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Caught by surprise near a supernova, you race away from the explosion in your spaceship, hoping to outrun the high-speed material ejected toward you. Your Lorentz factor gamma relative to the inertial reference frame of the local stars is 22.4. How long does that trip take according to you (in your reference frame)? |
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Answer» Solution :KEY IDEAS 1. We now want the time interval measured in a different REFERENCE frame- namely, yours. Thus, we need to TRANSFORM the data given in the reference frame of the stars to your frame. 2. The given path length of `9.00xx10^(16)`m, measured in the reference the two ends of the path are at rest in that frame. As observed from your reference frame, the stars. reference frame and those two ends of the path race past you at a relative speed of `v~~c`. 3. You measure a CONTRACTED length `L_(0)//gamma` for the path, not the proper length `L_(0)`. Calculations: We can now rewrite Eq. 36-19 as (time interval relative to you) `=("distance relative to you")/(c )=(L_(0)//gamma)/(c )`. Substituting known data, we find (time interval relative to you)`=((9.00xx10^(16)m)//22.4)/(299 792 458 m//s)` `=1.340xx10^(7)s=0.425y`. In part (a) we found that the flight takes 9.51 y in the reference frame of the stars. However, here we find that it takes only 0.425 y in your frame, due to the relative motion and the resulting contracted lengthof the path. |
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