Calculating Relative Time and Distance for Spaceship Catch-Up

[bethany555]

Homework Statement

Spaceship #1 moves with a velocity of .2c in the positive x direction of reference frame S. Spaceship #2, moving in the same direction with a speed of .6c is 3 x 10^9 m behind. At what times in reference frames S, and in the reference frame of ship #1, will 2 catch up with 1.

Homework Equations

The Attempt at a Solution

vrelevative = .6c-.2c / (1-.6(.2)) = .455c
Then, 10 light seconds / .455c = 22.0s

Or...is it done classically .6-.2 = .4, making it 25s?

For the second part, is it done like so?

beta = 1/(sqrt (1-.2)) = 1.02
Then t = 22/1.02 = 21.6s or if it was 25s, 25/1.02 = 24.5s

Can someone explain to me which of these two (if either) are correct? I am not sure if it is done classically or with relativity, so can you explain why it is either.

Thank you very much.

 

[tiny-tim]

welcome to pf!

hi bethany555! welcome to pf!

Or...is it done classically .6-.2 = .4, making it 25s?

yes … in any one frame, speed etc works just the way it should!

(you only need those pesky gammas :grumpy: if you're having to convert measurements from another frame first )

For the second part, is it done like so? …

(btw, that's called gamma, not beta … beta is v/c )

imo, it's always safest to use the original lorentz transformation formulas

(rather than use short-cut contraction formulas which usually work only for rigid separations)

in this case, you know that (in frame S) t = 25, x = … , so t' = … ?

 

[bethany555]

Thank you very much!

For the second part, would it be the following --

Vrelative = .455c (as shown above)

L = 10c seconds (sqrt(1-.45^2)) = 8.93

8.93/.455 = 19.6s

I'm not sure if I'm using the correct v for all parts, would you be able to tell me if I did this correct? Thank you again so much!

 

[tiny-tim]

(sqrt(1-.45^2))

no, your gamma should be for the relative speed between your two frames,

ie between S and the frame of the 1st ship (0.2 c)

and i really think you should use the t' = γt - γvx formula from the Lorentz transformation

 

[bethany555]

no, your gamma should be for the relative speed between your two frames,

ie between S and the frame of the 1st ship (0.2 c)

and i really think you should use the t' = γt - γvx formula from the Lorentz transformation

Thanks! So would it be 9.80cs / .455 = 21.5s?

When I use the Lorenz equation, I get
t' = (1/(sqrt(1-.2^2)) ) * 25 - (1/(sqrt(1-.2^2)) ) .2 * 10, I get 23.47?

 

[tiny-tim]

Thanks! So would it be 9.80cs / .455 = 21.5s?

what is this?

When I use the Lorenz equation, I get
t' = (1/(sqrt(1-.2^2)) ) * 25 - (1/(sqrt(1-.2^2)) ) .2 * 10, I get 23.47?

yes, that looks ok

(i haven't checked the actual figures)