Question

How fast must a rocket travel relative to the earth so that time in the rocket “slows down” to half its rate as measured by earth-based observers? Do present-day jet planes approach such speeds?

Short answer

u = 0.867c

Step-by-step solution

Concept of the time dilation equation

The time dilation equation is given as$\mathbf{∆}\mathbf{t}\mathbf{=}\mathbf{\gamma }\mathbf{∆}{\mathbf{t}}_{\mathbf{0}}\mathbf{/}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}}$where$\mathbf{∆}{\mathit{t}}_{\mathbf{0}}$ is the proper time in the rest frame,$\mathbf{∆}\mathbf{t}$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 speed of the jet plane.

We know that,$\Delta t=\frac{1}{2}\Delta t_0$ In order to calculate the u of the rocket, we will use the time dilation equation: $\Delta t=\gamma \Delta t_0$

$$\begin{gathered} y=\sqrt{1-u^2/c^2} \\ ∆t=∆t_0/\sqrt{1-u^2/c^2} \\ \sqrt{1-u^2/c^2}=∆t_0/∆t \end{gathered}$$

By squaring both sides:

$$\begin{gathered} 1-u^2/c^2={\left(\frac{1}{2}\right)}^2 \\ {u}^2/c^2=1-{\left(\frac{1}{2}\right)}^2 \end{gathered}$$

By taking square root for both sides:

$$\begin{gathered} \sqrt{u^2/c^2}=\sqrt{1-{\left(\frac{1}{2}\right)}^2} \\ \frac{u}{c}=0.867 \\ u=0.867c \end{gathered}$$

The fastest jet plane is around$1.09\times {10}^3\mathrm{m}/\mathrm{s}$

Thus, there is no jet plane nowadays approaches such speeds.