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Lets assume that the motions of the Moon around the Earth and of the Earth around the Sun are uniform circular. If $\omega_L, \omega_T$ are the respective angular speeds find $\omega_L/\omega_t$.

Since the motions are uniform circular we have $\omega = v/R$ with $R$ radius, hence $\omega_L/\omega_T=\frac{R_T}{R_L}\cdot \frac{v_L}{v_T}\simeq \frac{1.5}{4,0}\cdot 10^3 \cdot \frac{1,0}{3,0} = 125$.

If the planes of the motions are the same, is the angular speed of the Moon with respect to the Sun constant?

I think it is not. Let $r_1$ be the vector that connects the Sun and the Earth and $r_2$ be the vector that connects the Earth and the Moon, if $r_1 \perp r_2$ and we consider a rotation of an angle $\pi$ when $r_2$ is directed "as the motion of the earth" and when is "opposed to the motion of the earth", then the variation of the angle between the Sun and the Moon is bigger in the last case.

I've tried to prove my claim but I have some trouble. Is it correct so far? Is it sufficient my explanation? Otherwise can you give me a hint?

Note: sorry if I have not explained myself very well, English is not my native language.

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If the planes of the motions are the same, is the angular speed of the Moon with respect to the Sun constant?

No, it is not constant.

This is easiest to see by taking the parameters to the asymptotic regime where the answer is most clearly seen: if the claim were true, then it would be true for all parameters, and you can just find suitable parameter sets that make it ridiculous.

In particular, consider the regime where the Earth is far away and moving relatively slowly, but the Moon's orbit has been contracted around the Earth in order to increase its angular velocity so that it is going as fast as necessary. At some point in this process, as seen from the Sun, the Moon will start going retrograde, i.e. it will oscillate between going forwards and backwards in its orbit, with a slight bias towards forwards. This is plainly inconsistent with the claim, which is therefore false.

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