Specialy theory of relativity [Answer Check]

[AClass]

Homework Statement

a 30-year old female astronaut leaves her newborn child on Earth and goes on a round-trip voyage to a star that is 40light-years away in a spaceship traveling at 0.90c. what will be the ages of the astronaut and her son when she returns?

Homework Equations

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

The Attempt at a Solution

40 years (2) +(0.20)(40 years)= 88 years

The astronaut will be 30+88 years = 118 years old when she returns

Using the above equation.

$$\Delta$$ t = 118 years

v=0.90c

Obtained $$\Delta t_o$$ = 51.44 years

The astronaut's son will be 51 years old when she returns.

Wondering if I did this correct, I don't have any answers to check from It looks right to me.

 

[collinsmark]

40 years (2) +(0.20)(40 years)= 88 years

Ummm, I'm not following you there. :uhh:

The astronaut will be 30+88 years = 118 years old when she returns

Something went wrong again. You're not applying the right time to the right person.

You might want to just start in the child's frame of reference. Then use x = vt for that frame.

Using the above equation.

$$\Delta$$ t = 118 years

v=0.90c

Obtained $$\Delta t_o$$ = 51.44 years

The astronaut's son will be 51 years old when she returns.

No, that's not the right application of the formulas.

For starters, the astronaut will age the slowest.

 

[AClass]

40 years (2) +(0.20)(40 years)= 88 years
Means, the parent is on route for 80 years [40 Years there, then 40 years back], plus since she is traveling 10% slower than the speed of light 20% slower since she is going there and back.

I realize where I went wrong. I went to verify my notes.

The child will be 88 years old.

$$\Delta$$ t = 88 years [Earth Observer]

Using the equation,

Obtained $$\Delta t_o$$ = 38.36 Years

Therefore the child will be 88 years old, and the parent will be [30 years+38.36 years] 68 years old.

That should be right. Thanks for the help!

 

[collinsmark]

40 years (2) +(0.20)(40 years)= 88 years
Means, the parent is on route for 80 years [40 Years there, then 40 years back], plus since she is traveling 10% slower than the speed of light 20% slower since she is going there and back.

That's not the correct math. :uhh:

Please, start over. First calculate the total distance (in light years). Then write down the velocity (v = 0.90c). It's also helpful to note that c = 1 ly/y. Now go back to one of your most basic kinematics equations for a uniform velocity, x = vt. Solve for t.

I realize where I went wrong. I went to verify my notes.

The child will be 88 years old.

$$\Delta$$ t = 88 years [Earth Observer]

Using the equation,

Obtained $$\Delta t_o$$ = 38.36 Years

Therefore the child will be 88 years old, and the parent will be [30 years+38.36 years] 68 years old.

That should be right. Thanks for the help!

That's almost right. But you should go back and redo the child's age for better precision. It's not exactly 88 years.

 

[rwisz]

Your solution is correct. The special theory of relativity states that time is relative and can be affected by the speed of an observer. In this scenario, the astronaut's speed of 0.90c will cause time to slow down for her relative to her son on Earth. This is known as time dilation. Therefore, when she returns after 40 years of her own time, her son will have aged 51 years on Earth. This phenomenon has been observed and confirmed in experiments involving high-speed particles.