculate how long it takes (from the traveler's perspective) to accelerate to $$v_{\mathrm{max}}=0.93c$$, $$\tau_A$$.
2. Calculate how far the spaceship travels (from the perspective of an outside observer) during that time, $$x_A$$.
3. Calculate how long it takes the spaceship to travel the remaining distance at a speed of $$v_{\mathrm{max}}=0.93c$$, $$\tau_B$$.

With an acceleration $$a=1.5g$$, we can use some hyperbolic trigonometry and algebra to find $$\tau_A=\frac{c}{a}\mathrm{arctanh}^{-1}\left(\frac{v_{\mathrm{max}}}{c}\right)\approx1.07\;\text{years}$$ $$x_A=\frac{c^2}{a}\left(\cosh\left(\frac{a\tau_A}{c}\right)-1\right)\approx1.11\;\text{light-years}$$ The total distance to Tau Ceti is 13 light-years (I think more recent measurements have 12 light-years, but we'll use 13 here), so there is a distance $$x_B=11.89$$ light-years to go, which takes a time $$\tau_B=\frac{x_B}{v_{\mathrm{max}}}\sqrt{1-\frac{v_{\mathrm{max}}^2}{c^2}}\approx4.70\;\text{years}$$ for a total travel time, from the ship's perspective, of $$\tau_A+\tau_B=5.77$$ years.

Now, this assumes that the ship does not slow down before entering the system. If we assume it does decelerate, then step 3 instead becomes calculating the time it takes to reach the halfway point of the journey (which turns out to be 2.13 years after reaching maximum speed), add that to the time it takes to accelerate (1.07 years) and double the sum to account for the second half of the journey, making 6.40 years, which matches the answer L.Dutch's calculator gave.

• So not exactly “simple math…” Nov 29 '21 at 0:00
• @J.D.Ray Yeah, certainly not. Nov 29 '21 at 0:16