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Calculation of the speed of light

  1. Feb 7, 2014 #1
    1. The problem statement, all variables and given/known data

    The Apollo 11 astronauts set up a panel of efficient corner–cube retroreflectors on the Moon's surface (see figure below). The speed of light can be found by measuring the time interval required for a laser beam to travel from the Earth, reflect from the panel, and return to the Earth. Assume this interval is measured to be 2.51 s at a station where the Moon is at the zenith and take the center–to–center distance from the Earth to the Moon to be equal to 3.84 108 m. (The Moon's radius is 1.74 106 m, and the Earth's radius is 6.37 106 m.)

    (a) What is the measured speed of light?

    (b) Explain whether it is necessary to consider the sizes of the Earth and the Moon in your calculation.


    2. Relevant equations



    3. The attempt at a solution
    Why the measured speed of light is not equal to 2d/t?And I don't know why the size of the earth and the moon need to be taken into account?

    THANKS!!
     
  2. jcsd
  3. Feb 7, 2014 #2

    NascentOxygen

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    Both are rotating bodies?

    That seems rather small. :smile: Does m = metres?
     
    Last edited: Feb 7, 2014
  4. Feb 7, 2014 #3

    phyzguy

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    Does the laser beam travel from the center of the Earth to the center of the moon and back?
     
    Last edited: Feb 7, 2014
  5. Feb 7, 2014 #4
    I highly doubt the laser is fired from the Earth's core and it is also said that the panels are situated on the Moon's surface hence the given radiuses.
     
  6. Feb 7, 2014 #5

    HallsofIvy

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    Oddly enough, there was NO "figure below"! It might be important because there might be some important distance between reflectors on the moon. If that is not the case then the speed of light is, indeed, "2s/t" where s is the distance from the surface of the earth to the surface of the moon.
     
  7. Feb 7, 2014 #6

    BvU

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    Corner cubes are retroflectors like in car taillights. Size can be ignored.
    earth and moon radius amount to a 2% smaller d than the c-to-c distance. That's relevant if the time measurement suggests a 0.5% accuracy.
     
  8. Feb 8, 2014 #7
    Oh sorry.:shy:This is the photo.
    35-figure-08a.gif
     
  9. Feb 8, 2014 #8

    NascentOxygen

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    Presumably NASA went to all the expense of sending that reflector to the moon with the intention of demonstrating some impressive accuracy in taking measurements. For that money, I'd be expecting more than just 3 or 4 significant figures!
     
  10. Feb 8, 2014 #9

    BvU

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    Yes, but this is an exercise. The radii were given in 3 digits too.
     
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