Why the Moon always shows the same face

[Tony Hambro]

From Earth, we always see precisely the same half of the moon.
Isn't the simplest explanation for this odd fact that the Moon's core is not uniformly dense, and it's centre of gravity is situated some distance further way from us than its geometric centre but on a direct line extended beyond the centre of its face from the centre of gravity of the Earth? The heaviest bit will always be the furthest away.
If you stick a weight on one side of a tennis ball, put it in a bucket of water on a rope and swing it round fast, the weighted side of the floating ball will stay furthest way from you.
The common consensus explanation has to do with our tides but I don't buy that.

 

[Orodruin]

https://en.wikipedia.org/wiki/Tidal_locking

 

[Orodruin]

The common consensus explanation has to do with our tides but I don't buy that.

The consensus explanation builds on a basic understanding of classical Newtonian mechanics and can be shown mathematically from Newton's laws of motion. Your "explanation" is, forgive the expression, word sallad with no basis in Newtonian physics. Gravitation does not work like your example with the bucket and the tennis ball. That is based on a pressure gradient and potential energy in a rotating frame of reference with dissipation of energy to the surroundings.

In the end, it does not matter if you "buy" it or not. It is the correct explanation. It is up to you if you want to learn actual physics or base your understanding of the world upon your own imagination.

 

[sophiecentaur]

From Earth, we always see precisely the same half of the moon.
Isn't the simplest explanation for this odd fact that the Moon's core is not uniformly dense, and it's centre of gravity is situated some distance further way from us than its geometric centre but on a direct line extended beyond the centre of its face from the centre of gravity of the Earth? The heaviest bit will always be the furthest away.
If you stick a weight on one side of a tennis ball, put it in a bucket of water on a rope and swing it round fast, the weighted side of the floating ball will stay furthest way from you.
The common consensus explanation has to do with our tides but I don't buy that.

That's not the reason for the water in the bucket. There is a rope pulling the handle towards the person from an off-centre position on the bucket. The Moon has no rope and, left to itself, it would rotate at some other rate for ever. The 'real' reason is that parts of the Earth and Moon are free to flow a small amount as the other body's gravity pulls at it and acts as a Drag force between the two. That is a mechanism for loss of rotational energy and tends to dissipate differential rotational energy. Eventually the two will end up facing each other all the time. Because there has to be conservation of angular momentum, the actual separation will increase, eventually. But we are talking in terms of a verrrry long time. That link from @Orodruin will tell you the details. His explanation is perfectly correct but may need a bit of translation. I have given a Noddy Version for you. No matter if you didn't need it; someone else may find it useful.
PS the Moon rocks a bit from side to side so we actually see a bit more than half of it over a period of time.

 

[Janus]

PS the Moon rocks a bit from side to side so we actually see a bit more than half of it over a period of time.

It also "nods" up and down as its axis of rotation is perpendicular to the plane of its orbit around the Earth.

This video shows this libration.

 

[alantheastronomer]

From Earth, we always see precisely the same half of the moon.

Yes, look up - see the dark features that make up the Man in the Moon? When Soviet satellites first imaged the far side of the moon in the early 60's, they discovered that there were no similar features, only a heavily cratered surface. It turns out that those dark features were more than just a difference in coloration. After the Apollo missions it was learned that those dark areas, called Maria, consisted of a very dense mineral called basalt, while the rest of the crust was made of much lighter silica.

If you stick a weight on one side of a tennis ball, put it in a bucket of water on a rope and swing it round fast, the weighted side of the floating ball will stay furthest away from you. The common consensus explanation has to do with our tides but I don't buy that.

Yes, I can see why you'd be skeptical - if the gravity of the Earth acted on the Moon the same way that the rope acted on the weighted tennis ball, then the heavier side of the Moon should be facing away from the Earth, not towards it. Well, in the tennis ball example, it's not the rope that's exerting a force on the ball, it's a fictitious force from the curved motion called centrifugal force, which is directed outward, not inward. The weighted side of the ball, being heavier, having more inertia, is affected more by this outward directed force. The near side of the Moon however, is affected more by the inward force of gravity, and so faces towards the Earth, rather than away.

 

[Tony Hambro]

The consensus explanation builds on a basic understanding of classical Newtonian mechanics and can be shown mathematically from Newton's laws of motion. Your "explanation" is, forgive the expression, word sallad with no basis in Newtonian physics. Gravitation does not work like your example with the bucket and the tennis ball. That is based on a pressure gradient and potential energy in a rotating frame of reference with dissipation of energy to the surroundings.

In the end, it does not matter if you "buy" it or not. It is the correct explanation. It is up to you if you want to learn actual physics or base your understanding of the world upon your own imagination.

 

[Tony Hambro]

The explanation given in Wikipedia, that the bulge in the shape of the moon from the stronger gravity on the nearer side facing the Earth produces torque that eventually cancels rotation, may be true if the moon really does bulge enough. But if the moon’s centre of gravity is not in its geometric centre (who knows?), as with my weighted tennis ball in a bucket of water swung round on a rope, it would produce a similar effect, the face with the weight always being furthest away from the centre of rotation. So both views could be true. Incidentally neither explanation tells us why the moon/ball has no rotation around the axis linking the two centres of gravity. Any ideas?

 

[russ_watters]

The explanation given in Wikipedia, that the bulge in the shape of the moon from the stronger gravity on the nearer side facing the Earth produces torque that eventually cancels rotation, may be true if the moon really does bulge enough. But if the moon’s centre of gravity is not in its geometric centre (who knows?)...

Scientists know. The shape and mass distribution of the moon is very accurately known, and scientists have mathematically modeled its behavior. There are no open questions warranting speculation here.

...as with my weighted tennis ball in a bucket of water swung round on a rope, it would produce a similar effect, the face with the weight always being furthest away from the centre of rotation.

I'm sorry, but repeating it does not help. Moreover you've made your scenario overly complicated; you don't need to swing the bucket, just set it down on the floor and the tennis ball assumes the same orientation; center of gravity low due to gravity and buoyancy. The moon does not have buyoyancy.

So both views could be true.

Sorry, but no. You're going to have a lot of trouble learning science if you want to develop the answers yourself. It is much better to learn from the thousands of scientists who have spent hundreds of years figuring this stuff out.

Incidentally neither explanation tells us why the moon/ball has no rotation around the axis linking the two centres of gravity. Any ideas?

If the Moon was formed in an impact, the impact would impart a rotation in the direction of a motion...like bouncing a tennis ball at an angle does.

 

[PeterDonis]

the heavier side of the Moon should be facing away from the Earth, not towards it.

The near side of the Moon however, is affected more by the inward force of gravity, and so faces towards the Earth, rather than away.

You appear to be claiming that the near side of the Moon is heavier than the far side. That is not correct. And it is not necessary to account for the Moon keeping the same face to the Earth. Tidal locking accounts for that, as has already been pointed out several times.

 

[Orodruin]

This is not correct. The rope does exert a force on the ball.

The rope is not connected to the ball, it is connected to the bucket. The centripetal force on the ball is that from the pressure gradient of the water in the bucket.

 

[PeterDonis]

The rope is not connected to the ball, it is connected to the bucket.

Oops, you're right, I misread that in the OP.

 

[alantheastronomer]

You appear to be claiming that the near side of the Moon is heavier than the far side. That is not correct. And it is not necessary to account for the Moon keeping the same face to the Earth. Tidal locking accounts for that, as has already been pointed out several times.

The GRAIL spacecraft provides a map of lunar surface gravity that shows a clear asymmetry between the near and far sides. An uneven mass distribution is a facilitator of tidal locking, not meant as a substitute for it.

 

[PeterDonis]

The GRAIL spacecraft provides a map of lunar surface gravity that shows a clear asymmetry between the near and far sides.

Ah, that's right, I was forgetting about the mass concentrations beneath the maria, which are concentrated on the near side.

 

[sophiecentaur]

An uneven mass distribution is a facilitator of tidal locking, not meant as a substitute for it.

Yes. That's the point. If there is no internal friction, the Moon could rotate or swing back and forth like a pendulum for ever. To get a stable situation, you need damping.

 

[PeterDonis]

If there is no internal friction, the Moon could rotate or swing back and forth like a pendulum for ever. To get a stable situation, you need damping.

I don't see how an unequal mass distribution (a better term might be "unsymmetrical" since nobody expects the density to be the same everywhere) is required to get frictional damping. Even a perfectly symmetrical Moon won't stay that way under tidal forces; it will be distorted, and the distortion will vary with time as the Moon swings back and forth about the equilibrium position. This time-varying distortion will cause friction in the Moon's material and hence will damp the vibrations.

 

[sophiecentaur]

I don't see how an unequal mass distribution (a better term might be "unsymmetrical" since nobody expects the density to be the same everywhere) is required to get frictional damping. Even a perfectly symmetrical Moon won't stay that way under tidal forces; it will be distorted, and the distortion will vary with time as the Moon swings back and forth about the equilibrium position. This time-varying distortion will cause friction in the Moon's material and hence will damp the vibrations.

It's a matter of Gravitational Potential of an asymmetrical body* in different orientations and not just the separation of the CMs. PE is added and taken away and the KE changes accordingly - it's not uniform rotation. Tidal effects are due to the forces on the body as Energy changes cyclically. Where there's any movement (water or molten core) there can be losses.
In your response I think there's a hidden a truth(?) and that is there can be movement and losses, for any shape if there is any non-uniformity and some liquidity (if that's the right word.) A lop-sided distribution will presumably make the tidal losses faster.

*Do the sums for a dog bone shape and a nearby sphere. The GP when the bone lies along a radius and when at right angles to the radius are easy to calculate and the difference is apparent. (Bed time for me though!)

 

[PeterDonis]

It's a matter of Gravitational Potential of an asymmetrical body

No asymmetry of gravitational potential is required for tidal gravity to be present. A perfectly spherically symmetric, non-rotating Earth would have a perfectly spherically symmetric gravitational potential, but would still cause tidal gravity in the space around it. Of course the non-sphericity and rotation of the actual Earth add additional effects.

Tidal effects are due to the forces on the body as Energy changes cyclically.

No, they're due to tidal gravity, i.e., an extended body like the Moon, moving in the tidal gravity of the Earth, will be stretched in the radial direction and squeezed in the tangential directions. This will happen even if the Moon's orbit is perfectly circular and its kinetic and potential energies never change. Of course the non-circular orbit of the actual Moon means the tidal gravity it feels won't be exactly the same as this idealized example.

there can be movement and losses, for any shape if there is any non-uniformity and some liquidity (if that's the right word.)

"Non-uniformity" ("asymmetry" would be a better word for what I think you mean here) is required only in the sense that the tidal gravity will stretch the object in some directions and squeeze it in others. It is not required for either the Earth or the Moon themselves to have any asymmetry.

A lop-sided distribution will presumably make the tidal losses faster.

I haven't done the math but intuitively this seems right to me, yes.

 

[sophiecentaur]

How would tidal gravity operate? Whet mechanism would be at work?

 

[Orodruin]

Tidal forces are due to the gravitational acceleration changing from point to point, leading to a different acceleration of different parts of the same object (leading to internal stress and thereby deformation if the object is deformable). As long as the separation $d\vec x$ between two points is small, the difference in gravitational acceleration at the points is given by
$$
\delta \vec g = \vec g(\vec x + d\vec x) - \vec g (\vec x) \simeq d\vec x \cdot \nabla \vec g.
$$
In terms of the gravitational potential $\Phi$ such that $\vec g = - \nabla\Phi$, you would have
$$
\delta \vec g \simeq - \vec e_j dx^i \partial_i \partial_j\Phi = - (d\vec x \cdot \nabla) \nabla \Phi.
$$
Thus, tidal forces are best described by the rank 2 tensor that is the second derivative of the gravitational potential. You do not need any asymmetric objects to generate a potential with non-zero second derivatives. A point mass works just fine (although the object affected needs to be extended for the tidal forces to be noticeable).

 

[snorkack]

Try rotating a well-balanced bicycle wheel vs. rotating a poorly balanced one.
A well-balanced bicycle wheel set in motion would still slow down because of friction against hub and air, and eventually stop. But it would slow down and stop without any rocking back or forth in the end, and stop in a random position each time. Whereas a poorly balanced bicycle wheel would slow down and stop at the same specific position each time, and oscillate before stopping.
Further, since a well-balanced wheel stops at random position, even a small nudge will make a small but permanent movement, accumulating over time.
Moon acts like an ill-balanced wheel, not like a well-balanced one.

 

[PeterDonis]

How would tidal gravity operate? Whet mechanism would be at work?

What mechanism makes gravity in general work? In either Newtonian gravity or General Relativity, tidal gravity is no more mysterious than gravity in general.

 

[russ_watters]

Try rotating a well-balanced bicycle wheel vs. rotating a poorly balanced one.
A well-balanced bicycle wheel set in motion would still slow down because of friction against hub and air, and eventually stop. But it would slow down and stop without any rocking back or forth in the end, and stop in a random position each time. Whereas a poorly balanced bicycle wheel would slow down and stop at the same specific position each time, and oscillate before stopping.
Further, since a well-balanced wheel stops at random position, even a small nudge will make a small but permanent movement, accumulating over time.
Moon acts like an ill-balanced wheel, not like a well-balanced one.

The problem with this description (the OP's take) is that you have two separate forces: one holding the wheel up at its geometric center and another pulling down at the center of mass. The moon in this model (where we ignore tides) only has one force, pulling "down" (toward Earth) on the center of mass. There's no second force to provide the torque (couple).

There is no geometric bias with the tidal force. The fact that the moon is oriented this way is a coincidence or result of a different process such as biased meteor bombardment.

 

[jbriggs444]

Try rotating a well-balanced bicycle wheel vs. rotating a poorly balanced one.

For fairness, you need to either do this in a weightless environment (difficult) or with a rigidly mounted axle (easier). Do not hold it in your hands or use an upside-down bicycle. You do not want to compromise the experiment by having an energy dump into a partially damped axle.

You should find that a small motion imparted to an ill balanced wheel persists for a long time, just like a well balanced wheel. It is just that it is a pendulum motion rather than a rotary one.

 

[snorkack]

The problem with this description (the OP's take) is that you have two separate forces: one holding the wheel up at its geometric center and another pulling down at the center of mass. The moon in this model (where we ignore tides) only has one force, pulling "down" (toward Earth) on the center of mass. There's no second force to provide the torque (couple).

There is no geometric bias with the tidal force. The fact that the moon is oriented this way is a coincidence or result of a different process such as biased meteor bombardment.

Are you sure there is no "geometric bias"?

Tidal force is principally a difference between two forces. Centrifugal repulsion and gravitational attraction. They add to zero over the whole Moon, but since the repulsion increases with distance (proportional to it) and attraction decreases (with inverse square), the far side of Moon experiences more repulsion and near side more attraction. Both are thus away from centre of Moon.

Now what kind of asymmetry can cause Moon to librate?
Take the example of bicycle wheel again.
A well balanced wheel is a perfect circle with hub in its exact centre.
An ill-balanced wheel might be imbalanced in several ways.
It might be a perfect circle but with hub off circle. Then it would, on a road, bump once each turn - and when rotating on suspended hub, swing off the hub as a pendulum.
It might be elliptical, with hub at centre. A wheel like this would on the road bump twice each turn - but rotating on suspended hub, the two long ends of the elliptic wheel would balance each other exactly, so it would rotate freely then.
Or it might have higher order deviations. A square wheel would not roll on a road, but its corners would balance each other when hanging in the air.
In case of Moon, elliptical Moon would experience a torque. Two bulges would be attracted to near and far sides respectively.
But higher order asymmetry? Would a cubic moon turn one face (or edge or corner) to Earth, or would the perfect balance of pairs of opposing corners leave Moon freely rotating?

 

[PeterDonis]

Tidal force is principally a difference between two forces. Centrifugal repulsion and gravitational attraction.

No, it isn't. The difference between those two forces is the net "acceleration due to gravity" on a body in a rotating frame, not the tidal force. In the language you are using here, tidal force would be the "difference in the difference" from one point to another.

Also, tidal gravity is present in an inertial frame, where there is no centrifugal force at all, so thinking of it in terms of centrifugal force is not the best way to think of it. A much better way was given by @Orodruin in post #20.

what kind of asymmetry can cause Moon to librate?

No asymmetry in the Moon itself is necessary for the Moon to librate. The effects of the asymmetries due to the eccentricity and inclination of the Moon's orbit, plus the asymmetry between the Moon's period of rotation and the Earth's, are sufficient.

 

[sophiecentaur]

There is no geometric bias with the tidal force. The fact that the moon is oriented this way is a coincidence or result of a different process such as biased meteor bombardment.

I think it would be a good idea if someone specified what they actually mean by the terms "tidal gravity" and "tidal locking". Is it just the forces involved or is it to do with distortion of a body (and friction losses)? Clearly, it is the result (can be described in terms) of Newtonian Gravity. The forces on any distributed mass will be different in all parts of the mass (CM is just an approximation) and the geometry of the object will have an influence on those forces.
I am sure that any form of "locking" must be due to the loss of Energy due to distortion and friction and the tendency towards a minimum of Potential Energy in the system. It's not clear which parts of the argument are accepted or rejected by contributors to the thread so could I have a response to that last sentence?

 

[sophiecentaur]

No asymmetry in the Moon itself is necessary for the Moon to librate.

I haven't understood why a symmetrical, rigid body would have any potential minimum dependent on its orientation. I think we need to be defining terms more precisely as we may well not be disagreeing.

 

[jbriggs444]

I haven't understood why a symmetrical, rigid body would have any potential minimum dependent on its orientation. I think we need to be defining terms more precisely as we may well not be disagreeing.

In a gravitational field with a uniform gradient, it would not. However, the Earth's gravitational field is neither uniform nor does it have a uniform gradient.

Edit: I have to correct myself. A stick is symmetrical, but nonetheless has an orientation-dependent potential in a field with uniform gradient.

 

[PeterDonis]

I think it would be a good idea if someone specified what they actually mean by the terms "tidal gravity" and "tidal locking".

@Orodruin gave a good explanation of tidal gravity in post #20.

To understand tidal locking, consider the Earth-Moon system. Not only does each body exert tidal gravity on the other, the tidal gravity of each on the other changes with time, for a number of reasons:

The Earth is rotating much faster than the Moon revolves around it;

The Moon's orbit about the Earth is inclined to the Earth's equator;

The Moon's orbit about the Earth is not perfectly circular.

All three of these things cause the deformation of both the Earth and the Moon due to tidal gravity to change with time, which causes frictional damping and torques on both bodies. These effects act to slow the Earth's rotation, to make the Moon's orbit less inclined to the Earth's equator, and to make the Moon's orbit more circular. Also, conservation of angular momentum means that as the Earth's rotation slows, the Moon's average distance from the Earth increases.

At some point in the future, the Earth's rotation will have slowed to exactly match the Moon's rate of revolution about the Earth; the Moon's orbit will be in the Earth's equatorial plane; and the Moon's orbit will be perfectly circular. At that point, the deformation of each body due to the tidal gravity of the other will no longer be time-dependent; it will be "locked" in a certain configuration on each body. Once that point is reached, the orbital parameters will stop changing, and the system is said to be tidally locked.