Time Dialation-Theory of Relativity

[AlmonzoWilder]

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

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

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:
\Delta t_o = \Delta t \sqrt{1-v^2/c^2}
\Delta t_o =88yrs \Delta t \sqrt{1-(0.9c)^2/c^2}
\Deltat_o=8.756yrs

So age of astronaut is \Deltat_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.

 

[AlmonzoWilder]

Sorry about the long spaces in between the equations, I'm not quite sure what I did wrong.

 

[AlmonzoWilder]

Okay, looking over my equation on paper, I found a small error when I calculated the age of the astronaut, she would be 38.358 =37yrs, making her 75.4 yrs old, however this still does not solve my problem, any thoughts?
Thanks in advance!

 

[sylas]

In the equations, you are mixing up tex and php. The tex language, for example, uses x^y for a superscript... not the php "sup" tags that you get with the superscript button.

Here is a rewrite of your original post using the correct syntax for tex. If you click on an equation, you will see how it is written.

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

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

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:

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

\Delta t_o = 88 \text{yrs} \times \sqrt{1-(.90c)^2/c^2}

\Delta t_o = 8.756 \text{yrs}

So age of astronaut is Δt_o + her original age, making her 45.756, and her child 88.
​

Now you have a problem there. Travelling 40 light years at 0.9c takes 40/0.9 = 44.444 years, not 44.

The factor you multiply by is

\sqrt{1-0.9^2} = \sqrt{1 - 0.81} = \sqrt{0.19} = 0.436​

You originally had this wrong, but this factor looks correct now. But you should divide by 0.9, not multiply by 1.1, to get the time elapsed back on Earth.

Cheers -- sylas

PS. (Note that you can edit your post. Getting rid of those long limes would make the thread fit on the page better.)

 

[AlmonzoWilder]

Thanks, that looks better now, not perfect but better. :)
I get what you did, but I still do not understand why the child would be so much older than the mother when she returns...

 

[sylas]

Thanks, that looks better now, not perfect but better. :)
I get what you did, but I still do not understand why the child would be so much older than the mother when she returns...

Why not?

Of course, I know why not. You are used to time being something that passes at the same rate everywhere. It's what we are all used to. If we habitually traveled at close to the speed of light, this would not be hard to understand, because we would all be used to it. But because we move at velocities so much less than light speed, the relative nature of time seems strange... seems hard to understand.

But that's the way it is, all the same. The amount of time that passes between two events is not an absolute, but depends on how you move between those two events.

Cheers -- sylas