- #1

AlmonzoWilder

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## Homework Statement

A 30 year-old astronaut leaves her newborn child on Earth and goes on a round-trip voyage to a star that is 40 light-years away traveling in a spaceship that is traveling at 0.90

*c*What will the ages of the astronaut and her child be when she returns?

## Homework Equations

[tex]\Delta t_o = \Delta t \sqrt{1-v^2/c^2}[/tex]

## The Attempt at a Solution

Since spaceship is traveling at 0.90c, the trip will take 10% longer to reach star each way, therefore time to reach star as viewed by an observer(child) can be calculated as such:

t=40yrs+(0.20)x(40yrs)

t=88yrs

And then time for the astronaut can be calculated from that value for observed time:

[tex]\Delta t_o = \Delta t \sqrt{1-v^2/c^2}[/tex]

[tex]\Delta t_o =88yrs \Delta t \sqrt{1-(0.9c)^2/c^2}[/tex]

[tex]\Delta[/tex]t

_{o}=8.756yrs

So age of astronaut is [tex]\Delta[/tex]t

_{o}+ her original age, making her 45.756, and her child 88.

That's as far as I've gotten, I can't figure out what is done wrong, the numbers are right but if the astronaut perceives time on Earth as moving slower than that in the spaceship and yet she returns home to find her child almost twice as old as her, there has to be something wrong.

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