Question

# The distance from earth to the center of our galaxy is about 30,000 ly (1 ly...

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.)

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

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