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For example; is there any possibility for the moon to always have a side facing the Sun whilst orbiting Earth? And if so then what would the day cycle be like?

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    $\begingroup$ No, it's not possible because the tidal forces will inevitably be stronger from the planet than the star. If the tidal forces were stronger from the star then the Moon couldn't be in a stable orbit around the planet. If nobody else explains the math, I could give it a shot, but I tend to be a little clumsy with math and formulas. $\endgroup$ – userLTK Jun 15 '16 at 23:38
  • $\begingroup$ what a fascinating fun question! $\endgroup$ – Fattie Jun 18 '16 at 12:35
  • $\begingroup$ A body at one of the Lagrange points of a sun/planet system could keep the same face to the sun and the same face to the planet. Whether you would call such a body a "moon" is a valid question though. $\endgroup$ – Steve Linton Jul 22 '18 at 22:40
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I worked out the math, it's pretty straight forward. If we take the Hill sphere, which is an estimate for the furthest possible orbit, and I'm going to just run the math on circular orbits. Eliptical orbits are harder to put in resonance anyway.

the simple Hill Sphere formula. $3\frac{r^3}{a^3} = \frac{m}{M}$, where $a$ is the semi major axis planet to sun and $r$ is the distance moon to the planet, little $m$, mass of the planet, big $M$, mass of the star.

To calculate tidal forces, the force changes with the 3rd power of the radius, so for equal tidal force on a moon from both the star it's planet orbits and the planet it orbits, $\frac{r^3}{R^3} = \frac{m}{M}$, and that rather neatly puts the equal tidal forces outside the outer edge of the Hill Sphere by a factor of the cube root of 3, or about 1.44

The outer half of the Hill Sphere isn't long term stable anyway in real systems, so for the Moon to experience a stronger tidal force from the star it's planet orbits than it experiences from it's planet, it would need to be well outside the stable orbital region around the planet, by at least a multiple of 2.88.

There's no way for any moon in any system to be tidally locked to it's star and not it's planet.

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The moon is gravitationally locked to the Earth. Because gravitational field depends on distance, the Earth's gravity does not affect the moon uniformly: the farthest side is not attracted as strongly as the closest side, these are tidal effects. This causes the heaviest side of the moon to face the earth, i.e. the moon is tidally locked to the earth. The sun is much farther away (~360 times). That means that the difference in the gravitational field vector from the sun across the moon (i.e. tidal effects) is much lower. Therefore, if an object is gravitationally bound, it will be so to the object it orbits. If we were close enough to the sun that the moon would be affected more by the tidal effects of the sun than those of the Earth, we would be so close that the moon would no longer orbit the Earth because it would be pulled away by the sun.

So no, a moon will not be tidally locked to a star. However, if a moon were perfectly spherically symmetric (and solid), it would not be able to be tidally locked. Then, by extreme coincidence it could have a sidereal rotation period similar to the orbital period of its planet around the sun.

There are objects that are tidally locked to stars, for example Mercury. Mercury orbits the sun in 87 days, and it rotates about its axis in 87 days. That means that on one side of Mercury it is always day, and on the other side it is always night, there is no day-night cycle on Mercury. Mercury is actually in orbital resonance, see below.

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    $\begingroup$ Mercury is in a 3:2 orbital resonance with the sun. It has days and nights. A full day on Mercury is 2 Mercury years. $\endgroup$ – userLTK Jun 15 '16 at 23:40
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    $\begingroup$ And its orbital period is closer to 88 days (87.9691 according to Wikipedia). I'm not going to downvote your answer because the first two paragraphs are quite good, but you should correct that last paragraph. $\endgroup$ – Keith Thompson Jun 16 '16 at 2:09
  • $\begingroup$ Whoops, you're right! $\endgroup$ – Coen Jul 22 '18 at 12:48

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