<p>A belated answer, but neither of the existing answers properly explain this.</p>

<p>The proper explanation is simple. In Newtonian mechanics, tidal influences make all objects in retrograde orbits and those objects in prograde orbits below the equivalent of geosynchronous radius spiral inward. Only objects orbiting prograde above the equivalent of geosynchronous radius spiral outward. Our Moon orbits at about 385000 km, well above geosynchronous radius of 42164 km. Phobos orbits Mars with a semi-major axis of 9377 km, well below the areosynchronous radius of 20400 km. While our Moon is spiraling outward from Earth, Phobos is spiraling inward toward Mars.</p>

<p>Mutually tidally locked objects spiral neither inward nor outward in Newtonian mechanics. General relativity modifies this dynamic a bit, making mutually tidally locked objects spiral inward. </p>

CC BY-SA 3.0
2017-07-19T17:43:25.830Z
2017-07-19T17:43:25.830Z
David Hammen
<p>Other answers are right at explaining why tidal forces move Earth and Moon apart but they don't move apart a pair of black holes. However, I think it is also needed to explain why the phenomena making two black holes spiral inward don't make the Moon spiral inward to the Earth.</p>
<p>In fact, every pair of rotating masses radiate gravitational waves. What makes the difference is that only very large masses rotating very close to each other produce gravitational waves large enough to meaningfully affect those masses' orbits.</p>
<p>According to <a href="https://en.wikipedia.org/wiki/Gravitational_wave#Binaries" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Gravitational_wave#Binaries</a> the time that takes a pair of masses to fall into each other due to radiated gravitational waves is:</p>
<p><span class="math-container">$$t= \frac{5}{256}\, \frac{c^5}{G^3}\, \frac{r^4}{(m_1m_2)(m_1+m_2)}$$</span></p>
<p>Let's plug the masses of Earth and Moon and its distance into that equation (all data taken from Wikipedia in SI units):</p>
<pre><code>&gt; G &lt;- 6.674e-11
&gt; r &lt;- 384e6
&gt; mluna &lt;- 7.342e22
&gt; c &lt;- 299792458
&gt; mterra &lt;- 5.97237e24
&gt; (t &lt;- 5/256*c^5/G^3*r^4/(mterra*mluna)/(mterra+mluna))
[1] 1.304925e+33
</code></pre>
<p>That is, left alone, radiating gravitational waves would make the Moon crash into the Earth in 1.3<span class="math-container">$\times$</span>10<sup>33</sup> seconds, that is 4.13<span class="math-container">$\times$</span>10<sup>25</sup> years or 3<span class="math-container">$\times$</span>10<sup>15</sup> times the current age of the universe. In other words, the effect of radiating gravitational waves in the motion of Earth and Moon is so tiny — specially compared with other forces like tide ones — that we can't absolutely forget about it.</p>
<p>Just for comparison, two one solar mass neutron stars orbiting each other at the same distance of the Earth and the Moon would fall into each other in:</p>
<pre><code>&gt; msol &lt;- 1.9885e30
&gt; (t &lt;- 5/256*c^5/G^3*r^4/(msol*msol)/(msol+msol))
[1] 2.19985e+14
</code></pre>
<p>Which is just about 7 million years, showing that changing masses have a large effect on the outcome. As stated in the beginning, gravitational waves make pairs of star-sized objects spiral inwards, but they don't have noticeable effects on a satellite orbiting a planet.</p>

CC BY-SA 4.0
2019-03-22T17:01:05.367Z
Excellent answer with real numbers. Wish I could accept your answer if you'd replied earlier!
Gnubie
CC BY-SA 4.0
2019-03-24T13:30:06.367Z
@Gnubie This [Meta FAQ](https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work) says that "You may change which answer is accepted, or simply un-accept the answer, at any time." :-)
Chappo Hasn't Forgotten
CC BY-SA 4.0
2019-03-24T13:38:40.493Z
Thank you, for your appreciation, but I think a this question is in fact made of two questions. Other answers are right at answering one of them. Mine answers the other one. Therefore, the accepted answer isn't wrong. It's just one half of the answer and mine is the other half.
Pere
CC BY-SA 4.0
2019-03-21T14:55:40.150Z
2024-09-01T18:40:45.857Z
2024-09-01T18:40:45.857Z
Pere
<p>The recession rate of the Moon from the Earth is given as 38.04 mm per year, due to tidal forces, according to Wikipedia.  A good description (with diagrams) about how this occurs can be found <a href="http://www.themcdonalds.net/richard/astro/papers/602-tides-web.pdf" rel="nofollow noreferrer">here</a> (now sadly a <em>dead link</em>).</p>
<p>The orbital decay rate due to gravitational radiation can be determined by <span class="math-container">$$\frac{\mathrm{d}r}{\mathrm{d}t} = - \frac{64}{5}\, \frac{G^3}{c^5}\, \frac{(m_1m_2)(m_1+m_2)}{r^3}$$</span><a href="https://en.wikipedia.org/wiki/Gravitational_wave#Orbital_decay_from_gravitational_radiation" rel="nofollow noreferrer">Wikpedia:Gravitational wave</a></p>
<p>This works out to approximately 2nm per year, or 7 orders of magnitude smaller.  So, yes, both tidal recession and orbital decay via gravitational waves are expected to be happening, but tidal recession is a far larger effect.</p>

CC BY-SA 3.0
2016-02-13T14:35:30.217Z
But doesn't this contradict the fact from O. P.  that two mutually rotating black holes tend to come closer?
Astroynamicist
CC BY-SA 3.0
2016-02-13T21:46:17.350Z
Exactly! Why don't inspiraling black holes and neutron stars suffer from tidal forces too?
Gnubie
CC BY-SA 3.0
2018-03-05T16:38:56.403Z
I prefer this answer because it compares the magnitude of each effect instead of trying to explain either one. In other words, all orbiting pairs have the potential for both tidal forces and gravitation waves but the OP wants to know why one system is influenced by one effect over the other and you do that quite well here.
Kelly S. French
CC BY-SA 4.0
2016-02-13T13:30:48.590Z
2023-08-08T00:42:36.500Z
2023-08-08T00:42:36.500Z
user10789
<p>Here is how the tides move the moon away from the Earth:</p>
<p>The moon orbits the earth, and there is a difference in gravitational force between the the side of the Earth nearest the moon, and the side far from the moon.</p>
<p>This difference in force tends to pull the Earth into a oval shape with its long axis pointing towards the moon.</p>
<p>But the Earth is also spinning in the same direction as the moon orbits the Earth, and this spinning moves the axis of the oval forward, so the oval doesn't point towards the moon, but a little ahead of it. So there is a bump on the Earth, and it is permanently a little in front of the moon. This bump has mass and it pulls the moon towards it, so the moon is being pulled forwards. The moon pulls the Earth back. So the Earth's spin is slowed, but the moon gains energy and moves a little further from the Earth.</p>
<p>The moon has also been slowed down to the extent that the same side always faces Earth, and it will remain locked like this.</p>
<p>Gravitational waves, on the other hand, imply an emission of energy from the binary system, and as energy is lost, the black holes spiral in.</p>
<p>There are no tides on black holes because there is nothing there to be pulled into an oval shape. The event horizon is not a solid surface. The mass of a black hole is concentrated entirely at the singularity, there is no structure to be deformed into a bump.</p>

CC BY-SA 3.0
2016-02-14T21:49:30.740Z
There ARE tides on the black holes!
Anixx
CC BY-SA 3.0
2016-02-15T00:39:04.693Z
Black holes can produce tides on other objects (very power ones, rip you apart type tides) but a black hole has no matter to be tidally distorted. It has mass, but no matter.
James K
CC BY-SA 4.0
2024-09-03T11:14:34.133Z
The Earth-Moon system is also emitting gravitational waves. Because its mass is small, and its spacing wide compared to a binary black hole, the loss of energy is tiny, compared to a binary black hole. So the Moon continues to recede from the Earth, but this is very slightly slowed by gravitational radiation.
John Dallman
CC BY-SA 4.0
2024-09-04T09:12:18.603Z
Why does the moon gaining energy mean it moves away? Why does it not become warmer instead?
Parrotmaster
CC BY-SA 4.0
2024-09-04T21:44:42.120Z
It gains gravitational potential energy, not heat.  There is a mechanical process that pulls it forwards and outwards, not a thermodynamic process of heating.
James K
CC BY-SA 4.0
2024-11-20T02:57:35.120Z
@JohnDallman The perturbation with the least effect would be relativistic frame dragging from the rotating earth.
Miss Understands
CC BY-SA 4.0
2016-02-13T22:20:12.003Z
2024-09-01T05:02:42.487Z
2024-09-01T05:02:42.487Z
James K
<p>According to the accepted answer on <a href="https://astronomy.stackexchange.com/questions/131/is-the-moon-moving-further-away-from-earth-and-closer-to-the-sun-why?newreg=c97ffe82214b40f5bc69c703df67d44b">Is the moon moving further away from Earth and closer to the Sun? Why?</a>, the moon is receding from Earth because tidal forces and friction cause energy to be lost.</p>

<p>However, according to <a href="http://ligo.org/science/GW-Inspiral.php" rel="noreferrer">LIGO's website</a>,</p>

<blockquote>
  <p>As the two masses rotate around each other, their orbital distances
  decrease</p>
</blockquote>

<p>as they lose energy radiating <em>inspiral gravitational waves</em>.</p>

<p>Why do the bodies move apart in the Earth-moon case but closer together in the black hole case?</p>

<p>If both opposing phenomena are present but a different one is stronger in the two cases, what determines the fate of a system?</p>

CC BY-SA 3.0
2016-02-12T23:34:32.243Z
You may have misunderstood the answer to the linked question a bit.  The answer tells you that the tidal forces and friction transfers energy from the Earth (from the angular momentum from its rotation about its axis) and imparts it to the Moon (in the form of orbital angular momentum).  Energy isn't lost (pretty much ever).  It just gets converted and transported around.   The gravity waves take orbital angular momenta out of the binary system, which will cause them to move into a tighter orbit, though it is still reasonable to ask why the same Earth-Moon conversion doesn't compensate/apply.
zibadawa timmy
CC BY-SA 3.0
2016-02-13T00:07:00.683Z
@zibadawatimmy should be an answer
ProfRob
CC BY-SA 3.0
2016-02-13T01:18:22.803Z
@RobJeffries There's this technicality in my mind which I can't resolve that makes me not want to actually claim to have an answer: can there be enough rotational momentum from a spinning black hole, and can it be converted into orbital angular momentum (for the other black hole), at a rate fast enough to actually unbind the system (despite the gravity waves)?  I get that there's a finite amount of rotational momentum, so if the system remains bound eventually the gravity waves win out and an inspiral begins...but can it get unbound?  I can't figure out if that's an easy or hard question.
zibadawa timmy
CC BY-SA 3.0
2016-02-15T23:31:18.507Z
Got it! The reason the moon is receding is that it's revolving in the same direction that Earth rotates, and the Earth is not tidally locked to the moon. In the Neptune-Triton system, Triton has a retrograde orbit (unique among large moons) and so gets closer to Neptune instead -- from http://universetoday.com/56042/triton
Gnubie
<the-moon><gravity><black-hole><tidal-forces><gravitational-waves>
Why do Earth and moon move apart but binary black holes move closer?
Gnubie
# Why do Earth and moon move apart but binary black holes move closer?

(Asked by: username_0 on 2016-02-12)

According to the accepted answer on [Is the moon moving further away from Earth and closer to the Sun? Why?](https://astronomy.stackexchange.com/questions/131/is-the-moon-moving-further-away-from-earth-and-closer-to-the-sun-why?newreg=c97ffe82214b40f5bc69c703df67d44b), the moon is receding from Earth because tidal forces and friction cause energy to be lost.

However, according to [LIGO's website](http://ligo.org/science/GW-Inspiral.php),

> As the two masses rotate around each other, their orbital distances
> decrease

as they lose energy radiating *inspiral gravitational waves*.

Why do the bodies move apart in the Earth-moon case but closer together in the black hole case?

If both opposing phenomena are present but a different one is stronger in the two cases, what determines the fate of a system?

**Tags:** the-moon, gravity, black-hole, tidal-forces, gravitational-waves

> **Comments:**
> **username_1**: You may have misunderstood the answer to the linked question a bit.  The answer tells you that the tidal forces and friction transfers energy from the Earth (from the angular momentum from its rotation about its axis) and imparts it to the Moon (in the form of orbital angular momentum).  Energy isn't lost (pretty much ever).  It just gets converted and transported around.   The gravity waves take orbital angular momenta out of the binary system, which will cause them to move into a tighter orbit, though it is still reasonable to ask why the same Earth-Moon conversion doesn't compensate/apply.  
> **username_2**: @username_1 should be an answer  
> **username_1**: @RobJeffries There's this technicality in my mind which I can't resolve that makes me not want to actually claim to have an answer: can there be enough rotational momentum from a spinning black hole, and can it be converted into orbital angular momentum (for the other black hole), at a rate fast enough to actually unbind the system (despite the gravity waves)?  I get that there's a finite amount of rotational momentum, so if the system remains bound eventually the gravity waves win out and an inspiral begins...but can it get unbound?  I can't figure out if that's an easy or hard question.  
> **username_0**: Got it! The reason the moon is receding is that it's revolving in the same direction that Earth rotates, and the Earth is not tidally locked to the moon. In the Neptune-Triton system, Triton has a retrograde orbit (unique among large moons) and so gets closer to Neptune instead -- from http://universetoday.com/56042/triton  

---

## 4 answers

### Answer by username_9 on 2016-02-13. Score: 23 | Accepted answer

Here is how the tides move the moon away from the Earth:

The moon orbits the earth, and there is a difference in gravitational force between the the side of the Earth nearest the moon, and the side far from the moon.

This difference in force tends to pull the Earth into a oval shape with its long axis pointing towards the moon.

But the Earth is also spinning in the same direction as the moon orbits the Earth, and this spinning moves the axis of the oval forward, so the oval doesn't point towards the moon, but a little ahead of it. So there is a bump on the Earth, and it is permanently a little in front of the moon. This bump has mass and it pulls the moon towards it, so the moon is being pulled forwards. The moon pulls the Earth back. So the Earth's spin is slowed, but the moon gains energy and moves a little further from the Earth.

The moon has also been slowed down to the extent that the same side always faces Earth, and it will remain locked like this.

Gravitational waves, on the other hand, imply an emission of energy from the binary system, and as energy is lost, the black holes spiral in.

There are no tides on black holes because there is nothing there to be pulled into an oval shape. The event horizon is not a solid surface. The mass of a black hole is concentrated entirely at the singularity, there is no structure to be deformed into a bump.

> **Comments:**
> **username_10**: There ARE tides on the black holes!  
> **username_9**: Black holes can produce tides on other objects (very power ones, rip you apart type tides) but a black hole has no matter to be tidally distorted. It has mass, but no matter.  
> **username_11**: The Earth-Moon system is also emitting gravitational waves. Because its mass is small, and its spacing wide compared to a binary black hole, the loss of energy is tiny, compared to a binary black hole. So the Moon continues to recede from the Earth, but this is very slightly slowed by gravitational radiation.  
> **username_12**: Why does the moon gaining energy mean it moves away? Why does it not become warmer instead?  
> **username_9**: It gains gravitational potential energy, not heat.  There is a mechanical process that pulls it forwards and outwards, not a thermodynamic process of heating.  
> **username_13**: @username_11 The perturbation with the least effect would be relativistic frame dragging from the rotating earth.  

---

### Answer by username_3 on 2017-07-19. Score: 16

A belated answer, but neither of the existing answers properly explain this.

The proper explanation is simple. In Newtonian mechanics, tidal influences make all objects in retrograde orbits and those objects in prograde orbits below the equivalent of geosynchronous radius spiral inward. Only objects orbiting prograde above the equivalent of geosynchronous radius spiral outward. Our Moon orbits at about 385000 km, well above geosynchronous radius of 42164 km. Phobos orbits Mars with a semi-major axis of 9377 km, well below the areosynchronous radius of 20400 km. While our Moon is spiraling outward from Earth, Phobos is spiraling inward toward Mars.

Mutually tidally locked objects spiral neither inward nor outward in Newtonian mechanics. General relativity modifies this dynamic a bit, making mutually tidally locked objects spiral inward.


---

### Answer by username_4 on 2019-03-21. Score: 9

Other answers are right at explaining why tidal forces move Earth and Moon apart but they don't move apart a pair of black holes. However, I think it is also needed to explain why the phenomena making two black holes spiral inward don't make the Moon spiral inward to the Earth.

In fact, every pair of rotating masses radiate gravitational waves. What makes the difference is that only very large masses rotating very close to each other produce gravitational waves large enough to meaningfully affect those masses' orbits.

According to <https://en.wikipedia.org/wiki/Gravitational_wave#Binaries> the time that takes a pair of masses to fall into each other due to radiated gravitational waves is:

$$t= \frac{5}{256}\, \frac{c^5}{G^3}\, \frac{r^4}{(m\_1m\_2)(m\_1+m\_2)}$$

Let's plug the masses of Earth and Moon and its distance into that equation (all data taken from Wikipedia in SI units):

```
> G <- 6.674e-11
> r <- 384e6
> mluna <- 7.342e22
> c <- 299792458
> mterra <- 5.97237e24
> (t <- 5/256*c^5/G^3*r^4/(mterra*mluna)/(mterra+mluna))
[1] 1.304925e+33

```

That is, left alone, radiating gravitational waves would make the Moon crash into the Earth in 1.3$\times$1033 seconds, that is 4.13$\times$1025 years or 3$\times$1015 times the current age of the universe. In other words, the effect of radiating gravitational waves in the motion of Earth and Moon is so tiny — specially compared with other forces like tide ones — that we can't absolutely forget about it.

Just for comparison, two one solar mass neutron stars orbiting each other at the same distance of the Earth and the Moon would fall into each other in:

```
> msol <- 1.9885e30
> (t <- 5/256*c^5/G^3*r^4/(msol*msol)/(msol+msol))
[1] 2.19985e+14

```

Which is just about 7 million years, showing that changing masses have a large effect on the outcome. As stated in the beginning, gravitational waves make pairs of star-sized objects spiral inwards, but they don't have noticeable effects on a satellite orbiting a planet.

> **Comments:**
> **username_0**: Excellent answer with real numbers. Wish I could accept your answer if you'd replied earlier!  
> **username_5**: @username_0 This [Meta FAQ](https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work) says that "You may change which answer is accepted, or simply un-accept the answer, at any time." :-)  
> **username_4**: Thank you, for your appreciation, but I think a this question is in fact made of two questions. Other answers are right at answering one of them. Mine answers the other one. Therefore, the accepted answer isn't wrong. It's just one half of the answer and mine is the other half.  

---

### Answer by username_6 on 2016-02-13. Score: 7

The recession rate of the Moon from the Earth is given as 38.04 mm per year, due to tidal forces, according to Wikipedia. A good description (with diagrams) about how this occurs can be found [here](http://www.themcdonalds.net/richard/astro/papers/602-tides-web.pdf) (now sadly a *dead link*).

The orbital decay rate due to gravitational radiation can be determined by $$\frac{\mathrm{d}r}{\mathrm{d}t} = - \frac{64}{5}\, \frac{G^3}{c^5}\, \frac{(m\_1m\_2)(m\_1+m\_2)}{r^3}$$[Wikpedia:Gravitational wave](https://en.wikipedia.org/wiki/Gravitational_wave#Orbital_decay_from_gravitational_radiation)

This works out to approximately 2nm per year, or 7 orders of magnitude smaller. So, yes, both tidal recession and orbital decay via gravitational waves are expected to be happening, but tidal recession is a far larger effect.

> **Comments:**
> **username_7**: But doesn't this contradict the fact from O. P.  that two mutually rotating black holes tend to come closer?  
> **username_0**: Exactly! Why don't inspiraling black holes and neutron stars suffer from tidal forces too?  
> **username_8**: I prefer this answer because it compares the magnitude of each effect instead of trying to explain either one. In other words, all orbiting pairs have the potential for both tidal forces and gravitation waves but the OP wants to know why one system is influenced by one effect over the other and you do that quite well here.  

---

