How Does Relativity Affect Bullet Travel Time in a Moving Rocket?

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SUMMARY

The discussion focuses on calculating the time it takes for a bullet to reach the front of a moving rocket, specifically one with a rest length of 100 meters traveling at 0.6c relative to Earth. The bullet is fired at a velocity of 0.8c relative to the astronaut, resulting in a bullet speed of 0.946c relative to Earth observers. The length of the rocket is perceived as 80 meters by Earth observers and 60 meters by observers moving with the bullet. The correct formula for calculating time in the Earth frame is time = length/speed, specifically using the values 80 meters and the difference in speeds.

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dmayers94
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Homework Statement


A rocket of rest length 100 meters is moving at .6c relative to Earth and contains an astronaut at the tail end of the ship. The astronaut fires a bullet toward the front of the ship. The velocity of the bullet relative to the astronaut is 0.8c. How much time does it take the bullet to reach the front of the ship as measured by Earth observers? As measured by astronauts at rest with respect to the ship? As measured by observers moving with the bullet?

Homework Equations


t = t_0*gamma
L = L_0/gamma
time = length/speed

The Attempt at a Solution


There are more parts to this problem that I did not type in because I already solved them, and this gives some extra information that we could use. The bullet is moving .946c relative to Earth observers. The length of the ship according to the astronaut is 100 m, according to Earth observers it is 80 m, and according to observers in the frame of the bullet it is 60 m.

So I know that the time passed is the length of the ship divided by the speed of the bullet relative to the ship, but there are many speeds and multiple lengths of the ship, so I can't figure out which combinations to use. Is the time measured in the Earth frame 80/(.946c -.6c) or can I not simply take the difference of these relativistic speeds? Thanks for any help.
 
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dmayers94 said:
Is the time measured in the Earth frame 80/(.946c -.6c)... ?

Yes, that's correct. You can check this result by calculating the time using the Lorentz transformation equations.
 

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