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

I'm building an earthlike world that has a moon orbiting a gas giant. Is it possible for the moon to always be between the planet and the sun? Also, is it possible for a moon to rotate around its own axis?

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• It may be possible for the moon to never be in the shadow of the planet if it's in a sun synchronous orbit.
– Luke
May 8 '20 at 17:01

No, it's not possible for the moon to always be between the planet and the sun.

For the moon to be in a stable orbit around the planet, and always be in front of the sun, two things must be true (We'll ignore the situation of putting the Moon at the L1 Lagrange Point, it wouldn't be in orbit around the planet, and L1 is not long-term stable):

1. The Moon's sidereal orbital period around the planet must be equal to the planet's orbital period around the sun.
2. The Moon must be orbiting inside the Planet's Hill Sphere.

The Hill Sphere is the region of long-term stable orbits around the planet, which is based on the mass of the planet, the mass of the star being orbited, and the distance between the two. Its radius can be estimated as follows.

$$r_h=a_p\sqrt[3]{\frac{m_p}{3m_s}}$$

Where $$m_p$$ is the mass of the planet, $$m_s$$ is the mass of the star, and $$a_p$$ is the semimajor axis of the planets orbit, or the radius, in the circular orbit case we'll be using for simplicity.

The Orbital Period of an object around another object can be determined by the following formula.

$$t=2\pi\sqrt{\frac{a^3}{GM}}$$ Where $$a$$ is the semimajor axis of the orbit in question in meters, $$G$$ is the Newtonian Gravitational Constant, and $$M$$ is the mass of the body being orbited. This assumes that the satellite is negligible compared to the mass of the central body.

When I started fooling around with the numbers determined by these values in a Desmos Graph, a curious relation emarged, that I'll have to take some time to work through the derivation of:

Regardless of the mass I chose for the Sun and the Planet, regardless of the semimajor axis of the planet, The Period of the largest Moon orbit that could fit inside the Planet's Hill Sphere was always 55.7% of the Planet's orbital period.

So, no. You can't have a Moon with a long-term-stable orbit around a planet that keeps it between the planet and the sun.

• @novotny I do not have the math to calculate it, but my gut tells me (it is the fashion in the US!) that rather than a moon, a binary planet with equal sized partners could fulfill your conditions 1 and 2. The orbital calculators I found all make your assumption that satellite is negligible & I am not sure Hill sphere calculation works for equal mass orbiting partners. Apr 26 '20 at 20:18
• Here is a bounty in hopes that you will expand your answer to detail your use of the Desmos graph and also explore the case of a binary planet and orbital period. May 8 '20 at 1:17
• I've come up with a derivation of the curious relation, if you'd like to see it - it's really just applying Kepler's third law. cc @Willk. May 8 '20 at 2:10
• @HDE226868 - Go ahead and put up an answer, HDE. The bounty is for whomever provides sweet math education & you have a fine track record in that regard. May 8 '20 at 2:15
• @Willk Excellent, done. May 8 '20 at 2:22

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.

• /𝑇𝑀 and 𝑇𝑝 are the periods of the planet and the moon; 𝑎𝑀 and 𝑎𝑝 are their periods;/. I think T is the period. What is a? Circumference? May 8 '20 at 12:53
• @Willk Whoops, typo on my part - let me fix that. May 8 '20 at 13:55
• I should just plug in numbers and try to do the math. I may yet. But consider your scenario where Mp is close to Ms. That is like the situation in your question worldbuilding.stackexchange.com/questions/71971/…. If the ship plays the role of moon and "lunar period" is larger, can the orbiting ship orbit (the center of mass) so as to keep the black hole between itself and the super star? May 8 '20 at 18:14
• @Willk The big issue (so far) is that the expression for the Hill sphere assumes that $M_p\ll M_S$, so we would need to find a different equation for that equal-mass regime to think about that case. I'll have to think about it. May 9 '20 at 0:26
• @Willk Having done the algebra out, I believe there's no general solution because it would require finding roots of a fifth-order polynomial, and there's no algebraic formula for the roots of such a polynomial in the general case. May 11 '20 at 3:50

Not in practice, and yes, if you justify it.

If a moon is orbiting a planet, it has to go around the planet. To simplify, the movement around the planet is what keeps it from just falling onto the planet. Orbits are simple in principle, but those principles don't especially correspond to everyday ideas about moving objects. If you learned about orbits from films or TV, you likely have things to un-learn, because those sources tend to ignore how things actually work. The Wikipedia page about orbits is a good place to start.

There is a special case that looks as if it will solve your problem, but doesn't work in practice. That's the "L1" Lagrangian point. In that, the "moon" is not actually orbiting the planet. It's orbiting the sun, close enough to the planet that it gets dragged along by the planet's gravity, and is always approximately between the planet and the sun. The reason that this doesn't work in practice is that the position is unstable: the slightest disturbance of the moon's position, such as the gravity of another planet in the system, will set the moon drifting away from the L1 position.

Staying at L1 requires frequent course corrections. Humanity has several spacecraft at the L1 point between Earth and our Sun, but they all need to use small rockets ("thrusters") to stay there. The SOHO satellite is an example. However, any body big enough to retain an atmosphere and be habitable is too large for its position to be adjusted with any reasonable level of technology.

Moons always rotate, it's just that they usually do it in a time equal to the time they take to orbit their planet. This means they always show the same face to the planet, giving a false impression that they don't rotate. This is called "tidal locking," and it happens naturally with most moons.

To have a moon that is not tidally locked, you need some kind of explanation. The simplest way is to say that the moon collided with another sizable body, which can change its rotational speed and axis quite seriously, and provides a reason for exciting topography. You need to have this happen before life appeared on the moon, because such a collision is likely to kill all life on the moon.

• Tidal locking is a very slow process. It may take several billion years for a moon to be perfectly in sync with its planet. This means that you can justify having a moon not in sync with its planet just by saying that tidal locking has not done its job yet. Their is no need for a celestial bodies to crash. Apr 27 '20 at 8:42