Why is the moon drifting into space

[oldunion]

My astronomy book tried saying that the moon is spinning into outer space, away from earth,; it was blamed on tidal effects between the Earth and moon and the subsequent generation of frictional forces. I don't see it.

Also, from physics, if you look at a force diagram there should only be weight acting on the moon pulling it towards the earth, so there would be a net force towards earth. but the moon is drifting into space (confirmed by NASA experiments), so how is this possible. my physics teacher couldn't understand what i was saying.

 

[tony873004]

It's because the net force isn't exactly towards the Earth. The moon pulls tides on Earth. There is a high tide almost directly underneath the moon, and another high tide on the opposite side of the Earth. I say almost because as the Earth rotates, the buldge created by the moon is pulled forward slightly. This means that the Earth is not a perfect sphere, and the moon itself gets pulled mostly towards the Earth, but a tiny bit in the direction it is orbiting. This results in the moon's orbit gaining energy and that causes it to spiral out. The effect is very small, something like an inch per year, but builds up over millions of years since millions of inches is a lot of miles. The other thing that happens is that the Earth's rotation slows down a little bit. Again, a very small amount, but enough that over millions of years it adds up. Some theorize that the moon used to be a lot closer and the Earth's day was 12-18 hours long. In the distant future, after the moon spirals out more, and the Earth's rotation slows and becomes the same as the moon's orbital period, the moon will no longer spiral out. The system will be tidally locked. But other bad things (such as the Sun expandinging into a red giant and engulfing the Earth) are probably in store for Earth long before this has a chance of happening.

 

[Janus]

The tidal interaction works like this:

The Moon causes tidal bulges on the Earth. Ideally, these bulges would line up with the Moon. The Earth, however, turns on its axis and friction between the body of the Earth and the tidal bulges tends to drag the tidal bulges along. The Moon, meanwhile tries to keep the bulges aligned with itself. The resulting tug of war results in the tidal bulges "leading" the Moon by a bit.

So we have an action/reaction situation here, while the Moon tries to pull "back" on the bulges, the bulges pull "forward" on the Moon. If you push or pull forward on an object in orbit, you add to its orbital energy and it climbs into a orbit with a higher average altitude.

The other side is that while the friction between the Earth and tidal bulges pull them forward, it also pulls backward( against its rotation) on the Earth. This causes the Earth to lose rotational speed.

In effect, the 'canted forward' tidal bulges act as a mechanism to transfer angular momentum from the Earth to the Moon.

 

[SpaceTiger]

Also, from physics, if you look at a force diagram there should only be weight acting on the moon pulling it towards the earth, so there would be a net force towards earth.

You need to take into consideration that the moon and Earth aren't perfect spheres. They are actually kind of egg-shaped, with the long axis of the egg pointing towards the other body. Why? It has to do with the fact that each pulls more strongly on the close side of the other than the far side. The lack of sphericity means that you can no longer treat them as point masses (as I'm assuming you did in your diagram), so the problem gets a little more complicated. What ends up happening is described in more detail here.

The short version is that the moon got tidally "locked", so that it always has the same side facing us. Given enough time, the same thing will happen to the Earth (rotation period will match the moon's orbital period), so that must mean the Earth is spinning down. But if the Earth is spinning down, conservation of angular momentum dictates that something else must "spin up". It turns out that this something else is the moon's orbit. Over time, it gains angular momentum and moves away from the earth.

 

[tony873004]

An interesting side note is that if the Moon's orbital period were faster than the Earth's rotational period, the exact opposite would happen. The Earth's rotation would speed up and the Moon would spiral in. This is happening to Mars' moon, Phobos.

 

[Crosson]

This is the easy way to think about it:

The moon and the Earth together have a certain amount of energy. The tides are like friction, they slowly drain the Earth-Moon energy. This causes the moon to slow down, and so it drifts further out.

 

[tony873004]

This is the easy way to think about it:

The moon and the Earth together have a certain amount of energy. The tides are like friction, they slowly drain the Earth-Moon energy. This causes the moon to slow down, and so it drifts further out.

No, it causes the moon to speed up. If the moon slowed down it would drift inward. Friction only applies to the Earth part of it. It makes the Earth's rotation slow down.

 

[Janus]

This is the easy way to think about it:

The moon and the Earth together have a certain amount of energy. The tides are like friction, they slowly drain the Earth-Moon energy. This causes the moon to slow down, and so it drifts further out.

Easy, but wrong. If the Moon were losing energy it would fall into a lower orbit, not climb away.

While it is true that higher orbits mean lower orbital speeds and thus lower kinetic energies, The increased altitude of the orbits involve an increase of gravitational potential energy that is greater than the decrease in kinetic energy.

 

[SpaceTiger]

While it is true that higher orbits mean lower orbital speeds and thus lower kinetic energies, The increased altitude of the orbits involve an increase of gravitational potential energy that is greater than the decrease in kinetic energy.

An interesting way of looking at it is that gravitational systems have a negative specific heat. That is, if you add energy, the particles/bodies in the system will slow down (have a lower temperature).

 

[Uno Lee]

When we move a satellite to a higher orbit we must fire the rocket. Unless the moon is somehow already past the point of no return, there must be a force acting on it. Tidal patterns on Earth would, I think, slow the moon down-sort of like spinning a raw egg compared to a hard boiled one. This is a very good question. I think the moon is receding from the Earth at about 2.4 cm/year, now I wonder, is it accelerating?

 

[pervect]

When we move a satellite to a higher orbit we must fire the rocket. Unless the moon is somehow already past the point of no return, there must be a force acting on it. Tidal patterns on Earth would, I think, slow the moon down-sort of like spinning a raw egg compared to a hard boiled one. This is a very good question. I think the moon is receding from the Earth at about 2.4 cm/year, now I wonder, is it accelerating?

No, tidal forces act to speed the moon up, not slow it down, as several posters have already mentioned. However, the tidal forces act as a whole to dissipate energy, while conserving angular momentum. The Earth loses more energy by spinning down than the moon gains by climibing to a higher orbit - the difference powers the Earth's tides.

 

[Uno Lee]

Then, why are satellites crashing down to Earth unlike the moon which is getting flung out to space? Are they too light? Too close?

 

[selfAdjoint]

Then, why are satellites crashing down to Earth unlike the moon which is getting flung out to space? Are they too light? Too close?

Too close. Their orbits cut through the upper parts of the atmosphere, so they experience air drag, which slows them down.

 

[marlon]

I have always remembered this explanaition :

the Earth slows down due to the friction with the oceans (the tidal bulges). Part of this energy goes to a rise in potential energy of the earth-moon system because of the extra forces between the bulges and the moon (a torque is exterted between the bulges and the moon because both are NOT aligned). The other (less in magnitude) amount of energy goes to the rise in kinetic energy of the moon (due to conservation of angular momentum).

Now, normally the radius of the moon would lower because the kinetic energy is bigger, but in the same time the potential energy has become much greater and this can only happen if the radius of the moon-orbit has become larger.

the net effect yields a lunar orbit of bigger radius. It is a bit like the radius rises 5 km (because the potential energy rises) and lowers 2 meters (because the kientic energy rises). In the end both energy conservation and angular momentum conservation are respected.

I have one question though : how do we know for sure that more energy from the 'slowing down' of the Earth goes to the rise in potential energy (coming from the force between the moon and the tidal bulges) and less energy goes to the rise in kinetic energy of the moon ? Does anyone have the calculations to prove this ?

regards
marlon

 

[Phobos]

Remember that an orbit is a balacing of tangential velocity and gravity between the two objects. If you increase the velocity of the moon (as Janus described), then it can overcome the gravitational attraction more and move further away from the Earth (move to a higher orbit).

Of course, this is not free energy...increasing the moon's tangential velocity is at the expense of decreasing the Earth's rotational speed.

 

[tony873004]

On a kinda related subject, how strong is the Earth's grip on the Moon's tidal buldge? I don't imagine its very strong, and that if the Moon did have a period other than sychronous, that it would take a long time for Earth to slow it down to sychronous again.


If the Moon had a slow rotation as viewed from Earth so that from Earth we witnessed it rotate once a year (ie in 6 months from now the far side of the moon was facing us, and 6 months later, the familiar side was facing us again), how long would it take for the Earth to put the brakes on the Moon's rotation and make it sychronous again?

How much energy would it take to make the moon rotate, as viewed from Earth, no matter how slowly, so that the far side was facing us at times? IE: If an astronaut drove a moon buggy in a westward direction on the Moon's equator and traveled around the Moon once, how massive would the moonbuggy have to be to impart a rotation on the moon that exceeded the strength of Earth's grip on the tidal buldge and break the sychronous rotation for at least one rotation?

 

[marlon]

Remember that an orbit is a balacing of tangential velocity and gravity between the two objects. If you increase the velocity of the moon (as Janus described), then it can overcome the gravitational attraction more and move further away from the Earth (move to a higher orbit).

Of course, this is not free energy...increasing the moon's tangential velocity is at the expense of decreasing the Earth's rotational speed.

Yes, but total energy needs to be conserved so increasing the kinetic energy of the moon will indeed result in a higher orbit. However if the radius has increased then so has the potential energy between the moon and the earth. So the increase in both kinetic and potential energy must come from the decrease in rotational energy of the earth. What i am wondering about is what calculation shows us how much of the Earth's rotational energy goes to the increase in kinetic energy and potential energy of the moon ?


marlon

 

[BobG]

Yes, but total energy needs to be conserved so increasing the kinetic energy of the moon will indeed result in a higher orbit. However if the radius has increased then so has the potential energy between the moon and the earth. So the increase in both kinetic and potential energy must come from the decrease in rotational energy of the earth. What i am wondering about is what calculation shows us how much of the Earth's rotational energy goes to the increase in kinetic energy and potential energy of the moon ?


marlon

You're counting the same increase twice. You add energy to the Moon's orbit, period.

Theoretically, there's two ways you could do that. You could change the potential energy by instantly teleporting the Moon to a new location (okay, that doesn't happen). Or, you could add kinetic energy.

Once you've added the energy, the balance between kinetic and potential energy is automatic. In other words, if you only accelerated the Moon once, the Moon would have to return back through that same point (the one where you accelerated it) every orbit. You're affecting the opposite side of your orbit - how far away the Moon will reach on the opposite side. As that distance increases, the balance shifts from kinetic energy towards potential energy.

In reality, the Moon is accelerated at every point in its orbit (not just one point). You see the end result of an infinite number of little accelerations, which is why it appears you added speed to wind up with a slower speed (for any given point in an orbit, you're looking at the opposite side of the orbit than where the acceleration took place).

 

[marlon]

Thanks BobG

marlon

 

[pervect]

Yes, but total energy needs to be conserved so increasing the kinetic energy of the moon will indeed result in a higher orbit. However if the radius has increased then so has the potential energy between the moon and the earth. So the increase in both kinetic and potential energy must come from the decrease in rotational energy of the earth. What i am wondering about is what calculation shows us how much of the Earth's rotational energy goes to the increase in kinetic energy and potential energy of the moon ?


marlon

There's a theorem called the virial theorem which says that the kinetic energy of an orbiting body with an inverse square law force law is -1/2 the potential energy.

(The potential energy is negative, the kinetic energy is positive, that's why the factor is negative).

So when the moon goes to a higher orbit, if you increase the potential energy by 2 units, making it less negative, the kinetic energy decreases by 1 unit, and the total energy increases by 1 unit.

See for instance Goldstein, "Classsical mechanics", pg 83-85.

For a sanity check, consider a circular orbit.

The potential energy is -GM/R

Setting the centripetal acceleration equal to the force of gravity gives
GMm/R^2 = mv^2/R, so mv^2=GM/R, thus .

5*m*v^2 = kinetic energy = GM/2R

which confirms the relationsip

kinetic energy = -1/2 potential energy

Note that the factor of -1/2 is specific to the inverse square force law.

 

[Lil`SciWizGirl]

I heard it's, because the sun very hot & the engery, and force emitted by the sun pulls the moon futher into Space. Does the clairify you question?

I glad to help answer any basic questions you have on Astronomy & basic Science! Contact me any Time!

Thank You!

 

[marlon]

I heard it's, because the sun very hot & the engery, and force emitted by the sun pulls the moon futher into Space. Does the clairify you question?

I glad to help answer any basic questions you have on Astronomy & basic Science! Contact me any Time!

Thank You!

This is completely incorrect. The main reason, as has been given many times here, is the fact that you put energy into the moon's orbit, which will extent it's radius (due to energy conservation) as a consequence. The energy comes from the slowing down of the Earth's rotation due to friction with the tidal bulges...

regards
marlon

 

[RoboSapien]

I am sorry . All of U have failed to tell me why is moon gaining height in orbit. Something as heavy as moon to do that would require fantastic amount of energy. Where is the energy comming form ?

By the way if Earth is slows down by moon then moon too should get slowed down.


Thanks for posting the this :

... I think the moon is receding from the Earth at about 2.4 cm/year, now I wonder, is it accelerating?

 

[SpaceTiger]

I am sorry . All of U have failed to tell me why is moon gaining height in orbit. Something as heavy as moon to do that would require fantastic amount of energy. Where is the energy comming from ?

The Earth's rotation. Read my original response. The Earth is spinning down, so the moon is gaining angular momentum and energy.

 

[DaveC426913]

I am sorry . All of U have failed to tell me why is moon gaining height in orbit. Something as heavy as moon to do that would require fantastic amount of energy. Where is the energy comming form ?

As a way of cutting past the many incomplete and wrong answers posted in addition to the correct ones, I'll try to summarize once again:



The Earth has a lot of angular momentum - and ultimately, that is where the Moon is getting its energy from - that's the answer to your question.

The Moon's gravitational pull causes bulges in the Earth's oceans.
The bulges try to stay pointed towards the Moon, but the Earths rotation pulls them ahead.
The bulges mean the gravitational pull on the Moon is slightly ahead of where it should be with a perfectly round Earth.
The net pull on the Moon is slightly forward, causing the Moon to speed up.
The sped up Moon, with its extra energy, moves into a higher orbit.

Where does the energy come from? It is bled off from the Earth's rotation.
Those tidal bulges cause friction. The water acts against the ocean floor and the continental shelves.
This slows the Earth's rotation.

By the way if Earth is slows down by moon then moon too should get slowed down.

Indeed it has! That is why it shows only one face. The Moon has slowed down to the point where all its rotational energy has been bled off (millions of years ago). It is now locked, one side facing the Earth. (It won't stop rotating with respect to a frame of reference outside the Earth, because it would actually have to find energy to counter-rotate.)

 

[pervect]

I am sorry . All of U have failed to tell me why is moon gaining height in orbit. Something as heavy as moon to do that would require fantastic amount of energy. Where is the energy comming form ?

By the way if Earth is slows down by moon then moon too should get slowed down.


Thanks for posting the this :

There's been a few wrong answers, but a lot of correct answers too, which have answered your question.

To provide yet another recap:

As various other people have said, the Earth is losing kinetic energy as its rotation speeds down. It's also losing angular momentum.

The moon is gaining some of the energy the kinetic energy the Earth loses, and all of the angular momentum the Earth loses.

Thus the spin angular momentum of the Earth is being changed into orbital angular momentum of the moon. This prcoess is sometimes known as spin-orbit coupling.

The total angular momentum of the Earth-moon system remains constant. I was going to give some formulas, but they got too messy and long. If you want to calculate the total angular momentum of the Earth-moon system you have to add together the

spin angular momentum of the earth
orbital angular momentum of the Earth around the earth-moon center of mass
spin angular momentum of the moon
orbital angular momentum of the moon around the earth-moon center of mass

The energy formulas are also somewhat long. ONe specific formula that might be of interest is that the energy of a rotating body is .5*I*w^2, where I is the moment of inertia of the body around it's spin axis, and w is the angular frequency at which the body rotates.

For the Earth-moon system, the total energy iof the system s the sum of the following:

the spin energy of the Earth
the spin energy of the Moon
the kinetic energy of the Earth due to it's orbital velocity around the COM
the kinetic energy of the Moon due to it's orbital velocity around the COM
the gravitational potential energy between the Earth and moon.

Angular momentum must stay constant - but energy can be dissipated in the form of heat, so it is allowed to decrease.

 

[Creator]

... consider a circular orbit.

The potential energy is -GM/R

.5*m*v^2 = kinetic energy = GM/2R

which confirms the relationsip

kinetic energy = -1/2 potential energy

Hmmm. Good point;
You are correct about the not so commonly realized relation between kinetic energy K and potential energy U of an orbital system; K=(1/2)U.

However,
U=-\frac{GMm}{r}

Also, converting kinetic energy K in terms of potential energy U may be helpful; so:

K= \frac{1}{2}\frac{GMm}{r}
It is positive since kinetic E is always positive.

And thus the total energy E is given by:

E= K + U = \frac{1}{2}\frac{GMm}{r}-\frac{GMm}{r}=-\frac{GMm}{2r}

This total energy then (last eqn.) is usually referred to as a 'constant' of motion, (the total angular momentum being the other). The negatve sign indicates a bound (closed) system.
However, we can see from inspection that if the moon's orbital radius r actually changes then we are left with a non conservative equation.. IOW, we realize something is missing.

The problem is that this idealistic conservation equation leaves out a critical component (not spoken of in first yr. astronomy), namely, spin. In general, spin angular momentum as well as the deformation of the planet must be taken into account.

As DaveC correctly pointed out, taking into account the Earth's deformation (tides) is necessary to correctly account for why the Earth's spin angular momentum gets transformed into the moon's orbital angular momentum, in other words, the spin-orbit coupling.

It seems to me that in Class. Gen. relativity a somewhat analogous situation arises. However, this is may not the place to address it.

Creator

--A clear conscience is the sign of a bad memory.--

 

[pervect]

The virial theorem as I described it and as developed in the textbook I referenced applies to point particles, which can't have any rotational energy.

You are of course right when you point out that the energy that propels the moon into a higher orbit comes from the Earth's rotation - I"ve made that same point myself in an earlier post, as have a number of other posters.

 

[Creator]

The virial theorem as I described it and as developed in the textbook I referenced applies to point particles, which can't have any rotational energy.

If that is the case then why are you trying to use it to explain rotation of extented bodies? ...The point is...that the logical extension of that relation (as I presented it) IS USED in many 1st year texts to explain circular orbital motion of extented bodies. However, it is an idealization which ignores spin related deformations.

You are of course right when you point out that the energy that propels the moon into a higher orbit comes from the Earth's rotation - I"ve made that same point myself in an earlier post, as have a number of other posters.

That's not the point I was making, and I think you well know it. I was pointing out that a planet's (non-relativistic) rotation, by itself, is not suffficient to boost the moon to a higher orbit; deformation of the planet (or inhomogenous densities) is required, a point which no one except DaveC specifically brought up. A spherically symmetric (homogenous) planet, even if it is rotating, will not suffice; (in non-relativistic motions).

Heres' the quote again:
"As DaveC correctly pointed out, taking into account the Earth's deformation (tides) is necessary to correctly account for why the Earth's spin angular momentum gets transformed into the moon's orbital angular momentum, in other words, the spin-orbit coupling."

Creator

Edited: Sorry; I just re-read & noticed Janus, Tony, & Space tiger (and possibly Marlon) also mentioned 'tidal bulges' as a mechanism to transfer energy.- my mistake.

 

[pervect]

If that is the case then why are you trying to use it to explain rotation of extented bodies? ...

You should read back in the thread a bit more closely - you'll see that my response with repsect to the virial theorem was not to the original poster or the original question, but to another poster with a different question.

The original thread had wound down at that point, (to be later re-opened by the orignal poster, who was apparently confused by the the chaff mixed in with the wheat.)

(Umm, that's an anology, there wasn't any actual wheat in the thread, nor any grain at all for that matter.)

 

[tony873004]

(from other thread)I hear that the moon is steadily moving away from the earth. Is it due to the suns gravity? Will it eventually go out of Earth orbit and circle the sun on its own? Would there be any chance of a posssible collision with Earth if this happened or would it leave Earth orbit at its closest approach to the sun if so?

If the Moon left the Earth, it almost certainly would have an Earth collision in its future.

But it won't ever leave. As it moves out, it also spins down the Earth. Once the Moon's period and the Earth's rotation are equal to each other, the Moon will stop receeding. And the Sun will probably go Red Giant long before that happens, engulfing both Earth and Moon.

 

[quetzalcoatl9]

If the Moon left the Earth, it almost certainly would have an Earth collision in its future.

But it won't ever leave. As it moves out, it also spins down the Earth. Once the Moon's period and the Earth's rotation are equal to each other, the Moon will stop receeding. And the Sun will probably go Red Giant long before that happens, engulfing both Earth and Moon.

i have read that, among other things, if the Earth were to suddenly disappear, that the moon would continue in orbit around the sun. and that supposedly this is unique to our moon: other planetary moons, if in the same situation, would fly off into space.

is this true?

 

[tony873004]

i have read that, among other things, if the Earth were to suddenly disappear, that the moon would continue in orbit around the sun. and that supposedly this is unique to our moon: other planetary moons, if in the same situation, would fly off into space.

is this true?

Every moon, including the Earth's would fly off into space relative to the position of their now-missing parent planet. Many would continue to orbit the Sun, many would escape the solar system, and many would enter planet-crossing orbits.

To determine what would happen, you have to compare the circular orbital velocity of the planet around the Sun, and see what effect adding or subtracting the moon's orbital velocity to the planet's orbital velocity.

If the moon's new solar orbital velocity exceeds solar escape velocity, then it will escape the solar system. Escape velocity is \sqrt{2}*V_{circ}.

In the case of the Earth's Moon, it travels around the Earth with an orbital velocity of about 1 km/s, while the Earth is traveling around the Sun with an orbital velocity of about 30 km/s. So relative to the Sun, the Moon is orbiting in the range of 29 - 31 km/s depending on where in its orbit it is. This is not a huge difference, so if the Earth disappeared, the Moon's new solar orbit would be very similar to its original solar orbit while it was a part of the Earth/Moon system.

Mars' moons would fare a little worse. Phobos and Deimos, with orbital velocities of 2.1 and 1.4 km/s are orbiting Mars faster than the Moon orbits the Earth. And Mars' orbital velocity at ~24km/s is a bit slower than Earth's. So the percentage change in orbital velocity experienced by the now-free Martian moons would be higher than that experienced by Earth's Moon. Depending on where they were in their orbits when Mars disappeared, and also depending on where Mars was in its solar orbit, since it's orbit is noticably elliptical causing its orbital velocity to vary between 21-27 km/s, Phobos could enter an Earth-crossing orbit.

The story changes for Jupiter. It's solar orbital velocity is only about 13 km/s. Escape velocity from the Sun is about 18 km/s at Jupiter's distance. But its innermost moons orbit it with velocities of 17 km/s for Io, to 8 km/s for Callisto. There are a few moons interior to Io which orbit even faster. So these moons will not happily orbit the Sun in similar orbits to the ones they enjoyed while Jupiter existed. They will either get flung out of the solar system, dropped into the Sun, or enter planet-crossing orbits. But Jupiter has many moons that orbit it at a great distance. For example, Eurydome has an orbital velocity of about 1.7 km/s. Its fate, and the fate of similar moons depends on where in their orbits they were when Jupiter disappeared.

The moons of Saturn, Uranus, and Neptune would have fates similar to Jupiter's moons. But Pluto's moon, Charon, would continue to orbit the Sun in a very similar orbit.

Here's a screen shot of a simulation of the solar system a few years after Jupiter's mass suddenly dropped to 0 kg.
http://orbitsimulator.com/orbiter/jup.GIF

 

[quetzalcoatl9]

thank you, that makes sense

 

[mavisgold]

.moon gravitational pull
↓
↑
.earth bulge from moons gravitational pull

.but because the Earth is rotating faster than the moon is orbiting

.the Earth bulge gets ahead of the moon


.moon position
. ↓
.↑ ←earth rotation
.earth bulge


.so the moon is pulled towards the Earth bulge increasing its speed
.← moon orbit
.→
.and the Earth bulge is pulled towards the moon slowing it down


do this help?
or just add confusion?

 

[sylas]

...

This thread is a bit over four years old.

 

[pegasus980]

.moon gravitational pull
↓
↑
.earth bulge from moons gravitational pull

.but because the Earth is rotating faster than the moon is orbiting

.the Earth bulge gets ahead of the moon


.moon position
. ↓
.↑ ←earth rotation
.earth bulge


.so the moon is pulled towards the Earth bulge increasing its speed
.← moon orbit
.→
.and the Earth bulge is pulled towards the moon slowing it down


do this help?
or just add confusion?

confusion for me.

 

[ideasrule]

I know this thread is 4 years old, but for people reading it, I have to mention one thing.

Where does the energy come from? It is bled off from the Earth's rotation.
Those tidal bulges cause friction. The water acts against the ocean floor and the continental shelves.
This slows the Earth's rotation.

No! Tidal bulges are not the tides that we see on the beach; Earth's oceans have negligible volume compared to the Earth itself.

Because the Moon's gravity is stronger on the side closer to the Earth than on the farther side, Earth is slightly elongated. This is the so-called tidal bulge; it's a bulge in the rock, not a bulge in the ocean.

 

[sylas]

I know this thread is 4 years old, but for people reading it, I have to mention one thing.



No! Tidal bulges are not the tides that we see on the beach; Earth's oceans have negligible volume compared to the Earth itself.

Because the Moon's gravity is stronger on the side closer to the Earth than on the farther side, Earth is slightly elongated. This is the so-called tidal bulge; it's a bulge in the rock, not a bulge in the ocean.

The original post to which you are responding is completely correct.

The bulge in the ocean IS a tidal bulge, and this IS the tides you see on the beach. You can also get a bulge in rock, but it is much much smaller. Which matters for for slowing rotation depends on the circumstances. For Earth, it is the ocean tides which are most significant for the drag that slows Earth's rotations and transfers energy to the Moon, leading to it moving gradually further away.

That the volume of the ocean is smaller is irrelevant. There's still a tidal bulge with the ocean, which corresponds to tides.

Cheers -- sylas

 

[D H]

The Earth tides are smaller, but not much smaller, than the ocean tides. Much smaller means smaller by at least an order of magnitude. The Earth tides are about 1/3 of the oceanic tides in magnitude. The "lossiness" of the Earth tides is much smaller than is that of the oceanic tides.

Right now, the lossiness of the oceanic tides is particularly high because the oceanic tides run into two north-south barriers: the Americas and Africa+Eurasia. This configuration is rather unusual. The average change in the Earth's rotation rate over the last billion years is smaller than the present rate. If the change was due to the Earth tides the change in length of day would have been a lot closer to constant.

 

[sylas]

The Earth tides are smaller, but not much smaller, than the ocean tides. Much smaller means smaller by at least an order of magnitude. The Earth tides are about 1/3 of the oceanic tides in magnitude. The "lossiness" of the Earth tides is much smaller than is that of the oceanic tides.

Thank you! I did not realize that they were that close in magnitude. I've learned something.

The rest is my understanding as well. Much more energy is dissipated in the oceanic tides. That's probably how I jumped too quickly to the bulge size. I've also been aware that energy dissipation from ocean tides varies with the locations of continents, and is particularly high in the present epoch.

Cheers -- sylas

 

[ideasrule]

I stand corrected (and interested). I really feel this should go in Wikipedia somewhere; do any of you have sources you could cite, DH and sylas?

 

[sylas]

I stand corrected (and interested). I really feel this should go in Wikipedia somewhere; do any of you have sources you could cite, DH and sylas?

The wikipedia article Tide seems pretty good, to me, at a quick glance. I am a registered editor on wikipedia, although I have done hardly anything there for a couple of years now. (Anyone can register; it is very easy to do, and has some advantages when you edit articles.)

There's a section on Dissipation, and that links to a more detailed article on Tidal acceleration, which covers this particular aspect of tides.

Cheers -- sylas