Special Relativity - time paradox question, not sure if it's right

[daleklama]

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 traveling 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 :)

 

[baddin]

Use more figures in your gamma factor, I got Y=1.000001205095829, then used t' = t/y = 86400/1.000001205095829 = 86399.90 seconds, so the difference should be about 0.1 seconds, seems correct because Y is so close to 1.

 

[harrylin]

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 traveling 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 :)

As it's the difference between 1 and 1-v²/c² that matters, even a small v does impact on the equation - γ is close to 1, but different from 1.

Thus, either solve as baddin suggested*, or solve as Einstein did in 1905 - he did not have a pocket calculator! Derive an approximate equation for the difference (easy math, simplify 1/√(1-x²) for x<<1). See the second half of §4 of http://www.fourmilab.ch/etexts/einstein/specrel/www/

*PS: I get a much smaller result both ways

 

[daleklama]

Thanks both!

My calculator doesn't show enough figures to give me a non-1 answer, so I'm going to do it the way harrylin suggested, cheers :)