Calculating Time Dilation in a Relativistic Spaceship Passage

[Little PBrane]

Homework Statement

Two fixed navigation beacons mark the approach lane to a star. The beacons are in line with the star and are 40 x 106 m apart. A spaceship approaches the star with a relative velocity of 0.30 c and passes the beacons. The passage of the ship between the beacons is timed by observers on the beacons. The time interval of the passage according to observers is closest to:

Homework Equations

The given distance between the star and the beacons=s=40*10^6 m

Velocity of the given spaceship=0.30 c

The passage of the ship between the beacons is timed by observers on the beacons

Δt=Δt0 /√(1 - v2/c2)

The Attempt at a Solution

=(s/v)/√(1 - v2/c2)=(40*10^6 m/0.30c)√(1 - 0.302)
=0.44/0.954=0.465 seconds=465 ms
I think this is wrong. Answer choices given are:
470, 420, 250, 440, 170
Am I thinking this right? For the observer the time should be larger not smaller. 444ms ((4*10^6)/0.3c) would be the time inside the spaceship, correct?

 

[Quinzio]

In the observers reference frame s=vt is always true.

 

[0ddbio]

The difficulty here is that you (or the problem) did not state in which reference fram the 40x10^{6}m is measured.

As Quinzio suggested you will not have to do any calculations to adjust the velocity of the ship... therefore if the 40x10^{6}m is given in the reference frame of the observers then this isn't even a relativistic problem.. you would just do d/v = t and be done with it..

However, I think it would make more sense if this was a relativistic calculation, so it's probably best if you assume that the distance 40x10^{6}m is actually measured in the reference frame of the ship.

So perhaps that will help you?
usually relativity problems are very good about specifying in what reference frame all measurements are made... but I guess not in this case.