Question

You fly $5000km$across the United States on an airliner at $250m/s.$You return two days later at the same speed.

a. Have you aged more or less than your friends at home?

b. By how much?

Hint: Use the binomial approximation

Short answer

$a.)$Because my proper time is smaller than the time in the Earth frame, I have aged less.

$b.)$In frame $S\to ∆t=4\times {10}^{4}s,$

In frame $S\text{'}\to ∆\tau =14ns$

Step-by-step solution

Part (a) Step 1: Given Information

We have given that,

Airliner takes of and attains $v=250m/s$.

Airliner returns to original position after $2$days.

We have to assume you've aged more or less than your family and friends.

Part (a) Step 2: Simplify

Because my proper time is smaller than the time in the Earth frame, I have aged less.

Part (b) Step 1: Given Information

We have given that,

Airliner takes of and attains $v=250m/s.$

Airliner returns to original position after $2$days.

We need to figure out how much is the Earth's frame and the airliner's frame.

Part (b) Step 2: Simplify

The Earth's frame is $S$.

The Airliner's frame is $S\text{'}.$

In the $S$frame:

$$\begin{gathered} ∆t=\frac{2\times 5\times {10}^{6}m}{250m/s} \\ ∆t=4\times {10}^{4}s \end{gathered}$$

In the frame$S\text{'}:$

$$\begin{gathered} ∆\tau =∆t{\left(1-\frac{v^2}{c^2}\right)}^{1/2} \\ ∆\tau =∆t\left(1-\frac{v^2}{2c^2}\right) \\ ∆\tau =∆t·\frac{v^2}{2c^2} \\ ∆\tau =(4\times {10}^{4}s)\frac{1}{2}\left(\frac{250m/s}{3\times {10}^{8}m}\right) \\ ∆\tau =1.4\times {10}^{-8}s \\ ∆\tau =14ns \end{gathered}$$