A person is travelling on a spaceship that moves at a speed of 5c/13 . Determine the time interval calculated by his friend on the earth between consecutive birthday celebrations of the person on the spaceship.

Step-by-step solution

The proper time interval between consecutive birthday celebrations for the person on the spaceship is t₀ = 1 year . The speed of the spaceship is v = 5c 13 . The time interval measured by the friend on Earth is given by the time dilation formula: t = t₀ 1 - v^2 c^2 Substituting the given values: t = 1 1 - ( 5 13 )^2 t = 1 1 - 25 169 t = 1 144 169 t = 1 12 13 = 13 12 year