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

AI Thread Summary
The discussion revolves around calculating the speed of a spaceship traveling to a star a certain distance away, factoring in relativistic effects like time dilation and length contraction. Participants explore the implications of the Lorentz factor and how it affects the equations used to determine velocity and distance. There is confusion regarding the application of length contraction, leading to complex polynomial equations that require the quadratic formula for solutions. The conversation emphasizes the need to clarify methods and ensure consistency in calculations. Overall, the thread highlights the challenges of applying relativistic physics to practical problems in space travel.
member 731016
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!
 
Physics news on Phys.org
(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)
 
  • Love
Likes member 731016
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} ?
 
Last edited:
  • Love
Likes member 731016
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.
 
  • Love
Likes member 731016
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.
 
Back
Top