Question

Earth turns on its axis about once every 24 hours. The Earth's equatorial radius is 6.38 x 10^6 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 kinematic equations of movement description for both linear and angular movement (include the description of the period)

The speed of a body at a point on the equator is given by

$v= \omega r$

Where,

$\omega$ = Angular Speed

r = Radius

The period of the earth is 24 hours or 84600s, and the radius of the earth is $6.38 * 10 ^ 6m$. If the angular velocity as a function of the period is described as

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

Replacing at the first equation we have that

$v = (\frac{2\pi}{T})R$

Replacing the values,

$v = (\frac{2\pi}{84600})(6.38*10^6)$

$v = 463.966 m/s$

Therefore the speed would go flying  off the Earth's inhabitants who live at the equator equator is 464m/s