We all know about the fact that tidal friction is slowly slowing down the Earth's rotation about its axis, and that subsequently the Moon is slowly drifting away, in order for the angular momentum of the Earth-Moon system to be conserved.

Now: the angular momentum of the Earth Moon system about the Earth's centre should be: $$ L_{E+M} = L_{\text{Earth}}^{\text{Rot.}} + L_{\text{Moon}} $$

The total angular momentum of the Moon about the centre of the Earth can be decomposed into $$ L_{\text{Moon}} = L_{\text{CoM Moon}}^{\text{about CoM Earth}} + L^{\text{about Moon's axis}} $$

SO : $$ L_{E+M} = L_{E}^{\text{Rot.}} + L_{M}^{\text{Rot.}} + L_{M}^{\text{Orbital}} $$

Tidal friction decreases $ L_{E}^{\text{Rot.}} $, and by conservation of $L_{E+M}$ the other two terms should increase.

Which term increases? Can the Moon just revolve faster around its axis without receding in its orbit?

And on a related note:

Physically, how can it be that tidal friction on Earth makes the Moon do something? I know it is because conservation of angular momentum, but I always try to use conservation laws as a conclusion more than as the primary explanation: does the water bulge accumulated on one side of the planet because of whatever exert an extra gravitational pull?

How much time does it take for this information to travel from the Earth to the Moon?

  • $\begingroup$ yeah that's what I am calling $L_M^{Rotational}$ $\endgroup$
    – SuperCiocia
    Sep 8, 2014 at 23:44
  • $\begingroup$ Although the tidal forces don't change the rotation of the moon directly, the rotation of the moon slows since the orbit slows and it remains tide locked. $\endgroup$
    – LDC3
    Sep 9, 2014 at 1:27

1 Answer 1


Physically, how can it be that tidal friction on Earth makes the Moon do something? I know it is because conservation of angular momentum,

No, conservation of angular momentum alone can't predict that one object will lose angular momentum and another will gain. It would be equally consistent with conservation of angular momentum if both stayed the same.

The changes occur because the earth's tidal bulges make opposite torques on the moon, and these torques don't exactly cancel. The lack of cancellation is because friction causes the bulges to be misaligned with the earth-moon axis, and also because the bulges are at unequal distances from the moon, as explained by the following diagram:

enter image description here

As the moon moves in its orbit, the bulge of the tides leads a little bit (because of drag on the earth's surface). Consequently, the bulge that is closer (and thus has a stronger force on the moon) is slowing the moon down a little bit; this force is not completely canceled out by the "leading" bulge on the other side, which is further away and therefore provides a weaker force.

This results in the dotted line (net force) drawn, which speeds the moon in its orbit. This actually causes it to move to a larger radius (per the comment by Logan R Kearsley). But there is no torque on the moon about its own axis of rotation based on these tides on earth.

Note also that this is a very idealized sketch. In reality, the motion of the oceans is not a simple bulge - it's not even close - but on average there is some more material "closer to the moon, and leading" as a result of these tidal effects. So while it's not simply "a bump in the ocean" (although my drawing makes it look that way), the effect is real.

  • 3
    $\begingroup$ As usual we should note that this pretty theory relies on some conditions that are not actually met on Earth, so that the real situation is complicated by resonances in oceanic basins. While the long-term trend is to slow the Earth and lift the moon the system has run in reverse during some geological epochs. It depends on where the land masses are. We have a user who is an expert in these matter, though I can't quite recall who at this time. $\endgroup$ Sep 9, 2014 at 0:38
  • $\begingroup$ @dmckee: Very interesting. Would love to learn more about this. Maybe it would make a good question of its own. $\endgroup$
    – user4552
    Sep 9, 2014 at 1:15
  • $\begingroup$ @BenCrowell Poke around the other tidal transfer questions. He left a string of comments a couple of months ago. I have a look and post some link here in a few minutes if I find them. $\endgroup$ Sep 9, 2014 at 1:16
  • 3
    $\begingroup$ Here's the biggie: physics.stackexchange.com/a/121858 $\endgroup$ Sep 9, 2014 at 1:17

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