A space ship travels (almost) at the speed of light. When it nearly reaches its destination it starts decreasing the speed. What would be the most appropriate deceleration for human body in such conditions?

Astronauts endure around 3G on lift-off, which is about 29.4 $m/s^{2}$, but only for a short time. To decelerate from speed of light to zero then it takes nearly 4 months. Is there any condition that astronauts could be put in, so they can bear the pressure for such a long time, like cryogenic freeze or something else?

  • $\begingroup$ @a4android Please don't answer question in comments. Make an answer to stun us all. $\endgroup$ Jul 4, 2019 at 2:51
  • $\begingroup$ 4 months isn't all that long if we're talking interstellar or is this interplanetary? Do these human bodies need to be natural or could they be cyborgs with nanobots instead of cells? Or maybe their cells have reinforced membranes? $\endgroup$
    – slOOP
    Jul 4, 2019 at 2:53
  • 2
    $\begingroup$ The premise of your question is somewhat incorrect, you seem to be using the Newtonian formula t = v/a to figure out the time t needed to decelerate from velocity v to 0 with acceleration a, but that formula becomes increasingly inaccurate the closer your starting v is to light speed. The correct relativistic formulas for a ship undergoing constant acceleration can be found on this page, and the equation giving v as a function of T (time onboard the ship) can be rearranged as T = (c/a) * atanh(v/c) $\endgroup$
    – Hypnosifl
    Jul 4, 2019 at 4:33
  • 1
    $\begingroup$ It simplifies things a bit if you use units like light-years for distance and years for time, in which case c = 1, and an acceleration of 9.8 m/s^2 becomes an acceleration of 1.03 ly/y^2, as mentioned on the relativistic rocket page. Then you could rewrite the expression as (1/(1.03 * g)) * atanh(v) where g is the acceleration in multiples of the acceleration felt at the surface of the Earth. You can plug that equation into the calculator here & pick values for g and v to get the onboard time, for ex. with g=3 and v=0.8 you get a time of 0.356 years. $\endgroup$
    – Hypnosifl
    Jul 4, 2019 at 4:36
  • 1
    $\begingroup$ Also, in your other question you mentioned wanting a time dilation factor of 2000/66 million, and I mentioned that this could be done with a velocity of 0.999999999540863177124 times light speed. If we plug that into the equation I gave in my previous comment and use the calculator I linked, it tells us the onboard time to decelerate from that speed to 0 (or accelerate from 0 to that speed) would be 10.774 years at acceleration g=1, 5.387 years at g=2, and 3.591 years at g=3. $\endgroup$
    – Hypnosifl
    Jul 4, 2019 at 5:16

2 Answers 2


If the spaceship was decelerating at one g, meaning a comfortable 9.8 $m/s^{2}$, then deceleration will take roughly one year shiptime. No need for special counter-acceleration measures. Plus, it provides convenient environment for astronauts to work and play in, while decelerating. Also, time to survey their destination prior to their arrival.

However, the spaceship will actually need to decelerate for the same length it took to accelerate to near-lightspeed. For preference, and the comfort of crew and passengers and any stowaways lurking the bowels of the starship, this should be one g (for reasons given above in one paragraph).

Accelerations and decelerations of one g are the most appropriate for spacecraft approaching lightspeed (assuming we can ignore the energetic requirements for such interstellar space travel, because they're astronomically mind-boggling).

  • 1
    $\begingroup$ +1 for mentioning the energy requirements - they make such ultra relativistic travel very unrealistic. $\endgroup$ Jul 4, 2019 at 3:18
  • $\begingroup$ @StephenG My sentiments exactly. Even modestly relativistic space travel is incredibly energy greedy. It's fine for science fiction stories, but for realistic spaceflight not so. Relativistic travel is a physical possibility, but a practical impossibility $\endgroup$
    – a4android
    Jul 4, 2019 at 7:38
  • $\begingroup$ If the target planet has a known gravity, the best deceleration may be to match that - or, if the gravity is higher than the Earth's, start with 1 g deceleration and slowly increase to the target planet's gravity.. $\endgroup$ Jul 4, 2019 at 7:43
  • $\begingroup$ @a4android - There are possibilities for getting things close to light speed without having to carry the fuel onboard, so you don't have to deal with the relativistic tsiolkovsky equation implying a huge ratio of fuel mass to payload mass. One idea is to aim an enormous laser array in the solar system at a ship with a solar sail, another is to magnetically accelerate a stream of small pellets to near light speed with something like a magnetic coilgun, which could push against a magnetic sail (magsail) on the ship. $\endgroup$
    – Hypnosifl
    Jul 4, 2019 at 14:30
  • $\begingroup$ (a magsail can also potentially be used for decelerating using particles in the interstellar medium without needing to have fuel onboard, see here and here) $\endgroup$
    – Hypnosifl
    Jul 4, 2019 at 14:37


han solo in the carbonite


"Freeze" your spacefarers in carbonite. How does carbonite work, you may ask?

Carbonite was a liquid substance that was made from carbon gas and could change into a solid through rapid freezing. Goods could be encased in carbonite for preservation, through a process known as carbon-freezing,1 Carbonite blocks could also be used to place people in hibernation.[3] Before the invention of the hyperdrive, some early spacers would use carbonite to endure long voyages. That technique, however, had brutal side effects, collectively referred to as hibernation sickness. https://starwars.fandom.com/wiki/Carbonite

(waves hands) There you go! They even used it for what you want to use it for.

Remember to make a funny face as the carbonite flows around you.

Do not freeze people in carbonite if you exist in a hard science world.

Do not confuse carbonite with the explosive of the same name.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .