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Consider watching an alien space ship at Alpha Centauri (at 4.5 light years away) through a telescope from earth. This space ship turns towards us, and starts travelling toward us at what appears to be a velocity of exactly 2 million kilometers per second, when measuring the distance every hour.

The space ship will arrive here in about 8 months.

How fast is the space ship really travelling?

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  • $\begingroup$ Note that this is entirely possible, according to relativity. $\endgroup$
    – frodeborli
    Feb 11, 2014 at 10:03
  • $\begingroup$ How do we measure the distance every hour ? $\endgroup$
    – Tom-Tom
    Feb 11, 2014 at 10:06
  • $\begingroup$ @V.Rossetto I don't know. The sonic screwdriver comes to mind, but it is not relevant for the answer. If it makes it easier, just consider watching it depart, and then land 8 months later and knowing that Alpha Centauri is about 4.5 light years away. $\endgroup$
    – frodeborli
    Feb 11, 2014 at 10:16
  • $\begingroup$ We see apparent velocities (much) faster than light all the time. A photon from Alpha Centauri reaches your eye at exactly the same instant you see it leave, making its apparent velocity infinite. $\endgroup$
    – WillO
    Feb 11, 2014 at 14:22
  • $\begingroup$ @WillO Yes. But the really interesting thing with this question is that for the aliens on that space ship, the trip only takes 8 months, I believe? In other words; they will only age by 8 months during their journey. $\endgroup$
    – frodeborli
    Feb 12, 2014 at 11:13

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Suppose the spaceship is moving at speed $v$ and covers a distance $d$. The light emitted when the spaceship left reaches us at a time:

$$ t_0 = \frac{d}{c} $$

while the time the spaceship reaches us is:

$$ t_1 = d/v $$

The apparent velocity is distance divided by time so:

$$\begin{align} v_a &= \frac{d}{d/v - d/c} \\ &= \frac{vc}{c - v} \end{align}$$

Your question asks for the real velocity, so we rearrange to get:

$$ v = \frac{cv_a}{c + v_a} $$

If you substitute your value for $v_a$ you get $2.603×10^8\text{ m/s}$.

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  • $\begingroup$ For many, a more intuitive way to think may be to think in terms of distance and journey time - taking into account that the journey took 8 months + 4.5 years. (4.5 light years) / ((4.5 years) + (8 months)) = 261 109 560 m / s. A slightly different number from what you got, it seems... $\endgroup$
    – frodeborli
    Feb 12, 2014 at 11:20
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    $\begingroup$ @frodeborli: the OP said velocity of exactly 2 million kilometers per second compared to The space ship will arrive here in about 8 months. The journey time is approximate and that's why I didn't use it. $\endgroup$ Feb 12, 2014 at 12:45
  • $\begingroup$ @GlenTheUdderboat: the difference in the 3rd decimal place is presumably just the value we used for $c$. I wanted to add the reference to your comment since you obviously did the calculation and I didn't want it to look as if I was claiming any originality for my answer. However I'll remove it now. $\endgroup$ Feb 14, 2014 at 8:17
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The answer to the apparent contradiction is that when you observe the spacecraft, it has long since left its starting point.

Suppose the spacecraft leaves at velocity $v=\beta c$ at time $t=0$ in the shared Earth / Alpha Centauri rest frame. Then the light emitted at that event will not reach earth until time $t=4\tfrac{6}{12}\:\mathrm{yr}$. If the spacecraft arrives on Earth eight months after that, at $t=5\tfrac{2}{12}\:\mathrm{yr}$, then its speed will be simply $$ v=\frac{\Delta x}{\Delta t}=\frac{4\tfrac{6}{12}}{5\tfrac{2}{12}}c=\tfrac{27}{31}c\approx0.87c=2.61\times10^8\:\mathrm{m}\:\mathrm{s}^{-1}, $$ which is of course slower than light, as the light got here first.

The spacecraft's apparent speed will, of course, be much faster, since we must observe it to transverse the $4.5 \:\mathrm{ly}$ in the eight months between our initial obervation and the spacecraft's arrival, which gives a speed of $2.61\times10^9\:\mathrm{m} \:\mathrm{s}^{-1} =6.75c$. John Rennie's answer provides a good explanation of how this number can be arrived at directly from the numbers above.

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