Finding speed of spaceship traveling to star ##a## light years away

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Homework Help Overview

The discussion revolves around calculating the speed of a spaceship traveling to a star located a light-year distance away, incorporating concepts from special relativity, such as time dilation and length contraction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the implications of length contraction on the distance to the star and how it affects the calculation of speed. There are attempts to derive equations involving the Lorentz factor and polynomial forms related to the speed of the spaceship.

Discussion Status

Several participants express confusion regarding the application of length contraction and the resulting equations. Some suggest numerical approaches to simplify the problem, while others seek clarification on specific expressions and methods presented in the discussion.

Contextual Notes

There are indications of varying interpretations of the problem setup, particularly concerning the reference frames and the assumptions about distances and speeds involved in the calculations.

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Homework Statement
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Relevant Equations
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For this problem,
1715728502105.png

My working,
(a) ##d = a~ly = ac~y##

##v = \frac{d}{\Delta t} = \frac{ac}{b} \frac{m}{s}##
(b) Lorentz factor is ##γ = \frac{1}{1 - \frac{a^2}{b^2}}## Thus time dilation is ##\Delta t = \frac{b}{1 - \frac{a^2}{b^2}} y##, however, I think my arugment is only valid if ##a >> b## in ##\frac{a}{b}## so ##v \approx c##.

Is that please correct?

Thanks!
 
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(a) The distance between the Earth and the star is contracted for the pilot. You should take it.
(b) The distance for the Earth / v of (a)
 
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anuttarasammyak said:
(a) The distance between the Earth and the star is contracted for the pilot. You should take it.
(b) The distance for the Earth / v of (a)
Thank you for your reply @anuttarasammyak !

So I'm slightly confused. If the distance is length contracted then we end up wiht a very ugly polynomial to solve which is from ##v = \frac{L_0}{(\sqrt{1 - \frac{v^2}{c^2}})\Delta t}##
Squaring both sides gives
This gives ##-\frac{v^4}{c^2} + v^2 - \frac{L^2_0}{\Delta t^2} = 0##, which I would need to use the quadratic formula to solve. Is that what you were intending?

Where I leave in terms of ##v## since otherwise it gets even more ugly.

Thanks!
 
anuttarasammyak said:
(a) The distance between the Earth and the star is contracted for the pilot. You should take it.
(b) The distance for the Earth / v of (a)
Here is a version of the problem with numbers in it. Maybe it would be easier to solve numerically then generalize symbolically.
1715753562977.png

Thanks!
 
ChiralSuperfields said:
So I'm slightly confused. If the distance is length contracted then we end up wiht a very ugly polynomial to solve which is from v=L0(1−v2c2)Δt
Squaring both sides gives
This gives −v4c2+v2−L02Δt2=0, which I would need to use the quadratic formula to solve. Is that what you were intending?
Though I do not follow you, how about
\sqrt{1-\beta^2}\ a=\beta \ c b
where
\beta=\frac{v}{c} ?
 
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anuttarasammyak said:
Though I do not follow you, how about
\sqrt{1-\beta^2}\ a=\beta \ c b
where
\beta=\frac{v}{c} ?
Thank you for your reply @anuttarasammyak ! Sorry, I'm still confused. Where did you get that expression from?

Thanks!
 
LHS is distance. RHS is speed of the star * time to meet. All in IFR of space pilot.
 
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anuttarasammyak said:
LHS is distance. RHS is speed of the star * time to meet. All in IFR of space pilot.
Thank you for your reply @anuttarasammyak ! Sorry I'm still confused. Does anybody please know whether my method in post #3 and #4 are correct?

Thanks!
 
You may calculate and check the v result of yours and mine. I hope they coincide.
 
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