Answer to Question #101661 in Mechanics | Relativity for AbdulRehman

Question

Of course, the earth also spins on its axis, so that neither city in 7-5 is
standing still... Estimate the relative velocity between the two cities, and
use this to estimate the size of the time dilation effect you would expect
from special relativity. Do we need to consider this effect as well?

Answer 1

Consider two cities: one is located at the equator (with latitude 0°) and the other at 54° and calculate their velocities. We know that a tangential speed can be found as

$$v=\frac{2\pi R}{T},$$

where

R - radius of the circumference with center at earth's axis of rotation that crosses the given city,

T - earth's period, 23.93 hours.

Now calculate the radii of these circumferences:

for the city at the equator: $R_1=R_{earth}=6375000\text{ m},$

for the city at 54°: $R_2=R_{earth}\cdot\text{cos}54^\circ=3747131\text{ m}.$

Thus:

$$v_1=\frac{2\pi 6375000}{23.93\cdot3600}=464\text{ m/s},\\ \space\\ v_2=\frac{2\pi 3747131}{23.93\cdot3600}=273\text{ m/s}.$$

Imagine that you're an observer sitting at the North pole. You do not move with regard to the two considered cities. Therefore, 1 year for you will be

$$t_1=1\sqrt{1-\bigg(\frac{464}{3\cdot10^8}\bigg)^2}=1\text{ year}$$

for the first city at the equator and

$$t_2=1\sqrt{1-\bigg(\frac{273}{3\cdot10^8}\bigg)^2}=1\text{ year}$$

for the city at 54°.

No need to consider time dilation.