# Special relativity question

• August 18th 2012, 02:46 PM
alexmahone
Special relativity question
A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.6c. The pursuit ship is traveling at a speed of 0.8c relative to Tatooine, in the same direction as the cruiser. What is the speed of the pursuit ship relative to the cruiser?
• August 18th 2012, 05:23 PM
HallsofIvy
Re: Special relativity question
I believe I answered this on another board and said that I agreed with the result you gave their. But perhaps we were both looking at this incorrectly (I'm not a physicist!). We both were doing this as $\frac{u+ v}{1+ \frac{uv}{c^2}}$. But relative the cruiser, Tatoine has a velocity of -.8c. So the "sum" of speeds should be $\frac{-.8c+ .6c}{1+ \frac{(-.8c}(.6c)}{c^2}}= \frac{-.2c}{1- .48}= -.38c$.
• August 18th 2012, 05:34 PM
GJA
Re: Special relativity question
Hey, guys.

I think the numerator should be reversed:

$u=.8c$ is the velocity of sprinter ship relative to Tatooine and

$v=-.6c$ is the velocity of Tatooine relative to the cruiser ship.
• August 18th 2012, 05:41 PM
HallsofIvy
Re: Special relativity question
If that were the case, the sprinter ship would see the cruiser ship with speed +.38c and it would not be gaining!
• August 18th 2012, 06:45 PM
GJA
Re: Special relativity question
The velocity is measured relative to the cruiser, so it should be gaining.
• August 19th 2012, 03:00 AM
alexmahone
Re: Special relativity question
Quote:

Originally Posted by HallsofIvy
I believe I answered this on another board and said that I agreed with the result you gave their. But perhaps we were both looking at this incorrectly (I'm not a physicist!). We both were doing this as $\frac{u+ v}{1+ \frac{uv}{c^2}}$. But relative the cruiser, Tatoine has a velocity of -.8c. So the "sum" of speeds should be $\frac{-.8c+ .6c}{1+ \frac{(-.8c}(.6c)}{c^2}}= \frac{-.2c}{1- .48}= -.38c$.

Actually, Doc Al on physicsforums spotted the sign error.