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Confusion with time dialation.

  1. May 26, 2012 #1
    Reading Brian Cox and Jeff Forshaw's book "Why does E=MC2". They mention that satellites speed up with time, but, then reading Wiki it says the crew of the ISS experience the slowing of time. Which one is correct? I'm slightly confused.
    Both experience a weaker gravitational pull and high velocities.
    I feel I have the answer but it's been some months since I read anything about this.
    Could it be the overall difference in altitude and speed?
    Many thanks


    http://en.wikipedia.org/wiki/Time_dilation
     
    Last edited: May 26, 2012
  2. jcsd
  3. May 26, 2012 #2
  4. May 26, 2012 #3
    The velocity based time dilation combined with the gravitational time dilation can be written as:

    [tex]\sqrt{1-\frac{v^2}{c^2}} * \sqrt{1-\frac{2GM}{rc^2}} [/tex]

    Since the orbital velocity of a satellite is given by:

    [tex]v = \sqrt{\frac{GM}{r}} [/tex]

    then the first equation can be rewritten as:

    [tex]\sqrt{1-\frac{GM}{rc^2}} * \sqrt{1-\frac{2GM}{rc^2}} [/tex]

    It can be seen for increasing radius the time dilation reduces due to increased height and due to reduced orbital velocity. Clocks on satellites with large orbits tick faster than clocks on satellites at lower orbits. Clocks on the surface of the Earth are moving much slower than the required orbital velocity at that radius and so tick faster than clocks on satellites with very low orbits. The ISS has a relatively low orbit (its radius is approximately 1.05 times the radius of the Earth), so clocks on the ISS are indeed ticking slower than clocks on the surface of the Earth. The speed up of clocks with increasing radius means that once an orbital radius is larger than 3 times the Earth surface radius (the break even point) the clocks on board a satellite are ticking faster than a clock on the Earth surface. The GPS satellites have an orbital radius of about 4.1 times the radius of the Earth so they are ticking faster.

    P.S. The above equations for the time dilation of an orbiting satellite can be fairly accurately approximated in this case by:

    [tex]\sqrt{1-\frac{3GM}{rc^2}} or \left(1-\frac{3GM}{2rc^2}\right) [/tex]
     
  5. May 26, 2012 #4
    Thank you, yuiop. You have hit the nail on the head.
    Since I posted the question, a few hours ago, I have been pondering over it and knew their had to be a explanation. From reading your post it looks like I was touching on the reason but couldn't quite grasp it.

    Thank you for your post also, Naty1.
     
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