According to the accepted answer on Is the moon moving further away from Earth and closer to the Sun? Why?, the moon is receding from Earth because tidal forces and friction cause energy to be lost.

However, according to LIGO's website,

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?

  • $\begingroup$ 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. $\endgroup$ Feb 12, 2016 at 23:34
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    $\begingroup$ @zibadawatimmy should be an answer $\endgroup$
    – ProfRob
    Feb 13, 2016 at 0:07
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    $\begingroup$ @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. $\endgroup$ Feb 13, 2016 at 1:18
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    $\begingroup$ 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 universetoday.com/56042/triton $\endgroup$
    – Gnubie
    Feb 15, 2016 at 23:31

4 Answers 4


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, 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 emmision 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.

  • $\begingroup$ There ARE tides on the black holes! $\endgroup$
    – Anixx
    Feb 14, 2016 at 21:49
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    $\begingroup$ 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. $\endgroup$
    – James K
    Feb 15, 2016 at 0:39

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.


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 (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

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.

  • $\begingroup$ But doesn't this contradict the fact from O. P. that two mutually rotating black holes tend to come closer? $\endgroup$ Feb 13, 2016 at 14:35
  • $\begingroup$ Exactly! Why don't inspiraling black holes and neutron stars suffer from tidal forces too? $\endgroup$
    – Gnubie
    Feb 13, 2016 at 21:46
  • $\begingroup$ 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. $\endgroup$ Mar 5, 2018 at 16:38

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 phenomenons 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*10^33 seconds, that is 4.13*10^25 years or 3*10^15 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 to spiral inwards but they don't have noticeable effects on a satellite orbiting a planet.

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    $\begingroup$ Excellent answer with real numbers. Wish I could accept your answer if you'd replied earlier! $\endgroup$
    – Gnubie
    Mar 22, 2019 at 17:01
  • $\begingroup$ @Gnubie This Meta FAQ says that "You may change which answer is accepted, or simply un-accept the answer, at any time." :-) $\endgroup$ Mar 24, 2019 at 13:30
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    $\begingroup$ 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. $\endgroup$
    – Pere
    Mar 24, 2019 at 13:38

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