## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

According to passengers on the spaceship, the passage takes $~~11.2~hours$
We can find the speed of the ship: $v = \frac{d}{t}$ $v = \frac{10~light~hours}{15~hours}$ $v = \frac{2}{3}~c$ We can find $\gamma$: $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-\frac{(2c/3)^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-\frac{4}{9}}}$ $\gamma = 1.34$ Let $t'$ be the time measured on earth. We can find the time that passes according to passengers on the spaceship: $t' = \gamma~t$ $t = \frac{t'}{\gamma}$ $t = \frac{15~h}{1.34}$ $t = 11.2~hours$ According to passengers on the spaceship, the passage takes $~~11.2~hours$