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    I am rather stuck on this.:confused:

    A star is at rest relative to Earth and 27c y from Earth. A spaceship is making a trip from Earth to the Star at such a speed that the trip from Earth to the star takes 12y according to clocks on the spaceship. At what speed relative to Earth, must the spaceship travel.

    The answer is 0.91c.
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    I get v = 0.90c. Not sure why there's a small difference. I did it like this:

    The S' frame is the frame moving with the spaceship and so in the S' frame the velocity of the spaceship is zero. In this S' frame the time to make the trip is 12 (years). The S frame is the earth frame, in which the time for the spaceship to travel to the sun is 27 and the velocity of the spaceship is v. We now use the fact that the spacetime interval is conserved to write:

    c^2{\delta}t'^2 = c^2{\delta}t^2-{\delta}x^2

    so

    c^2{\delta}t'^2 = c^2{\delta}t^2-v^2{\delta}t^2

    Substituting numbers

    c^2(12^2) = c^2(27^2)-v^2(27^2)


    v^2 = \frac{c^2(27^2 - 12^2)}{27^2} 



v = 0.90c
 
 
 
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