Hahahaha, I'm sorry. Right, ##v^{pilot}>c## so the number would be imaginary. The velocity that goes into gamma, at least on my notes, is the velocity seen from the earth.
Okay, then I think that ##\displaystyle\frac{t}{t'}=\frac{1}{\gamma}={\sqrt{1-{(v^{pilot}/c)}^{2}}}##. Am I right? Thank you again, wabbit. I'm learning more from this exercise than from my professor.
Oh! okay, I understand it now. I obtained
Oh, okay, I understand it. On the section a) I obtained ##v^{pilot}=7.81\cdot 10^{12}## m/s. From here, to solve b), I was thinking on use ##\displaystyle v^{earth}=\frac{v^{pilot}\cdot t'}{t}=\frac{7.81\cdot 10^{12}\cdot t'}{t}##, with ##t'=1209600##s...
Yes, that's what I've noticed too. If we do ## v^{pilot}=1000 light years/2 weeks##, we can't use the obtained value to find ## v^{earth} ##, because ## v^{pilot} ## is not a well-calculated velocity. It doesn't matter?
Hi, thanks for your answer. Then are you saying that the value ##v^{pilot}=d/t'## can be used to find ##v^{earth}## using that ##v^{pilot}=d/t\cdot t/t'=v\cdot t/t'##? Thank you.
Hi phinds. I think that the paragraph of Star Wars on the exercise has nothing to do there. It's only saying that on the exercise, if you don't know what relativity is, it could happen to obtain velocities >c. In fact, on the a) section we obtain a v>c.
Thank you for your answer.
Hi! First of all, sorry for my english, I'll try to do my best translating the exercise. I've been working on it this evening but I'm having some troubles:
Problem: On the Star Wars film, there's a starship traveling at a velocity of 5c. This is nonsense, but we can assume this if we suppose...