<p>There is the famous case of the i five Lagrange points in celestial mechanics, where an astronomical object can remain in the same posiiton relative to two much more massive worlds.</p>
<p>Part One:  L2, or maybe L2 or L3?</p>
<p>See illustratins of the Lagrange points:</p>
<p><a href="https://en.wikipedia.org/wiki/Lagrange_point#/media/File:Lagrange_points_simple.svg" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Lagrange_point#/media/File:Lagrange_points_simple.svg</a></p>
<p><a href="https://en.wikipedia.org/wiki/List_of_objects_at_Lagrange_points#/media/File:Lagrange_very_massive.svg" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/List_of_objects_at_Lagrange_points#/media/File:Lagrange_very_massive.svg</a></p>
<p>One of those points, the L2 point, is on a line between the two larger objects, beyond the smaller of the two large objects.</p>
<p>As the smaller object orbits around the larger objects, it travels at the orbital speed for its distance.  All objects orbiting farther out from the largest object will orbit at slower orbital speeds.  So if some third object farther out is in a specific position relative to the seond object, the second obect body will move ahead of the third object, and eventually catch up with the third object and pass it, over and over again.</p>
<p>Unless the third object is in the L2 position on the line betweenthe first and second object.  At the L2 positionthe gravitational attraction of the first and second objects will combine to keep the third object orbiting in the L2 point, a specific point on th elinte between first and second objects, and beyond the 2nd object.</p>
<p>So if the gravitational force of the second object on the third objet is strong enough, the tidal forces - the difference between  the attactions on the sides of the third object closer and farther from the second object - could eventually tidally lock the third object to the second object.</p>
<p>Thus one side of the third object will always be pointed toward the second object, and since the third object will always be on the opposite sid of the second object from thefirst object, that side of the third object will always be pointed at the first object also.</p>
<p>Thus some people might think that the third object, objectively tidally locked to the secod object, is alos tidally locked to the first object as well.</p>
<p>I guess that is at least one way such a situation cuuld happen.</p>
<p>The first object could be a star, and the scond object could be a planet orbiting around the star, and the third object could be a moon orbiting the planet.  If the moon orbits around the planet in the the prograde directon,the driction that the planet rotates around its axis in, and if the moon orbits the planet beyond the geosychronous orbital distance. the titdal interactins between the palenet and the moon will push the orbit ouf the moon farther from the planet while also slowing down the rotatin rates of the planet and of the moon.</p>
<p>and if the object orbits closer to the first object than the second object is, there   is a L1 point where the third object would stay on the line between the first and secnd objects.</p>
<p>Since the moon will be much less massive than the planet, the tidal interactions should tidally lock the moon to the planet long (maybe many billions of years) before the planet becomes tidlly locked ot the moon.</p>
<p>And that is the present situation with the Sun, the Earth, and the Moon; the Earth orbits the Sun, the Moon orbits the Earth, and only the Moon is tidally locked, to the Earth.</p>
<p>The Moon's orbit is elliptical, and it has a semi-major axis of 384,399  kilometers (it gets abut 12,000 kilometers closer to and farther from Earth than that semi-major axis).</p>
<blockquote>
<p>Like L1, L2 is about 1.5 million kilometers or 0.01 au from Earth. An example of a spacecraft at L2 is the James Webb Space Telescope, designed to operate near the Earth–Sun L2.<a href="https://en.wikipedia.org/wiki/Planetary_habitability#Size" rel="nofollow noreferrer">3</a> Earlier examples include the Wilkinson Microwave Anisotropy Probe and its successor, Planck.</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Lagrange_point#L2_point" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Lagrange_point#L2_point</a></p>
<p>And if the Moon was still expanding its orbit, eventually the Mooon's orbit would reach the L2 point about 1,115,601 kilometers farther from Earth.   And the Moon's orbit is very slightly increasing.</p>
<blockquote>
<p>Measurements from laser reflectors left during the Apollo missions (lunar ranging experiments) have found that the Moon's distance increases by 38 mm (1.5 in) per year (roughly the rate at which human fingernails grow).[168][169][170] Atomic clocks show that Earth's day lengthens by about 17 microseconds every year,[171][172][173] slowly increasing the rate at which UTC is adjusted by leap seconds.</p>
<p>This tidal drag makes the rotation of Earth and the orbital period of the Moon very slowly match. This matching first results in tidally locking the lighter body of the orbital system, as already the case with the Moon. Eventually, after 50 billion years,[174] also the Earth would be made to always face the Moon with the same side. This would complete the mutual tidal locking of Earth and the Moon, matching the length of Earth's day to the then also significantly increased lunar month and the Moon's day, and suspending the Moon over one meridian (comparable to the Pluto-Charon system). However, the Sun will become a red giant engulfing the Earth-Moon system long before the latter occurs.[175][176]</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Moon#Influence_on_orbit_and_rotation" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Moon#Influence_on_orbit_and_rotation</a></p>
<p>So at 38 mm per year, it would take the Moon 26.315789 years to move one meter farther from Earth, and 26,315.789 years to move one kilometer farther from Earth.  Thus it would take the Moon's orbit 26,315,789 years to move one thousand kilometers farther from Earth, and 26,315,789,470 years to move 1,000,000 kiloemters farther from Earth.</p>
<p>And if an object in Earth's L1 positon becomes tidlly locked to the Earth, with one side always facing Earth and the other side always facing away fromt he Earth, that other side always facing away from the Earth would aways face the Sun.  Thus the third object in Earth's L1 postion could be considered to be be tidally locked to both the Sun and the Moon.</p>
<p>Except that the Sun is predicted to become a red giant in only 5,000,000,000 and will probably engulf and destroy the Earth and the Moon when the Moon's orbit shoudl have moved only about 100,000 kilometers frather from Earth. And as the Moon moves farther from Earth, the rate at which it moves away should slow down.</p>
<blockquote>
<p>The Hill sphere of an astronomical body is the region in which it dominates the attraction of satellites. To be retained by a planet, a moon must have an orbit that lies within the planet's Hill sphere. That moon would, in turn, have a Hill sphere of its own. Any object within that distance would tend to become a satellite of the moon, rather than of the planet itself.</p>
</blockquote>
<p>[https://en.wikipedia.org/wiki/Hill_sphere<a href="https://en.wikipedia.org/wiki/Hill_sphere" rel="nofollow noreferrer">6</a></p>
<blockquote>
<p>The Hill sphere for Earth thus extends out to about 1.5 million km (0.01 AU). The Moon's orbit, at a distance of 0.384 million km from Earth, is comfortably within the gravitational sphere of influence of Earth and it is therefore not at risk of being pulled into an independent orbit around the Sun. All stable satellites of the Earth (those within the Earth's Hill sphere) must have an orbital period shorter than seven months.</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Hill_sphere#Formula_and_examples" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Hill_sphere#Formula_and_examples</a></p>
<p>And I don't think that it is a coincidence that Earth's Hill Sphere extends to about the distances of its L1 and L2 points.</p>
<p>However:</p>
<blockquote>
<p>The Hill sphere is only an approximation, and other forces (such as radiation pressure or the Yarkovsky effect) can eventually perturb an object out of the sphere. This third object should also be of small enough mass that it introduces no additional complications through its own gravity. Detailed numerical calculations show that orbits at or just within the Hill sphere are not stable in the long term; it appears that stable satellite orbits exist only inside 1/2 to 1/3 of the Hill radius.</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Hill_sphere#Formula_and_examples" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Hill_sphere#Formula_and_examples</a></p>
<p>So when the moon's orbit increases to about 500,000 or 750,000 kilometers, the Moon will pass out out the zone of true stability.  Eventually it will wander out of Earth orbit and take an independent orbit around the Sun. And after a probably very long time the Moon would crash into the Sun, the Earth, or another planet, or be ejected from the Solar system.</p>
<p>What about an object in Earth's L1 point which started out as an independent planet orbiting the Sun and was captured into Earth's L1 position?</p>
<blockquote>
<p>The triangular points (L4 and L5) are stable equilibria, provided that the ratio of M1/M2 is greater than 24.96.[note 1] This is the case for the Sun–Earth system, the Sun–Jupiter system, and, by a smaller margin, the Earth–Moon system. When a body at these points is perturbed, it moves away from the point, but the factor opposite of that which is increased or decreased by the perturbation (either gravity or angular momentum-induced speed) will also increase or decrease, bending the object's path into a stable, kidney bean-shaped orbit around the point (as seen in the corotating frame of reference).<a href="https://en.wikipedia.org/wiki/Habitability_of_red_dwarf_systems#Tidal_effects" rel="nofollow noreferrer">8</a></p>
<p>The points L1, L2, and L3 are positions of unstable equilibrium. Any object orbiting at L1, L2, or L3 will tend to fall out of orbit; it is therefore rare to find natural objects there, and spacecraft inhabiting these areas must employ station keeping in order to maintain their position.</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Lagrange_point#Stability" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Lagrange_point#Stability</a></p>
<p>So an object in a L1, L2, or L3 point relative to two other objects will tend to wander out of the Lagrange Point and take an independent orbit around the larger object.  And tht orbit will be so unstable that it will eventually crash into a star or planet or be ejected from the star system.</p>
<p>So any system where a third object is in a L1, L2, or L3 point relative to two other objects will be unstable, and the third object will probably leave the Lagrange point before it becomes tidally locked to either body.</p>
<p>Part Two:  L4 or L5.</p>
<p>Where are the L4 and L5 points?</p>
<p>They are in the same orbit as the second and smaller object, but 60 dgrees separate from the second object.  The L4 point is 60 degrees ahead of the second object and the L5 point is 60 degrees behind the second object.</p>
<p>There many known objects in the L4 and L5 points of various solar solar sytem objects, so obviously such &quot;trojan&quot; orbits can sometimes be stable for millions or billions of years.</p>
<p>An object in a L4 or L5 point will be equally distant from the first and most massive object in the system and from the second and less massive object in the system.  Except that they do tend to oscillate around their Lagrange points by some degrees before returning to them.</p>
<p>If the first and second objects had idnetical mass, they would have identical tidal braking effects on the rotation of the third object at equal distance from them (not counting the aforementined wandering frm the exact Lagrange point).</p>
<p>However, the first object must be a number of times more massive than the second object, for the trojan obit to be stable, and thus it must have a much gstronger tidal braking effect on the the third object.  So the third object would become tidally locked to the first object (probably a star) long before becoming tidally locked to the second object (probably aplanet).</p>
<p>Of course if the third object constantly stays in the same position relative to the first object and the second object, when it stops rotationg relative to the first object it should stop rotating relative to the the second object. it that case the third object will be as good as tidally locked to the second object.</p>
<p>Since the directions from the third object tothe first and second objects will be sixty degrees apart, the circumference of the surface of the third object can be divided into six segments sixty degrees wide (if the third object is large enough to be spherical in shape).<br />
In two segements totalling 180 degrees both the first and the second objects will always be visible.  In the third segment of 60 degees only the second object will be visible.  In the 4th and 5th segments totalling another 180 degrees neither the first nor the second object will ever be visible. And in the sixth segement of 60 degrees only the first object will be visible.</p>
<p>We know that the tidal fprce of a planet on its moon can be strong enough to tidally lock the moon to the planet.  The Moon is tidally locked to the Earth, and dozens of other other moons in the solar system are tidally locked to their planets.</p>
<p>So if a planet is orbited by a moon which has smaller moon the larger moon L 4 or L5 point, it is perfectly possible for the tidal forces of the planet to tidally lock one or probably both of the to the panet.  And and any moon tidally to locked to its planet would also keept one side aways facing the othr moon 60degrees away on the same orbit.  Thus two moons sharing an orbit around a planet can be tidally locked to the planet, whichhas the effect of making them tidally locked to eachother.</p>
<p>Can a star have a tidal force strong enough to tidally locall a planet or asteroid oriting it, so tht twoobjects sharing the same orbit around the star would become tidall locked to the star, and thus in a sense to each other?</p>
<p>Yes.</p>
<p>A planet has to orbit within the circumstellar habitable zone, or Goldilocks zone, of its star to have the correct temperatures for liqud water on its surface and be potentially habital for life.  And different stars with different masses have Goldilocks zones at difference distances from them.  But if a star's Goldilocks zone  is so clos to the star that a planet in it becomes tidally locked to the star, the planet would have ahot side always facing the star and and a cold dark side always facing away from the star. And such a situation might make the planet uninhabitable.</p>
<p>So discussions of planetary habitability discuss whether a star would tidally lock a planet in its Goldilocks zone (or closer to the star).  Calculations shsw that planets in the habitable zones of main sequence stars less massive than some of the spectral class K stars and all of the spectral class M stars would probably be tidally locked to theeir stars, which might possibly ake them uninhabitable.</p>
<p>There are discusssion of this on pages 68 to 75 of <em>Habitable Panets for Man</em>, Stephen H. Dole, 1964.</p>
<p><a href="https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf" rel="nofollow noreferrer">https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf</a></p>
<p>And:</p>
<p><a href="https://en.wikipedia.org/wiki/Planetary_habitability#Size" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Planetary_habitability#Size</a></p>
<p>And:</p>
<p><a href="https://en.wikipedia.org/wiki/Habitability_of_red_dwarf_systems#Tidal_effects" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Habitability_of_red_dwarf_systems#Tidal_effects</a></p>
<p>Andof course it is perfectly possible for planets orbit their stars closer to their stars than the Goldiocks zone, and thus to have a higher probability of becoming tidally locked than planets in the Goldilocks zone.  Many exoplants have been detected orbiting their stars much closer than the Goldilocks zones of those stars and thus having much higher teperatures than a habitable planet could have - temperatures tens, hundreds, and even thousands of degrees Kelvin above the temperatures where water could be liquid.</p>
<p>And in many cases those very hot p;anets should be tidally locked to their stars.</p>
<p>Thus a situation where a giant planet and a terrestrial type planet orbit a star in the same orbit and both are tidally locked to the star, and thus in a sense to each other, is perfectly possible.</p>

CC BY-SA 4.0
2022-08-27T23:09:18.320Z
While L2 itself is not a stable position, the answer to [this answer to *Are some Halo Orbits actually Stable? (stable orbits about unstable Lagrange points)*](https://space.stackexchange.com/a/21910/12102) is "yes" and links to [this answer to *Is it possible to have stable orbits around Lagrange point 1?*](https://physics.stackexchange.com/a/338685/83380) as well.
uhoh
CC BY-SA 4.0
2022-08-27T22:01:30.793Z
2022-08-27T22:01:30.793Z
M. A. Golding
<p>Could a planet be tidally locked to both the star it is orbiting and a moon orbiting the planet at the same time? I feel like it wouldn't be possible, because I think it would cause the moon to crash into the planet, but maybe there's a way for it to work?</p>

CC BY-SA 4.0
2022-08-26T14:09:33.290Z
Think about it: where would the moon have to remain if the planet were locked to the star?
Carl Witthoft
CC BY-SA 4.0
2022-08-27T09:58:11.103Z
If tidal locking means absolutely locked, then it's hard/impossible. But if it means that the rotational motion of the body around its axis is locked into some kind of *synchrony* with the other two objects (i.e. some kind of resonance), I think it's certainly possible, but it would require some special circumstances and like most things may not be stable for millions of years. See for example [Just how "locked" are resonant-chains of exoplanets thought to be? (e.g. K2-138 and TOI-178)](https://astronomy.stackexchange.com/q/23980/7982)
uhoh
CC BY-SA 4.0
2022-08-28T20:21:48.063Z
I am missing the problem here.  You could have a planet go around its star in say 30 days and locked so one side always faces the star.  And, it could have a moon that would go around it, in 30 days and locked to the planet and hence to the star too. So, what is the problem here?
eshaya
<tidal-locking>
Can a celestial body be tidally locked to multiple other bodies at the same time?
Nzall
# Can a celestial body be tidally locked to multiple other bodies at the same time?

(Asked by: username_0 on 2022-08-26)

Could a planet be tidally locked to both the star it is orbiting and a moon orbiting the planet at the same time? I feel like it wouldn't be possible, because I think it would cause the moon to crash into the planet, but maybe there's a way for it to work?

**Tags:** tidal-locking

> **Comments:**
> **username_1**: Think about it: where would the moon have to remain if the planet were locked to the star?  
> **username_2**: If tidal locking means absolutely locked, then it's hard/impossible. But if it means that the rotational motion of the body around its axis is locked into some kind of *synchrony* with the other two objects (i.e. some kind of resonance), I think it's certainly possible, but it would require some special circumstances and like most things may not be stable for millions of years. See for example [Just how "locked" are resonant-chains of exoplanets thought to be? (e.g. K2-138 and TOI-178)](https://astronomy.stackexchange.com/q/23980/7982)  
> **username_3**: I am missing the problem here.  You could have a planet go around its star in say 30 days and locked so one side always faces the star.  And, it could have a moon that would go around it, in 30 days and locked to the planet and hence to the star too. So, what is the problem here?  

---

## 1 answers

### Answer by username_4 on 2022-08-27. Score: 3

There is the famous case of the i five Lagrange points in celestial mechanics, where an astronomical object can remain in the same posiiton relative to two much more massive worlds.

Part One: L2, or maybe L2 or L3?

See illustratins of the Lagrange points:

<https://en.wikipedia.org/wiki/Lagrange_point#/media/File:Lagrange_points_simple.svg>

<https://en.wikipedia.org/wiki/List_of_objects_at_Lagrange_points#/media/File:Lagrange_very_massive.svg>

One of those points, the L2 point, is on a line between the two larger objects, beyond the smaller of the two large objects.

As the smaller object orbits around the larger objects, it travels at the orbital speed for its distance. All objects orbiting farther out from the largest object will orbit at slower orbital speeds. So if some third object farther out is in a specific position relative to the seond object, the second obect body will move ahead of the third object, and eventually catch up with the third object and pass it, over and over again.

Unless the third object is in the L2 position on the line betweenthe first and second object. At the L2 positionthe gravitational attraction of the first and second objects will combine to keep the third object orbiting in the L2 point, a specific point on th elinte between first and second objects, and beyond the 2nd object.

So if the gravitational force of the second object on the third objet is strong enough, the tidal forces - the difference between the attactions on the sides of the third object closer and farther from the second object - could eventually tidally lock the third object to the second object.

Thus one side of the third object will always be pointed toward the second object, and since the third object will always be on the opposite sid of the second object from thefirst object, that side of the third object will always be pointed at the first object also.

Thus some people might think that the third object, objectively tidally locked to the secod object, is alos tidally locked to the first object as well.

I guess that is at least one way such a situation cuuld happen.

The first object could be a star, and the scond object could be a planet orbiting around the star, and the third object could be a moon orbiting the planet. If the moon orbits around the planet in the the prograde directon,the driction that the planet rotates around its axis in, and if the moon orbits the planet beyond the geosychronous orbital distance. the titdal interactins between the palenet and the moon will push the orbit ouf the moon farther from the planet while also slowing down the rotatin rates of the planet and of the moon.

and if the object orbits closer to the first object than the second object is, there is a L1 point where the third object would stay on the line between the first and secnd objects.

Since the moon will be much less massive than the planet, the tidal interactions should tidally lock the moon to the planet long (maybe many billions of years) before the planet becomes tidlly locked ot the moon.

And that is the present situation with the Sun, the Earth, and the Moon; the Earth orbits the Sun, the Moon orbits the Earth, and only the Moon is tidally locked, to the Earth.

The Moon's orbit is elliptical, and it has a semi-major axis of 384,399 kilometers (it gets abut 12,000 kilometers closer to and farther from Earth than that semi-major axis).

> Like L1, L2 is about 1.5 million kilometers or 0.01 au from Earth. An example of a spacecraft at L2 is the James Webb Space Telescope, designed to operate near the Earth–Sun L2.[3](https://en.wikipedia.org/wiki/Planetary_habitability#Size) Earlier examples include the Wilkinson Microwave Anisotropy Probe and its successor, Planck.

<https://en.wikipedia.org/wiki/Lagrange_point#L2_point>

And if the Moon was still expanding its orbit, eventually the Mooon's orbit would reach the L2 point about 1,115,601 kilometers farther from Earth. And the Moon's orbit is very slightly increasing.

> Measurements from laser reflectors left during the Apollo missions (lunar ranging experiments) have found that the Moon's distance increases by 38 mm (1.5 in) per year (roughly the rate at which human fingernails grow).[168][169][170] Atomic clocks show that Earth's day lengthens by about 17 microseconds every year,[171][172][173] slowly increasing the rate at which UTC is adjusted by leap seconds.
>
> This tidal drag makes the rotation of Earth and the orbital period of the Moon very slowly match. This matching first results in tidally locking the lighter body of the orbital system, as already the case with the Moon. Eventually, after 50 billion years,[174] also the Earth would be made to always face the Moon with the same side. This would complete the mutual tidal locking of Earth and the Moon, matching the length of Earth's day to the then also significantly increased lunar month and the Moon's day, and suspending the Moon over one meridian (comparable to the Pluto-Charon system). However, the Sun will become a red giant engulfing the Earth-Moon system long before the latter occurs.[175][176]

<https://en.wikipedia.org/wiki/Moon#Influence_on_orbit_and_rotation>

So at 38 mm per year, it would take the Moon 26.315789 years to move one meter farther from Earth, and 26,315.789 years to move one kilometer farther from Earth. Thus it would take the Moon's orbit 26,315,789 years to move one thousand kilometers farther from Earth, and 26,315,789,470 years to move 1,000,000 kiloemters farther from Earth.

And if an object in Earth's L1 positon becomes tidlly locked to the Earth, with one side always facing Earth and the other side always facing away fromt he Earth, that other side always facing away from the Earth would aways face the Sun. Thus the third object in Earth's L1 postion could be considered to be be tidally locked to both the Sun and the Moon.

Except that the Sun is predicted to become a red giant in only 5,000,000,000 and will probably engulf and destroy the Earth and the Moon when the Moon's orbit shoudl have moved only about 100,000 kilometers frather from Earth. And as the Moon moves farther from Earth, the rate at which it moves away should slow down.

> The Hill sphere of an astronomical body is the region in which it dominates the attraction of satellites. To be retained by a planet, a moon must have an orbit that lies within the planet's Hill sphere. That moon would, in turn, have a Hill sphere of its own. Any object within that distance would tend to become a satellite of the moon, rather than of the planet itself.

[https://en.wikipedia.org/wiki/Hill\_sphere[6](https://en.wikipedia.org/wiki/Hill_sphere)

> The Hill sphere for Earth thus extends out to about 1.5 million km (0.01 AU). The Moon's orbit, at a distance of 0.384 million km from Earth, is comfortably within the gravitational sphere of influence of Earth and it is therefore not at risk of being pulled into an independent orbit around the Sun. All stable satellites of the Earth (those within the Earth's Hill sphere) must have an orbital period shorter than seven months.

<https://en.wikipedia.org/wiki/Hill_sphere#Formula_and_examples>

And I don't think that it is a coincidence that Earth's Hill Sphere extends to about the distances of its L1 and L2 points.

However:

> The Hill sphere is only an approximation, and other forces (such as radiation pressure or the Yarkovsky effect) can eventually perturb an object out of the sphere. This third object should also be of small enough mass that it introduces no additional complications through its own gravity. Detailed numerical calculations show that orbits at or just within the Hill sphere are not stable in the long term; it appears that stable satellite orbits exist only inside 1/2 to 1/3 of the Hill radius.

<https://en.wikipedia.org/wiki/Hill_sphere#Formula_and_examples>

So when the moon's orbit increases to about 500,000 or 750,000 kilometers, the Moon will pass out out the zone of true stability. Eventually it will wander out of Earth orbit and take an independent orbit around the Sun. And after a probably very long time the Moon would crash into the Sun, the Earth, or another planet, or be ejected from the Solar system.

What about an object in Earth's L1 point which started out as an independent planet orbiting the Sun and was captured into Earth's L1 position?

> The triangular points (L4 and L5) are stable equilibria, provided that the ratio of M1/M2 is greater than 24.96.[note 1] This is the case for the Sun–Earth system, the Sun–Jupiter system, and, by a smaller margin, the Earth–Moon system. When a body at these points is perturbed, it moves away from the point, but the factor opposite of that which is increased or decreased by the perturbation (either gravity or angular momentum-induced speed) will also increase or decrease, bending the object's path into a stable, kidney bean-shaped orbit around the point (as seen in the corotating frame of reference).[8](https://en.wikipedia.org/wiki/Habitability_of_red_dwarf_systems#Tidal_effects)
>
> The points L1, L2, and L3 are positions of unstable equilibrium. Any object orbiting at L1, L2, or L3 will tend to fall out of orbit; it is therefore rare to find natural objects there, and spacecraft inhabiting these areas must employ station keeping in order to maintain their position.

<https://en.wikipedia.org/wiki/Lagrange_point#Stability>

So an object in a L1, L2, or L3 point relative to two other objects will tend to wander out of the Lagrange Point and take an independent orbit around the larger object. And tht orbit will be so unstable that it will eventually crash into a star or planet or be ejected from the star system.

So any system where a third object is in a L1, L2, or L3 point relative to two other objects will be unstable, and the third object will probably leave the Lagrange point before it becomes tidally locked to either body.

Part Two: L4 or L5.

Where are the L4 and L5 points?

They are in the same orbit as the second and smaller object, but 60 dgrees separate from the second object. The L4 point is 60 degrees ahead of the second object and the L5 point is 60 degrees behind the second object.

There many known objects in the L4 and L5 points of various solar solar sytem objects, so obviously such "trojan" orbits can sometimes be stable for millions or billions of years.

An object in a L4 or L5 point will be equally distant from the first and most massive object in the system and from the second and less massive object in the system. Except that they do tend to oscillate around their Lagrange points by some degrees before returning to them.

If the first and second objects had idnetical mass, they would have identical tidal braking effects on the rotation of the third object at equal distance from them (not counting the aforementined wandering frm the exact Lagrange point).

However, the first object must be a number of times more massive than the second object, for the trojan obit to be stable, and thus it must have a much gstronger tidal braking effect on the the third object. So the third object would become tidally locked to the first object (probably a star) long before becoming tidally locked to the second object (probably aplanet).

Of course if the third object constantly stays in the same position relative to the first object and the second object, when it stops rotationg relative to the first object it should stop rotating relative to the the second object. it that case the third object will be as good as tidally locked to the second object.

Since the directions from the third object tothe first and second objects will be sixty degrees apart, the circumference of the surface of the third object can be divided into six segments sixty degrees wide (if the third object is large enough to be spherical in shape).  
In two segements totalling 180 degrees both the first and the second objects will always be visible. In the third segment of 60 degees only the second object will be visible. In the 4th and 5th segments totalling another 180 degrees neither the first nor the second object will ever be visible. And in the sixth segement of 60 degrees only the first object will be visible.

We know that the tidal fprce of a planet on its moon can be strong enough to tidally lock the moon to the planet. The Moon is tidally locked to the Earth, and dozens of other other moons in the solar system are tidally locked to their planets.

So if a planet is orbited by a moon which has smaller moon the larger moon L 4 or L5 point, it is perfectly possible for the tidal forces of the planet to tidally lock one or probably both of the to the panet. And and any moon tidally to locked to its planet would also keept one side aways facing the othr moon 60degrees away on the same orbit. Thus two moons sharing an orbit around a planet can be tidally locked to the planet, whichhas the effect of making them tidally locked to eachother.

Can a star have a tidal force strong enough to tidally locall a planet or asteroid oriting it, so tht twoobjects sharing the same orbit around the star would become tidall locked to the star, and thus in a sense to each other?

Yes.

A planet has to orbit within the circumstellar habitable zone, or Goldilocks zone, of its star to have the correct temperatures for liqud water on its surface and be potentially habital for life. And different stars with different masses have Goldilocks zones at difference distances from them. But if a star's Goldilocks zone is so clos to the star that a planet in it becomes tidally locked to the star, the planet would have ahot side always facing the star and and a cold dark side always facing away from the star. And such a situation might make the planet uninhabitable.

So discussions of planetary habitability discuss whether a star would tidally lock a planet in its Goldilocks zone (or closer to the star). Calculations shsw that planets in the habitable zones of main sequence stars less massive than some of the spectral class K stars and all of the spectral class M stars would probably be tidally locked to theeir stars, which might possibly ake them uninhabitable.

There are discusssion of this on pages 68 to 75 of *Habitable Panets for Man*, Stephen H. Dole, 1964.

<https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf>

And:

<https://en.wikipedia.org/wiki/Planetary_habitability#Size>

And:

<https://en.wikipedia.org/wiki/Habitability_of_red_dwarf_systems#Tidal_effects>

Andof course it is perfectly possible for planets orbit their stars closer to their stars than the Goldiocks zone, and thus to have a higher probability of becoming tidally locked than planets in the Goldilocks zone. Many exoplants have been detected orbiting their stars much closer than the Goldilocks zones of those stars and thus having much higher teperatures than a habitable planet could have - temperatures tens, hundreds, and even thousands of degrees Kelvin above the temperatures where water could be liquid.

And in many cases those very hot p;anets should be tidally locked to their stars.

Thus a situation where a giant planet and a terrestrial type planet orbit a star in the same orbit and both are tidally locked to the star, and thus in a sense to each other, is perfectly possible.

> **Comments:**
> **username_2**: While L2 itself is not a stable position, the answer to [this answer to *Are some Halo Orbits actually Stable? (stable orbits about unstable Lagrange points)*](https://space.stackexchange.com/a/21910/12102) is "yes" and links to [this answer to *Is it possible to have stable orbits around Lagrange point 1?*](https://physics.stackexchange.com/a/338685/83380) as well.  

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