\[0.9c\]

, 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.