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?

Respuesta :

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

[tex]\omega = \frac{2\pi}{T}[/tex]

Where,

[tex]\omega =[/tex]Angular velocity

T = Period

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

[tex]v = \omega r[/tex]

Where,

r = Radius

Our values are given as,

[tex]T = 24 hours[/tex]

[tex]T = 24hours (\frac{3600s}{1 hour})[/tex]

[tex]T = 86400s[/tex]

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

[tex]6.38*10^6m[/tex]

Usando la ecuación de la velocidad angular entonces tenemos que

[tex]\omega = \frac{2\pi}{T}[/tex]

[tex]\omega = \frac{2\pi}{86400}[/tex]

[tex]\omega = 7.272*10^{-5}rad/s[/tex]

Then the linear velocity would be,

[tex]v = \omega *r[/tex]

x[tex]v = \omega *r[/tex]

[tex]v= 463.96m/s[/tex]

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