<p>Short Answer:</p>
<p>There is an inner limit to how clase a palnet could orbit to its star and keep a moonin orbit around it.  But I don't know how to calculate it.  As far as I know there is no outer limit to how far a planet could be from its star and have moons.  A rogue planet in interstellar space far from a star could have moons.</p>
<p>Long Answer:</p>
<p>Every planet whch orbits a star has an inner and outer distance between which any moons would have to orbit.</p>
<p>The inner distance partially depends on the characteristics of the specific moon.  It is call the Roche radius or Roche limit.  Any object, that is held together mostly by its gravity, like a star, planet, or moon, which passes with the Roche limit of a more massive object will be torn apart by tidal forces.  The Roch elimit depends of the relative densiites of the two bodies and whether the smaller object is rigid or fluid.</p>
<p>A moon can not form out of smaller objects if it is within the Roche limit of its planet, and if a moon passes within the Roche limit of its planet it will break up.  So there should be no moons within the Roche limits of their planets.</p>
<p>Formulas for calculating the Roche limits of astronomical bodies can be found at:</p>
<p><a href="https://en.wikipedia.org/wiki/Roche_limit%5B1%5D" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Roche_limit[1]</a></p>
<p>The outer limit at which a moon can orbit a planet is call the Hill radius or Hill sphere.  The size of the hill sphere depends on the masses of the planet and its star and the distance between them, since the stronger the gravity of the star is relative to the planet, the smaller the Hill sphere of the planet will be.</p>
<p>There are formulas for calculating an object's Hill Sphere at:</p>
<p><a href="https://en.wikipedia.org/wiki/Hill_sphere%5B2%5D" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Hill_sphere[2]</a></p>
<p>However, it turns out that a moon orbiting in the outer parts of a Hill sphere will not have an orbit stable for long periods of astronomical time.</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. The region of stability for retrograde orbits at a large distance from the primary is larger than the region for prograde orbits at a large distance from the primary. This was thought to explain the preponderance of retrograde moons around Jupiter; however, Saturn has a more even mix of retrograde/prograde moons so the reasons are more complicated.<a href="https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability" rel="nofollow noreferrer">3</a></p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability%5B3%5D" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability[3]</a></p>
<p>The Hill sphere of Earth extends to about 1,500,000 kilometers, so the region of truely stable orbits around Earth extends only to about 500,000 to 750,000 kilometers from Earth.</p>
<p>The Roche limit for Earth is 1.49 radii for rigid objects and 2.88 radii for fluid objects. Since the radius of Earth is 6,371 kilometers, the Roche limit for rigid objects is 9,492.79 kilometers and for fluid objects is 18,348.48 Kilometers.</p>
<p>The larger and more massive a planet is, the greater the radii of its Roche limit and Hill sphere will be.  The farther a planet is from its star, the greter the radii of its Hill sphere will be.</p>
<p>Mercury and Venus are less massive than the Earth, and so their Roche limits are smaller, which is good for not destroying close moons..  They are closer to the Sun than the Earth, so the Sun's gravity is stronger where they orbit, and since they are less massive than Earth, their mases and distances from the Sun combine to make their HIll spheres much smaller than Earth's, which is bad for keeping moons.</p>
<p>The Hill sphere of Mercury has a radius of only 175,300 kilometers, so the true region of stability should have an outer edge at only 58,432 to 87,650 kilometers.</p>
<p>The Hill Spere of Venus is 1,004,200 kilometers, so the true region of stability should have an outer edge at only 334,733.3 to 502,100 kilometers, much larger than Mercury's but smaller than Earth's.</p>
<p>But there ae other problems with Mercury and Venus having moons.  The orbit of a moona round tis planet will chang eover time, as the moon moves closer to or farther from the planet.  So if a moon starts out between the Roche limit and the Hill sphere of its planet, it could move out of the zone where stable orbits are possible.</p>
<p>If a planet and its moon form together, the moon will have a prograde orbit.  Most of the objects in the Solar System rotate in the same direction that they orbit the sun in, whch is called a prograde orbit.  If a moon forms with its planet, what is called a regular moon, it will orbit the planet in the same direction as the planet rotates, which will be prograde relative to the planet.  Since most planets rotate in a prograde direction relative to their orbit, most of the regular moons formed with their planets orbit in a prograde direction relative to the plant's rbit around the Sun.</p>
<p>Planets can also capture passing objects and make them their moons.  The capture process can result in either a prograde orbit, or an orbit in the opposite direction, a retrograde orbit.  Therer are a number of captured moons in the Solar System, some with prograd eorbits and some with retrograde orbits.</p>
<blockquote>
<p>All retrograde satellites experience tidal deceleration to some degree. The only satellite in the Solar System for which this effect is non-negligible is Neptune's moon Triton. All the other retrograde satellites are on distant orbits and tidal forces between them and the planet are negligible.</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Retrograde_and_prograde_motion#Natural_satellites_and_rings%5B4%5D" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Retrograde_and_prograde_motion#Natural_satellites_and_rings[4]</a></p>
<blockquote>
<p>Triton's revolution around Neptune has become a nearly perfect circle with an eccentricity of almost zero. Viscoelastic damping from tides alone is not thought to be capable of circularizing Triton's orbit in the time since the origin of the system, and gas drag from a prograde debris disc is likely to have played a substantial role.<a href="https://en.wikipedia.org/wiki/Triton_(moon)#Orbit_and_rotation" rel="nofollow noreferrer">4</a> Tidal interactions also cause Triton's orbit, which is already closer to Neptune than the Moon's is to Earth, to gradually decay further; predictions are that 3.6 billion years from now, Triton will pass within Neptune's Roche limit.[25] This will result in either a collision with Neptune's atmosphere or the breakup of Triton, forming a new ring system similar to that found around Saturn.[25]</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Triton_(moon)#Orbit_and_rotation%5B5%5D" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Triton_(moon)#Orbit_and_rotation[5]</a></p>
<p>So if Mercury or Venus captured a moon in an orbit retrograde to their rotation, that moon would gradually spiral down to within the Roche limit and be destroyed after millions or billions of years.</p>
<p>Moons in prograde orbits around their planets will either move away from the planets or move in toward their planets.</p>
<p>If a moon is in a prograde orbit above the synchronous orbit level, it will move away from the planet due to tidal acceleration. Thus, after millions or billions or psosibly trillions of years, it might pass outside the Hill sphere of the planet and go into orbit around their star.  MOs tof the moons is the Solar System orbit farther than the synchronous orbits of their planets, and so a gradually receeding form them.</p>
<p><a href="https://en.wikipedia.org/wiki/Tidal_acceleration#Other_cases_of_tidal_acceleration%5B6%5D" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Tidal_acceleration#Other_cases_of_tidal_acceleration[6]</a></p>
<p>Some moons orbit their planets below the synchronous orbit level.  Their orbital periods are less than one day of the planet. Those moons experience tidal deceleration and slowly spiral inward toward their planets, eventually reaching the Roche limits of their planets and breaking up.</p>
<p><a href="https://en.wikipedia.org/wiki/Tidal_acceleration#Tidal_deceleration%5B7%5D" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Tidal_acceleration#Tidal_deceleration[7]</a></p>
<p>The sidereal days, the period it takes them to rotate 360 degrees with respect to the disant stars, of Mercury and Venus are 58.646 Earth days and 243.0226 Earth days respectively.  So any moons orbiting Mercury and Venus at the synchrnous orbits would be orbiting very far from their planets.</p>
<blockquote>
<p>Mercury and Venus are believed to have no satellites chiefly because any hypothetical satellite would have suffered deceleration long ago and crashed into the planets due to the very slow rotation speeds of both planets; in addition, Venus also has retrograde rotation.</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Tidal_acceleration#Tidal_deceleration%5B7%5D" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Tidal_acceleration#Tidal_deceleration[7]</a></p>
<p>For a long time astronomers believed that Mercury, and possibly Venus, were tidally locked to the Sun, with rotation periods equal to their oribital periods, so that a day would be equal toa year, and one side of theplanet would always face the Sun and the other side always face away from the Sun. That is called a 1:1 resonance.  It is now known that Mercury and Venus do not have 1:1 resonances, but their rotation periods are very long compared to their years of 87.97 Earth days and 224.7 Earth days.</p>
<p>When the  gravity of a star is too strong at the orbit of a planet - which depends on the mas sof hte star and the orbital distance of the planet from the star - that planet will become tidally locked to the Star, either with a 1:1 resonance or a resonance with simple integers, such as the 3:2 resonance of the planet Mercury - Mercury rotates three times in two Mercurian years - or some other simple resonance.</p>
<p>So if a planet is too deep in the gravity well of its star, the planet's rotation will be slowed down a lot, and the synchronous orbit of the planet will be very far from the planet, possbily beyond the Hill sphere.  So a moon outside the Hill spehre would be lost into space, and a moon below the synchronous orbit would have a decaying orbit and eventually reach the roche limit of the planet.</p>
<p>So if the mass of a star is known, the distance at which a planet would have its day greatly slowed down, thus making it hard or impossible for it to keep its moon(s) can be calculated.</p>
<p>And of course the shorter years of planets orbiting close to their stars is another problem for any potential moons.</p>
<blockquote>
<p>The longest possible length of a satellite’s day compatible with Hill stability has been shown to be about P∗p/9, P∗p being the planet’s orbital period about the star (Kipping 2009a)</p>
</blockquote>
<p><a href="https://arxiv.org/ftp/arxiv/papers/1209/1209.5323.pdf%5B8%5D" rel="nofollow noreferrer">https://arxiv.org/ftp/arxiv/papers/1209/1209.5323.pdf[8]</a></p>
<p>And the source is given as:</p>
<p><a href="https://academic.oup.com/mnras/article/392/1/181/1071655%5B9%5D" rel="nofollow noreferrer">https://academic.oup.com/mnras/article/392/1/181/1071655[9]</a></p>
<p>So this claims that a mooon can not have a stable orbit around its planets unless the orbital of the moon is less than one nineth (0.111111) theorbital period of the planet around the star.</p>
<p>The greater the mass of a star, the shorter the year of a lanet oriting it at a specific distance will be, since the increased gravity of the star means the planet will have to orbit faster to stay in orbit.  The closer a planet orbits to a star, the faster it will have to orbit, due to the stronger gravity at tht distance, and the shorter its year will be.</p>
<p>So the closer a planet orbit sits star, and the more massive the star is, the shorter will be the year of the planet.</p>
<p>The shortest orbital period or year of any known exoplanet orbiting a main sequence star is that of K2-137b, I think, whichis only 4.31 hours long.</p>
<p><a href="https://academic.oup.com/mnras/article/474/4/5523/4604789%5B10%5D" rel="nofollow noreferrer">https://academic.oup.com/mnras/article/474/4/5523/4604789[10]</a></p>
<p>So if a moon of K2-137b had a stable orbt it around it, that orbit would proably have to be less than about 0.4788 hours long. Which should be far below the Roche limit of K2-137b and probably also far below the surface of K2-137b.</p>
<p>So it should be possible to create a formula showing how close a planet could get to a star of a specific mass and still retain a moon.</p>
<p>Of course there is no outer limit to how far away a planet with a moon could get from its star.</p>
<p>.</p>

CC BY-SA 4.0
2021-06-02T02:08:48.087Z
I gave the question a +1 but was I really looking for math formulas.
William
CC BY-SA 4.0
2021-06-01T19:48:05.280Z
2021-06-01T19:48:05.280Z
M. A. Golding
<p>There are several factors determining the inner limit to moons. Perhaps the simplest is that it needs to stay inside <a href="https://en.wikipedia.org/wiki/Hill_sphere" rel="nofollow noreferrer">the Hill sphere</a>, the region around the planet where the planet's gravity dominates over the sun's. If the planet's orbit has semi-major axis <span class="math-container">$a$</span> and eccentricity <span class="math-container">$e$</span> the farthest the moon can orbit is <span class="math-container">$$r_H \approx a(1-e)\sqrt[3]{\frac{m}{3M}}$$</span> where <span class="math-container">$m$</span> is the planet mass and <span class="math-container">$M$</span> the sun mass.</p>
<p>The closest a satellite can orbit a planet is the <a href="https://en.wikipedia.org/wiki/Roche_limit" rel="nofollow noreferrer">Roche limit</a>, <span class="math-container">$$r_R = r_m\sqrt[3]{\frac{2m}{m_m}}$$</span> where <span class="math-container">$r_m$</span> and <span class="math-container">$m_m$</span> is the radius and mass of the moon. Equating <span class="math-container">$r_H=r_R$</span> and assuming <span class="math-container">$e=0$</span> to get the minimal <span class="math-container">$a$</span> where a moon is possible gives
<span class="math-container">$$ a_{min}= r_m\sqrt[3]{\frac{6M}{m_m}}.$$</span></p>
<p>For a Moon with <span class="math-container">$r_m=1737.4$</span> km and <span class="math-container">$M/m_m=27090711$</span> (e.g. our moon), this is 0.006 AU (948,179 km), 1.36 solar radii! This is still just barely outside the Roche limit for an Earth-sized planet relative to the sun.</p>
<p>(See <a href="https://academic.oup.com/mnras/article/406/3/1918/978194?login=true" rel="nofollow noreferrer">(Donnison 2010)</a> for a more careful estimation for the full three-body problem applied to moons. <a href="https://academic.oup.com/mnras/article/373/3/1227/1063626" rel="nofollow noreferrer">(Domingos, Winter &amp; Yokoyama 2006)</a> found the rough limits <span class="math-container">$a_{crit}\approx 0.4895(1-1.0305e_{planet}-0.2738e_{sat})r_H$</span> for prograde satellites and about twice this limit for retograde satellites.)</p>
<p>However, while this shows that <em>in principle</em> you can have moons extremely close to stars, in practice they are not going to occur.</p>
<p>The most obvious problem is that very close planets will become tidally locked to the sun, and that will make the moon spiral inward since it will dissipate orbital energy through tidal deformation of the planet. The effect becomes bigger for heavier satellites. <a href="https://iopscience.iop.org/article/10.1086/341477/pdf" rel="nofollow noreferrer">(Barnes &amp; O'Brien 2002)</a> calculate the following allowed region in a 4.6 Gy old system around a 1 Jupiter mass planet:
<a href="https://i.sstatic.net/G3N5e.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/G3N5e.png" alt="Fig 2 from (Barnes &amp; O'Brien 2002)" /></a>
The curve scales as <span class="math-container">$m_m \propto a^{13/2} m^{8/3}r^{-5}$</span>; for an Earth-like primary the corresponding masses have to be 2.7% of the Jupiter case (although the different tidal properties of the planets makes this estimate somewhat dodgy).</p>
<p>There are other destabilizing factors for small bodies close to the sun such as <a href="https://en.wikipedia.org/wiki/Yarkovsky_effect" rel="nofollow noreferrer">the Yarkovsky effect</a>. So while a tiny moon can reside close to the sun, it will likely not remain there long.</p>
<p>The opposite question, if there is an outer limit to planets holding satellites, presumably can be answered negatively. Clearly there are less and less disturbances the further you go outward, and the only issue is if a planet can accrete or capture a satellite. <a href="https://en.wikipedia.org/wiki/Kuiper_belt#Satellites" rel="nofollow noreferrer">Given the common presence of satellites around trans-Neptunian objects</a> this seems fairly common.</p>

CC BY-SA 4.0
2021-06-02T08:35:30.610Z
I'm a little surprised that the size of the moon (its radius) matters (you did set e=0). Can't one assume it's a point particle and come up with a similar estimate for a? Why does $r_m$ matter?
Buck Thorn
CC BY-SA 4.0
2021-06-02T12:57:35.057Z
@BuckThorn The Roche limit formula factors in the relative density of both bodies. Hence you need mass and volume (and volume derives from radius for a roughly spherical body).
Tonny
CC BY-SA 4.0
2021-06-02T13:37:51.853Z
Note that the Barnes & O'Brien 2002 paper assumes that even large moons are still much smaller than the planet, such that the tidal torque exerted on the planet by the moon is negligible compared to solar tides. I'm not sure if (or to what extent) their conclusions can be applied to systems where lunar and solar tides are comparable in magnitude, like the Earth–Moon system. In particular, off the top of my head I cannot say for sure that there couldn't be some critical moon size above which the system would become more stable again.
Ilmari Karonen
CC BY-SA 4.0
2021-06-03T07:37:27.903Z
Your "Domingos, Winter & Yokoyama 2006" link appears to have some (expired) session information in it.
Peter - Reinstate Monica
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2021-06-03T12:35:42.143Z
@Peter-ReinstateMonica - Oops, replaced the link with a stable one.
Anders Sandberg
CC BY-SA 4.0
2021-06-01T20:51:34.060Z
2021-06-03T12:35:15.920Z
2021-06-03T12:35:15.920Z
Anders Sandberg
<p>Mercury and Venus are theorized to have no moons because they are so <a href="https://image.gsfc.nasa.gov/poetry/venus/q328.html" rel="noreferrer">close to the sun</a>.</p>
<p>Is there a theoretical distance in which moons tend to exist based on simulations?</p>

CC BY-SA 4.0
2021-06-01T17:07:02.767Z
You want to take into account the Hill Sphere of the body and check its long-term stability via an nbody simulation
WarpPrime
<natural-satellites><mathematics>
At what distance from the Sun can planetary moons exist?
William
# At what distance from the Sun can planetary moons exist?

(Asked by: username_0 on 2021-06-01)

Mercury and Venus are theorized to have no moons because they are so [close to the sun](https://image.gsfc.nasa.gov/poetry/venus/q328.html).

Is there a theoretical distance in which moons tend to exist based on simulations?

**Tags:** natural-satellites, mathematics

> **Comments:**
> **username_1**: You want to take into account the Hill Sphere of the body and check its long-term stability via an nbody simulation  

---

## 2 answers

### Answer by username_3 on 2021-06-01. Score: 20 | Accepted answer

There are several factors determining the inner limit to moons. Perhaps the simplest is that it needs to stay inside [the Hill sphere](https://en.wikipedia.org/wiki/Hill_sphere), the region around the planet where the planet's gravity dominates over the sun's. If the planet's orbit has semi-major axis $a$ and eccentricity $e$ the farthest the moon can orbit is $$r\_H \approx a(1-e)\sqrt[3]{\frac{m}{3M}}$$ where $m$ is the planet mass and $M$ the sun mass.

The closest a satellite can orbit a planet is the [Roche limit](https://en.wikipedia.org/wiki/Roche_limit), $$r\_R = r\_m\sqrt[3]{\frac{2m}{m\_m}}$$ where $r\_m$ and $m\_m$ is the radius and mass of the moon. Equating $r\_H=r\_R$ and assuming $e=0$ to get the minimal $a$ where a moon is possible gives
$$ a\_{min}= r\_m\sqrt[3]{\frac{6M}{m\_m}}.$$

For a Moon with $r\_m=1737.4$ km and $M/m\_m=27090711$ (e.g. our moon), this is 0.006 AU (948,179 km), 1.36 solar radii! This is still just barely outside the Roche limit for an Earth-sized planet relative to the sun.

(See [(Donnison 2010)](https://academic.oup.com/mnras/article/406/3/1918/978194?login=true) for a more careful estimation for the full three-body problem applied to moons. [(Domingos, Winter & Yokoyama 2006)](https://academic.oup.com/mnras/article/373/3/1227/1063626) found the rough limits $a\_{crit}\approx 0.4895(1-1.0305e\_{planet}-0.2738e\_{sat})r\_H$ for prograde satellites and about twice this limit for retograde satellites.)

However, while this shows that *in principle* you can have moons extremely close to stars, in practice they are not going to occur.

The most obvious problem is that very close planets will become tidally locked to the sun, and that will make the moon spiral inward since it will dissipate orbital energy through tidal deformation of the planet. The effect becomes bigger for heavier satellites. [(Barnes & O'Brien 2002)](https://iopscience.iop.org/article/10.1086/341477/pdf) calculate the following allowed region in a 4.6 Gy old system around a 1 Jupiter mass planet:
[![Fig 2 from (Barnes & O'Brien 2002)](https://i.sstatic.net/G3N5e.png)](https://i.sstatic.net/G3N5e.png)
The curve scales as $m\_m \propto a^{13/2} m^{8/3}r^{-5}$; for an Earth-like primary the corresponding masses have to be 2.7% of the Jupiter case (although the different tidal properties of the planets makes this estimate somewhat dodgy).

There are other destabilizing factors for small bodies close to the sun such as [the Yarkovsky effect](https://en.wikipedia.org/wiki/Yarkovsky_effect). So while a tiny moon can reside close to the sun, it will likely not remain there long.

The opposite question, if there is an outer limit to planets holding satellites, presumably can be answered negatively. Clearly there are less and less disturbances the further you go outward, and the only issue is if a planet can accrete or capture a satellite. [Given the common presence of satellites around trans-Neptunian objects](https://en.wikipedia.org/wiki/Kuiper_belt#Satellites) this seems fairly common.

> **Comments:**
> **username_4**: I'm a little surprised that the size of the moon (its radius) matters (you did set e=0). Can't one assume it's a point particle and come up with a similar estimate for a? Why does $r_m$ matter?  
> **username_5**: @username_4 The Roche limit formula factors in the relative density of both bodies. Hence you need mass and volume (and volume derives from radius for a roughly spherical body).  
> **username_6**: Note that the Barnes & O'Brien 2002 paper assumes that even large moons are still much smaller than the planet, such that the tidal torque exerted on the planet by the moon is negligible compared to solar tides. I'm not sure if (or to what extent) their conclusions can be applied to systems where lunar and solar tides are comparable in magnitude, like the Earth–Moon system. In particular, off the top of my head I cannot say for sure that there couldn't be some critical moon size above which the system would become more stable again.  
> **username_7**: Your "Domingos, Winter & Yokoyama 2006" link appears to have some (expired) session information in it.  
> **username_3**: @username_7-ReinstateMonica - Oops, replaced the link with a stable one.  

---

### Answer by username_2 on 2021-06-01. Score: 4

Short Answer:

There is an inner limit to how clase a palnet could orbit to its star and keep a moonin orbit around it. But I don't know how to calculate it. As far as I know there is no outer limit to how far a planet could be from its star and have moons. A rogue planet in interstellar space far from a star could have moons.

Long Answer:

Every planet whch orbits a star has an inner and outer distance between which any moons would have to orbit.

The inner distance partially depends on the characteristics of the specific moon. It is call the Roche radius or Roche limit. Any object, that is held together mostly by its gravity, like a star, planet, or moon, which passes with the Roche limit of a more massive object will be torn apart by tidal forces. The Roch elimit depends of the relative densiites of the two bodies and whether the smaller object is rigid or fluid.

A moon can not form out of smaller objects if it is within the Roche limit of its planet, and if a moon passes within the Roche limit of its planet it will break up. So there should be no moons within the Roche limits of their planets.

Formulas for calculating the Roche limits of astronomical bodies can be found at:

[https://en.wikipedia.org/wiki/Roche\_limit[1]](https://en.wikipedia.org/wiki/Roche_limit%5B1%5D)

The outer limit at which a moon can orbit a planet is call the Hill radius or Hill sphere. The size of the hill sphere depends on the masses of the planet and its star and the distance between them, since the stronger the gravity of the star is relative to the planet, the smaller the Hill sphere of the planet will be.

There are formulas for calculating an object's Hill Sphere at:

[https://en.wikipedia.org/wiki/Hill\_sphere[2]](https://en.wikipedia.org/wiki/Hill_sphere%5B2%5D)

However, it turns out that a moon orbiting in the outer parts of a Hill sphere will not have an orbit stable for long periods of astronomical time.

> 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. The region of stability for retrograde orbits at a large distance from the primary is larger than the region for prograde orbits at a large distance from the primary. This was thought to explain the preponderance of retrograde moons around Jupiter; however, Saturn has a more even mix of retrograde/prograde moons so the reasons are more complicated.[3](https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability)

[https://en.wikipedia.org/wiki/Hill\_sphere#True\_region\_of\_stability[3]](https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability%5B3%5D)

The Hill sphere of Earth extends to about 1,500,000 kilometers, so the region of truely stable orbits around Earth extends only to about 500,000 to 750,000 kilometers from Earth.

The Roche limit for Earth is 1.49 radii for rigid objects and 2.88 radii for fluid objects. Since the radius of Earth is 6,371 kilometers, the Roche limit for rigid objects is 9,492.79 kilometers and for fluid objects is 18,348.48 Kilometers.

The larger and more massive a planet is, the greater the radii of its Roche limit and Hill sphere will be. The farther a planet is from its star, the greter the radii of its Hill sphere will be.

Mercury and Venus are less massive than the Earth, and so their Roche limits are smaller, which is good for not destroying close moons.. They are closer to the Sun than the Earth, so the Sun's gravity is stronger where they orbit, and since they are less massive than Earth, their mases and distances from the Sun combine to make their HIll spheres much smaller than Earth's, which is bad for keeping moons.

The Hill sphere of Mercury has a radius of only 175,300 kilometers, so the true region of stability should have an outer edge at only 58,432 to 87,650 kilometers.

The Hill Spere of Venus is 1,004,200 kilometers, so the true region of stability should have an outer edge at only 334,733.3 to 502,100 kilometers, much larger than Mercury's but smaller than Earth's.

But there ae other problems with Mercury and Venus having moons. The orbit of a moona round tis planet will chang eover time, as the moon moves closer to or farther from the planet. So if a moon starts out between the Roche limit and the Hill sphere of its planet, it could move out of the zone where stable orbits are possible.

If a planet and its moon form together, the moon will have a prograde orbit. Most of the objects in the Solar System rotate in the same direction that they orbit the sun in, whch is called a prograde orbit. If a moon forms with its planet, what is called a regular moon, it will orbit the planet in the same direction as the planet rotates, which will be prograde relative to the planet. Since most planets rotate in a prograde direction relative to their orbit, most of the regular moons formed with their planets orbit in a prograde direction relative to the plant's rbit around the Sun.

Planets can also capture passing objects and make them their moons. The capture process can result in either a prograde orbit, or an orbit in the opposite direction, a retrograde orbit. Therer are a number of captured moons in the Solar System, some with prograd eorbits and some with retrograde orbits.

> All retrograde satellites experience tidal deceleration to some degree. The only satellite in the Solar System for which this effect is non-negligible is Neptune's moon Triton. All the other retrograde satellites are on distant orbits and tidal forces between them and the planet are negligible.

[https://en.wikipedia.org/wiki/Retrograde\_and\_prograde\_motion#Natural\_satellites\_and\_rings[4]](https://en.wikipedia.org/wiki/Retrograde_and_prograde_motion#Natural_satellites_and_rings%5B4%5D)

> Triton's revolution around Neptune has become a nearly perfect circle with an eccentricity of almost zero. Viscoelastic damping from tides alone is not thought to be capable of circularizing Triton's orbit in the time since the origin of the system, and gas drag from a prograde debris disc is likely to have played a substantial role.[4](https://en.wikipedia.org/wiki/Triton_(moon)#Orbit_and_rotation) Tidal interactions also cause Triton's orbit, which is already closer to Neptune than the Moon's is to Earth, to gradually decay further; predictions are that 3.6 billion years from now, Triton will pass within Neptune's Roche limit.[25] This will result in either a collision with Neptune's atmosphere or the breakup of Triton, forming a new ring system similar to that found around Saturn.[25]

[https://en.wikipedia.org/wiki/Triton\_(moon)#Orbit\_and\_rotation[5]](https://en.wikipedia.org/wiki/Triton_(moon)#Orbit_and_rotation%5B5%5D)

So if Mercury or Venus captured a moon in an orbit retrograde to their rotation, that moon would gradually spiral down to within the Roche limit and be destroyed after millions or billions of years.

Moons in prograde orbits around their planets will either move away from the planets or move in toward their planets.

If a moon is in a prograde orbit above the synchronous orbit level, it will move away from the planet due to tidal acceleration. Thus, after millions or billions or psosibly trillions of years, it might pass outside the Hill sphere of the planet and go into orbit around their star. MOs tof the moons is the Solar System orbit farther than the synchronous orbits of their planets, and so a gradually receeding form them.

[https://en.wikipedia.org/wiki/Tidal\_acceleration#Other\_cases\_of\_tidal\_acceleration[6]](https://en.wikipedia.org/wiki/Tidal_acceleration#Other_cases_of_tidal_acceleration%5B6%5D)

Some moons orbit their planets below the synchronous orbit level. Their orbital periods are less than one day of the planet. Those moons experience tidal deceleration and slowly spiral inward toward their planets, eventually reaching the Roche limits of their planets and breaking up.

[https://en.wikipedia.org/wiki/Tidal\_acceleration#Tidal\_deceleration[7]](https://en.wikipedia.org/wiki/Tidal_acceleration#Tidal_deceleration%5B7%5D)

The sidereal days, the period it takes them to rotate 360 degrees with respect to the disant stars, of Mercury and Venus are 58.646 Earth days and 243.0226 Earth days respectively. So any moons orbiting Mercury and Venus at the synchrnous orbits would be orbiting very far from their planets.

> Mercury and Venus are believed to have no satellites chiefly because any hypothetical satellite would have suffered deceleration long ago and crashed into the planets due to the very slow rotation speeds of both planets; in addition, Venus also has retrograde rotation.

[https://en.wikipedia.org/wiki/Tidal\_acceleration#Tidal\_deceleration[7]](https://en.wikipedia.org/wiki/Tidal_acceleration#Tidal_deceleration%5B7%5D)

For a long time astronomers believed that Mercury, and possibly Venus, were tidally locked to the Sun, with rotation periods equal to their oribital periods, so that a day would be equal toa year, and one side of theplanet would always face the Sun and the other side always face away from the Sun. That is called a 1:1 resonance. It is now known that Mercury and Venus do not have 1:1 resonances, but their rotation periods are very long compared to their years of 87.97 Earth days and 224.7 Earth days.

When the gravity of a star is too strong at the orbit of a planet - which depends on the mas sof hte star and the orbital distance of the planet from the star - that planet will become tidally locked to the Star, either with a 1:1 resonance or a resonance with simple integers, such as the 3:2 resonance of the planet Mercury - Mercury rotates three times in two Mercurian years - or some other simple resonance.

So if a planet is too deep in the gravity well of its star, the planet's rotation will be slowed down a lot, and the synchronous orbit of the planet will be very far from the planet, possbily beyond the Hill sphere. So a moon outside the Hill spehre would be lost into space, and a moon below the synchronous orbit would have a decaying orbit and eventually reach the roche limit of the planet.

So if the mass of a star is known, the distance at which a planet would have its day greatly slowed down, thus making it hard or impossible for it to keep its moon(s) can be calculated.

And of course the shorter years of planets orbiting close to their stars is another problem for any potential moons.

> The longest possible length of a satellite’s day compatible with Hill stability has been shown to be about P∗p/9, P∗p being the planet’s orbital period about the star (Kipping 2009a)

[https://arxiv.org/ftp/arxiv/papers/1209/1209.5323.pdf[8]](https://arxiv.org/ftp/arxiv/papers/1209/1209.5323.pdf%5B8%5D)

And the source is given as:

[https://academic.oup.com/mnras/article/392/1/181/1071655[9]](https://academic.oup.com/mnras/article/392/1/181/1071655%5B9%5D)

So this claims that a mooon can not have a stable orbit around its planets unless the orbital of the moon is less than one nineth (0.111111) theorbital period of the planet around the star.

The greater the mass of a star, the shorter the year of a lanet oriting it at a specific distance will be, since the increased gravity of the star means the planet will have to orbit faster to stay in orbit. The closer a planet orbits to a star, the faster it will have to orbit, due to the stronger gravity at tht distance, and the shorter its year will be.

So the closer a planet orbit sits star, and the more massive the star is, the shorter will be the year of the planet.

The shortest orbital period or year of any known exoplanet orbiting a main sequence star is that of K2-137b, I think, whichis only 4.31 hours long.

[https://academic.oup.com/mnras/article/474/4/5523/4604789[10]](https://academic.oup.com/mnras/article/474/4/5523/4604789%5B10%5D)

So if a moon of K2-137b had a stable orbt it around it, that orbit would proably have to be less than about 0.4788 hours long. Which should be far below the Roche limit of K2-137b and probably also far below the surface of K2-137b.

So it should be possible to create a formula showing how close a planet could get to a star of a specific mass and still retain a moon.

Of course there is no outer limit to how far away a planet with a moon could get from its star.

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> **Comments:**
> **username_0**: I gave the question a +1 but was I really looking for math formulas.  

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