The Earth can be approximated as a sphere of uniform density, rotating on its axis once a day. The mass of the Earth is 5.97×10^24kg , the radius of the Earth is 6.38×10^6m , and the period of rotation for the Earth is 24.0hrs .

I found in a previous problem that I = 9.72×10^37 kg⋅m^2

For this problem, I know I have to eventually use the equation
KErot = (1/2)(I)(omega)^2

I'm thinking first I need to find omega then use the above eqution... But I'm not sure which equation to use to find omega and what values to plug in?

Question

The Earth can be approximated as a sphere of uniform density, rotating on its axis once a day. The mass of the Earth is 5.97×10^24kg , the radius of the Earth is 6.38×10^6m , and the period of rotation for the Earth is 24.0hrs .

I found in a previous problem that 	I = 9.72×10^37 kg⋅m^2

For this problem, I know I have to eventually use the equation 
KErot =  (1/2)(I)(omega)^2

I'm thinking first I need to find omega then use the above eqution... But I'm not sure which equation to use to find omega and what values to plug in?

anonymous · Answer

Okay, I figured out omega. 

$$\omega=\frac{ 2 \Pi }{ T }$$
$$\omega=7.27\times10^{-27}$$

Let me plug this into the main equation to see if I get it...

anonymous · Answer

Yup!
(1/2)(9.72x10^23)(7.27x10^-27) = 2.57x10^29