**an observer's point of view in frame S** **I HAVE ANSWERED PART A AND PART B. PLEASE HELP ME WITH PART C**
An armada of spaceships that is 1.00 ly long (in its rest frame) moves with speed .800c relative to a ground station in frame S. A messenger travels from the rear of the armada to the front with a speed of .950c relative to S. How long does the trip take as measured in:
(a) the messenger's rest frame?
(b) the armada's rest frame?
(c) an observer's point of view in frame S?
All the normal non-relativistic equations work provided you take all quantities in the same frame of reference.
Part (a)
In the messenger's frame, we can use s=vt to work out the trip time. But before we can do that, we need to know how far the armada moves as its front moves to the messenger (who is stationary in this frame), and how fast the armada moves in the same frame.
The speed of the armada relative to the messenger is given by the velocity addition formula.
w = (v - u)/(1 - uv/c^2)
where v is the messenger's speed relative to S, u is the armada's speed relative to S, and w is the armada's speed relative to the messenger.
We get
w = (0.950c - 0.800c)/(1 - (0.950)(0.800)) = 0.625c
The length of the armada L in the messenger's frame, is:
L = L_0 / \gamma
where
\gamma = (1)/(√(1-(w/c)^2)) = 1.28
So
L = 1.00 ly / 1.28 = 0.78 ly
The time taken, in this frame, is:
t = L/w = 0.78 ly / (0.625 ly/yr) = 1.25 years
Part (b)
We could follow the same procedure as in part (a) here. You should try that. But, there's another way to do this - use the time dilation formula between the messenger's frame and the armada's frame.
The messenger's elapsed time is 1.25 years, and the messenger is moving relative to the armada, so the armada measures a longer time, by a factor of \gamma (given above).
The time taken is 1.60 years, in this frame.
(c)??????
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Last edited by olyviab; Mar 31st 2010 at 08:01 PM.
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