<p>You can think about this in terms of energy conservation.   The orbiting moon has a certain amount of kinetic and potential energy.  The spinning planet has a certain amount of kinetic energy in its rotation.  The kinetic energy of the rotation will be transferred to the kinetic and potential energy of the moon, until such time as the orbit period of the moon equals the rotation time of the planet.  This distance depends on the masses and rate of rotation of the bodies, It doesn't depend on the mass and distance to the star.  (Here I've adopted a simple model in which the tides don't dissipate energy. That they do dissipate energy changes the values of the numerical calculation, but not the reasoning that the limiting distance isn't equal to the Hill sphere)</p>
<p>The Hill sphere depends on the mass of the planet, star and their distance. It doesn't depend on the rotation rate of the planet.</p>
<p>So you see that if there was no star, the moon would move further away until no more energy was being transferred.  This distance might be larger than the radius of the Hill sphere.  If that's the case then tidal forces will push the moon out of orbit.</p>
<p>There's nothing special about the Hill sphere for the matter of tides.  Yes, tidal forces decrease rapidly with distance, but while the planet is spinning faster than the orbit of the moon, the moon will raise tides on the planet, the rotation of the planet will pull these tides ahead of the moon, and the moon will be accelerated, causing the moon to move further.  There's no reason why the maximum distance the moon will be pushed to must be equal to or less than the Hill sphere's radius.</p>
<p>For the Earth-Moon system, Wikipedia gives an estimate that the maximum distance the moon could reach is when it has an orbital period of 47 days, corresponding to an orbital distance of 610000km, which is still well inside the Hill Sphere relative to the sun (1.5 million km).  But by the time the moon reaches that distance, the sun will be long gone.</p>

CC BY-SA 4.0
2023-11-19T13:29:59.730Z
The Hill Sphere also depends on the mass and size of the third gravitational body (e.g., the Sun). If there was no star (or other somewhat nearby third gravitational body), how could there be a Hill Sphere?
David Hammen
CC BY-SA 4.0
2023-11-19T16:29:04.657Z
I say that "The Hill sphere depends on the mass of the planet, star and their distance"
James K
CC BY-SA 4.0
2023-11-19T10:44:38.187Z
2023-11-19T10:49:53.510Z
2023-11-19T10:49:53.510Z
James K
<p>While I agree with everything said in <a href="https://astronomy.stackexchange.com/a/55343/30705">the answer by James K</a>, there is an important detail to be added to this story.</p>
<p>Numerical analysis by <a href="https://www.nature.com/articles/nature01622" rel="nofollow noreferrer">Astakhov et al. (2003)</a> and <a href="https://academic.oup.com/mnras/article/373/3/1227/1063626" rel="nofollow noreferrer">Domingos et al. (2006)</a> indicated that the orbits too closely approaching the Hill radius become unstable in the long term, while those confined to a smaller domain, the <em>reduced Hill sphere</em>, remain stable. Therefore, when analysing if a tidally receding moon can reach a <em>stable</em> synchronous configuration, we should check if synchronism is attained within the reduced Hill sphere, not just within the ordinary Hill sphere.</p>
<p>The radius of the reduced Hill sphere (`the reduced Hill radius') for a circular (<span class="math-container">$e_m=0$</span>) orbit of the moon is
<span class="math-container">$$
   r^{\,'}_H=B\,a_p\,(1-b\,e_p)\,\left(\frac{M_p}{3\,M_{\ast}} \right)^{1/3}
 $$</span>
and depends on the direction of the orbit:
<span class="math-container">$$
   B\;=\;0.49\;\,,\;\;b\,=1.03\;\,,\quad\mbox{for prograde-orbiting moons}\,;\;
$$</span>
<span class="math-container">$$
   B\;=\;0.93\;,\;\;b\,=1.08\;\,,\quad\mbox{for retrograde-orbiting moons}\,.
$$</span>
Here <span class="math-container">$a_p$</span> and <span class="math-container">$e_p$</span> and are the planet's semimajor axis and eccentricity, while <span class="math-container">$M_p$</span> and <span class="math-container">$M_\ast$</span> are the planetary and stellar masses.</p>
<p><a href="https://astronomy.stackexchange.com/a/50836/30705">This calculation</a> demonstrates that, for the Moon, synchronism is established <em>barely</em> below the reduced Hill radius.</p>

CC BY-SA 4.0
2024-05-12T01:35:24.250Z
2024-05-12T01:35:24.250Z
Michael_1812
<blockquote>
<p>Would the planet's tidal forces seriously cause its moon's orbit to expand so much that it's outside its Hill Sphere? That seems counterintuitive to me. Tidal forces drop by the distance cubed, and as the moon got farther away, the tidal forces that caused it to drift away in the first place would get exponentially weaker.</p>
</blockquote>
<p>The two concepts, Hill Sphere versus tidal acceleration, have nothing to do with one another. The Hill Sphere is an approximation of where a very small body can stably orbit a not quite so small body in the presence of a much more massive third body. For example, the Earth orbits the Sun, and Moon orbits the Earth (but with significant perturbations from the Sun). The Hill Sphere is concerned with the radial gravitational acceleration of the smallest body toward the intermediate sized body versus that of the radial gravitational acceleration of the smallest body toward the larger but more remote third body.</p>
<p>Tidal acceleration addresses the fact that the gravitational acceleration toward the central body (e.g., the Moon toward the Earth) is not exactly radial. The Moon raises tides on the Earth, both in the oceans and the in Earth as a whole. These tides more or less rotate with the Earth. Because the Earth is rotating, the tidal gravitational influences on the Moon on average lead the Moon's orbital position. This results in a secular transverse (non-radial) component of the Moon toward the Earth. Note well: This transverse acceleration has nothing to do with the Sun.</p>

CC BY-SA 4.0
2023-11-19T18:39:20.220Z
Okay..... no one claimed they're the same. But the reason a moon might drift outside its planet's Hill Sphere is because of the tidal forces at play, which get exponentially weaker as the moon gets farther away.
user267545
CC BY-SA 4.0
2023-11-21T05:46:31.640Z
@user267545 While the Earth's tidal influence on the Moon does drop as roughly the inverse of the cube of distance, that's not an exponential decay. That said, the Moon's initial recession rate was initially rather quick due to extremely strong tides; the Moon formed at about six Earth radii from the center of the Earth. As the Moon receded,  the recession rate stabilized to a lower rate, but then it rose! The current rate is anomalously high (much higher than the average over the last 650 million years) due to resonances between the Moon and the oceans, particularly the North Atlantic.
David Hammen
CC BY-SA 4.0
2023-11-19T15:47:39.493Z
2023-11-19T15:47:39.493Z
David Hammen
<p>If a moon orbits in the same direction as its planet's rotation (i.e. a prograde orbit), and its orbital period is longer than its planet's rotational period, then tidal forces would cause the moon to slowly drift away. This is true for the vast majority of moons in the Solar System, and for every large round moon bar Triton.</p>
<p>The moon's orbit would continue expanding, and the planet's rotation would continue slowing down, until both bodies are tidally locked to each other. This is what's expected to happen to the Earth-Moon system if it is not engulfed by the Sun as a red giant.</p>
<p>However, what if the moon just can't tidally lock its primary? Whether that's because the planet is hundreds of thousands of times more massive than the moon, or because its rotation is too fast to be slowed down in time.</p>
<p>Would the planet's tidal forces seriously cause its moon's orbit to expand so much that it's outside its Hill Sphere? That seems counterintuitive to me. Tidal forces drop by the distance <em>cubed</em>, and as the moon got farther away, the tidal forces that caused it to drift away in the first place would get exponentially weaker.</p>
<p>I've heard the titular claim multiple times, but to me it seems more likely that a moon would asymptotically approach a distance within the Hill Sphere that it can't be pushed farther out of, as the tidal forces would become negligible.</p>

<orbital-mechanics><tidal-forces><hill-sphere>
Would a planet's tidal forces really push away a moon out of its Hill Sphere?
user267545
# Would a planet's tidal forces really push away a moon out of its Hill Sphere?

(Asked by: username_0 on 2023-11-19)

If a moon orbits in the same direction as its planet's rotation (i.e. a prograde orbit), and its orbital period is longer than its planet's rotational period, then tidal forces would cause the moon to slowly drift away. This is true for the vast majority of moons in the Solar System, and for every large round moon bar Triton.

The moon's orbit would continue expanding, and the planet's rotation would continue slowing down, until both bodies are tidally locked to each other. This is what's expected to happen to the Earth-Moon system if it is not engulfed by the Sun as a red giant.

However, what if the moon just can't tidally lock its primary? Whether that's because the planet is hundreds of thousands of times more massive than the moon, or because its rotation is too fast to be slowed down in time.

Would the planet's tidal forces seriously cause its moon's orbit to expand so much that it's outside its Hill Sphere? That seems counterintuitive to me. Tidal forces drop by the distance *cubed*, and as the moon got farther away, the tidal forces that caused it to drift away in the first place would get exponentially weaker.

I've heard the titular claim multiple times, but to me it seems more likely that a moon would asymptotically approach a distance within the Hill Sphere that it can't be pushed farther out of, as the tidal forces would become negligible.

**Tags:** orbital-mechanics, tidal-forces, hill-sphere

---

## 3 answers

### Answer by username_1 on 2023-11-19. Score: 3 | Accepted answer

You can think about this in terms of energy conservation. The orbiting moon has a certain amount of kinetic and potential energy. The spinning planet has a certain amount of kinetic energy in its rotation. The kinetic energy of the rotation will be transferred to the kinetic and potential energy of the moon, until such time as the orbit period of the moon equals the rotation time of the planet. This distance depends on the masses and rate of rotation of the bodies, It doesn't depend on the mass and distance to the star. (Here I've adopted a simple model in which the tides don't dissipate energy. That they do dissipate energy changes the values of the numerical calculation, but not the reasoning that the limiting distance isn't equal to the Hill sphere)

The Hill sphere depends on the mass of the planet, star and their distance. It doesn't depend on the rotation rate of the planet.

So you see that if there was no star, the moon would move further away until no more energy was being transferred. This distance might be larger than the radius of the Hill sphere. If that's the case then tidal forces will push the moon out of orbit.

There's nothing special about the Hill sphere for the matter of tides. Yes, tidal forces decrease rapidly with distance, but while the planet is spinning faster than the orbit of the moon, the moon will raise tides on the planet, the rotation of the planet will pull these tides ahead of the moon, and the moon will be accelerated, causing the moon to move further. There's no reason why the maximum distance the moon will be pushed to must be equal to or less than the Hill sphere's radius.

For the Earth-Moon system, Wikipedia gives an estimate that the maximum distance the moon could reach is when it has an orbital period of 47 days, corresponding to an orbital distance of 610000km, which is still well inside the Hill Sphere relative to the sun (1.5 million km). But by the time the moon reaches that distance, the sun will be long gone.

> **Comments:**
> **username_2**: The Hill Sphere also depends on the mass and size of the third gravitational body (e.g., the Sun). If there was no star (or other somewhat nearby third gravitational body), how could there be a Hill Sphere?  
> **username_1**: I say that "The Hill sphere depends on the mass of the planet, star and their distance"  

---

### Answer by username_3 on 2024-05-12. Score: 0

While I agree with everything said in [the answer by James K](https://astronomy.stackexchange.com/a/55343/30705), there is an important detail to be added to this story.

Numerical analysis by [Astakhov et al. (2003)](https://www.nature.com/articles/nature01622) and [Domingos et al. (2006)](https://academic.oup.com/mnras/article/373/3/1227/1063626) indicated that the orbits too closely approaching the Hill radius become unstable in the long term, while those confined to a smaller domain, the *reduced Hill sphere*, remain stable. Therefore, when analysing if a tidally receding moon can reach a *stable* synchronous configuration, we should check if synchronism is attained within the reduced Hill sphere, not just within the ordinary Hill sphere.

The radius of the reduced Hill sphere (`the reduced Hill radius') for a circular ($e\_m=0$) orbit of the moon is
$$
r^{\,'}\_H=B\,a\_p\,(1-b\,e\_p)\,\left(\frac{M\_p}{3\,M\_{\ast}} \right)^{1/3}
$$
and depends on the direction of the orbit:
$$
B\;=\;0.49\;\,,\;\;b\,=1.03\;\,,\quad\mbox{for prograde-orbiting moons}\,;\;
$$
$$
B\;=\;0.93\;,\;\;b\,=1.08\;\,,\quad\mbox{for retrograde-orbiting moons}\,.
$$
Here $a\_p$ and $e\_p$ and are the planet's semimajor axis and eccentricity, while $M\_p$ and $M\_\ast$ are the planetary and stellar masses.

[This calculation](https://astronomy.stackexchange.com/a/50836/30705) demonstrates that, for the Moon, synchronism is established *barely* below the reduced Hill radius.


---

### Answer by username_2 on 2023-11-19. Score: 0

> Would the planet's tidal forces seriously cause its moon's orbit to expand so much that it's outside its Hill Sphere? That seems counterintuitive to me. Tidal forces drop by the distance cubed, and as the moon got farther away, the tidal forces that caused it to drift away in the first place would get exponentially weaker.

The two concepts, Hill Sphere versus tidal acceleration, have nothing to do with one another. The Hill Sphere is an approximation of where a very small body can stably orbit a not quite so small body in the presence of a much more massive third body. For example, the Earth orbits the Sun, and Moon orbits the Earth (but with significant perturbations from the Sun). The Hill Sphere is concerned with the radial gravitational acceleration of the smallest body toward the intermediate sized body versus that of the radial gravitational acceleration of the smallest body toward the larger but more remote third body.

Tidal acceleration addresses the fact that the gravitational acceleration toward the central body (e.g., the Moon toward the Earth) is not exactly radial. The Moon raises tides on the Earth, both in the oceans and the in Earth as a whole. These tides more or less rotate with the Earth. Because the Earth is rotating, the tidal gravitational influences on the Moon on average lead the Moon's orbital position. This results in a secular transverse (non-radial) component of the Moon toward the Earth. Note well: This transverse acceleration has nothing to do with the Sun.

> **Comments:**
> **username_0**: Okay..... no one claimed they're the same. But the reason a moon might drift outside its planet's Hill Sphere is because of the tidal forces at play, which get exponentially weaker as the moon gets farther away.  
> **username_2**: @user267545 While the Earth's tidal influence on the Moon does drop as roughly the inverse of the cube of distance, that's not an exponential decay. That said, the Moon's initial recession rate was initially rather quick due to extremely strong tides; the Moon formed at about six Earth radii from the center of the Earth. As the Moon receded,  the recession rate stabilized to a lower rate, but then it rose! The current rate is anomalously high (much higher than the average over the last 650 million years) due to resonances between the Moon and the oceans, particularly the North Atlantic.  

---

