A ship, earth, and a missile. Which time gests dilated and why?

[HelloDrChewz]

Homework Statement

"A hostile spacecraft approaches your outpost on a distant planet at a speed of 0.3c. When it is at a distance of 3.0*10^{7} km you launch a missile at the enemy with a speed of 0.5c.

How much time will elapse before the missile hits the enemy spacecraft as seen in the rest frame of the enemy spacecraft ?"

Homework Equations

x' = γ(x-vt)
t' = γ(t-\frac{vx}{c^{2}})
U' = \frac{U-v}{1-\frac{uv}{c^{2}}}

The Attempt at a Solution

This was a test question that I got wrong. I figured if I found the velocity of the missile with respect to the ship, then divided the distance by the velocity, I'd have the time. What I forgot to account for was length contraction. But, it seems to me that this is essentially the twin paradox. How does one determine which length is contracted when in either reference frame the other is flying toward it with v=0.3c. The scenarios are symmetric.
I tried to do the lorentz transformations to find the distance in the moving reference frame, but found the time coordinates were different for those two points.
Am I overlooking something here? Is there a logic behind picking one to be contracted over the other?
When I ask my professor, he tells me that he defined it as proper length. But his choice seems to me to be entirely arbitrary, and not explicitly defined in the problem, so there must be something I'm not getting, right?
When I'd first approached my professor about this, I told him that if he sees a contracted length due to him being in a moving reference frame, then according to him, we would see a contracted length because we're in a moving reference frame (according to him). Then you get in an endless contraction loop. To this my professor said, you can't change reference frames. But I thought that's what these Lorentz transformations were for! What does that mean when they say you can't change reference frames?

 

[vela]

The distance to the ship, $3.0\times 10^7\text{ km}$ is measured in the rest frame of the outpost, so it isn't length-contracted in that frame. That's what your professor meant when he said it was the proper length. It's the distance as measure in the designated rest frame. The ship, which is moving relative to the outpost, will see a smaller spatial separation because of length contraction.

 

[HelloDrChewz]

yea, but my question is, why? If there is indeed no preferred reference frame, where does this asymmetric answer come from? If it's because he shot the missile, then let us consider the instant before the missile was shot. At that point you have a ship approaching a planet (and conversely, a planet approaching a ship). An observer on the planet measures the distance between them at an instant to be A meters, while an observer on the ship measures the distance between them at the same instant to be B meters. Where is the asymmetry? What am I missing?

 

[vela]

I think it would be helpful if you drew space-time diagrams for both the planet's rest frame and the ship's rest frame.

The reason behind your confusion is manifest in this sentence: "An observer on the planet measures the distance between them at an instant to be A meters, while an observer on the ship measures the distance between them at the same instant to be B meters." There is no same instant for the observer on the planet and the observer on the ship. Simultaneity is observer dependent.

The problem tells you that the ship and planet are separated by $D=3.0\times 10^7\text{ km}$ as measured in the rest frame of the planet. Unprimed coordinate correspond to the planet's rest frame, and primed coordinates, to the ship's rest frame. So say this means that at t=0, x_ship=0 and x_planet=D. In the planet's rest frame, the launch of the missile also happens at t=0. Now when you say "same instant" in the ship's frame, what do you mean? Is it when the ship passes through x=0 or when the missile is launched from x=D at t=0? Those two events aren't simultaneous as far as observers on the ship are concerned.