# Another two ships relativity question

1. Dec 14, 2011

### nearc

I wanted to double check my calculations an interpretation of relativity. Thanks in advance.

Two identical ships leave the same point at the same time traveling .999C and travel a 100 light years each but leave on paths that are 10 degrees apart. If the passengers where to compare their time spent traveling to an outside observer they would only show that 4.47 years [1] have passed compared to the 100 years from the external frame of reference. However, since they have traveled 100 light years on their two different vectors the two ships are now 17.4 light years apart [2].

[1] let c = 1 and v = .999, then 100 years * (1-.999^2)^.5 = 4.47 years
[2] sin 5 = opp/hyp = x/100, x = 100 * sin 5 = 8.72, so 5 degrees gives 8.72 light year distance and then 10 degrees would be 17.4 light years distance between the two ships after traveling 100 light years.

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2. Dec 14, 2011

### joris_pixie

don't have my calculator here but I think your logic is correct

3. Dec 17, 2011

### nearc

thanks for all who have replied or viewed my question, but can i get a little more confirmation that this is correct? as a teacher i don't want spread misinformation. do i need to clarify anything? do i need to account for length contraction?

4. Dec 17, 2011

### Fredrik

Staff Emeritus
I get the same results. No need to consider length contraction if you're only a) comparing the time coordinate in an inertial coordinate system in which the ships both have speed 0.999c, with the times measured by the ships, and b) talking about the coordinate distance between the two ships in that inertial coordinate system.

However, you need to be very careful if you start talking about such things as one ship's idea about the distance to the other ship or what time it is on the other ship, because of relativity of simultaneity.