the Apollo 10 spacecraft traveled at 11 km/s just prior o its re-entry, setting a speed record for humans. Suppose advanced technology raises the speed record by a factor of 100, to 1100 km/s. if a time travel of 1 hour elapsed on a ground-based clock, by how many seconds would the ship clock differ from one hour?
To calculate the time difference between the ship clock and the ground-based clock, we need to take into account the time dilation effect predicted by Einstein's theory of relativity. According to this theory, time runs slower for an object moving at high speeds relative to an observer at rest.
The time dilation formula is given by:
Δt' = Δt * sqrt(1 - (v^2 / c^2))
Where:
Δt' is the perceived time on the moving clock (ship clock in this case).
Δt is the time on the stationary clock (ground-based clock in this case).
v is the velocity of the moving object relative to the stationary observer.
c is the speed of light in a vacuum (299,792 km/s).
In this scenario, the velocity of the spacecraft (v) just prior to re-entry is 1100 km/s. We want to find the time difference (Δt' - Δt) when 1 hour (3600 seconds) elapses on the ground-based clock.
Plugging in the values into the equation, we have:
Δt' = 3600 * sqrt(1 - ((1100 km/s)^2 / (299792 km/s)^2))
Now, let's calculate the time difference using this formula:
Δt' = 3600 * sqrt(1 - (1100^2 / 299792^2))
≈ 3600 * sqrt(1 - (1210000 / 89875558464))
≈ 3600 * sqrt(1 - 0.000013482)
≈ 3600 * sqrt(0.999986518)
Using a calculator, we find that:
Δt' ≈ 3600 * 0.999993259
≈ 35999.75 seconds
Therefore, the ship clock would differ from one hour by approximately 35999.75 seconds, which is equivalent to approximately 10 hours and 6 minutes.