91Ó°ÊÓ

(a)

The dilation equation states that $t=\frac{t_0}{\sqrt{1-{\mathrm{β}}^2}}$, where $t$ is the time in stationary reference frame, $t_0$ is the time in moving frame.

Here, $t_0=30$ and it is also known that the time taken to travel distance $d$with velocity $v$is $t=\frac{d}{v}$.

So, the equation becomes $\frac{d}{v}=\frac{t_0}{\sqrt{1-{\mathrm{β}}^2}}$. Substitute the values in the formula as follows:

$$\begin{gathered} \frac{d}{v}=\frac{t_0}{\sqrt{1-{\mathrm{β}}^2}} \\ \frac{23000ly}{v}=\frac{30y}{\sqrt{1-{\mathrm{β}}^2}} \\ \frac{23000cy}{30y}=\frac{v}{\sqrt{1-{\mathrm{β}}^2}} \\ \frac{v}{c}=\left(\frac{30}{23000}\right)\sqrt{1-{\mathrm{β}}^2} \end{gathered}$$

Solve the above equation as follows:

$$\begin{gathered} \mathrm{β}=\left(\frac{3}{2300}\right)\sqrt{1-{\mathrm{β}}^2} \\ {\mathrm{β}}^2=\left(\frac{3}{2300}\right)\sqrt{1-{\mathrm{β}}^2} \\ 1=\left(1+1.7×{10}^{-6}\right){\mathrm{β}}^2 \\ \mathrm{β}=0.99999915 \end{gathered}$$

Thus, the speed parameter has to be equal to $\mathrm{β}=0.99999915$.

The formula that has to be used to find the length of the Galaxy passed is $l=l_0\sqrt{1-{\mathrm{β}}^2}$.

Substitute role="math" localid="1663140492479" $l_0=23000ly$and$\mathrm{β}=0.99999915$ :

$$\begin{gathered} l=l_0\sqrt{1-{\mathrm{β}}^2} \\ l=23000\sqrt{1-{\left(0.99999915\right)}^2} \\ l=29.88ly \end{gathered}$$

Thus, the length of the Galaxy would pass by you during the trip is $29.998ly$.