Mystery of the Moon's Rotation: Investigating Comet Impacts

[quietrain]

why does the moon keep rotating about its own axis?

especially after so many comets have hit it?

 

[JaredJames]

Because the objects impacting aren't large enough to affect it.

They certainly do change the angular momentum of the moon, but in such a small way it's hardly noticeable.

 

[K^2]

More importantly, Moon is tidally locked to Earth, meaning that Moon's rotation around own axis matches rotation around Earth. If something like meteorite impacts would affect Moon's rotation, tidal forces due to Earth's gravity would compensate for it.

 

[quietrain]

but the craters are so huge and so many!
and i rmb reading about something that says a meteorite impact the size of a city= many times of TNT or something

it is so much more lethal that a nuclear bomb.

from wiki
"Based on crater formation rates determined from the Earth's closest celestial partner, the Moon, astrogeologists have determined that during the last 600 million years, the Earth has been struck by 60 objects of a diameter of 5 km (3 mi) or more. The smallest of these impactors would release the equivalent of ten million megatons of TNT and leave a crater 95 km (60 mi) across. For comparison, the largest nuclear weapon ever detonated, the Tsar Bomba, had a yield of 50 megatons."

i think that is enough energy to move the moon?

also if the gravitational force of Earth on moon is around ~10²⁰N
then what is the force of impact about? anyone knows how to calculate? or have some figures?

thanks!

 

[quietrain]

More importantly, Moon is tidally locked to Earth, meaning that Moon's rotation around own axis matches rotation around Earth. If something like meteorite impacts would affect Moon's rotation, tidal forces due to Earth's gravity would compensate for it.

i don't really understand why there are tidal forces on the moon when there aren't any water?

from wiki
http://en.wikipedia.org/wiki/Tidal_force
"it is the difference between the force exerted by the third body on the second and the force exerted by the third body on the first."

what do they mean?

 

[JaredJames]

but the craters are so huge and so many!
and i rmb reading about something that says a meteorite impact the size of a city= many times of TNT or something

it is so much more lethal that a nuclear bomb.

from wiki
"Based on crater formation rates determined from the Earth's closest celestial partner, the Moon, astrogeologists have determined that during the last 600 million years, the Earth has been struck by 60 objects of a diameter of 5 km (3 mi) or more. The smallest of these impactors would release the equivalent of ten million megatons of TNT and leave a crater 95 km (60 mi) across. For comparison, the largest nuclear weapon ever detonated, the Tsar Bomba, had a yield of 50 megatons."

Let's put it another way - the moon is still here after all of those collisions. Hence, they are not powerful enough to make much difference.

 

[K^2]

i don't really understand why there are tidal forces on the moon when there aren't any water?

Tidal forces act on all matter. Water simply moves a lot more than everything else, so we can observe the tides. The Moon itself is slightly stretched due to tidal forces. That little stretch is sufficient for Earth's gravity to "grab" at it, and force the same end of the Moon to always point at Earth.

The Moon is doing the same to Earth, but Earth being significantly more massive, and Moon's gravity being weaker, the combined pull of Moon and Sun has only slowed down Earth's rotation enough to increase Earth's day by a few hours since the Moon has formed. (I'm going with Giant Impact Hypothesis for this one.)

 

[quietrain]

Let's put it another way - the moon is still here after all of those collisions. Hence, they are not powerful enough to make much difference.

is it possible to tell me about how much force N is required to cause a disturbance to the moon's rotation?

 

[quietrain]

Tidal forces act on all matter. Water simply moves a lot more than everything else, so we can observe the tides. The Moon itself is slightly stretched due to tidal forces. That little stretch is sufficient for Earth's gravity to "grab" at it, and force the same end of the Moon to always point at Earth.

The Moon is doing the same to Earth, but Earth being significantly more massive, and Moon's gravity being weaker, the combined pull of Moon and Sun has only slowed down Earth's rotation enough to increase Earth's day by a few hours since the Moon has formed. (I'm going with Giant Impact Hypothesis for this one.)

oh... so that's why the same face of moon keeps facing us?

let me go read more about tidal forces... kinda confusing

 

[K^2]

is it possible to tell me about how much force N is required to cause a disturbance to the moon's rotation?

It's not about the force. It's about momentum. Any amount of net force, applied over long enough period of time, would make a difference. But impact force from asteroids is very brief.

 

[quietrain]

oh...

but the speed of impact is very high right?

so around what values of momentum will it affect the moon?

 

[JaredJames]

Well every impact transfers momentum to the moon - technically, every impact affects it.

The problem is that the amount imparted on the moon by an asteroid is tiny compared to the Earth-Moon systems overall momentum. Think of it like a fly hitting a truck windshield - lots of mess, but the truck doesn't go skidding off the road.

After enough small impacts it will have the same result as a few large ones. What that number is I'm not entirely sure.

 

[D H]

Tidal forces act on all matter. Water simply moves a lot more than everything else, so we can observe the tides. The Moon itself is slightly stretched due to tidal forces.

The Earth itself (along with the oceans) is subject to tidal forces from the Moon and the Sun. These are called Earth tides. While both the Moon and the Sun raise solid body tides on the Earth, the solid body tides on the Moon are almost solely due to the Earth.

That little stretch is sufficient for Earth's gravity to "grab" at it, and force the same end of the Moon to always point at Earth.

Another factor at play is the fact that the Moon's center of mass is not located at the Moon's geometric center. The center of mass is about about 2 km away from the center of figure, in the direction of the Earth.

The Moon is doing the same to Earth, but Earth being significantly more massive, and Moon's gravity being weaker, the combined pull of Moon and Sun has only slowed down Earth's rotation enough to increase Earth's day by a few hours since the Moon has formed. (I'm going with Giant Impact Hypothesis for this one.)

Per the giant impact hypothesis, the Moon formed from the orbiting debris field at a distance of the about six Earth radii from the center of the Earth. The length of an Earth day at the time the Moon coalesced from this debris was only about six hours or so. In other words, the Moon has slowed the Earth's rotation rate by about 75%.

 

[K^2]

Another factor at play is the fact that the Moon's center of mass is not located at the Moon's geometric center. The center of mass is about about 2 km away from the center of figure, in the direction of the Earth.

I'm not sure how you figure this affects the situation. Center of mass is precisely the point where the tidal force is zero by definition. Not to mention the fact that any forces acting on center of mass cannot affect the torque around the center of mass, again, by definition.

What matters here is mass distribution. Fact that mass is distributed further from center on the radial line than in tangential direction is what marks a stable equilibrium with respect to tides. The amount of mass on near side and on the far side from center of mass is the same, again, by definition of center of mass.

Fact that geometrical center differs from gravitational center is peculiar, however. I did not know that. I wonder what caused that.

 

[D H]

I'm not sure how you figure this affects the situation. Center of mass is precisely the point where the tidal force is zero by definition.

Not at all. You can do physics in any frame you want. The torque on some object is zero at whatever point you choose as the origin of the object. Choosing the center of mass as the origin often (but not always) results in simpler equations of motion because the translational and rotational equations of motion become uncoupled. If you wanted to one could model the Moon's state (position, velocity, orientation, rotation rate) with respect to the Moon's geometric center as opposed to with respect to the Moon's center of mass. However, this would mean that the translational state and rotational state would couple to one another, and in a rather nasty way. It is easier to model the state with respect to the Moon's center of mass.

So, how does this off-center center of mass play into the equations of motion when the origin of the object is chosen to be the center of mass? Imagine a thin spherical shell of mass m_s and radius of r_s connected rigidly (by massless struts) to some point mass m_p that is located some distance r_p from the center of the spherical shell. Now compute the inertia tensor about the center of mass. Even though the object appears to be spherical, the inertial tensor about the center of mass will not be that of a spherical mass distribution (some constant times the identity matrix).

The Moon is not quite spherical; it has a small not non-zero k₂ Love number. The non-spherical nature of the Moon's inertia tensor results in part from the permanent tides and in part from the off-center CoM.

 

[Dale]

but the craters are so huge and so many!
and i rmb reading about something that says a meteorite impact the size of a city= many times of TNT or something

it is so much more lethal that a nuclear bomb.

from wiki
"Based on crater formation rates determined from the Earth's closest celestial partner, the Moon, astrogeologists have determined that during the last 600 million years, the Earth has been struck by 60 objects of a diameter of 5 km (3 mi) or more. The smallest of these impactors would release the equivalent of ten million megatons of TNT and leave a crater 95 km (60 mi) across. For comparison, the largest nuclear weapon ever detonated, the Tsar Bomba, had a yield of 50 megatons."

i think that is enough energy to move the moon?

Let's do some rough estimates.

The moon's moment of inertia is about 9E34 kg m².

The angular velocity of the moon is about 3E-6 rad/s corresponding to a rotational KE of 4E23 J. Because it is tidally locked with the Earth the next available angular velocity is about 6E-6 rad/s corresponding to a rotational KE of 1.6E24 J. So the difference in energy is about 1.2E24 J.

On the other hand 10 million megatons of TNT is only 4E22 J. So the energy in such a collision is only about 3% of the necessary energy to move the moon up to the next tidal resonance mode.

Furthermore, the energy is probably not as important as the momentum because there are so many places for the energy to go. You could do a similar analysis for the momentum.

 

[K^2]

Not at all. You can do physics in any frame you want. The torque on some object is zero at whatever point you choose as the origin of the object. Choosing the center of mass as the origin often (but not always) results in simpler equations of motion because the translational and rotational equations of motion become uncoupled.

Which is exactly the point. CoM of the Moon is the part that follows simple orbital motion. When talking about tidal forces, it is the only relevant pivot.

So, how does this off-center center of mass play into the equations of motion when the origin of the object is chosen to be the center of mass? Imagine a thin spherical shell of mass m_s and radius of r_s connected rigidly (by massless struts) to some point mass m_p that is located some distance r_p from the center of the spherical shell. Now compute the inertia tensor about the center of mass. Even though the object appears to be spherical, the inertial tensor about the center of mass will not be that of a spherical mass distribution (some constant times the identity matrix).

And would you like me to construct an example where the center of mass of a spherical object is off geometrical center, yet the object possesses fully degenerate inertia moments?

Yes, broken symmetry is going to nearly guarantee that the symmetry in the tensor will also be broken, and I_r will be different than either I_t. But whether it will be lower or higher is still up to density distribution.

On the other hand, knowing that planetary/lunar surface is very near equipotential, simply because forces involved are way outside of any material's critical strength, I can tell you right away that a tidally locked body will be elongated radially with respect to its orbit around parent body, regardless of whether geometrical and gravitational centers coincide.

 

[D H]

Let's do some rough estimates.

The moon's moment of inertia is about 9E34 kg m².

The angular velocity of the moon is about 3E-6 rad/s corresponding to a rotational KE of 4E23 J. Because it is tidally locked with the Earth the next available angular velocity is about 6E-6 rad/s corresponding to a rotational KE of 1.6E24 J. So the difference in energy is about 1.2E24 J.

That is the 2:1 spin-orbit resonance. You missed a couple of more likely scenarios here. One is a 3:2 spin-orbit resonance, which requires a smaller jolt of energy. Another possibility, one that requires a much, much smaller jolt, would be just enough to get the Moon rotating at a slightly different rate, only to be later drawn back into a 1:1 spin-orbit resonance after a half a revolution. The current far side would be the new near side.

 

[D H]

On the other hand, knowing that planetary/lunar surface is very near equipotential, simply because forces involved are way outside of any material's critical strength, I can tell you right away that a tidally locked body will be elongated radially with respect to its orbit around parent body, regardless of whether geometrical and gravitational centers coincide.

The Moon's surface is not an equipotential surface, nor is it in hydrostatic equilibrium. It was an equipotential surface after coalescing. The Moon was also considerably hotter and more plastic way back then than it is now. This made for a short relaxation time in response to deviations from hydrostatic equilibrium and deviations from the geoid. At some point (conjectured to be ~3 billion years ago) the Moon "froze" into its current shape. The Moon's k₂ tidal Love number is now about 0.026. Compare that to the 0.30 value for the Earth. The Moon is now very stiff and has an extremely long (billions of years) relaxation time.

 

[Dale]

That is the 2:1 spin-orbit resonance. You missed a couple of more likely scenarios here. One is a 3:2 spin-orbit resonance, which requires a smaller jolt of energy.

Oops, I didn't think of half-integer resonances.

Another possibility, one that requires a much, much smaller jolt, would be just enough to get the Moon rotating at a slightly different rate, only to be later drawn back into a 1:1 spin-orbit resonance after a half a revolution.

Yes, but I was assuming that we were considering impacts that were far enough in the past for all transients to have already died down.

 

[quietrain]

suddenly i don't understand anything... :(

except maybe what dale said about not enough energy to affect the moon ...

oh well, thanks everyone ..

 

[K^2]

The Moon's surface is not an equipotential surface, nor is it in hydrostatic equilibrium. It was an equipotential surface after coalescing. The Moon was also considerably hotter and more plastic way back then than it is now. This made for a short relaxation time in response to deviations from hydrostatic equilibrium and deviations from the geoid. At some point (conjectured to be ~3 billion years ago) the Moon "froze" into its current shape. The Moon's k₂ tidal Love number is now about 0.026. Compare that to the 0.30 value for the Earth. The Moon is now very stiff and has an extremely long (billions of years) relaxation time.

Now you are being silly. "Froze"? What sort of a material you are imagining that can resist these sort of forces? Its viscosity is certainly significantly higher, but it is still for all intents and purposes a liquid. Small deviations from equipotential might take a long time to disperse, but compared to the effect of elongated shape of the quipotential surface, these minor deviations are going to be minor corrections to Moon's dynamics.

And again, what does shifted CoM have to do with any of it?

 

[rcgldr]

Just in case this point was missed, the moon's orbit isn't quite circular and people on the Earth can see about 59% of the moons surface instead of just 50% of the moon surface if there was no variation in the part of the moon facing the Earth during the moons orbit.

 

[D H]

Now you are being silly. "Froze"?

Yes, "froze". The Moon is a lot more ellipsoidal than it would be were it in hydrostatic equilibrium. If the Moon was in hydrostatic equilibrium the tidal bulge would only be 10 meters. The bulge is nearly three orders of magnitude larger than that, ~770 km on the near side. The common explanation is that the Moon obtained its shape a long, long time ago and has retained it ever since.

D. E. Smith and M. T. Zuber, "Inferences About the Earth Moon from Gravity and Topography," Origin of the Earth and Moon Conference (1998),
http://www.lpi.usra.edu/meetings/origin98/pdf/4075.pdf

On its simplest level the present shape of the Moon is slightly flattened by 2.2 ± 0.2 km while its gravity field, represented by an equipotential surface, is flattened only ~0.5 km. The hydrostatic component to the flattening arising from the Moon’s present-day rotation contributes only 7 m. This difference between the topographic shape of the Moon and the shape of its gravitational equipotential has frequently been explained as the “memory” of an earlier Moon that was rotating faster and had a correspondingly larger hydrostatic flattening.​

The paper concludes with (emphasis mine)

Thus, it seems that both the observed flattening and the ellipticity of the equator require a Moon that was much closer to the Earth, at a distance of 13 to 16 R_Earth, when the Moon “froze in” its present shape.​

Here is another paper. There are many, many more. That the Moon is not in hydrostatic equilibrium is a well-known fact.

M. Lefftz and H. Legros, "Some remarks on the non-hydrostatic form of the Moon," Physics of the Earth and Planetary Interiors, 76:3-4, 317-327
http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V6S-470WC04-3B&_user=2148702&_coverDate=03%2F31%2F1993&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_acct=C000056362&_version=1&_urlVersion=0&_userid=2148702&md5=33d4c1c6ae6be4632af398a8f353b9bf&searchtype=a

Finally, we show that the prevailing non-hydrostatic form of the Moon can be explained by an internal load (which appeared 3 Ga ago) located between the upper and the lower parts of the lunar lithosphere, which creates viscoelastic surface deformations in the selenoid that are not completely relaxed at present.​

What sort of a material you are imagining that can resist these sort of forces? Its viscosity is certainly significantly higher, but it is still for all intents and purposes a liquid.

Liquid? And you called me silly? Even the bulk of the Earth is not "liquid". While the outer core is liquid, the inner core is solid, the mantle is solid, and the crust is solid. Not a rigid solid, but solid. It is a deformable solid, so plastic might be a better term. This is evidenced by the Earth potential k₂ Love number, 0.39. Compare that to Jupiter's moon Io, which has a potential k₂ Love number of 1.2, and to our Moon, which has a potential k₂ Love number of about 0.021 (see below). I previously reported in post #19 a value of 0.026, which apparently is an older estimate. Published estimates vary, but they are all very small.

The potential k₂ Love number is a measure of the elasticity of a body with regard to potential (gravitational, tidal, rotational). It is a unitless parameter that can range from 0 for a perfectly rigid body to 1.5 for an incompressible liquid. Io is close to an incompressible liquid. Our Moon is close to being a rigid body. This paper addresses recent measurements of the Moon's potential k₂ Love number:

S. Goossens and K. Matsumoto, "Lunar degree 2 potential Love number determination from satellite tracking data," Geophysical Research Letters, 35:K02204 (2008).
http://www.agu.org/journals/ABS/2008/2007GL031960.shtml

This Lunar and Planetary Science Conference paper by Goossens and Matsumoto presents the same material but is not behind a paywall:
S. Goossens and K. Matsumoto, "Determination of the Lunar k₂ Love Number from Satellite Tracking Data,"
http://www.lpi.usra.edu/meetings/lpsc2008/pdf/1536.pdf

And again, what does shifted CoM have to do with any of it?

From the first cited paper,

The present shape of the lunar equator is dominated by an ellipticity (2, 2 terms) of ~770 m, coupled with a 1.7 km degree 1 term that represents the offset of the center of figure of the Moon from its center of mass.​

 

[D H]

Yes, but I was assuming that we were considering impacts that were far enough in the past for all transients to have already died down.

Mercury is in a 3:2 spin-orbit resonance precisely because its orbit is quite eccentric. Because the Moon's orbit is nearly circular, any transient would have died down to the Moon once again being in a 1:1 spin-orbit resonance.

Because the Moon is fairly rigid (see my previous post), a small "whack" would, over the long term, have zero effect. The tidal torques would make the Moon settle back to original orientation. That the Moon is not perfectly rigid and that its orbit is not perfectly spherical means that perturbations in its rotation caused by that small whack will damp out rather quickly (geologically quickly).

By way of analogy, think of a car on a road over a series of undulating hills, with the car at rest in one of the troughs between hilltops. If the road was frictionless, a small push on the car would result in the car climbing partway up a hill, then rolling back down, then climbing the hill behind, etc. This oscillatory motion would be undamped if the road was frictionless. A slightly harder push would make the car climb higher but still oscillate in its local valley. Give the car a good hard push and the car would climb hill after hill after hill. Now suppose the road is soft sand. This will damp out the small oscillations from a slight push. If you push the car just hard enough to climb over the first hill it will settle down in the next valley.

The tidal torques resulting from the Moon's ellipsoidal shape give rise to a hilly landscape with respect to deviations from a tidally-locked configuration. That the Moon is not perfectly rigid means this hilly landscape is made of soft sand. A good solid "whack" from an asteroid hitting the Moon could have made the Moon flip by 180 degrees. The distribution of craters on the Moon suggest that this did happen at some time in the past.

M. Wieczorek and M. Le Feuvre, "Did a large impact reorient the Moon?", Icarus, 200:2, 358-366 (2009)
e-print: http://peer.ccsd.cnrs.fr/docs/00/51/72/48/PDF/PEER_stage2_10.1016%252Fj.icarus.2008.12.017.pdf
Lunar and Planetary Science Conference paper: http://www.lpi.usra.edu/meetings/lpsc2009/pdf/1554.pdf

Lay article:
Marianne Dyson, "The New Face of the Moon", Ad Astra 22:3, 38-41 (2010)
http://www.nss.org/adastra/volume22/AdAstra_MoonFace_2010.pdf

 

[K^2]

What sort of a material you are imagining that can resist these sort of forces? Its viscosity is certainly significantly higher, but it is still for all intents and purposes a liquid.

Liquid? And you called me silly? Even the bulk of the Earth is not "liquid". While the outer core is liquid, the inner core is solid, the mantle is solid, and the crust is solid. Not a rigid solid, but solid.

I hope I do not need to explain the difference between solid as a phase and solid as a mechanical property. Even an ordered crystal, due to defects, can flow. Let alone poly-crystal structures. Under sufficient stress, any object in its solid phase is still modeled as a liquid in terms of its mechanical properties. Granted, it is a very, very viscous liquid, but it is most certainly not a rigid body.

From the first cited paper,
The present shape of the lunar equator is dominated by an ellipticity (2, 2 terms) of ~770 m, coupled with a 1.7 km degree 1 term that represents the offset of the center of figure of the Moon from its center of mass.

That still says absolutely nothing. Which part of this tells me that eccentricity has anything to do with the Moon's inertia tensor? Again, a perfectly spherical object with shifted center of mass can have fully degenerate moment of inertia. Do you need me to construct an example, or can you see that?



The rest is interesting. I'll have to run the numbers. The difference in the actual bulge and the equipotential surface is only as significant as the gradient in the effective potential. I do not think the gradient can be strong, as the forces required to maintain such a bulge under significant gradient would be enormous. But in light of your citations, I'm going to need actual numbers to back that up. I'm going to go come up with some.