# Homework Help: Where did I go wrong in this special relativity problem?

1. Aug 19, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
Star A and Star B are at rest relative to Earth. Star A is 27 c*y from Earth, and as viewed from Earth, Star B is located beyond Star A.

a) A spaceship is making a trip from Earth to Star A at a speed such that the trip from Earth to Star A takes 12 years according to clocks on the spaceship. At what speed, relative to earth, must the spaceship travel?

b) Upon reaching Star A, the spaceship speeds up and departs for Star B at a speed such that the gamma factor is twice that of part a. The trip from Star A to star B takes 5 years according to the spaceship time. How far, in c*y, is Star B from Star A in the rest frame of Earth and the two stars?

2. Relevant equations
tpγ=t
γ=(1/(1-(v^2/c^2)^1/2)

3. The attempt at a solution
I got part a correct. i took the equation tpγ=t and multiplied both sides by v, then set the right side equal to the given distance between Earth and Star A and solved for v.

The answer is .91c confirmed by the back of my book.

For part b, the gamma factor is doubled, so I solved for the value of gamma in part one with the now known velocity of .91c and multiplied it by 2, giving me 4.823. I then used this new gamma to solve for the new velocity, and got v=.97826c.
I used similar method as part one with the tpγ=t equation. first used the 5 year proper time and the new gamma factor to get the time according to earths frame, then multiplied that amount of time by the velocity of .97826c. I end up getting 23.5 c*y when the back of my book says 22 c*y.
;-(

2. Aug 19, 2013

### Staff: Mentor

I got roughly the same answer that you got. If they rounded the velocity in part a to 0.9, and then rounded the result in part b to two significant figures, they would have obtained 22c. I don't know whether this is what they did or not.