Solving the Time Dilation Problem: Insights from a 1971 Experiment

[KyleS4562]

In 1971 four portable atomic clocks were flown around the world in jet aircraft, two east bound and two westbound, to test the times dilation predictions of relativity. a) If the westbound plane flew at an average speed of 1500 km/h relative to the surface, how long would it have to fly for the clock on board to lose 1s relative to the reference clock on the ground? b) In the actual experiment the plane circumflew Earth once and the observed discrepancy of the clocks was 273ns. What was the plane's average speed?

For the first part, I convereted 1500km/h to 416.6m/s, then put it terms of c, or 1.389x10^-6c. I then took the standard equation t'/(sqrt(1-(1.389x10^-6c)^2)/c^2) = t.
I then made t-t'=1, solved for t'=-1+t and put that into the equation.
Solving, I got 1x10^12 s, or 31,688 years. Is this a reasonable answer?

I think my method would be wrong, because I followed similar steps to get b. After removing the t' prime from the equation as I did in part a, I set t = 40075160m/v, 40075160m being the circumference of the earth. My equation was as follows:
(1/(sqrt(1-(v^2/c^2)))*(-273x10^-9s + 40075160/v) = 40075160/v

I solved for v, but got 1226.2 m/s, which is a speed I don't believe we've even held for a sustained flight now, let alone in 1971. Where am I going wrong?

 

[jbsteele]

For part a, the answer is not reasonable. You are assuming that the speed of the plane is constant, which it is not in reality. You have also assumed that the time dilation is due only to the speed of the plane, which is not true. Time dilation also depends on altitude and gravitational potential as well. For part b, you have not taken into account the time dilation due to altitude and gravitational potential. The actual equation used to calculate the velocity of the plane would be t'/(sqrt(1-(v^2/c^2))*(1+g/c^2)) = t, where g is the acceleration due to gravity. Taking this into account should give you a more accurate result.