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Special relativity problem

  1. Jan 17, 2015 #1
    1. The problem statement, all variables and given/known data
    An astronaut went to sun from earth by a rocket. According to the clock in the rocket, the traveling time was 5 minute. What was the velocity of the rocket?
    ( earth is 8 light minute away from sun)

    2. Relevant equations
    [itex]t = \frac {t_0}{\sqrt {1 - \frac{v^2}{c^2}}}[/itex]


    3. The attempt at a solution
    According to the reference frame of earth, the distance between sun and earth is, [itex]L_0 = 8\cdot 60 \cdot c [/itex]
    Let, the time required to travel this distance be [itex]t_0[/itex] [according to earth];
    So, [itex]t_0 = 5 \cdot 60 \cdot \sqrt {1 - \frac{v^2}{c^2}}[/itex]
    and the velocity, [itex] v = \frac {L_0}{t_0} = \frac { 8\cdot 60 \cdot c }{ 5 \cdot 60 \cdot \sqrt {1 - \frac{v^2}{c^2}}} = \frac { 8c }{ 5 \sqrt {1 - \frac{v^2}{c^2}}} [/itex]
    [itex]v^2 = \frac {64c^2}{25(1 - \frac{v^2}{c^2})}[/itex]
    [itex]v^2 \cdot 25(1 - \frac{v^2}{c^2}) = 64c^2[/itex]
    Simplifying, [itex]\frac{25}{c^2} (v^2)^2 - 25 (v^2) +64c^2 = 0 [/itex]
    But, the solution of this equation is complex number.
     
  2. jcsd
  3. Jan 17, 2015 #2

    Orodruin

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    You are doing the time dilation in the wrong way. ##t_0## is the proper time, i.e., the time for the astronaut and should be smaller than the time in the Earth reference frame.
     
  4. Jan 17, 2015 #3
    Look, we are calculating the distance according to earth. So, shouldn't we take the time according to earth?
     
  5. Jan 17, 2015 #4

    Orodruin

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    Yes, and you are not. The quantity ##t_0## is the time elapsed for the astronaut.
     
  6. Jan 17, 2015 #5
    I got it. We have to calculate the time interval between two events : 1. The rocket starts from earth , 2. The rocket reaches the sun. And these two events occured at the 'same location' in the reference frame of rocket. So, the time measured from the rocket is to be taken as the proper time.
    There is another way to solve the problem. According to earth, both the earth and the sun are stationary (as it is described for this problem), and according to the rocket they are moving. So, the distance between the sun and earth is contracted in the reference frame of the rocket. So, ## L = L_0 \sqrt { 1 - \frac{v^2}{c^2}}##
     
  7. Jan 18, 2015 #6

    Orodruin

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    That's more like it. :)
     
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