A Breakdown in Simultaneity

[Gymnos]

Hi, I'm stuck on a couple problems from my Mastering Physics homework.

#1
There are three space ships (each 10m in length,) each traveling 90m apart from the center spaceship while traveling at a constant speed of .9c. They're entering into an asteroid (215m in length) where some people hope to capture the ships by closing them inside. The locals would like the two ends to close simultaneously in their rest frame, so the front trapdoor will be set to spring .72 microseconds after the first enemy spacecraft passes whereas the rear trapdoor will spring the instant it gets the signal.

How close to the rear of the asteroid will the first enemy spacecraft be when the trapdoors close?


Perhaps L = L$$_{0}\sqrt{1-(v/c)^{2}}$$ is relevant in this situation.


I tried multiplying .72 microseconds by .9c to get the distance traveled by the first craft during that time, but it returned an incorrect answer of 194. I also tried other combinations, such as adding 10m (in case I hadn't correctly interpreted the wording, and the event activated as soon as the nose of the ship passed the entrance of the asteroid.) That was wrong, too.

For #2
What is the angle $$\theta$$ that these lines make to the ct axis in the asteroid frame?

I think that's the correct image.

I have no idea how to go about this one, to be honest. I'm not sure if I even understand what the axes represent.

 

[anathema]

Hello there,

For problem #1, you are on the right track by using the equation L = L_{0}\sqrt{1-(v/c)^{2}}. However, you also need to take into account the length of the asteroid and the distance between the three spaceships.

Since the ships are traveling at a constant speed of 0.9c, the time it takes for the first ship to reach the asteroid is given by t = d/v, where d is the distance between the ship and the asteroid. In this case, d = 90m.

Now, using the time t = 0.72 microseconds, we can calculate the distance traveled by the first ship during that time, which is 0.72 microseconds * 0.9c = 194.4m. This means that the first ship will be 194.4m away from the rear of the asteroid when the trapdoors close.

For problem #2, the axes represent the ct axis (time) and the x axis (space). The angle \theta is the angle between the x axis and the lines connecting the two spaceships and the asteroid.

To solve this problem, you can use the equation tan\theta = x/t, where x is the distance between the two spaceships and t is the time it takes for the first ship to reach the asteroid (which we calculated in problem #1).

I hope this helps. Good luck with your homework!