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I solved exercise 6.1 of "Gravitation" by Misner, Thorne and Wheeler and reached the impressive result that it takes only 11 years of proper time to reach the center of the Milky Way (~30,000 light years away) if you accelerate at one earth-gravity (103 cm/sec2) for half the trip and decelerate at the same rate for the remaining half.

What I did was to simply use the formula x = g-1 cosh gτ. If I got relativistic units right, then g = 1 (in light-years) and x = 30,000, which yields τ = 11 years. But since these accelerations are quite mundane (lower than that of a Formula 1 race car), I suspect I went wrong somewhere. Did I?

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    $\begingroup$ The calcuations on this page agree with yours. It's a reputable site as well, so I think you're correct. $\endgroup$
    – Michael Seifert
    Commented Jan 7, 2023 at 13:38
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    $\begingroup$ @MichaelSeifert Baez says 20 because of deceleration, which doesn't agree with OP's 11? $\endgroup$
    – Kyle Kanos
    Commented Jan 7, 2023 at 13:40
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    $\begingroup$ @KyleKanos: Good point. If the goal is to arrive at the center of the galaxy, then you should use $x = 15,000$ ly and then double the result. But the order of magnitude is correct, which is (I think) the main point that the OP was skeptical of. $\endgroup$
    – Michael Seifert
    Commented Jan 7, 2023 at 13:45
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    $\begingroup$ I leave this comment after voting to reopen this question to explain why. It is a question starting from a calculation and the following doubt of a mistake. But it should be clear to everybody, but closing-vote-addicted people, that there is a genuine conceptual/quantitative issue related to not grasping the effect of a continuous $1g$ acceleration for a long time. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jan 7, 2023 at 14:28
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    $\begingroup$ Guys, you are amazing. @MichaelSeifert, I checked the page you mentioned, it really clarifies everything! $\endgroup$
    – Guilherme Bahia
    Commented Jan 7, 2023 at 17:36

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