Question

Earth turns on its axis about once every 24 hours. The Earth's equatorial radius is 6.38 x 106 m. If some catastrophe caused Earth to suddenly come to a screeching halt, with what speed would Earth's inhabitants who live at the equator go flying off Earth's surface?

Answer 1

To solve this problem it is necessary to apply the concepts related to the Period of a body and the relationship between angular velocity and linear velocity.

The angular velocity as a function of the period is described as

$\omega = \frac{2\pi}{T}$

Where,

$\omega =$Angular velocity

T = Period

At the same time the relationship between Angular velocity and linear velocity is described by the equation.

$v = \omega r$

Where,

r = Radius

Our values are given as,

$T = 24 hours$

$T = 24hours (\frac{3600s}{1 hour})$

$T = 86400s$

We also know that the radius of the earth (r) is approximately

$6.38*10^6m$

Usando la ecuación de la velocidad angular entonces tenemos que

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{86400}$

$\omega = 7.272*10^{-5}rad/s$

Then the linear velocity would be,

$v = \omega *r$

x$v = \omega *r$

$v= 463.96m/s$

The speed would Earth's inhabitants who live at the equator go flying off Earth's surface is  463.96