Special Relativity and Composition Law for Velocities with respect to an observer watching two objects moving away from a central point

So I understand that special relativity is all about the frame of reference and there is a lot to do with time dilation and how space-time is warped at velocities near $c$. So my question is what about an observer on Earth, watching two ships accelerate, in opposite directions, away from the observer at say 1g. As they pass a velocity between the observer on Earth and a ship of .51 , cant the observer on earth measure both of them going at .51c? At which point, with respect to the observer, they are moving away from each other at 1.02c? Now special relativity says the two ships would be moving away from each other at $(v+u)/[1+(vu/c^2)]$, but how does that relate to the stationary observer?

Thanks for your help on this a co-worker and I were unable to find another reference.

• Spacetime is not "warped at velocities near $c$" for approximately 7,000 reasons, the first of which is that spacetime does not have a velocity. – WillO Mar 27 '16 at 4:34
• As @AndreaDiBiagio notes, the limitation is only on two objects observing each other, not on a third party observer observing two objects. In that case, the two individual limits are c, so the total limit is 2*c. No contradiction since there's no physical object moving faster than c, just the measurement of a distance. – barrycarter Mar 28 '16 at 2:35

What you are saying is right. Another situation in which something similar happens is when you consider two proton beams in the LHC, travelling at $99.999\%$ the speed of light collide, the distance between them decreases at $199.998\%$ of $c$ in the lab frame.
Basically that equation tells you: if you observe that there are two bodies, one moving at $u$ and the other at $v$ (either towards or away from each other), then the speed of either, in the frame of the other is given by tha formula.