Tidal effects on Earth's rotation and moon's orbit

[Buzz Bloom]

Issues about two topics were discussed in another thread.

https://www.physicsforums.com/threads/do-solar-tides-affect-Earth's-orbit.933073/​

The topics are: how do the lunar tides (1) cause the Earth's rotation to slow down, and (2) cause the moon to move away from the Earth. This thread is intended to be more specific.

@Drakkith cited (post #8) an on-topic 1996 Science article

http://science.sciencemag.org/content/sci/273/5271/100.full.pdf​

whose abstract sounds very interesting, and which I hope to read sometime over the next few weeks.

Drakkith also quoted from a Wikipedia article on tidal acceleration:

The mass of the Moon is sufficiently large, and it is sufficiently close, to raise tides in the matter of Earth. In particular, the water of the oceans bulges out towards and away from the Moon. The average tidal bulge is synchronized with the Moon's orbit, and Earth rotates under this tidal bulge in just over a day. However, Earth's rotation drags the position of the tidal bulge ahead of the position directly under the Moon. As a consequence, there exists a substantial amount of mass in the bulge that is offset from the line through the centers of Earth and the Moon. Because of this offset, a portion of the gravitational pull between Earth's tidal bulges and the Moon is not parallel to the Earth–Moon line, i.e. there exists a torque between Earth and the Moon. This boosts the Moon in its orbit, and slows the rotation of Earth.​

In post #7 I confessed that I did not understand this logic. There was some further discussion about this sub-topic in that thread which still left me somewhat confused, so I decided to start this more specific thread.

I think the following hypothetical model will be helpful in explaining my current understanding by simplifying the dynamics.

Consider a spinning solid oblate spheroid with the mass of the Earth without any oceans. The shape of the oblateness leads to every point on the surface having the same local downward pull as every other surface point. Near the equator the bulge causes a larger gravitational downward pull than at other surface points, but the spin of the Earth cause a centrifugal force which reduces the downward pull so the the net pull is the same as the gravitational pull at the poles. Now assume that a mass of water equal to that of the oceans is spread over the Earth. There are no surface land masses, and a single worldwide ocean. The depth of the water will be greatest at the equator and least at the poles due to the centrifugal force of the Earth's spin. The is the "nominal" shape of this Earth model in the absence of any moon.

Now assume a non-rotating spherical moon with a circular orbit in the same plane as the nominal Earth's equator. This is the nominal moon. Assume that the sidereal day period of the nominal Earth's spin (west to east) is the same as for the real Earth. Assume that the nominal moon's sidereal orbital period (appearing east to west from an observer on the Earth) is the same as for the real moon. The nominal moon will cause tidal bulges on the nominal Earth's oceans.

Below is my new interpretation of the Wikipedia quote which is different that the interpretation I gave in the other thread because I think I now have an improved understanding based on the discussion in the other thread.

The start of the interpretation

The spin of the Earth will cause the tidal bulges to be a bit east of the line connecting the centers of the Earth and Moon. This is because the bulge moves west to east with the spin of the Earth.​

Question

If an instantaneous snapshot is taken, what is the gravitational and rotational dynamics explanation of the snapshot not showing that the bulge is exactly aligned with the line connecting the centers of Earth and moon?​

A thought process for a partial answer

Why does it matter that the Earth is spinning. Suppose it wasn't and the Earth's shape was perfectly spherical without a moon. Then since the nominal moon is revolving about the nominal Earth, the tidal bulge moves (more-or-less) with the moon around the Earth from east to west. I am guessing that it must matter, and for this case, the alignment of the bulge would be (almost?) exactly with the line connecting the centers of Earth and moon. The difference with the spinning Earth must have something to do with the water's viscosity and inertia. As the Earth spins, the budge is not able to fall instantaneously or as fast as needed as the bulge moves eastward from the moon-Earth line. When the Earth does not spin, the bulge moves much slower with the moon's motion, so it has much more time to fall to it's proper level.​

Is this explanation correct?
Some more interpretation

Since the bulge is slightly to the east of the Earth-moon line, the bulge can pull the moon towards the east, which is in the directing the moon is moving (west to east) in it's orbit. This will increase the moon's orbital angular momentum. The is an identical bulge on the side of the Earth away from the moon which will pull the moon Westward However, this force is slightly weaker since the distance is slightly farther away from the moon by the diameter of the earth. The net then is a pull eastward.​

Is this explanation correct?

Now the part of the interpretation that I have the most trouble with.

While the net of the two bulges is pulling on the moon eastward, the moon is pulling on the two bulges with a net westward pull, thereby slowing the rotation of the earth.​

Question

How does the moon's pull on the water bulges result in a net torque on the solid and more massive part of the Earth?​

Thoughts about the answer

Somehow (?) the water's friction against the solid part of the Earth acts to create a torque on the solid part. However, this seems quite strange. The increase in the moon's orbital angular momentum must exactly match the reduction in the Earth's rotational angular momentum. But, the torque based on the liquid-solid friction depends on the nature of the liquid and solid, while the pull on the moon does not depend on this.​

All comments are welcome.

Regards,
Buzz

 

[andrewkirk]

Why does it matter that the Earth is spinning.

I think it doesn't matter. What matters is the difference in angular velocity $\omega$ between the moon's rotation and the Earth's. If $\omega_E-\omega_M>0$ (which is the case for this Earth and this moon) then the bulge will be ahead of the line between the centres of mass. That will create a torque that retards the Earth and speeds up the moon. Conversely, if the diff is negative, the bulge will be behind the line connecting the centres of mass, creating a torque that speeds up the Earth and retards the moon.

Either way, the force will act to make the rotation rates converge towards equality.

 

[Buzz Bloom]

I think it doesn't matter.

Hi andrew:

I think there is a misunderstanding.
How far is the distance on the surface of the Earth between (a) the peak of the bulge facing towards the moon and (b) I_EM the intersection of the Earth-moon line with the surface? In the "normal" model I described this should be a constant difference along the equator given ω_E and ω_M. This distance should be proportional to

|ω_E - ω_M|.​

When the Earth rotates, the distance should be much larger (factor of ~ 28) than when the Earth does not rotate. This is because

|ω_E / ω_M| ~= 28.​

This large difference should matter for the reason I discussed about the speed with which the water bulge responds to moving away from the point I_EM.

Regards,
Buzz

 

[andrewkirk]

When the Earth rotates, the distance should be much larger (factor of ~ 28) than when the Earth does not rotate. This is because
|ω_E / ω_M| ~= 28.

I can't think of any reason why the absolute rotation rate would make a difference, at least at first order. Absolute rotation imparts centrifugal force on the water, but that is applied symmetrically around the equator, and so should not materially change the location of the bulge peak. In contrast, the difference in rotation rates directly affects the lag of the water bulge behind the relative position of the moon.

What mechanism did you have in mind in which the absolute rotation rate or the ratio of rotation rates, rather than the difference in rates, would be the first-order driver of the phenomenon?

 

[Buzz Bloom]

What mechanism did you have in mind in which the absolute rotation rate or the ratio of rotation rates, rather than the difference in rates, would be the first-order driver of the phenomenon?

Hi Andrew:

The question that this relates to is

what is the gravitational and rotational dynamics explanation of the snapshot not showing that the bulge is exactly aligned with the line connecting the centers of Earth and moon?

The explanation I give, as my guess, is that the I_EM point moves around the equator (east to west) at a

speed = ω_E - ω_M,​

where a positive value is east to west, and a negative is west to east. The absolute value is the absolute speed, or just "speed" for short, and the sign gives the direction of movement along the equator.
When ω_E = 0 the speed ω_M is approximately 1/28 of the speed ω_E - ω_M ~= ω_E.
The question, "Why is the rotation important?" becomes, "Why is the speed of the I_EM point important?" or "Why does a large difference in the speed make a difference in the dynamics?" My guess about this is:

The difference with the spinning Earth must have something to do with the water's viscosity and inertia. As the Earth spins, the budge is not able to fall instantaneously or as fast as needed as the bulge moves eastward from the moon-Earth line. When the Earth does not spin, the bulge moves much slower with the moon's motion, so it has much more time to fall to it's proper level.

This also implies that the distance between the I_EM point and the peak of the bulge depends on the speed along the equator of the I_EM point. My guess is also that the distance is as a first approximation roughly proportional to the I_EM point speed. For large speeds, a second order term may be needed to get the same degree of approximate accuracy in the calculated distance value. The rate of change in the moon's velocity is also approximately (first order) proportional to this distance value.

I hope this answers the question you asked me in your post.

Regards,
Buzz

 

[stefan r]

Consider a spinning solid oblate spheroid... The shape of the oblateness leads to every point on the surface having the same local downward pull as every other surface point... The depth of the water will be greatest at the equator and least at the poles...

No,your set up cancels itself. Water would be the same depth. If you had a rotating sphere then the equator would be much deeper.

...Suppose it wasn't [spinning] and the Earth's shape was perfectly spherical without a moon. Then since the nominal moon is revolving about the nominal Earth, the tidal bulge moves (more-or-less) with the moon around the Earth from east to west.

That should read "around the Earth from west to east."

 

[Buzz Bloom]

Water would be the same depth.

Hi stefan:

Thank you much for pointing out my two careless mistakes.

Do you have any thoughts about the issue I raised about the following which assumes the "nominal'" Earth and moon.?

Somehow (?) the water's friction against the solid part of the Earth acts to create a torque on the solid part. However, this seems quite strange. The increase in the moon's orbital angular momentum must exactly match the reduction in the Earth's rotational angular momentum. But, the torque based on the liquid-solid friction depends on the nature of the liquid and solid, while the pull on the moon does not depend on this.

Regard,
Buzz

 

[Janus]

Hi stefan:

Thank you much for pointing out my two careless mistakes.

Do you have any thoughts about the issue I raised about the following which assumes the "nominal'" Earth and moon.?Regard,
Buzz

The loss of angular momentum of the Earth does not exactly equal that gained by the Moon. The Earth loses significantly more momentum than the Moon gains. The difference is made up by the generation and emission of heat radiation due to the friction between the tidal waters and the Earth. Only ~1/30 of the rotational energy lost by the Earth ends up being transferred to the Moon, the rest is lost as waste heat.

Earlier, you brought up the case where the Earth was not rotating. In this scenario, the tidal bulge would still travel around the Earth following the Moon, There would still be drag between the tidal bulge and the body of the Earth (If the Earth was perfectly spherical and smooth, this would be minimal), and the bulge would lag the Moon by a tad. This would result in a torque acting on the Earth that would tend to spin it up in the direction the Moon is orbiting, and the gravitational interaction between the trailing bulge and the Moon would rob it of angular momentum, causing it to fall into a lower orbit.
Again the transfer would not be 100% efficient and the Earth would not gain as much angular momentum as the Moon lost.

Now since the Moons orbital period will shorten as it moves in closer, the difference between the Earth's rotation and the Moons orbit will grow, causing more of a lag between Moon and tidal bulges, so the Moon will just continue to spiral in until it reaches the Roche limit and breaks up.

You don't even need the Earth to be covered by water. The Earth is large enough for its shape to be primarily determined by gravity and thus is subject to having its shape slightly altered by the Moon's tidal forces. These tidal bulges are much, much smaller than the ocean tides, but would still participate in transferring angular momentum between the Earth and Moon.

 

[Drakkith]

Now the part of the interpretation that I have the most trouble with.
While the net of the two bulges is pulling on the moon eastward, the moon is pulling on the two bulges with a net westward pull, thereby slowing the rotation of the earth.Question
How does the moon's pull on the water bulges result in a net torque on the solid and more massive part of the Earth?

I'm guessing friction plus the non-frictional force exerted on coastlines, underwater hills, etc. I don't know much about the details of this mechanism.

 

[Buzz Bloom]

The loss of angular momentum of the Earth does not exactly equal that gained by the Moon. The Earth loses significantly more momentum than the Moon gains. The difference is made up by the generation and emission of heat radiation due to the friction between the tidal waters and the Earth. Only ~1/30 of the rotational energy lost by the Earth ends up being transferred to the Moon, the rest is lost as waste heat.

Hi Janus:

Thank you very much for your excellent answer. This makes everything I was confused about clear, except for one detail. How does the heat radiation carry angular momentum?

Regards,
Buzz

 

[stefan r]

Do you have any thoughts about the issue I raised about the following which assumes the "nominal'" Earth and moon.?

I had just accepted the tide and rotation effects because of conservation of momentum and conservation of energy. The tide is dissipating energy as heat the energy has to come from something. That something is Earth's rotation. Momentum is also conserved so the moon must be picking up momentum. The equations have to balance. You are looking for the mechanics/kinetics of how that plays out. You do not need to know the detailed mechanism to accept a final result.

I am not sure if the tidal effect comes from the gravitational pull from a bulge or from the decreased pull of the depression. The water may also be displacing the rock. Earth experiences tidal effects on solid and liquid rock, water, and air. The moon is tidal locked so we can be fairly confident that tides dissipate energy without an ocean.

The tires on your car are dissipating something like 10 to 20% of the torque your engine produces. Inflating your tires will get better gas mileage. Over inflating will ruin the treads and also make you lose traction but still gets better mileage . The energy is lost in the bending and flexing tire. The road also bends. Asphalt roads reduce mileage more than concrete. Driving on sand, snow, or wet surfaces would be worse. A deflated tire at rest will make contact with the road at different angles from the over inflated tire at rest. The engine will increase torque to compensate for the losses. I am not sure how/if that changes the angle of forces that a car applies to the pavement while traveling at constant high speeds. If you put the car in neutral and coast to a stop there will be some forward force applied to the road. The deflated car will transfer the same momentum to the road but it will make the transfer occur faster.

The loss of angular momentum of the Earth does not exactly equal that gained by the Moon. The Earth loses significantly more momentum than the Moon gains. The difference is made up by the generation and emission of heat radiation due to the friction between the tidal waters and the Earth. Only ~1/30 of the rotational energy lost by the Earth ends up being transferred to the Moon, the rest is lost as waste heat..

You are mixing momentum and energy. A rifle will recoil with more momentum than the bullet carries. Some gas goes the same direction as the bullet. The bullet impact does a lot more damage than the recoil. A rifle with a spiky stock that was fired away from the shoulder would hurt a lot but it would not have the energy of a bullet.

By conservation of momentum the moon has to be picking up the angular momentum that Earth lost. You need some other object or particle in the mix. Tidal heating comes from the energy difference.