The distance from earth to the center of the galaxy is about 22500 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 0.9995c. According to a clock on board the spaceship, how long will it take to make the trip? Express your answer in years (1 yr = 3.16 107 s).
_______ years
WORK:
The distance from earth to the center of the galaxy is d = 22500 ly
= 22500 *9.47 1015 m
speed of space craft is v = 0.9995c
let time measured by an observer on earth is Ģt = d/v = 22500 *9.47 1015 m/ 0.9995c
= 710.6*109sec
now we have calculate time taken to make the trip According to a clock on board the spaceship
thet is Ģt'.
we know that ƒ¢t = ƒ¢t'/�ã[1-(v/c)2]
we know that ƒ¢t' = ƒ¢t*�ã[1-(v/c)2] = 710.6*109sec*�ã1-(0.995)2 = 710.6*106sec
= 71.06*107sec
given that 1 yr = 3.16 107 s
then 71.06*107sec = [71.06*107sec/ 3.16 107 s] years
= 23.68 years THE ANSWER IS STILL WRONG. WHERE IS THE ERROR?
71.06*107sec converted to years in the Google calculator is not what you got.
1 answer