Our Milky Way galaxy is $100,000ly$in diameter. A spaceship crossing the galaxy measures the galaxy’s diameter to be a mere $1.0ly$.

a. What is the spacecraft’s speed, as a fraction of $c$, relative to the galaxy?

b. How long is the crossing time as measured in the galaxy’s reference frame?

We have given that our Milky Way galaxy is $100,000ly$in diameter.

A spaceship crossing the galaxy measures the galaxy’s diameter to be a mere $1.0ly$.

We have to find the spacecraft’s speed, as a fraction of $c$, relative to the galaxy.

The correct length of the galaxy in this case is $100,000$light years.

To find the fraction of the speed of light, use the appropriate length equation.

localid="1649845082660" $$\begin{gathered} L=\frac{L_{\rho }}{\gamma }=L_{\rho }\sqrt{1-\frac{v^2}{c^2}} \\ {\left(\frac{L}{L_{\rho }}\right)}^2=1-\frac{v^2}{c^2} \\ {v}^2=c^2\left(1-{\left(\frac{L}{L_{\rho }}\right)}^2\right) \\ v=\sqrt{\left(1-{\left(\frac{L}{L_{\rho }}\right)}^2\right)}c=\sqrt{\left(-1{\left(\frac{1.0ly}{100000ly}\right)}^2\right)}=0.999999999c \end{gathered}$$

We have given that our Milky Way galaxy is $100,000ly$in diameter.

A spaceship crossing the galaxy measures the galaxy’s diameter to be a mere $1.0ly$.

We have to find the time in the galaxy’s reference frame.

Time in the galaxy’s reference frame is

$t=\frac{d}{v}$

$$\begin{gathered} =\frac{(100000ly)(9.46\times {10}^{15}m/ly)}{(0.999999999)(3.00\times {10}^{8}m/s} \\ =3153333336487s \\ =100,000\text{years} \end{gathered}$$

Therefore, time in the galaxy’s reference frame is$100,000\text{years}$.