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It is a simple fact that the moon causes tides in the earth. But these tides seem to be a violation of the conservation of energy.

Why exactly would this be? Well think of it the way I am thinking of it, the ocean is a bowl of water ok. Now the “moon” is right above that bowl of water and the water is attracted to it causing it to retreat from the sides of the bowl and bulge more up towards that little “moon”. As the water retreats, it rubs with the sides of the container, thus causing friction. As the moon swings around, and goes “away” the water in the bowl goes back to its original state causing friction in the process.

Now with this image can you better visualize why I think the moon causing the tides is a violation of the conservation of energy? And can you also explain why it may not be?

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    $\begingroup$ Well, the moon’s orbit is changing because of it, so no - you have to consider the whole system. $\endgroup$
    – Jon Custer
    Commented Sep 25, 2019 at 1:05
  • $\begingroup$ oh how is the orbit changing? is it very minuscule? $\endgroup$
    – Locomotive 38
    Commented Sep 25, 2019 at 1:18
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    $\begingroup$ Take a look at [chegg.com/homework-help/…, which explains how tidal friction affects the Earth's rotation. $\endgroup$
    – S. McGrew
    Commented Sep 25, 2019 at 1:29
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    $\begingroup$ Since you asked a previous question where you suggested that planetary accretion violates the conservation of energy, I want to mention that, as far as I know, physicists recognize only one scenario that may possibly violate energy conservation, and that is the expansion of the entire universe. For any other scenario, the answer is going to get boring: there is no violation, because we can define a conserved quantity that we call energy. In the other question, I explained that this is true for Newtonian gravity because it is a so-called conservative force. $\endgroup$
    – G. Smith
    Commented Sep 25, 2019 at 1:56
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/77606/2451 , physics.stackexchange.com/q/6400/2451 , physics.stackexchange.com/q/221430/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Sep 27, 2019 at 17:39

4 Answers 4

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Energy for tides comes from the Earth's rotation. Tidal drag is actually slowing Earth's rotation, making the days longer, but very gradually. Just as the Moon is tidally locked to Earth, Earth would eventually become tidally locked to the Moon, but it would take billions of years. As Earth spins much faster than the Moon orbits it, the tidal bulge stays slightly ahead of the Moon, this pulls on the Moon, accelerating it slightly, causing it to move farther from Earth. you might want to read https://en.wikipedia.org/wiki/Tidal_acceleration

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  • $\begingroup$ First off isn't it the moon that causes tides. $\endgroup$
    – Locomotive 38
    Commented Sep 25, 2019 at 13:10
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    $\begingroup$ @Locomotive 38: The moon causes the larger tidal bulge, the earths rotation causes the bulge to move around the earth. There are also solar tides but they are only inches, not feet. and yes there would be a tidal bulge in a locked planet , but it would not move on the planet $\endgroup$
    – Adrian Howard
    Commented Sep 25, 2019 at 13:21
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    $\begingroup$ @Locomotive 38: Yes if Earths rotation were over a month instead of 24 hours, then tidal drag would speed it up, If you read my link it explains tidal acceleration and deceleration. Also it is not alleged it is accepted, and clearly the case. $\endgroup$
    – Adrian Howard
    Commented Sep 26, 2019 at 19:09
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    $\begingroup$ @Locomotive 38: if a moon orbits faster than its planet spins then the planets tidal bulge is behind the moon in its orbit. This speeds the planetary spin, but slows the moons orbit, the planets extra rotational energy comes from slowing the moon. $\endgroup$
    – Adrian Howard
    Commented Sep 26, 2019 at 23:14
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    $\begingroup$ @Locomotive 38: you should read en.wikipedia.org/wiki/Orbit_of_the_Moon#Tidal_evolution for a better explanation $\endgroup$
    – Adrian Howard
    Commented Sep 26, 2019 at 23:27
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Conceptually the answer is simple- the heat energy created by the frictional effects associated with tides is offset by a loss of rotational energy, as the friction slows the rotation of the Earth. The impact on day-length is small, but will eventually cause the rotation to reduce to the point at which the day length equals the period of orbit of the moon.

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Considering the interaction of the tidal bulge with the kinetic and potential energy compartments in the earth-moon system, how can we understand/rationalize these transfers of energies as obeying the conservation of energy?

The moon creates a tidal bulge on the earth which drags across the ocean floor. Tidal friction between the flowing bulge and irregularities in the ocean floor generates heat (i.e., randomized kinetic energy). It is because of the tidal friction that the tidal bulge remains at a leading angle/offset to the moon-earth axis, which slows the earth's rotation and increases the moon's orbital speed (with the associated change in gravitational potential energy).

The loss of energy to the heat produced by tidal friction continues decreasing the total kinetic and potential energies of the earth-moon system until these two bodies reach a locked state (i.e., when the rate of the earth's rotation equals the orbital rate of the moon).

The compartments of kinetic and potential energy in the earth-moon system include 1) the angular kinetic energy of the earth’s rotation, 2) the moon's orbital kinetic energy, 3) the gravitational potential energy between the moon and earth (which includes the gravitational potential energy of the tidal bulge).

The moon creates a tidal bulge by its gravitational force acting upon the earth’s oceans. The earth rotates faster than the moon orbits, but if there were no tidal friction, the tidal bulge would track with the earth-moon-center axis.

But because of tidal friction and the faster rotation of the earth than the orbital velocity of the moon, the tidal bulge is always offset ahead of the orbiting earth-moon-center axis. As a result, the angle between the earth-moon-center and the tidal bulge causes an acceleration of the moon’s orbital velocity (with a corresponding increase in its orbital radius), and a deceleration of the earth’s rotational angular velocity (with a corresponding lengthening of the day). See Hyperphysics reference.

If the fluid mass of the tidal bulge were a frictionless/lossless superfluid, then the transit of the tidal bulge the earth’s circumference would track directly under the moon, and there would be no change in angular velocity of either the rotation of the earth or orbit of the moon. See Wikipedia reference.

But in the real-world earth covered with ocean water, the rate of heat loss to tidal friction equals the rate of decrement in the total of all potential energy and kinetic energies of the moon-earth system.

More specifically:

A) The gravitational pull on the moon by the tidal bulge accelerates the moon, causing it to go to a higher/larger radius orbit. (Note: at this higher orbit, although its orbital speed increases, its angular velocity decreases, as per Kepler's laws.) This larger orbital radius increases the gravitational potential energy of the earth-moon system.

B) The gravitational force applied to the tidal bulge also acts between the moon and earth and it decelerates the earth’s rotation.

C) The key energy-conservation concept is that without tidal friction, there would be no tidal bulge offset (i.e., the tidal bulge would track with the earth-moon axis, and there would be no change in the rate of the earth's rotation or the moon's orbital velocity).

But, because of the ongoing conversion of the tidal bulge’s fluid velocity into heat, the tidal bulge is offset from the earth-moon axis. And, the angle of the tidal bulge to the earth-moon axis unavoidably, and continuously decreases the total-sum of the earth-moon kinetic and potential systems, by 1) decreasing the earth’s rotational energy, 2) increasing the moon’s orbital kinetic energy, and 3) increasing the moon’s gravitational potential energy.

In Summary: If the tidal bulge moved around the earth's circumference with a lossless/frictionless flow, then the tidal bulge would track without offset to the earth-moon axis. Thus, by comparing the effects produced by ocean-water tidal-bulge with a frictionless superfluid, we can easily identify the effects produced by tidal friction.

The heat loss of tidal friction is produced by molecular velocity-randomization due to 1) the collisions between the tidal-bulge water molecules with ocean floor irregularities and 2) the shear forces of water's viscosity.

The earth's rotation is opposed, and slowed, by the force of these molecular collisions, and in the process, the organized energy of rotational kinetic energy converts into the disorganized kinetic energy of heat.

The connection between heat loss by tidal friction and the slowing of the earth's rotation is hidden within the complexity of these inelastic collisions at the molecular level.

Conservation of energy requires that the sum of the earth-moon's rotational, orbital, and potential energies decreases by the amount of energy lost to heat from the earth's rotational energy.

The medium of conversion between the earth's rotational energy and the heated ocean water is the force of tidal friction acting between the ocean floor and the tidal bulge and between the sheer planes of viscous laminar flow.

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The principle of conservation of energy is that energy is allowed to change forms. The gravitational energy being depleted due to friction (and presumably converted into heat in your rubbing analogy) is precisely this principle.

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