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 [tex]2\pi[/tex] and the Period.
[tex]\omega = \frac{2\pi }{T}[/tex]
Here,
T = Time period
[tex]\omega[/tex] = 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
[tex]a_E = \omega^2 R[/tex]
Here,
[tex]a_E[/tex] = Radial acceleration of the object on the ground at the Earth's equator
R = Radius of the Earth
Replacing,
[tex]a_E = \frac{2\pi }{T} (R)[/tex]
The period of the Earth is 24Hours and the radius was previously given, then
[tex]a_E = (\frac{2\pi }{24hours(\frac{3600s}{1hour})})^2 (6380km(\frac{1000m}{1km}))[/tex]
[tex]a_E = 3.37*10^{-2} m/s^2[/tex]
The radial acceleration of the object on the ground at the earth's equator is [tex]3.37*10^{-2}m/s^2[/tex]
