Question

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its search light blink on for 0.150 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 ms. (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth, expressed as a fraction of the speed of light c?

Short answer

A)$\Delta t_0=0.012s$b) u=0.997c

Step-by-step solution

Concept of the time dilation equation.

The time dilation equation is given as$\mathbf{\Delta t}\mathbf{=}{\mathbf{\gamma \Delta t}}_{\mathbf{0}}\mathbf{/}\mathbf{1}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}$where is the proper time in the rest frame,$\mathbf{∆}\mathit{t}$is the time interval measured in the second frame of reference u is the speed of the second frame with respect to the rest frame, c is the speed of light

Calculate the proper time and the speed of spacecraft

The searchlight which blinked on is the searchlight of the spacecraft, so the event happened in the first officer frame of reference. Thus, the proper time${\mathrm{\Delta t}}_0=12\mathrm{ms}=0.012\mathrm{s}$ is the time that was measured by the first officer on the spacecraft.

$$\begin{gathered} 1\mathrm{s}=1000\mathrm{ms} \\ ∆{\mathrm{t}}_0=\frac{12}{1000}=0.012\mathrm{s} \end{gathered}$$

Thus,${\mathrm{\Delta t}}_0=0.012\mathrm{s}$

By the time dilation equation,

role="math" localid="1664086479811" $$\begin{gathered} ∆\mathrm{t}=\mathrm{\gamma }∆{\mathrm{t}}_0 \\ ∆{\mathrm{t}}_0=\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}∆\mathrm{t} \\ 0.150=\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(\mathrm{u}\right)}^2}{{\left(\mathrm{c}\right)}^2}}}}\times 0.012 \\ =0.997\mathrm{c} \end{gathered}$$

Therefore, u=0.997c