# Parallel Axis Theorem and Kinetic Energy of the Earth

I'm looking for a little clarification on the parallel axis theorem and computing the total kinetic energy of the Earth in it's orbit round the Sun.

So the Parallel Axis Theorem states that $$I = I_{CM} + Md^2$$ and I'm wondering if we could us this in a calculation of the Earth's total kinetic energy.

My book states that:

$$K_{tot} = K_{trans} + K_{rot} = \frac {1}{2}\bigg(Mr_{CM}^2 + I_{CM} \bigg)\omega^2$$ for a rigid body whose rotation rate is the same as the rotation rate about the center of mass. Here $$r_{CM}$$ is the distance from the axis of rotation to the rigid body.

It sounds like the answer to my question is "no" because the Earth rotates on its own axis at a different rate to it's angular speed round the Sun. Does that sound correct?

And so if we can't use this approach, do we simply compute $$K_{rot}$$ for the Earth spinning on its axis and then determine the linear speed of the Earth in it's orbit round the Sun and just compute $$K_{trans} = \frac {1}{2}Mv^2$$? Then $$K_{tot} = K_{trans} + K_{rot}$$?

• Yes, your final idea is correct. – Aaron Stevens Jun 26 at 4:35