Question

At what speed, as a fraction of $c,$ would a round-trip astronaut “lose” $\frac{1}{25}$ of the elapsed time shown on her watch?

Short answer

At speed of $0.28c,$a round-trip astronaut “lose”$\frac{1}{25}$ of the elapsed time shown on her watch.

Step-by-step solution

Given Information

We have given that the rate at which a round-trip astronaut loses $\frac{1}{25}$of the time on her watch.

Simplify

We have: $∆t=\gamma ∆t\text{'}.$

Now, we use binomial approximation for $\gamma :$

localid="1649531467250" $$\begin{gathered} \frac{∆t}{∆t\text{'}}\approx \frac{v^2}{2c^2} \\ \frac{∆t}{∆t\text{'}}=\frac{v^2}{2c^2} \\ {v}^2=2c^2\left(\frac{∆t}{∆t\text{'}}\right) \\ v=c\sqrt{2\left(\frac{∆t}{∆t\text{'}}\right)} \\ v=c\sqrt{2\left(\frac{1}{25}\right)} \\ v=0.28c \end{gathered}$$