Question

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?

Answer 1

Put T = 1 hour. Then in the rest frame of an observer on the ground, the spacecraft will travel distance of v T, during a time of T, this means that the space-time interval between two points on the trajectory a time T apart is given (in c = 1 units)by

s^2 = T^2 - v^2 T^2 = (1-v^2)T^2

The space-time interval is an invariant, it will yield the same value if evaluated in the rest frame of the spacecraft. In that frame the spacecraft is at rest, while the time between the events is what we want to evaluate, let's call this T'. We then have that:

s^2 = T'^2.

We thus see that:

T' = sqrt(1-v^2) T

For small v we can expand this as:

sqrt(1-v^2) = 1 - 1/2 v^2

The ship's clock thus differs from one hour by:

-T/2 v^2

Restoring c can be done by recognizing that in c = 1 units you are free to multiply and divide by whatever powers of c you like because it is equal to 1 anyway, but of course, one particular choice will coincide with an equation that in SI units would also be dimensionally correct. So, we need to divide by c^2:

-T/2 v^2/c^2 = -2.4*10^(-2) s