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

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SUMMARY

The discussion focuses on calculating the speed of a spaceship traveling to a star located ##a## light years away, incorporating concepts of special relativity, specifically time dilation and length contraction. Participants utilize equations such as ##v = \frac{d}{\Delta t}## and the Lorentz factor ##\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}## to derive relationships between distance, speed, and time. The complexity of the resulting polynomial equations, particularly when applying length contraction, is highlighted, with suggestions to solve numerically for clarity. The conversation emphasizes the need for precise mathematical expressions and the importance of verifying calculations.

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  • Familiarity with algebraic manipulation and solving polynomial equations.
  • Knowledge of the Lorentz factor and its application in relativistic physics.
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Homework Statement
Please see below
Relevant Equations
Please see below
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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