Question

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?

Short answer

(a) The speed of the spaceship relative to the galaxy is $0.999999999c$.

(b) The time in the galaxy's reference frame is $100,000$years.

Step-by-step solution

Part (a) Step 1: Given Information

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.

Part (a) Step 2: Simplify

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

Part (b) Step 1: Given Information

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.

Part (b) Step 2: Simplify

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$100,000\text{years}$.