Traveling on a spaceship to Alpha Centauri A, 4.37 light years away, at a velocity of 0.92 c,(adsbygoogle = window.adsbygoogle || []).push({});

if you just calculate the time by dividing the distance by the velocity:

[itex] t = \frac{d}{v} = 1.496 x 10^8 sec = 4.74 years[/itex]

Is this the time from the Earth reference frame?

Then the time on the spaceship is:

[itex] t' = t \sqrt{ 1 - \frac{v^2}{c^2}} = 1.86 years[/itex]

This is less than the time it takes light to reach Earth. Using length contraction on the spaceship, then determining the time to travel the shorter distance results in this same time.

OR, the time of 4.74 years is the time onboard the spaceship and the time on Earth is 12.09 years.

I'm confused!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Time dilation to go a specific distance

**Physics Forums | Science Articles, Homework Help, Discussion**