Special relativity problem

  • #1

Homework Statement


A spaceship sets out from earth to cross the galaxy, which has diameter D=6e20, in T=35 years (as measured on ship). What velocity does it need to travel at? And then find time viewed on earth, distance viewed by spaceship.

Homework Equations

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The Attempt at a Solution



So I don't think I can do this simply with the lorentz factor because of the element of distance when the final time measurement is taken, right? As in
Let l' be the length viewed by the ship, l=D the length viewed on earth, t' = T the time on ship and t the time on earth.
Then [tex]\frac{l'}{v} = T[/tex] and [tex]l' = \frac{D}{\gamma}[/tex] so just combine and solve?

Should I be using the Lorentz transformations here instead?
[tex]T = \gamma (t - \frac{vD}{c^2})[/tex]
[tex]x' = \gamma (D - vT)[/tex]

But that's a huge mess to solve...

Are either of these approaches correct? If not what am I missing?
 

Answers and Replies

  • #2
Andrew Mason
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Homework Statement


A spaceship sets out from earth to cross the galaxy, which has diameter D=6e20, in T=35 years (as measured on ship). What velocity does it need to travel at? And then find time viewed on earth, distance viewed by spaceship.

...
So I don't think I can do this simply with the lorentz factor because of the element of distance when the final time measurement is taken, right?
You can just use the Lorentz factor. There is no change in measuring position in the spaceship frame. The events (beginning and ending of journey) occur at the same position in the spaceship frame.

As in
Let l' be the length viewed by the ship, l=D the length viewed on earth, t' = T the time on ship and t the time on earth.
Then [tex]\frac{l'}{v} = T[/tex] and [tex]l' = \frac{D}{\gamma}[/tex] so just combine and solve?
That's right, which means that:

[tex]T = D/v\gamma[/tex]

You have to work that out to determine v. Its not that difficult.

AM
 

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