Finding the relativistic speed of a spacecraft

AI Thread Summary
A spacecraft traveling to a star 1 lightyear away takes 1 year in its own frame, prompting a calculation of its speed and the journey duration for an Earth observer. The initial calculation suggested a speed of 25% of the speed of light, but the algebra was incorrect. Upon reevaluation, the correct relationship shows that if 1 - beta^2 = beta^2, then beta equals 1/sqrt(2), indicating a more accurate speed. Consequently, the time experienced on Earth for the journey would be 2 years. The discussion highlights the importance of careful algebra in relativistic physics calculations.
insipiens
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Homework Statement


If a spacecraft is traveling to a star which is located at a distance of 1 lightyear and it would take the spacecraft 1 year to reach the star in its own frame, how fast would the spacecraft actually fly? Also, how long would the journey take for an observer on Earth (t_e)?
t_sc = 1 y
L_e = 1 ly
t_e = ?

Homework Equations


gamma = 1/sqrt(1-v^2/c^2)
beta = v^2/c^2
t_sc = t_e / gamma
L_sc = L_e /gamma

The Attempt at a Solution


I would start by saying that the L_e = v * t_e (v = the velocity). After throwing everything around, I get that the speed should be 25% of the speed of light, which I don't actually think is correct.
My calculations so far:
L_e = t_e * v = t_sc *v *gamma
L_e / t_sc = c = v * gamma
1/gamma^2 = beta^2
1-beta^2 = beta^2 --> beta = 1/4
Hence the time on Earth should be 4 years.
Thanks for everyone willing to help me!
Cheers,
Merlin
 
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Hi @insipiens and welcome to PF.

Your method is correct, but the algebra is not.
insipiens said:
beta = v^2/c^2
You mean beta2=v2/c2 otherwise
insipiens said:
1/gamma^2 = beta^2
would not be correct. This is a minor point. More importantly if 1-beta^2 = beta^2, what is beta2? What is beta?
 
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Thank you, you are absolutely right. I guess I was just too quick in my head and that is why I thought that sqrt(1/2)=1/4.
Now, if 1-beta^2 = beta^2, then
1/2 = beta^2 and
beta = 1/sqrt(2)
That also sounds much more reasonable. Thanks again!
 
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