2 identical clocks are synchronised in a lab on the equator. One clock is carried around the equator in 24 hours at a constant speed. Given that the radius of the Earth is 6.4 x 10 ^ 6 m, find the difference between the times registered by 2 clocks when the travelling clock returns to the lab.
t' = t / γ
γ = 1 / (√(1 - v^2 / c^2))
The Attempt at a Solution
First I found the circumference of the Earth using c = 2 pi r
Found the circumference of the Earth to be 40212385 m.
The clock takes 24 hours to go around - 86400 seconds
so the speed the clock goes around the earth at is distance / time = circumference / time = 465 m/s.
I used the gamma equation to find the Lorentz factor
γ = 1 / (√(1 - v^2 / c^2)) = 1.
(the v = 465 m/s was too small to impact the equation?)
So then, subbing my Lorentz factor into the time dilation formula t' = t/y
t' = t/1 = t, so there's no time difference?
That doesn't sound right at all, I've done something stupid, if anyone can correct me that would be great :)