special relativity - Relativistic addition of velocities of spaceships

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 $.5c-.3c = .2c$.

Answer

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 $1.5c$!

See here http://en.wikipedia.org/wiki/Relativistic_velocities#Composition_of_velocities

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

In our case, we get

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