Question

The center of our Milky Way galaxy is about $\mathbf{23000}\mathit{l}\mathit{y}$ away. (a) To eight significant figures, at what constant speed parameter would you need to travel exactly (measured in the Galaxy frame) in exactly $\mathbf{23000}\mathit{l}\mathit{y}$ (measured in your frame)? (b) Measured in your frame and in light-years, what length of the Galaxy would pass by you during the trip?

Short answer

1. The speed parameter is $0.99999915$.
2. The length of the Galaxy would pass by you during the trip is$29ly$ .

Step-by-step solution

The dilation equation 

(a)

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

The speed parameter

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 $\frac{d}{v}=\frac{t_0}{\sqrt{1-{\beta }^2}}$. Substitute the values in the formula as follows:

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

Solve the above equation as follows:

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

Thus, the speed parameter has to be equal to $\beta =0.99999915$.

The length of the Galaxy passed during the trip 

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

Substitute role="math" localid="1663140492479" $l_0=23000ly$and$\beta =0.99999915$ :

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

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