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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.