# The Moon's orbital plane

1. Aug 30, 2009

### Leo Klem

According to Newton's law, the attraction exerted by the Sun on the Moon is greater than twice the attraction exerted by the Earth on its satellite, whatever the Moon's position during its motion.
Nevertheless, the Moon's orbital plane around the Earth forms a 5.13 degree angle with the ecliptic's plane. The texts of astronomy I could so far access do either omit explaining this fact or clearly admit that current perturbation analysis does not provide any complete and clear explanation on the topic. It seems to be one of the anomalies detected within the solar system.
Somebody might remark that it's a "neglible" aspect, whereas the fact seems to imply that the Moon orbits the Earth irrespective of the solar attraction. In this connection, there is also to consider that the centre of mass of the Earth-Moon system does also orbit the Sun out of the ecliptic, and not according to any plane curve as instead required by the dynamics of central motions.
My question: is there any updated study relevant to the issue?

2. Aug 30, 2009

### D H

Staff Emeritus
Both statements are true. So what? It appears that you think the first statement means the Moon should not be orbiting the Earth.
The Sun/Earth/Moon system comprises three bodies. There is no nice, simple closed form solution to the general three body problem, even for three point masses. There are several other bodies in the solar system besides the Sun, Earth, and Moon. One has to resort to numerical techniques. These numerical techniques could predict the state of the solar system at any point in the future -- if we knew the exact current state (mass, position, and velocity) of all of the bodies in the solar system and if those bodies were point masses and if the solar system was isolated gravitationally from the rest of the universe and if our current best model of gravitation represents exactly how gravity works. Taking these ifs one point at a time,
• We do not know the exact state of the solar system. Our measurements of the masses, positions, and velocities of the solar system bodies are quite accurate, but they are not perfect. The solar system is a chaotic system, which means the future state is highly sensitive to errors in the current state.

• The solar system bodies are not point masses. That the Earth is not a point mass is particular important when it comes to the Moon's orbit about the Earth. The Earth's rotational angular momentum is being transferred to the Moon's orbit via the tides. The rate at which this angular momentum transfer occurs depends strongly on the configuration of the Earth. Right now, the Americas and Eurasia+Africa form two north-south barriers to the tides. This makes the angular momentum transfer considerably higher right now than the average rate over the last billion years or so.

• The solar system is a part of the Milky Way. Nearby stars can perturb the orbits of the planets. Because the solar system is chaotic, even the smallest perturbation will, over time, grow to become huge. To accurately predict the state of the solar system for billions of years means we would have to know and be able to predict the motions of the stars with which the solar system bodies will interact over that time frame.

• Finally, does general relativity truly explain exactly how things work? Probably not. The fervent hope of physicists who are working on things like string theory or quantum gravity is that it is not.

3. Aug 30, 2009

### Leo Klem

It doesn't seem an explanation to me. By the way, due to its relative size, the Earth-Moon system might be viewed as a unique body with respect to the Sun's mass and distance. The point, however, is that the Earth-Moon system's center of mass doesn't describe a plane orbit, whereas it is actually assumed that the Earth's centre runs on the ecliptic's plane. In addition, there is to note that the Moon's orbit around the Earth lies on a plane too.
This “anomaly” could – for example - be considered in association with the planet-satellite system of Saturn, where all the planet’s rings and satellites lie on the planet’s equatorial plane, i.e., with 26.731 degree inclination with respect to the planet’s orbital plane. This fact shows that even the combined strength of the gravitational influence of Sun and Jupiter (the nearest planet to Saturn, whose mass is 2.5 times the mass of the other planets altogether) doesn’t alter the amazing regularity of co-planar orbits of Saturn’s satellites.

Thus, if the huge complication of the overall astronomic/cosmic system, as recalled by you, can provide a satisfactory justification for the “macro-anomalies” I've pointed out, do you thing - for instance - of "the Pioneer's micro-anomaly" as of a case that is worth so much theoretical concern?

4. Aug 30, 2009

### D H

Staff Emeritus
An explanation of what? What do you perceive to be the problem?

Note well: My crackpot meter is starting to go offscale high.

5. Aug 30, 2009

### Staff: Mentor

Why does a pendulum on earth swing back and forth when it should be orbiting the sun?

6. Sep 3, 2009

### Krisia

I think there is an error. According to Newton’s law, the Earth’s gravity strength undergone by the Moon is approximately 4.5 times that exerted by the Sun. The Earth’s gravity should prevails over the Sun’s (Hill sphere) up to a distance of about 1.5 million km. (The Earth-Moon mean distance is 384,400 km, and the radius of the Moon’s Hill sphere should be about 60,000 km).
Nevertheless, the non-coplanar orbits of natural-regular satellites with respect to the planets’ orbital planes is an intriguing issue. It seems that natural-regular satellites (at variance with “stone-satellites”) prefer keeping their orbits co-planar to the respective planet’s equatorial plane (Uranus being the most spectacular example), except the Moon whose orbital plane has 5.22 degree inclination to the ecliptic and 23.53 degree inclination to the Earth’s equatorial plane.
Natural-regular satellites, which show spherical/spheroid shapes, seem strangely suggesting (against Newton’s law) that their own gravity fields differ from those determined (?) by “stone-satellites”.

7. Sep 3, 2009

### D H

Staff Emeritus
Leo was asking about the gravitational acceleration undergone by the Moon, not the Earth. His statement regarding the strength of the gravitational force of the Sun and Earth on the Moon was correct.

Lunar gravitational acceleration toward the Sun: $$\frac{\mu_{\text{sun}}}{(1\,\text{AU})^2} \approx 5.9\,\text{mm}/\text{s}^2$$
Lunar gravitational acceleration toward the Earth: $$\frac{\mu_{\text{earth}}}{(384,400\,\text{km})^2} \approx 2.7\,\text{mm}/\text{s}^2$$

The implication of Leo's comments are incorrect. While he never made his objection clear, he certainly thinks that we don't know why the Moon orbits the Earth.

Now you are on the right track regarding why the Moon does orbit the Earth, even though the gravitational force of the Sun on the Moon is greater than that of the Earth.

The Hill sphere is one measure of the boundary beyond which an orbit will no longer be stable. An alternative is Laplace's sphere of gravitational influence. Both the Hill sphere and sphere of influence look at a restricted three body problem from the perspective of where the perturbative gravitational force from the primary body equals the gravitational force from the secondary body. They differ in that the Hill sphere looks at things from the perspective of a rotating frame in which the third body is stationary while the sphere of influence looks at things from the perspective of a non-rotating frame.

Orbits will most likely become unstable well inside the Hill sphere / sphere of influence. The reason for this is that these surfaces mark the boundaries where the perturbative force from the primary equals the force from the secondary. Calling one force that is equal to another a perturbation is a bit of a misnomer. Perturbation forces should be small. An object at the edge of either sphere almost certainly will escape the secondary body. A more conservative estimate of where orbits become unstable is a third to a half of the Hill sphere / sphere of influence. Note well: the Moon is well inside these more conservative boundaries.

No mystery here either. The planets are not spherical bodies. They are instead oblate spheroids due to their rotation about their axes. On top of that, their rotation axes precess due to gravitational torque exerted by the Sun (and the Moon in the case of the Earth). The dynamical oblateness (J2) couples with this precession to keep moons that orbit close to a planet in that planet's equatorial plane.

8. Sep 3, 2009

### Leo Klem

Thank you Krisia, thank you DH. I owe both of you - and other readers - my apologies, since I’m afraid that Krisia is right. I made a blunder in copying figures from my pocket calculator.
Sun’s and Earth’s gravity accelerations exerted on the Moon are 5.94607*10-4m/s2, and 2.70017*10-3m/s2, respectively. The ratio of the latter to the former is approximately 4.54, as per Krisia’s statement. DH has perhaps incurred a copying mistake similar to mine. Sorry.

9. Sep 3, 2009

### D H

Staff Emeritus
You were right the first time, Leo. The gravitational force on the Moon by the Sun is indeed more than twice that exerted by the Earth.

By the Sun: http://www.google.com/search?q=G*(solar+mass)/(1AU)^2

Now let's look at the things from the perspective of an Earth-centered versus Sun-centered frame. From the perspective of an Earth-centered frame, the third body acceleration of the Moon toward the Sun is about 0.03 mm/s^2.

A rule of thumb is that perturbative effects are an order of magnitude smaller than primary effects. Since the third body acceleration of the Moon toward the Sun is about two orders of magnitude smaller than the acceleration of the Moon toward the Earth, this third body acceleration can definitely be categorized as a perturbative effect.

On the other hand, from the perspective of a Sun-centered frame, the third body acceleration of the Moon toward the Earth is about 2.7 mm/s^2, which is about half of the acceleration of the Moon toward the Sun. The third body acceleration of the Moon toward the Earth from a Sun-centerered perspective does not qualify as a perturbative effect. The effect is quite significant.

Google calculator results again:
Third body effect, Earth-centered: http://www.google.com/search?q=G*(solar+mass)+*(1/(1AU-384e3km)^2-1/(1AU)^2)
Third body effect, Sun-centered: http://www.google.com/search?q=G*(earth+mass)+*(1/(384e3km)^2-1/(1AU)^2)

Last edited: Sep 3, 2009
10. Sep 3, 2009

### Count Iblis

Also note that what matters is the fictitious tidal force exerted by Sun on the Earth-Moon system, not the real gravitational force. The Earth-Moon system is a freely falling system as it orbits the Sun. You have to work in this accelerating reference frame to see what the influence of the Sun is.

11. Sep 3, 2009

### D H

Staff Emeritus
That is exactly what I calculated in post #9, Count. See the google calculator links. I also calculated the third body effect in the accelerating Sun-centered frame. To recapitulate, the third body effect (aka tidal gravity) from the perspective of an Earth-centered frame of the Sun on the Moon is small compared to the primary gravitational force exerted by the Earth; this third body effect definitively qualifies as perturbative. The third body effect from the perspective of a Sun-centered frame of the Earth on the Moon is of the same order of magnitude as the primary gravitational force exerted by the Sun; this does not qualify as perturbative.

In short, one can look at the Moon as being in orbit around the Earth with the Sun being but a perturbative effect on that Earth-centered orbit or one can look at the Moon as being in orbit around the Sun but with the Earth as having something that is much more than a perturbative influence on that Sun-centered orbit.

12. Sep 3, 2009

### Leo Klem

I should definitively avoid calculations by means of pocket calculator, especially when I deem the calculation easy. The remark from Krisia made me doubt, and my doubt led me to a further mistake. Yes DH, I've verified my initial statement was correct, and I can also share in principle the conclusion quoted above.
According to others' perturbation analysis, however, it seems that there is not sufficient stuff to explain the orbit of the center of mass of the Earth-Moon system around the Sun. If the Earth's orbit lies on the ecliptic plane, then the center of the Earth-Moon mass cannot lie on a plane because of the Moon's orbit inclination to the ecliptic. Which (from the Newtonian point of view of central motion mechanics) makes things more complicated than desired, so that also the possible explanations become too elaborate.
In my opinion, the "three-body or multi-body problem" in Newtonian gravitation should be viewed as the true limit of the theory, rather than a question of analytical complexity. It seems the case to remind ourselves of the way in which Newton's gravitation law was derived, to accept the idea that explanations by use of that law have insuperable limits, as it was made evident by the problem of the perihelion precession.
Let me close here by thanking all of you.

13. Sep 3, 2009

### D H

Staff Emeritus
The answer to your concern is simple: The Earth's orbit does not lie on the ecliptic plane. It oscillates around the ecliptic plane, primarily due to the Moon.

Regarding the rest of your post: Just because a differential equation does not have a nice, clean solution in the elementary functions does not mean that the differential equation is not soluble. A solution can be found to any degree of accuracy desired.

Even if it did have a solution in the elementary functions as does the two body problem, that does not mean the exact answer is going to pop out like magic. Suppose you know the orbital elements for some body at some point in time. To find the cartesian state at some time other than the epoch, you need to
• Compute the mean motion, which if you are doing using normal computer/calculator arithmetic is not exact,
• Compute the mean anomaly, which once again is not exact,
• Compute the true anomaly, which requires numerical estimation techniques,
• Compute the sine and cosine of the true anomaly, which once again requires numerical estimation techniques,
• Compute the cartesian coordinates, which once again is not exact.

There are ways to get extra precision beyond that on your calculator, but the result will not be exact. It can, however, with a bit of computation horsepower be made as precise as one desires. That of course assumes you know the orbital elements exactly.

Which you don't.