Proxima b is an Earth-like (i.e. rocky) planet, in orbit around the star Proxima Centauri, approximately 4.243 light years from Sol, our home star.

If I had a space vehicle capable of accelerating continuously at 9.81 m/s2 (i.e. 1g), how long would it take to reach Proxima b?

For the purposes of this question, assume that our vehicle departs from low Earth orbit and the goal is to achieve orbital capture of Proxima b, so our vehicle would probably need to turn around and decelerate with respect to the target at some point during its journey to achieve capture velocity.

Also, for the purpose of the question, I'm asking about the apparent travelling time from the frame of reference of a passenger on our vehicle. That said, it would also be interesting to know the apparent travel time and characteristics for observers on Earth and Proxima b.

Thanks.

  • 1
    Given the continuous acceleration and the (I'm assuming) continuous deceleration, one would have to spin the ship at the halfway mark and start decelerating. from about half way. – Edlothiad Feb 9 at 7:21
  • That's what I would imagine, but my understanding of the physics involved in such a long journey, and the effects of approaching light speed, are limited. – Mark Micallef Feb 9 at 7:27
  • Well, that was just a gross over-simplification, but effectively it's like that (I haven't answered because I don't know, I like your question and just wanted to add my $0.02) – Edlothiad Feb 9 at 7:35
  • 2
    If you just want a quick answer there's a calculator: nathangeffen.webfactional.com/spacetravel/spacetravel.php – Blake Walsh Feb 9 at 9:36
  • 1
    @uhoh: It seems a bit odd to say "it's just math/physics with decorations", since of course all space exploration equations are straightforward math/physics problems with decorations. The important point is that they arise in situations that are very unusual outside of space. And, as it happens, acceleration with relativistic compensation is very unusual outside of space. The only other field where that is of any relevance would be particle physics. – Nathan Tuggy Feb 9 at 15:27

5.8 years

According to this page, distance travelled under constant force acceleration, even up to relativistic speeds, is calculated by:

$s(t) = c(\frac{m_0c}{F})(\sqrt{1 + (\frac{F}{m_0c})^2t^2} - 1)$

...and since...

$\frac{F}{m_0} = a = g \approx 10 m/s^2$

...and since we accelerate half-way and decelerate halfway we calculate the time to travel half the distance and then double that. Half the distance is...

$s=300\cdot10^6m/s\cdot60s/min\cdot60min/h\cdot24h/day\cdot365day/y\cdot2.12y \approx 20\cdot10^{15}m$

...so from this we get...

$20\cdot10^{15} = 3\cdot10^8\cdot\frac{3\cdot10^8}{10}\cdot(\sqrt{1 + \frac{10\cdot10}{3\cdot10^8\cdot3\cdot10^8}t^2} - 1)$

...which gives us...

$t = 92\cdot10^6s = 2.9y$

Double this and you get a total traveltime of 5.8 years.

  • 4.2 light years in 5.6 years? Thats one hell of a low estimate – Cursed Feb 9 at 13:36
  • 1
    Do we know if those ~6 years are Earth or ship time? – Diego Sánchez Feb 9 at 14:09
  • 2
    This is correct and the 6 years are Earth time. Ship time would be quite a bit less, probably around 3 years. This mission profile does have a few issues though -- apart from needing a ridiculously efficient engine (you might do it with an antimatter powered radiation drive and a high mass ratio) every dust grain that hit you would go off like a nuclear explosion. – Steve Linton Feb 9 at 14:12
  • 2
    Now all we need to do is develop a propulsion system that can thrust continuously for ~6 years with 1 g acceleration. Simple. – John Bode Feb 9 at 15:26
  • 1
    @JohnBode Even a 0.01g propulsion system would be super interesting, as long as it can keep that up continuously, as it would let us explore our own neighbourhood. And maybe... just maybe... something could. – MichaelK Feb 9 at 15:33

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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