Question

The distance from earth to the center of our galaxy is about 30,000 ly (1 ly = 1 light year = 9.47 × 1015 m), as measured by an earth-based observer. A spaceship is to make this journey at a speed of 0.9990c. According to a clock on board the spaceship, how long will it take to make the trip? Express your answer in years. (1 yr is equal to 3.16 × 107 s.)

Answer 1

D = distance from earth to center of galaxy = 30,000 ly = 30,000 (9.47 x 10¹⁵) m = 28.41 x 10¹⁹ m

v = speed of the spaceship = 0.9999 c = (0.9999) (3 x 10⁸) = 2.9997 x 10⁸ m/s

t = time taken by spaceship as observed by the observer on earth

t = D/v

t = (28.41 x 10¹⁹) /(2.9997 x 10⁸)

t = 9.47 x 10¹¹ sec

t' = time observed on spaceship

using the equation

t' = t /sqrt(1 - (v/c)²)

t' = (9.47 x 10¹¹ ) /sqrt(1 - (0.9999 c/c)²)

t' = 1.34 x 10¹⁰ sec

t' = (1.34 x 10¹⁰ sec ) (1 yr/(3.16 x 10⁷) sec)

t' = 424 yrs