An atomic clock moves at 1000 km/h for 1.75 hours as measured by an identical clock on the Earth. At the end of the 1.75 hour interval, how many nanoseconds slow will the moving clock be compared with the Earth-based clock?Hint: The binomial expansion allows the approximationΔtp=1−v2c2−−−−−√Δt≃(1−v22c2)Δt which can be used to find the differenceΔtp−Δt
Question
An atomic clock moves at 1000 km/h for 1.75 hours as measured by an identical clock on the Earth. At the end of the 1.75 hour interval, how many nanoseconds slow will the moving clock be compared with the Earth-based clock?Hint: The binomial expansion allows the approximationΔtp=1−v2c2−−−−−√Δt≃(1−v22c2)Δt which can be used to find the differenceΔtp−Δt
Solution
To solve this problem, we need to use the time dilation formula from special relativity, which is given by the equation:
Δtp = Δt / sqrt(1 - v^2 / c^2)
where:
- Δtp is the time measured by the moving clock (the atomic clock in this case)
- Δt is the time measured by the stationary clock (the Earth-based clock in this case)
- v is the velocity of the moving clock
- c is the speed of light
Given that the atomic clock moves at 1000 km/h for 1.75 hours, we first need to convert these values to SI units. Therefore, v = 1000 km/h = 277.78 m/s and Δt = 1.75 hours = 6300 s.
The speed of light c is approximately 3 x 10^8 m/s.
Substituting these values into the equation, we get:
Δtp = 6300 / sqrt(1 - (277.78)^2 / (3 x 10^8)^2)
Solving this equation gives us the time measured by the moving clock.
To find how many nanoseconds slow the moving clock is compared with the Earth-based clock, we subtract Δt from Δtp and convert the result to nanoseconds.
Let's calculate it:
Δtp = 6300 / sqrt(1 - (277.78)^2 / (3 x 10^8)^2) = 6300.000000012 s
The difference in time is then Δtp - Δt = 6300.000000012 s - 6300 s = 0.000000012 s.
Converting this to nanoseconds (1 s = 10^9 ns), we get:
0.000000012 s * 10^9 ns/s = 12 ns
Therefore, the moving clock is 12 nanoseconds slow compared with the Earth-based clock.
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