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27

You are right, that would be weird if the Moon speeds up and slows down this way to always show the exact same side to the Earth. That's why it doesn't. At some point in the orbit the Moon's rotation (its phase) lags behind and at some point it is too fast. It is (pretty much) constant and does not adjust to the varying orbital velocity. That's (one reason) ...

16

As the moon orbits Earth, tidal forces slow down the Earth's rotation by 2 milliseconds per century. Eventually, in tens of billions of years, the Earth and Moon would achieve a double tidal lock, where both are stuck with one side facing the other as they orbit the Earth-Moon barycenter. In 7.5 billion years, the Sun will expand past the Earth's current ...

11

The Moon indeed "wobbles" about in its orbit because it goes around the Earth in an ellipse and not a circle. From our point of view, it wobbles a little back and forth such that over a lunar cycle, we end up seeing about 59% of its surface. The effect is referred to as lunar libration. You can find a number of videos that show how the Moon looks to us ...

11

The question is interesting, but I suspect the answer is that the Moon will never show its "far side" to the Earth, because there are differences between the side that faces us and the far side that suggest there is something quite permanent about its orientation. So while the rotation was locking, it either settled into a state of minimum ...

9

This is a physical result: The change in rotation rate necessary to tidally lock a body B to a larger body A is caused by the torque applied by A's gravity on bulges it has induced on B by tidal forces. Stellar bodies get tidally locked with time. You even get graphs like there. On how long it generally takes, from Wikipedia $$t_{\text{lock}} \approx \... 9 The Earth-facing side of the Moon doesn’t change currently due to any of the forces on it by other bodies (excepting librations). From Gladman et. al.: “tidal dissipation in the satellite drives it to a state where […] the spin axis of a satellite in a generalized Cassini state will appear stationary.” The S_2 Cassini state is one of the two possible ... 8 With a more eccentric orbit, tidal control of the rotation does not lead to a librating lock. Mercury is in a 3:2 spin-orbit resonance. Hyperion's spin is chaotic: it gets a quasi-random kick each close approach to Saturn. So, your question is a good one, and the phenomena are rich. The Moon is merely an example of the simplest case. 6 You're quite right. The Earth is (nearly) stationary in the Moon's sky. (I say "nearly" because the Moon is in a slightly elliptical orbit around the Earth, but rotates perfectly smoothly. This means that the Earth's motion through the Lunar sky is a bit faster when the Moon is at perigee and a bit slower when it's at apogee. Because the Moon is tide-... 6 Hot objects radiate heat. The hotter something is, the faster it radiates heat. The side of the planet that faces the sun would heat up until the rate at which it is absorbing heat from the star is equal to the rate at which it is radiating heat. Once it reaches this point it will be stable, and neither heat up more or cool down. As a simple analogy, ... 6 Your scenario isn't stable. A simple way to explain this is to imagine that the planet's orbit each other at the same rate they orbit the star (your scenario has them orbiting even slower). At the same rate of orbit, the synodic period essentially approaches infinity. (see diagram of the Moon's synodic orbit). When this happens, the inner planet is at ... 6 Really, it's just because the tidal locking timescale is so long for Earth:$$t\propto\frac{a^6m_{s}}{m_{p}^2R_s^3}$$where a is semi-major axis, m_s is the mass of the secondary object, m_p is the mass of the primary, and R_s is the radius of the secondary. If we compare the Sun-Earth system to the Earth-Moon system, we see$$\frac{a_1}{a_2}\...

5

This is called "physical libration". The Moon’s physical librations and determination of its free modes (2011) estimates it using the Apollo mission retroreflectors. Their result is very small. They note that the "damping times for these two modes are estimated as $2$x$10^4$ and $2$x$10^6$ years", so it is a mystery why it is even as large as their small ...

5

So that's question 1: Is it really possible for tidal acceleration to throw a satellite into space before the planet is tidally locked, and if so, does either objects' size impact it, or just fluidity and relative periods? During the locking process, the angular momentum of one is fed into the other until the periods match. The final destination ...

5

The tidal locking timescale depends on several factors: $$\tau_{lock} \approx \frac{0.4 \omega_0 a^5 m Q}{3 G M^2 k_2 r^3}$$ such as the initial spin rate $\omega_0$, the semimajor axis $a$, the mass $m$, the solar mass $M$, the radius $r$ and various dissipation parameters $Q$ and $k_2$. Two planets that merely differ in $a$ will have the inner one lock ...

5

From the cited article, The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth. I suspect that this is the statement that is causing the OP to be confused, and rightly so. This statement is incorrect. Whether "speed of rotation" means orbital velocity or angular velocity, both are ...

5

tl;dr: An observer on the "Moon side" would see only half the phases during the fortnight-long night: from waxing half-moon to waning half-moon. They would also see the Moon during the day (early morning, late afternoon), as we currently can, but they would never see the New Moon or crescent moons at night since these would be only visible during the day. ...

5

I think what you are envisioning is having one pole always pointed towards the primary star. So the planet would rotate about that pole, but then the direction of the pole would change over the course of a year to keep the pole pointing towards a star. This phenomenon of the direction of the pole changing is called "precession". The Earth does it over a ...

4

Some confusion may have arisen in parts of the literature due to the facts that (a) the moon's mean rate of angular motion (and revolution speed) does appear to be speeding up if it is measured by mean solar time or by sidereal time, but also (b) the mean rate appears to be slowing down if it is measured by an atomic or dynamical time-scale. That is ...

4

You're asking if the temperature of the side facing the star will increase without bound. Almost anytime you're talking about a real, physical system, "without bound" just doesn't happen. Think about it - if the temperature was always increasing, eventually the planet will be hotter than the star its orbiting. At that point, why would it continue to get ...

4

You missed the the most important factor, which is, how close the sun is and as a result of the distance to the sun and relative mass, the Hill sphere and the approximate distance where the stable orbital distance ends. The angular momentum to rate the moon gets pushed away from the planet is an interesting combination of formulas, and could probably be ...

4

Wikipedia gives the formula $$t_{\text{lock}} \approx \frac{\omega a^6 I Q}{3 G m_p^2 k_2 R^5}$$ where $\omega$ is the initial spin rate expressed in radians per second, $a$ is the semi-major axis of the motion of the satellite around the central body (given by the average of the periapsis and apoapsis distances), $I\approx 0.4 m_s R^2$ is the moment of ...

4

There's no back and forth motion on the way towards tidal locking. There's kind of two ways to look this kind of "settling", there's overshooting and correction, basically a back and forth, and there's a very gradual force that grows smaller, basically moving towards zero but not overshooting. I realize that's just re-writing your question, but it helps ...

4

Let us begin with the basics. If an exterior perturber generates the potential W, then an originally spherical planet gets distorted. Owing to this distortion, its potential acquires an increment U which is usually called the additional tidal potential or, simply, the tidal potential. Now let us fix a point on the surface of the planet. Let $\gamma$ be the ...

4

The planet would reach an equilibrium where the amount of heat absorbed is the same as the amount of heat radiated. If there is no way to transfer heat on the planet (no conduction, no atmosphere), then that condition must apply locally. The flux radiated from a blackbody surface (in W/m$^2$) is given by $\sigma T^4$, where $\sigma$ is Stefan's constant and ...

4

why doesn't Earth's leading tidal bulge (encircled in the green circle 1) pull on the moon's tidal bulge It does.... leading to a force, and consequently a torque There's nothing here that will give an ongoing torque. Because of the Earth's high rotational speed and lag in response to forces, the earth's bulge stays displaced from the direction of net ...

3

I'm not exactly sure what you're asking, cause you touch on a few related points. The reason planets around red-dwarf stars are thought to be tidally locked is because the tidal force is comparatively much greater for the habitable zone. Take, for example, a star with half the mass of our sun. It would have, roughly speaking, 1/16th the solar output, so ...

3

The simple answer to your question is that Mercury is not tidally locked. You may have seen old books (before 1965) that said it was tidally locked, because it was once assumed to be so. Alternatively, as zephyr said, your source may have been referring to the 3:2 resonance, but that is also not really the same thing.

3

It's not tidally locked like the moon is because it is in a 3:2 resonance with the sun. It rotates three times for every two orbits it makes. So it isn't considered a tidal lock because it means they usually need to be in a 1:1 resonance. I think you were referring to Wikipedia, where it said Mercury was in a tidal lock with the sun. A 3:2 resonance would ...

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