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Relativistic Bullet - perpendicular (Posted on 2005-01-17) Difficulty: 3 of 5
Suppose you're traveling on a space ship at 9/10 the speed of light (.9c). You have a high-powered rifle that shoots bullets at the same speed. Suppose you shoot the bullet perpendicular to your direction of travel.

It appears that the bullet would travel at a 45-degree angle (northeast, if the ship is traveling north and the bullet is shot eastward), at about 1.2728c which is faster than light. Why is this wrong, and what would the actual speed and direction be?

See The Solution Submitted by Ken Haley    
Rating: 4.5000 (6 votes)

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Some Thoughts First thoughts. | Comment 3 of 15 |

The problem as stated is badly formatted. In classic ("Newtonian" )physics, it is possible   to get away with saying that a vehicle is travelling at "speed" (actually velocity) v without mentioning the reference frame, but it is more difficult to do so in relativistic ("Einsteinian") physics.


Restating the problem: A vehicle, V, passes a planet, P, going north (parallel to the planet's axis) with a measured velocity of v(V)=.9c. At the moment of closest approach, it ejects a missle, M directly away from the planet ("east"). From the vehicle's reference frame, the missile appears to tavel "east" (in a direction perpendicular to the vehicle) at a velocity v(M)= .9c.


Classic physics would predict that from the refernce frame of the planet, the velocity of the missile would be the vector sum of the two velocities, which would resolve to a northeast velocity of .9c √2 = 1.2728c, but this can't be the velocity under Relativity, since it exceeds c. What is the velocity predicted by Relativity.


It is known that if two velocities are parallel, there sum under classic physics is v(1) + v(2), but under Relativity, it is [v(1) +v(2)]/[1 + v(1)v(2)/c²]. In other words, the velocity predicted by classic physics is reduced by a factor of 1/[1+v(1)v(2)/c²]


I assume that the reduction factor in this case becomes tan θ/[1+v(V)v(M)/c²], where θ is the observed angle from north of the missle's trajectory. The question then becomes is the observed angle still 45°? Or if not why not?


Assuming that θ is still 45°, the velocity then becomes tan 45°(.9c√2)/ (1+.81)= [1.2728c/√2]/1.81 =0.78133


Edited to correct trig function and add last paragraph. and then again to correct formatting problems that crept in.
Edited on January 18, 2005, 1:53 am
  Posted by TomM on 2005-01-18 01:20:30

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