The Earth's rotational kinetic energy is found from:

$\overline{){{\mathbf{K}}}_{\mathbf{r}\mathbf{o}\mathbf{t}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I}}{{\mathbf{\omega}}}^{{\mathbf{2}}}}$, where I is the Earth's moment of inertia and ω is the angular velocity of the Earth.

**(a)**

The moment of inertial of Earth can be considered to be the same as that of a solid sphere.

So we have:

$\overline{){{\mathbf{I}}}_{\mathbf{e}\mathbf{a}\mathbf{r}\mathbf{t}\mathbf{h}}{\mathbf{=}}\frac{\mathbf{2}}{\mathbf{5}}{\mathbf{M}}{{\mathbf{r}}}^{{\mathbf{2}}}}$, where M is the mass of the Earth and r is the radius of the Earth.

We know that M = 6 × 10^{24} kg

(a) Calculate the rotational kinetic energy of Earth on its axis.

(b) What is the rotational kinetic energy of Earth in its orbit around the Sun?

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