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jbriggs444
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Is the equation correct? Let us try to justify it. Note that it is your job to justify the equations that you write down. It is not our job to reverse engineer them.You have a contribution from the rotational kinetic energy of the earth about its own axis. This from its [sidereal?] rotation rate of ##\omega_1##.Lotto said:And when we get back to my original question - the total kinetic energy of Earth - what would it be?
I would say that it is similar to the pendulum, so ##E_k=\frac 12 J_{CM} {\omega_1}^2+\frac 12 (J_{CM}+ma^2){\omega_2}^2##, where that angular velocities correspond to velocities about Earth's own axis of rotation and a rotation about the Sun, ##a## is a distance of Earth's centre of mass from the Sun.
Is this equation correct?
So far, so good.
Then you have ##\frac{1}{2}J_\text{CM}\omega_2^2##. No. That is not right at all. You've already accounted for the rotational kinetic energy of the Earth about its own center of mass. Counting it again would be double dipping.
[Counting it once with rotation rate corresponding to a solar day and then adjusting with the orbital rate about the sun to obtain a sidereal day would also be wrong. It makes my head hurt to think about it. The rotation rate is squared. Corrections would not add linearly]
Finally, you have a term ##\frac{1}{2}ma^2\omega_2^2##. On the face of it, this is silly. Kinetic energy depends on velocity, not acceleration. But maybe some terms cancel out helpfully. So let us see where this leads...
Let us try dimensional analysis first. We are after a result in units of energy: ##kg \ m^2 / sec^2##. You've proposed a term with units of mass times acceleration squared. That's ##kg \ m^2 / sec^4##. No, that cannot possibly be correct.
[You could get a correct formulation such as ##\frac{1}{2}maR## by exploiting the fact that ##a = \frac{v^2}{R}##. That is where I halfway thought you were going].
You want the second term to be the translational kinetic energy of an Earth-like object orbitting the sun at the orbital velocity of the Earth while maintaining a fixed orientation with respect to the distant stars.
We could get that using ##E = \frac{1}{2}mv^2## where ##m## is the mass of the Earth and ##v## is its orbital speed. But since you seem to prefer using the angular velocity (##\omega_2##) of the Earth in its orbit about the sun, we can write ##v## as ##\omega_2 R##. Where ##R## is the radius of the Earth's orbit. That would give us:$$\text{KE}_\text{translation} = \frac{1}{2}mv^2 = \frac{1}{2}m\omega_2^2R^2$$
For consistency, we could write the rotational kinetic energy of a hypothetical Earth of uniform density as $$\text{KE}_\text{rotation} = \frac{1}{2}I\omega_1^2 = \frac{1}{2}\frac{2}{5}mr^2\omega_1^2 = \frac{1}{5}mr^2\omega_1^2$$