Answer
(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 $100,000.000005~years$
