# Relativistic addition of velocities of spaceships [duplicate]

If Spaceship 1 is traveling at speed $.5c$ relative to Earth, and Spaceship 2 is traveling at speed $.3c$ relative to Earth in the same direction, what does Spaceship 2 see Spaceship 1's speed as?

I don't think it is quite as simple as $.5c-.3c = .2c$.

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## marked as duplicate by David Z♦Mar 14 '15 at 17:17

Your thinking, as far as it goes, is correct. However, the relativistic velocity addition formula is so well known and so easy to find (e.g., hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html), I simply cannot imagine any serious student of physics asking this question here. – Alfred Centauri Apr 10 '14 at 23:09
@AlfredCentauri To be fair it was asked by 'Fool'. – user6972 Apr 11 '14 at 0:31
What makes you think I know physics? Beginners have to start somewhere! – fool Apr 11 '14 at 0:58
@Fool, as I wrote "I cannot imagine any serious student of physics asking this question here". What I think you know about physics is irrelevant. – Alfred Centauri Apr 11 '14 at 2:51
Okay, but judging from the intention it seemed like you were addressing me. Thx for the post though. – fool Apr 11 '14 at 6:34

So, you are correct, that it is not quite as simple as that.

Have you studied Special Relativity (SR) yet?

One of the most fundamental ideas is the limiting speed of light. Nothing moves faster than c, and this means that velocities must add differently.

Think of the reverse, if you were on the rocket-ship traveling at $0.5c$ and shot a laser-beam at $c$, you certainly won't get the laser traveling at $1.5c$!

So, we need $$u' = \frac{u-v}{1- \frac{uv}{c^2}}$$

In our case, we get

$$u' = \frac{0.5c-0.3c}{1- \frac{0.5c \times 0.3c}{c^2}} = \frac{0.2c}{1-0.15} = 0.235c$$

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