closest star to our solar system is Alpha Centauri, which is
4.30 light years away. A spaceship with a constant speed of 0.8c
relative to the Earth travels toward the star. What distance does
the space ship travel according to a passenger on the ship?
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The length contraction factor is
γ = 1 / √ ( 1 – v² / c² )
γ = 1 √ ( 1 – ( .8c )² / c² )
γ = 1 √ ( 1 – .8² )
γ = 1.67
The distance traveled, from the point of view of the moving spaceship, is
( 4.30 LY ) / 1.67 = 2.6 LY
The length will be contracted to shorter than its proper length.
L = L0 / gamma
= L0 sqrt (1 – (v/c)^2)
They give you L0 = 4.3 c years, and v/c = 0.8. Plugnchug.
I don’t think that the question was meant the way rs interpreted it, but after re-reading it, the interpretion that ducks the question isn’t wrong. I’d have to give that answer full credit.
If the passenger starts the trip INSIDE the spaceship and ends the trip INSIDE the spaceship, how far has he moved RELATIVE to the spaceship?