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 10^{15}) m = 28.41 x 10^{19} m

v = speed of the spaceship = 0.9999 c = (0.9999) (3 x
10^{8}) = 2.9997 x 10^{8} m/s

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

t = D/v

t = (28.41 x 10^{19}) /(2.9997 x 10^{8})

t = 9.47 x 10^{11} sec

t' = time observed on spaceship

using the equation

t' = t /sqrt(1 - (v/c)^{2})

t' = (9.47 x 10^{11} ) /sqrt(1 - (0.9999
c/c)^{2})

t' = 1.34 x 10^{10} sec

t' = (1.34 x 10^{10} sec ) (1 yr/(3.16 x 10^{7})
sec)

t' = 424 yrs

The distance from earth to the center of our galaxy is about
28,000 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.)

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