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Rotating a laser at Moon so that spot has v>c

  1. Sep 14, 2014 #1
    1. The problem statement, all variables and given/known data
    You shine a powerful laser onto to the surface of the Moon from Earth (Earth-Moon distance is 384,000 km or 3.84E8 m). About how fast must the laser pointer rotate (in degrees per second) for the spot on the Moon to move with velocity v>c? Does this violate Special Relativity?


    2. Relevant equations
    No equations were given.


    3. The attempt at a solution
    This problem is in the Special Relativity section of a Modern Physics class. At first glance, I had absolutely no idea where to start. My professor didn't give me any equations for this type of problem. So, here is my bad attempt (probably wrong):

    I started by thinking of a relation between the angle (θ) of each incremental rotation and the distance (d) the spot moves on the Moon. If you make the distance from Earth to the Moon the horizontal, and you rotate the laser at an angle θ from the horizontal, you get a right triangle (assuming the Moon's surface is flat) with opposite=d and adjacent=Earth-Moon distance. So, if you take the tangent of the angle, you get:

    tanθ = d/(3.84E8)

    So the distance the spot moves on the Moon after a rotation θ is:

    d = (3.84E8)tanθ

    I then assumed that the velocity of the spot on the Moon is given by:

    v = d/t, where t is the time the spot takes to move that distance (or, equivalently, the time it takes for the laser to rotate by θ)

    At this point I calculated that the time it takes for the light to move from the laser pointer to the Moon is:

    t = (Earth-Moon distance)/(speed of light) = (3.84E8)/(3E8) = 1.28 s

    If you make the time interval between each incremental rotation 1.28 s, then you get the following equation for the velocity of the spot:

    v = d/t = ((3.84E8)tanθ)/(1.28) = c*tanθ

    If you want v>c, then:

    c*tanθ>c

    tanθ>1

    θ> 45°

    But, I don't know how to interpret my answer (i.e. 45°/s or 45°/1.28s).

    More than that though, my method is probably completely wrong anyway and I need help.

    Can someone please show me how to do this problem? I'm completely lost.

    Thank you
     
  2. jcsd
  3. Sep 14, 2014 #2
    I'll explain the "does this violate special relativity?" first.

    It doesn't because the laser is ejecting 1 proton at a time, this happens so fast it seems like a continuous beam when you move it. However if the angles of 2 photons are different over a long distance, they can end up being far apart. So its like 2 different events.
    ---------------------
    This is how I would set it up, kind of a weird question because it is impossible for there to be a answer because the photon travels at the speed of light and can't go faster and if you have a right triangle with a photon on the side and the hypotenuse traveling with the same speed, they wouldn't meet up and the side photon would always be covering more distance. And the spot on the moon is going to be stationary because that is only one photon and in another sense, light moving with only a x direction will always be more than light with a x and a y direction. BUT I will assume the speed for the laser to travel is instantaneous. Maybe my logic is flawed as well haha.:tongue:


    I'm going to say for the first part to get the speed to travel faster than light, the distance between the 2 points has to be greater than 3.0x108m. The distance from Earth to the moon is 384,000,000m. So we draw a right triangle. The y being distance from Earth to Moon and the x being 3.0x108m distance.

    From this you can find the angle and it will already be in degree/second because we used 3.0x108m (c*1s).

    However this value (degree/s) is equal to the speed of light, therefore you need to indicate that it has to be greater than this value.
    :biggrin:
     
    Last edited: Sep 14, 2014
  4. Sep 14, 2014 #3
    I agree, this question is very weird:P.

    Oh, ok! When I do what you said, I now get a rotation speed of 37.926°/s (for it to equal light speed) so anything higher does yield a speed greater than light:biggrin:. Thank you for clarifying this! My previous answers were 45°/s and 35°/s but the first was too high and the second was too low, and so it makes sense for it to be 37.926°/s.

    Its always the small stuff that brings me so much confusion. I sure hope its right:D! I really hate it when professors don't even talk about this stuff and then expect us to figure it out.

    I really appreciate the explanation:D!

    Thanks again!
     
  5. Sep 14, 2014 #4
    Glad it helped :D
     
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