Is it possible for a moon to stay on the same side of its planet relative to the sun?

This answer is meant as a supplement to notovny's. I agree with their conclusions (the scenario is impossible because of the instability of this Lagrange point, and the fact that the Hill sphere is too small), and I just want to derive the "curious relation" they came up with.

We start with Kepler's third law. $T_M$ and $T_p$ are the periods of the planet and the moon; $a_M$ and $a_p$ are their semi-major axes; $M_p$ and $M_S$ are the masses of the planet and the start. Let's write out Kepler's third law for both the orbit of the moon and the orbit of the planet: $$T_M^2=\frac{4\pi^2}{GM_p}a_M^3,\quad T_p^2=\frac{4\pi^2}{GM_S}a_p^3$$ If we assume the moon is in its outermost orbit, we have $$a_M=a_p\sqrt[3]{\frac{M_p}{3M_S}}$$ Now we substitute and our first equation is $$T_M^2=\frac{4\pi^2}{GM_p}a_p^3\frac{M_p}{3M_S}$$ Finally, we divide by the equation for the planet's period: $$\frac{T_M}{T_p}=\frac{M_S}{M_p}\frac{M_p}{3M_S}$$ and so $T_M\approx0.58T_p$, which is the result notovny found. It's interesting to think about this in the case of a binary planet ($M_p\approx M_M$) or a binary star ($M_S\approx M_p$). Kepler's third law is easy to modify for both of those cases. However, the derivation of the Hill radius requires that $M_p\ll M_S$, and that the Hill radius $R_H\ll a_p$. If we get rid of that requirement, then I believe a general solution would require finding the roots of a fifth-order polynomial in $x\equiv R_H/a_p$, which unfortunately has no general solution. For particular values of $M_p$ and $M_S$, we may be able to find solutions, but we'd need to look at them on a case-by-case basis.

posted over 4 years ago

HDE 226868‭

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