Relativity of simultaneity (astronaut and simultaneous explosions.)

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Gator93
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Homework Statement



An astronaut is space-traveling from planet X to planet Y at a speed of 0.17c. When he is precisely halfway between the planets, a distance of 1 light-hour from each, nuclear devices are detonated. The explosions are simultaneous in the astronaut's frame. What is the difference in time of arrival of the flashes from the explosions as observed by the astronaut?

v = 0.17c = 51000000 m/s
x = 2(1.079E12)
t' = 0 is simultaneous?
t = ?

Homework Equations



t = (gamma)(t' + (vx'/c^2))

gamma = 1/(sqrt (1-B^2))

The Attempt at a Solution



1/(sqrt (1-0.17^2))
1/sqrt(0.9711)
= 1.014770943

(1.014770943)(0 + (51000000*2.158505698E12)/(9E16)
= 20.687 min?

Now I'm not seeing how the units follow through on this without taking it from sec to min, but I've arrived at this answer a few other ways... namely multple equations and setting them equal to see when the first flash would encounter the ship and also when the second flash would catch up to the ship, and taking the difference.

So, I was giving this as a multiple choice problem.
A) 14 min B) 0 min C) 56 min D) 28 min

There is not an E) other... but wouldn't put it past him... then again I'm not confident I'm doing it correctly.
 
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Welcome to PF!

Hi Gator93! Welcome to PF! :smile:
Gator93 said:
An astronaut is space-traveling from planet X to planet Y at a speed of 0.17c. When he is precisely halfway between the planets, a distance of 1 light-hour from each, nuclear devices are detonated. The explosions are simultaneous in the astronaut's frame. What is the difference in time of arrival of the flashes from the explosions as observed by the astronaut?

You're making this too complicated :redface:

first rewrite the question in the astronaut's frame …

what does it say? :smile: