Question

Calculate the radial acceleration (in m/s2) of an object on the ground at the earth's equator, turning with the planet. The radius of the earth is R = 6380 km.

Answer 1

To solve this problem we will rely on the kinematic equations of angular motion. For this case we have that the angular velocity is equivalent to the change between the proportion $2\pi$ and the Period.

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

Here,

T = Time period

$\omega$ = Angular velocity of the object on the ground at the Earth's equator

Now the angular acceleration of the object on the ground at the Earth's equator is

$a_E = \omega^2 R$

Here,

$a_E$ = Radial acceleration of the object on the ground at the Earth's equator

R = Radius of the Earth

Replacing,

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

The period of the Earth is 24Hours and the radius was previously given, then

$a_E = (\frac{2\pi }{24hours(\frac{3600s}{1hour})})^2 (6380km(\frac{1000m}{1km}))$

$a_E = 3.37*10^{-2} m/s^2$

The radial acceleration of the object on the ground at the earth's equator is $3.37*10^{-2}m/s^2$