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## Homework Statement

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.

## Homework Equations

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 :)