Into Outer Space

Classical Mechanics Level 3

Suppose that each human lives for 2.5 × 1 0 9 2.5\times10^9 seconds. Given that c = 3 × 1 0 8 m / s c=3\times10^8 m/s , what is the distance, in kilometers that a person can travel in a lifetime on a spaceship traveling at a speed of 0.5 c 0.5c ?


The answer is 4.33E+14.

Alex Li by Alex Li , 22, USA

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1 solution

Alex Li
Apr 13, 2015

The time is dilated by a factor of 1 1 − v 2 c 2 = 1 1 − 0. 5 2 = 1.1547 \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}=\sqrt{\frac{1}{1-0.5^2}}=1.1547 . The person can therefore travel ( 1.5 × 1 0 8 ) × ( 2.5 × 1 0 9 ) × 1.1547 = 4.33 × 1 0 17 (1.5\times10^8)\times(2.5\times10^9)\times1.1547=4.33\times10^{17} meters or 4.33 × 1 0 14 \boxed{4.33\times10^{14}} km.

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