where L' is the length measured by the observer and L is the proper length of the object.

L' = L / 2.29 = 0.436L

γ = 1 / sqrt(1 - v^2/c^2)

Problem 2.1 asks students to calculate the Lorentz factor for an object moving at 0.6c relative to an observer. Using the equation above, we can plug in the values:

γ = 1 / sqrt(1 - (0.6c)^2/c^2) = 1 / sqrt(1 - 0.36) = 1 / sqrt(0.64) = 1 / 0.8 = 1.25

This means that the astronaut will experience time passing 1.67 times slower than the observer on Earth.