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 $K_{tot} = K_{trans} + K_{rot}$?

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

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