# Chapter 36 - Relativity - Exercises and Problems - Page 1060: 24

(a) The speed of the spaceship relative to the galaxy is $~~0.99999999995~c$ (b) The crossing time measured in the galaxy's reference frame is $100,000.000005~years$

#### Work Step by Step

(a) Let $L_0 = 100,000~ly$ Let $L = 1~ly$ We can find the speed $v$ of the spaceship relative to the galaxy: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $\frac{L}{L_0} = \sqrt{1-\frac{v^2}{c^2}}$ $(\frac{L}{L_0})^2 = 1-\frac{v^2}{c^2}$ $\frac{v^2}{c^2} = 1- (\frac{L}{L_0})^2$ $v^2 = [1- (\frac{L}{L_0})^2]~c^2$ $v = \sqrt{1- (\frac{L}{L_0})^2}~c$ $v = \sqrt{1- (\frac{1~ly}{100,000~ly})^2}~c$ $v = \sqrt{0.9999999999}~c$ $v = 0.99999999995~c$ The speed of the spaceship relative to the galaxy is $~~0.99999999995~c$ (b) We can find the crossing time measured in the galaxy's reference frame: $t = \frac{d}{v}$ $t = \frac{100,000~ly}{0.99999999995~c}$ $t = 100,000.000005~ly/c$ $t = 100,000.000005~years$ The crossing time measured in the galaxy's reference frame is $100,000.000005~years$

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