A spaceship is moving relative to the Earth at a constant velocity with a speed

Question

A spaceship is moving relative to the Earth at a constant velocity with a speed v=0.997000c. (Hints: light years are a distance and one light year is the distance light travels in one year. One light year equals to 9.459 x 1015 m. A spaceship travelling at v=0.997000c will move a distance of 0.997000 light years in one year in the Earth frame).
a> If, in the Earth frame, the length of a moving spaceship is found to be L=3.00m (in the direction of motion), what is the length of the spaceship in the frame of your spaceship?
b> Assuming the front of the spaceship passes x=0.00 at t=0.00 in the Earth frame, when does the front of the spaceship pass x=1.00 x 107 m in the Earth frame?

c> How much time elapses in the spaceship frame, when, in the Earth frame the spaceship travels the distance x= 1.00 x 107 m?

d> The spaceship, travels at this speed, in a straight line, to a star 20.0 light years away from the Earth (as measured in the Earth frame). This is a one way trip - you can neglect the amount of time it takes to accelerate and decelerate because the spaceship flies by the Earth and then flies by the star 20.0 light years away. How long does this trip take in the Earth frame? How long does this trip take in the spaceship frame?

e> In the spaceship's inertial frame (i.e. a frame co-moving with the moving spaceship at v=0.997000c), what is the distance, in light years, from the Earth to the star in part "d"?

Explanation / Answer

A spaceship is moving relative to the Earth at a constant velocity with a speed v=0.997000c. (Hints: light years are a distance and one light year is the distance light travels in one year. One light year equals to 9.459 x 1015 m. A spaceship travelling at v=0.997000c will move a distance of 0.997000 light years in one year in the Earth frame). a> we have that L=L0/gamma. where gamma=1/sqrt(1-0.997^2) so that L0=38.76m b> Assuming the front of the spaceship passes x=0.00 at t=0.00 in the Earth frame, when does the front of the spaceship pass x=1.00 x 107 m in the Earth frame? it is x/v=1e7/0.997c=0.0334336 c> How much time elapses in the spaceship frame, when, in the Earth frame the spaceship travels the distance ?x= 1.00 x 107 m? we have that t0=t'/gamma. where t'=0.0334336. so that t0=2.58782e-3(s) d> The spaceship, travels at this speed, in a straight line, to a star 20.0 light years away from the Earth (as measured in the Earth frame). This is a one way trip - you can neglect the amount of time it takes to accelerate and decelerate because the spaceship flies by the Earth and then flies by the star 20.0 light years away. How long does this trip take in the Earth frame? How long does this trip take in the spaceship frame? a. t'=20c/0.997c=20.06(years). b. we have that t0=t'/gamma=1.553(years) e> In the spaceship's inertial frame (i.e. a frame co-moving with the moving spaceship at v=0.997000c), what is the distance, in light years, from the Earth to the star in part "d"? it is 1.553*0.997*c=1.548c=1.548(light years)

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