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

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

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!

 

[anuttarasammyak]

(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)

 

[member 731016]

(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!

 

[member 731016]

(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.

Thanks!

 

[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=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
$$\beta=\frac{v}{c}$$ ?

 

[member 731016]

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!

 

[anuttarasammyak]

LHS is distance. RHS is speed of the star * time to meet. All in IFR of space pilot.

 

[member 731016]

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!

 

[anuttarasammyak]

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