A.T. said:
It does matter how fast they go, and even which direction they go around.
This is correct, but a bit terse.
Maximeroom said:
Thanks a lot Nugatory. They didn't seem to mention that in the documentary video's I watched.
So basically, the 0.134 seconds difference is only correct if the speed you travel = c.
I know it's technically impossible to travel faster than c because of the mass, but I heard, IF you can, you'd be able to go back in time.
According to that formula that's not possible.
http://www.mathsisfun.com/data/func...800&ymin=-0.2122&ymax=0.4413&aval=2.000&uni=0
Anyway , my problem is solved :p
Thanks guys!
Unfortunately the problem is not correctly solved yet :(, for the original poster has misinterpreted the answers.
It takes about .133 seconds for light to travel around the Earth's equator. But that number isn't the answer to the OPs question.
Ignoring the time dilation and other effects due to GR, we can consider the time dilation depends on the bodies total velocity in the special-relativity--like Earth centered inertial frame. We conclude the time dilation of the clock moving in the same direction as the Earth's rotation caries it is greater than the time dilation of the clock moving in the opposing direction.
Now we can ask - what happens in the limit of low and high velocities. The high velocity limit is that the proper time on the carried clock approaches zero, in the limit as the velocity approaches infinity it takes no proper time to travel around the Earth. So let's look at the low velocity limit which should be more interesting.
If we take a series expansion of the well-known SR time dilation formula ##1/ \sqrt{1-( \frac{u+v}{c} )^2} ## as a series in v , with u being some additional velocity due to the Earth's rotation, we get
##\approx \frac{1}{\sqrt{1-u^2/c^2}} - \frac{uv}{c^2}## when u << cIn this low velocity limit , if we let ##r_e## be the radius of the Earth and ##T_e## be the period of rotation of the Earth (a sidereal day,if one wishes to be precise), and multiply the difference in time dilation by the trip time, we get for the total discrepancy
\frac{ \left( 2 \pi r_e \right)^2} {c^2 T_e} = (.133 sec)^2 / (1 sidereal day)
which turns out to be about 200 nanoseconds if I've calculated it right. I'd check the number (also the formulae) against other sources before considering the problem "solved" though.
Wiki might be helpful,
http://en.wikipedia.org/wiki/Sagnac_effect discusses some of the issues, but they don't ask the same question the OriginalPoster does, so the answer doesn't apply directly .
