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According to general relativity, speed is relative, so for example if you are running at 20 km/h and a car passes you at 30 km/h, the runner is actually moving at 50 km/h relative to the car.

Now imagine that 2 spaceships are traveling both at 99% the speed of light, and they pass each other in a linear fashion. Could it be argued that from 1 spaceships perspective, you are traveling at 198% the speed of light relative to the other spaceship, according to special relativity, yet violating the constant speed of light?.

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marked as duplicate by jinawee, Pulsar, dmckee Dec 13 '13 at 20:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ This is discussed in every single introduction to relativity. Every one. $\endgroup$ – dmckee Dec 13 '13 at 20:28
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The relativity theory goes further and implements rules how to add up speeds. In the case you described, one spceship will percieve the other one as being nearly the speed of light, but not above. The formula that has to be used to add up speeds correctly in special relativity in the described case is: $$s = \frac{v+u}{1+(vu/c^2)}$$ where $u$ and $v$ denote the speed of of each spaceship and $s$ denotes to percieved sum. With the given speeds this yields $s=0.99995\cdot c$, if I am correct.

For additional information you may consult the thread linked by Kyle, where an analogous problem was already discussed.

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