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 :)

 

[paranormal]

Your attempt at solving the problem is correct, but there is a small mistake in your calculation. The Lorentz factor should be calculated using the velocity of the clock relative to the speed of light, not the speed of the clock relative to the circumference of the Earth. The correct Lorentz factor is actually very close to 1 (approximately 1.000000000000000000014), so your final calculation of t' = t/1 is almost correct. However, due to the extremely high precision required for these calculations, there will still be a very small difference in the times registered by the two clocks. This difference, while negligible, is a result of the time dilation effect predicted by special relativity.