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Relativistic Addition of Velocities

  • #1
15
1
This question comes from a previous years exam as practice for my upcoming.


Homework Statement


Two spaceships are launched from Earth, going in opposite directions. Eventually, both spaceships have a velocity of 0.75c (where c is the speed of light), each in their respective directions. A confused relative of yours points out that, if you take the point of view of one of the spaceships, it would seem that the Earth would be moving away at 0.75c and the other ship would be moving away at 1.5c. Explain what is wrong with this argument, and calculate the actual speed of the other ship in the ship reference frame using special relativity.

Homework Equations


ux = (u'x + v)/(1+ vux/c2)

The Attempt at a Solution



Well, first off the problem with the original argument is that special relativity states that nothing can exceed the speed of light. (1.50c > 1.00c)

I'm pretty sure I know how to do this, my main issue is defining u'x.
I have it as u'x = 0.75c

Using u'x = 0.75c we have:
and v = 0.75c

(0.75c + 0.75c)/(1 + (0.75c)2/c2)
= 1.50c/1.5625
=0.96c

Does this make sense for the "actual speed of the other ship in the ship reference frame using special relativity"? (This wording is very poor as well, I'm 99% sure it means the speed of ship 1, as seen from ship 2).
 

Answers and Replies

  • #2
236
16
I'm pretty sure I know how to do this, my main issue is defining u'x.
I have it as u'x = 0.75c
I'm 99% sure it means the speed of ship 1, as seen from ship 2).
Yes, you're correct.
 

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