Question

Can't seem to get this right :/

The distance to Alpha Centauri is 4,3 light years. How fast would a space ship have to travel to get there in 10 years, according to the crew?

Answer 1

use

V = s/t

(in metre and seconds)
for s
s = 4.3 light years is equal to 4.3 x 12 x 30 x 24x x60x60x300000 metre
t = 10 year = 10 x 12x30x24x60x60 seconds

put the data into:
v=s/t

Answer 2

Hmm, that doesn't seem to take time dilation into account. The answer *should* be 0,395c

Answer 3

oh i'm sorry.., for relativistik

\[\delta t' = \frac{ \delta t }{ \sqrt{1-\frac{ v ^{2} }{ c ^{2} }} }\]

Answer 4

10 years wrt the crew?

Answer 5

10 years according to the crew

Answer 6

But, at 4.3 light-years away, getting there would be extremely difficult. ... grain of sand, then Alpha Centauri would be over 10 kilometers – or about 6 miles – away. ... Its distance of 4.3 light-years equals 25.6 trillion miles away – nearly ... and the space ship needs to go 1/2 speed of light about 17,600 mph or 28,300 kph

Answer 7

0,395 is not correct bro

Answer 8

y=1 year
Substituted 10y for \(\Delta t'\) in \(\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}\)
assuming that the ship is going at constant velocity,
and the planet isn't moving,
\(v^2 = \frac{(4.3yc)^2}{(\Delta t)^2}\)
Solving the quadratic above gives \(\Delta t\) = 8.7 years or 4.9 years(NA).
Well, times does go longer when you're near light speed so that was reasonable.
Which gives velocity wrt to Earth of 0.49c.
Anything I missed?

Answer 9

The calc that gerry gave gives 0.43c wrt Earth. However, taking that the time will be shorter than what the crew feels, shouldn't it be larger than 0.43c?

Answer 10

4.3 light years is the distance measured from earth. That distance should be shorter for the crew because of length contraction. Ie, the answer 0.395c should, it seems, be for a shorter distance than 4.3.

Answer 11

Hey, you need to use the lorentz transformation for this.\[t'=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}\]Instead of t you put 4.3/v and what you have to do now is plug the values and isolate v. I got 0.395c with that.

Answer 12

Because the transformation takes into acount the distance, the dilatation of time doesn't and instead of trying to combine dilatation of time and contraction of space, its easier to use the transformation because it is actually where all those come from.