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Still trouble with time dilation

  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). If I'm doing the time dilation factor right, it would be either 1 or the square root of a negative number, so I'm not sure what to do with that.

    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

    kuruman

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    Did you use a value for c less than 25,000 miles per hour? Unless you made a mistake in your calculation, that's the only way to get the square root of a negative number.
     
  4. Aug 31, 2009 #3
    I converted c=3e8 m/s to c=6.711e8 mi/h and when I put that into the equation, I get 1. Because I was getting 1, I looked at the class notes from this morning and saw that the professor omitted c and just used sqrt(1-(v^2)). When I do that, I get the negative obviously because v^2 is greater than 1.
     
  5. Aug 31, 2009 #4

    sylas

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    You get 1 because the dilation factor is so tiny. You need at least 11 significant figures to see the difference.

    The professor uses sqrt(1-(v^2)) because he is using units in which c is equal to 1. In these units, the velocity of the spacecraft is about 3.7*10-5.

    For small values of v, you can use
    [tex]\frac{1}{\sqrt{1-v^2}} \approx 1 + v^2/2 \approx 1 + 6.86*10^{-10}[/tex]​

    Cheers -- sylas
     
  6. Sep 1, 2009 #5
    Where does the 6.86*10^-10 come from? Wouldn't I use 1+(v^2)/2=1+25,000^2/2=3.12*10^8.
     
  7. Sep 1, 2009 #6
    Someone please help. I've gotten no sleep because of this problem. I just don't see/ understand what to do at this point in the problem.
     
  8. Sep 1, 2009 #7

    Doc Al

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    Note that sylas's (and your professor's) v is really v/c. So find 1+(v/c)^2/2
     
  9. Sep 1, 2009 #8
    When I use v/c, I get 1 because v is so much smaller than c. Is that the answer; the same as Earth's frame?
     
  10. Sep 1, 2009 #9

    Doc Al

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    Staff: Mentor

    No. You're asked to find the time difference, which you can find using the formula sylas gave you. (Even though it's small.)

    You should have no problem calculating v/c, then (v/c)^2/2.

    For example: v = 25000 miles/hour = 11176 m/s = 1.12 x 104 m/s.
    c = 3 x 108 m/s, so v/c = what?
     
  11. Sep 1, 2009 #10
    Ok, (v/c)^2/2= 6.96889e-10, so the answer to my problem is 1+6.96889e-10
     
  12. Sep 1, 2009 #11

    sylas

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    It's a start. You've got a gamma factor.

    As a matter of general complaint about the question; it appears you are only considering the time difference from velocity; but given that the trip moves in and out of the gravitational fields, and given that most of it is in free fall, the special relativity calculation is really not adequate. But I don't think that is your fault.

    I presume the question is intended to let you use a special relativity calculation. If so, you want the gamma factor -- which you have just calculated correctly -- and you have to use that somehow to get a time difference.

    What else would you need, along with the time dilation factor, to get the difference in two clocks?

    Cheers -- sylas
     
  13. Sep 1, 2009 #12
    I would need the distance from the moon to the Earth which is 3.84*10^8 m and t=L/v=3.84*10^8/1.12*10^4=3.44*10^4 s. Then to find t', I would use t'=t/gamma and subtract t' from t? But then, I still end up with 3.44*10^4 s, which is t.
     
  14. Sep 1, 2009 #13

    sylas

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    That's a reasonable approach -- if we ignore all the issues with gravity and general relativity.

    You are still seeming to do your calculations with a limited precision calculator and getting hung up because the effect is so much smaller than the total elapsed time.

    Can you express the problem in algebra first? In this problem, using the approximation for the time dilation factor of 1 + (v/c)2/2 will give you the right answer. You've got the total time for the trip t. What is the value of t' - t, not as a number, but as an expression involving v?

    Cheers -- sylas

    Postscript. by the way... the approximation I am using can also be rearranged to be:
    [tex]\sqrt{1-(v/c)^2} \approx 1 - (v/c)^2/2[/tex]​
    which may be more convenient.
     
    Last edited: Sep 1, 2009
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