## Basic Time Analysis of Space Travel

It is 4.37 light years to Alpha Centauri A, the nearest star to us, apart from the Sun. Suppose an astronaut were to travel there and back at
$0.9c$
, where
$c$
is the speed of light. How much older would he be on arriving back on Earth?
According to an observer on Earth, the astronaut takes
$2 \times \frac{4.37}{0.9}=9.711$
years to three decimal places. Now use the time dilation formula
$t_0=t \sqrt{1-v^2/c^2}$
, where
$t$
is the time interval for the observer left on Earth
$t_0$
is the time interval for the astronaut.
$t_0=t \sqrt{1-v^2/c^2}=9.711 \times \sqrt{1-0.9^2} =4.23$
years to two decimal places.