Daniel B. answered 05/23/21
A retired computer professional to teach math, physics
In general, suppose you have a spaceship going at velocity v relative to Earth.
For an observer on Earth, clocks in the spaceship's frame of reference will appear to run slow by the factor
√(1 - (v/c)²)
So the observer on Earth needs to wait 1/√(1 - 0.93²) = 2.72 hours to observe clocks in the
spaceship's frame of reference to advance 1 hour.
The situation is symmetrical: The astronaut sees the Earth's clocks as running slow
and it takes her 2.72 hours to see the Earth's clocks advance 1 hour.