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Time Dilation and gravity

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data

    The Apollo astronauts returned from the moon under the Earth's gravitational force and reached speeds almost 25,000 mi/h WRT Earth. Assuming (incorrectly) they had this speed for the entire trip from the moon to Earth, what was the time difference for the trip between their clocks and clocks on Earth?

    Given:
    Velocity of Apollow WRT Earth: 25,000 mi/h

    Assumed (I'm not sure if I'm actually supposed to use or assume this):
    Distance from Earth to the moon: 238, 857 mi

    2. Relevant equations

    I believe I have solved for t (Earth's frame), but I'm having a problem solving for t' (Apollo's frame).

    3. The attempt at a solution

    Earth's Frame:

    t=L/v=238,857 mi/ 25000 mi/h=9.55428 h

    Apollo's Frame:

    t'=t/ɣ=9.55428 h*√(1-(25,000/c)²)=this doesn't work
     
  2. jcsd
  3. Aug 31, 2009 #2

    Redbelly98

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    Hint: what is the time dilation factor at this speed?
     
  4. Aug 31, 2009 #3
    The way I have it set up, the time dilation factor is less than 1, am I doing something wrong?
     
  5. Aug 31, 2009 #4

    Redbelly98

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    No, that's okay. The astronauts clocks run slower than Earth's clocks, and will show less time elapsed.

    So what number do you get for the time dilation factor?

    Also, I noticed you used 238,857 miles. Do the astronauts just go to the moon (total distance = 238,857 mi.), or do they return to Earth as well?

    p.s. I'm logging off soon, good luck!
     
  6. Aug 31, 2009 #5
    From the wording of the question, it looks like it only wants the distance from the Moon to the Earth. That is the problem that I am having. According to my calculator, I have two choices. I can take it to be 1 or the square root of a negative number, which is what is messing me up.
     
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