The moon is in tidal lock with the earth, but a long time ago it was not. As the moon became tidally locked with the earth, its angular momentum changed and the delta went into it's orbit and possibly the rotation of the earth.

Tidal locking can also lock the central body, which would cause the earth to rotate once per lunar orbit. While this is not the case in the earth/moon system, this does happen when the two bodies have comparable mass and enough time has passed.

  • In the latter case, how could the delta in angular momentum of the central body possibly be transferred to its satellite?

I would assume that a tangential force is required to achieve this and the only force I can think of which is capable of reaching from the central body to its satellite would be gravity. But if the central body is a sphere, its gravitational field would show rotational symmetry and I cannot see how that could possibly exert a tangential force on the moon's orbit.

Now the tidal forces cause a deformation of the central body, so it is not a perfect sphere and it could indeed affect the orbit of its satellite.

However, the deformation would depend on the central body's rigidness. The deformation also causes the central body to loose rotational energy and this is often claimed to be the cause of tidal locking. However, a perfectly elastic body would not loose any energy due to its deformation, it would not become tidally locked (if that "brake" explanation is correct) but its gravitational influence on its satellite's orbit would still exist. So it is hard to believe that the distortion of the symmetry of its gravitational field changes the satellite's orbit.

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    $\begingroup$ There is no difference between the central body and the satellite in the Kepler problem. They both orbit around their common center of mass and angular momentum transfer is transfer from orbital angular momentum to rotational angular momentum and vice versa. There are no perfectly elastic bodies and planets and moons are certainly nowhere close to being perfectly elastic. $\endgroup$
    – CuriousOne
    Commented Apr 24, 2016 at 6:30

2 Answers 2


I am certain that mathematical analysis of tidal locking has been done many times but I have failed to find such an analysis where the mechanism for the transfer of angular momentum to spin angular momentum is included which is the question which has been asked. Perhaps someone is able to produce a reference or an analysis?

Having experienced on a number of occasions how counter-intuitive rotational dynamics can be I will not be surprised to find that my answer is flawed.

The period of revolution of the Moon about its own axis is the same as its period of revolution about the Earth.
This is due to tidal locking.

Diagram $(1)$ (adapted from the diagram in Wikipedia article "Tidal force") has the Earth somewhere below the diagram and shows the Earth's gravity differential field at the surface of the Moon which causes the Moon to change its shape.
Under the influence of these differential forces the Moon changes its shape to something like the greatly exaggerated green ellipsoid shown in diagram $(2)$.
There is no net torque on the Moon due to the Earth and so its period of revolution about its axis stays the same.

enter image description here

The Wikipedia article Tidal locking has a diagram $(2)$ which shows what the differential forces on the Moon would have been like if the Moon's period of revolution about its axis was greater than its period of revolution about the Earth.
There is a net torque on the Moon which reduces its spin angular momentum.
Ignoring the influence of the Sun and the rest of the Solar System as there are no external torques on the Moon-Earth system the loss in spin angular momentum of the Moon must increase the angular momentum of the system somewhere else. It is the orbital angular momentum of the Moon which increases as does, to a lesser extent, the spin angular momentum of the Earth.

Both Wikipedia articles are very informative.

The question is, "Where is the force that increases the orbital angular momentum?"

I think that the answer to the original question is, the force $F$ in diagram $(5)$?

The diagrams I have drawn are gross exaggerations of what actually happens. Any smaller and the force $F$ would be difficult to identify. I have also only considered a situation where all the mass is distributed in a plane and ignored any angular momentum components which are not perpendicular to that plane.

enter image description here

In diagram $(3)$ the Earth $E$ is at the bottom and the Moon's centre of mass $M$ is orbiting the Earth with an angular speed $\omega$.
The Moon is rotating about its centre of mass with an angular speed $\Omega (>\omega)$.

$F_c$ is the gravitational attraction due to the Earth on a particle at the centre of mass of the Moon.
$F_f$ and $F_n$ are the gravitational attractive forces acting on particles at points $A$ and $B$ on the Moon.

The differential forces acting on the particles at $A$ and $B$ are $F_f^\prime$ and $F_f^\prime$.
The diagram shows that these forces, which are responsible for reducing the spin angular momentum of the Moon, are not parallel to one another nor do they have the same magnitude.
Hence there is also a net force $F$ which acts at the centre of mass of the Moon.

Diagram $(4)$ illustrate the fact that for a body which is symmetrical about the $EM$ axis there is no torque.
If there is asymmetry about the axis $EM$ as in diagram $(2)$ then tidal locking is possible.

Diagram $5$ shows the couples (in green and blue) acting on the Moon which reduce the Moon's spin angular momentum and the net force $F$ (in red) which increases the Moon's orbital angular momentum about the Earth.

  • $\begingroup$ IIUC your reasoning is entirely based on gravitational forces. But how about this "brake" effect? The energy it takes to deform a body a certain amount depends on the body's material (elasticity). The gravitational effect of the deformation however, only depends on its density. Hence an earth made of clay should tidal-lock more quickly than an earth made of an elastic material, because it looses more energy. I cannot see how the other body (moon) can know about this difference and adjust its orbit accordingly. $\endgroup$ Commented Apr 25, 2016 at 17:11

If there is energy transfer between bodies, I believe it would be vastly insignificant compared to the heat dissipation, that is the real responsible for tidal lockings. When moon was spinning relative to Earth, its core was liquid, and most angular momentum energy was lost as heat due to viscous dissipation, keeping the core liquid as long as it was spinning relative to Earth.

You said that "a perfectly elastic body would not loose any energy due to its deformation", and that is true, but a perfectly elastic body is a very poor model for a planet-like body. A planet is more like a raw egg: try spinning one and see how quickly it stops compared to a boiled egg.

Same thing with Earth, the angular momentum it looses due to Moon's tidal forces makes the heat that helps to keep the core liquid.

  • $\begingroup$ Okay, I understand the energy thing. It's just that a moving object assumes constant velocity unless it is exposed to a force. I believe the same is true for a rotating object. If you want it to rotate faster or slower, you need to apply a force. I believe, there is a force, and that it can be explained by looking at the picture from a certain angle. A force, akin to the correolis force or tidal forces. $\endgroup$ Commented Jul 15, 2017 at 19:51
  • $\begingroup$ I believe it is easier if you look from the frame of reference of the body: there is a ever changing (rotating) tidal force field over its full volume, inducing movement in the fluid core. This fluid core thus transfer its movement to the outer solid shell via viscosity, that will eventually bring the full mass to rotate with the force field. This process is different from a full solid body because viscous momentum transfer is dissipative. $\endgroup$
    – lvella
    Commented Jul 17, 2017 at 20:30

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