Given the recent news about the discovery of an "Earthlike" planet orbiting Alpha Centauri (our nearest stellar neighbour) it got me wondering just how fast would spaceship have to travel to be able to reach Alpha Centauri within a person's lifetime (say 60 years)?

The reasoning is that even if we sent an unmanned probe - assuming that it would be a one way trip - the journey time would have to be sufficiently short to keep people's interest in the mission so that when it actually arrived and sent back data there would be somebody home to receive the data.

The added complication is that the craft would have to slow down sufficiently so that it could at least enter orbit around the star (I'm not going to suggest that it manages to orbit one of the planets!).

I suppose what I'm really interested in how this speed compares to the speeds we've managed to attain so far and thus get an idea of how much technology must advance for us to be able to even think of achieving this.

I know about relativistic effects due to travelling at a high percentage of the speed of light - but I'm not really interested in that here.


2 Answers 2


If you are not interested in relativistic effects, the answer to your question is easy to workout. According to Wikipedia, Alpha Centauri is 4.24 ly away (4.0114x$10^{16}\mathrm{m}$). So to get there in 60 years ($1892160000\mathrm{s}$).

So your non-relativistic answer is

$v = \frac{d}{t} = \frac{4.0114 \times 10^{16}}{1892160000} = 21200000 \mathrm{m}\,\mathrm{s}^{-1}$.

This is 21200 $\mathrm{km}\,\mathrm{s}^{−1}$. The fastest recored space flight was 24,791Mph which is around 11$\mathrm{km}\,\mathrm{s}^{−1}$ which is 0.05% of 21200$\mathrm{km}\,\mathrm{s}^{−1}$. This means we have to be able to get spaceships to travel 2,000 times faster than the fastest current spaceship.

Note, I believe satellites in geostationary orbits do $\approx 17\mathrm{km}\,\mathrm{s}^{−1}$.

Edit. The relativistic calculation can be found here.

  • $\begingroup$ Does this take into account the deceleration needed to enter orbit. Yes it's an average speed, but not the top speed. $\endgroup$
    – ChrisF
    Oct 17, 2012 at 11:13
  • $\begingroup$ This assumes that the space craft is at this constant speed instantaniously. To take into account the effects of acceleration to this speed would require a slightly more complex calculation... $\endgroup$
    – MoonKnight
    Oct 17, 2012 at 11:14
  • 1
    $\begingroup$ Note. The full special relativistic calculation with effects of acceleration has been answered here. $\endgroup$
    – MoonKnight
    Sep 23, 2013 at 10:11

The distance between Earth and Alpha Centauri is $4.4\,\text{ly}$.

Dividing by $60\,\text{years}$ it's approximately $22000\,\text{km/s}$.

The relativistic factor, (I mean $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$) for this is almost $1$.

If we take a constant acceleration of $2g$ (it's possible) it would take only $320\,\text{hours}$ to reach this speed (and, same amount to stop). It total that's only $28\,\text{days}$. Negligibly small in compare with $60\,\text{years}$.

  • $\begingroup$ Does this take into account the acceleration and deceleration needed to enter orbit? $\endgroup$
    – ChrisF
    Oct 17, 2012 at 11:12
  • $\begingroup$ I've added some calculation for constant $2g$ accelaration. $\endgroup$
    – m0nhawk
    Oct 17, 2012 at 11:21
  • $\begingroup$ How much time would've passed for an observer left on earth? $\endgroup$ Oct 17, 2012 at 12:22
  • $\begingroup$ The relativistic factor is 1.003, if more precise. No more than a 60 and a half years. $\endgroup$
    – m0nhawk
    Oct 17, 2012 at 12:49
  • $\begingroup$ @ChrisF: I would expect the time needed to enter an orbit would be much less than 60 years. And therefore with correct calculations, it should be possible to decelerate the spaceship to stop somewhere in the vicinity of the target planet. $\endgroup$
    – Lie Ryan
    Oct 17, 2012 at 15:24

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