1. Apr 12, 2010

### SAT2400

1. The problem statement, all variables and given/known data
Imagine an astronaut on a trip to Siris, which is 8 light years from Earth. On arrival at Siris,
the astronaut finds that the trip lasted 6 years. If the trip was made at a constant speed of .8c, how can the 8 light year distance be reconciled with the 6 year duration?

2. Relevant equations
E=mc^2

3. The attempt at a solution
I don't know how to solve this question....8 light years-> 6 years. 3/.8c??

2. Apr 12, 2010

### tiny-tim

Hi SAT2400!

(try using the X2 tag just above the Reply box )
(e = mc2 has nothing to do with it … and i think it's Sirius )

the astronaut's measurement of the time (on his own clock) is 6 years

what is his measurement of the distance?

3. Apr 12, 2010

### seagulloftime

The relevant equations are:

γ=1/root(1-v2/c2)
t'=γt
x=γx'

The 6 years is measured from his frame of reference, you need to measure the distance in the same frame as well.

4. Apr 15, 2010

### SAT2400

hmm, can you explain more easily/??:(

Thanks !!

5. Apr 15, 2010

### tiny-tim

Hi SAT2400!

Do you know the Lorentz transformation equations?

(you didn't mention them in your "Relevant equations")

If not, look them up in your book.

6. Apr 15, 2010

### SAT2400

root of( 1-v^2/c^2) ...is this right??

hmm...could you please explain more in detail??

Sorry,,but I still don't know how to solve this question....:(

7. Apr 15, 2010

### tiny-tim

Sort-of …

but how are you going to use it?

8. Apr 17, 2010

### SAT2400

I still have no idea.......T_T

If I knew how to do this,,I would have not come to this website.......

SO...

THank you very much!

9. Apr 18, 2010

### tiny-tim

Use x and t for the coordinates in the Earth's frame (so Sirius is at x = 8, for all t).

Use x' and t' for the coordinates in the astronaut's frame.

Start the trip at (0,0) in both frames.

What do you get?

10. Apr 18, 2010

### Matterwave

To the people traveling on the rocket, the length between Earth and Sirius is contracted to: $$L=L_0\sqrt{1-\frac{v^2}{c^2}}$$ The time that they measure is then simply: $$t=L/v$$