# An astronaut leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. The astronaut's destination is a star system 14.4 light-years away (one light-year is the distance...

An astronaut leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. The astronaut's destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year.) According to the astronaut, how long does the trip take?

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Each physical body has *its own* rest frame, the one at which *this* body is at rest.

The rest frames of an astronaut and a spaceship are almost the same (an astronaut cannot move very fast inside a ship). So there is no reason to distinguish between an astronaut and a spaceship.

Also, we must assume that Earth and a star have a negligible speed relative to each other, so their rest frames are almost non-moving relative to each other. In such a case, an observer on Earth is almost at rest relative to a star, and the distance between Earth and a star, as measured from Earth, is the largest among all frames of reference (a so-called proper length).

For a moving observer--an astronaut, for example--this distance is shorter by Lorentz factor `sqrt(1-v^2/c^2).` So "according to the astronaut," which is the same as "at the rest frame of an astronaut," the distance will be less and the duration of the flight for him will be less.

So your friend's calculation is correct *for an observer on Earth*. Your formula gives correct **distance ***for an astronaut,* but to find the time we have to divide it by the speed, `t=(14.4 * sqrt(1-0.96^2))/0.96 = 4.2` (years).