What is the duration according to the crew of the spaceship?
The time t2 recorded in the spaceship related to t1 is shortened by γ , the Lorentz factor.
t2 = t1/γ = t(1 − β2)1/2 = t(1 − 0.52)1/2 = 0.866 t
= 0.866 × 2,250 = 1,948s
