In a fictional future, spaceships may fight each other at relativistic speeds. C

Question

In a fictional future, spaceships may fight each other at relativistic speeds. Consider the following situation: the USS Enterprise is directly in front of a Klingon Warbird(both spaceship names), with both ships traveling at 0.8c. You are on a third ship and observe that the Enterprise and the Warbird fire photon torpedoes at each other simultaneously in your frame when you are halfway between the ships, whcih are 10 light-seconds in your frame. You know that a photon torpedo travels at 0.99c relative to the object that fired it.

a) According to the Enterprise, who fired first?

b) How fast are the two torpedoes traveling in your frame?

c) Photon torpedoes can be armed with a timed detonator. How long should the Enterprise set the detonator for in order to hit the Klingon ship?

Explanation / Answer

a)
Since the enterprise travels with a speed 0.8 c relative to our ship and we measured the firing of the torpedoes to be simultaneous ,the firing will not be simultaneous in any of the moving reference frames .Thus according to the USS enterprise ,Warbird fired first
b)
The torpedoes velocity relative to the warships is
u' = 0.99 c
The velocity of the warship is v = 0.85 c
The velocity of torpedo with respect to our ship is u
u = u' + v / (1 + u' v / c2)
u = 0.99 c + 0.85 c / ( 1 + (0.99 c x 0.85 c / c2) )
u =   0.9992 c
c)
The distance between the spaceships as measured by us = 2 x 10 light seconds = 2 x 30 x 108 m = 6 x 109 m
The distance measured by the moving USS is
x = d / sqrt (1 / (1- v2 /c2) )
x = 6 x 109 m x sqrt (1- 0.852 c2 / c2)
x = 3.16 x 109 m
The relative velocity of the torpedo is 0.99 c
The time needed to cover the distance of 3.16 x 109 m is
t = x / v = 3.16 x 109 / 0.99 c =  10.64 s

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