1
$\begingroup$

I've been wondering about this for a while. So let's assume a ship accelerates at 1 million G for an extremely short period of time, such that its total velocity change was (say) 1cm/s. Would the ship or its passengers suffer damage, and if so how much? Is there any level of velocity change where immense accelerations are safe/survivable?

$\endgroup$
3
$\begingroup$

Using $g\approx10\,\mathrm{m/s^2}$ for the standard acceleration of gravity on Earth, so $a=10^7\,\mathrm{m/s^2}$ and $$\Delta v=a\Delta t$$ we get $\Delta t=1\times10^{-9}\,\mathrm{s}$, that is, the extremely short period is about 1 nanosecond.

Plugging that into $$\Delta s=\frac12a\Delta t^2$$ we get that the object moves 5 picometres, 0.1 the Bohr radius of the hydrogen atom. I don't think the ship or its passengers will notice that tiny movement. ;)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Nice answer. +1 from me. $\endgroup$ – Gert Jun 24 at 14:59
  • $\begingroup$ yes, clear and simple, +1 from me too. $\endgroup$ – niels nielsen Jun 24 at 16:21
  • $\begingroup$ Maybe their displacement during the period of acceleration would be only 0.1 Bohr radius of the hydrogen atom, but, if the ship velocity suddenly changed by 1 cm/sec, they would certainly feel this.. $\endgroup$ – Chet Miller Jun 25 at 14:56
  • 1
    $\begingroup$ We don't need to do that. All we need to do is to jump to 1 cm/sec in a relatively short time, like say 1 millisecond. Assuming a kinetic friction coefficient of, say, 0.2, our feet would have applied to them a frictional force to accelerate at only 0.2g = 2 m/sec^2. So for our bodies to reach final velocity, it would require about 5 msec. Our bodies (the bottom of our feet) would experience a frictional force of 0.2mg horizontally, or nominally 200 N, over this time interval. $\endgroup$ – Chet Miller Jun 25 at 21:32
  • 1
    $\begingroup$ @Chet You make good points. I'll give this further thought, and invite input from others on this interesting question. $\endgroup$ – PM 2Ring Jun 27 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.