Mass of Moon in a Non-Equilibrium Earth-Moon System

The two body equilibrium problem is relatively trivial. It would occur when the mutual attractive gravitational forces were exactly counterbalances by the centripetal forces of each body around their center of mass. If the centripetal force is greater then the attractive forces they crash into each other, less--they fly apart.

The three body problem; however, is much more difficult And since the Earth/Moon system is in effect a three body problem (the third body is the sun) it really isn't in equilibrium and the moon is very slowly moving away from the Earth.

Lagrange, determined the stable solutions for the three body problems, and determined five exact solutions (known L1 through L5), and when 19th century astronomers looked at the L3 and L4 points in the Jupiter Sun system they found what are known as the Trojan asteroids.

Rap

The Earth and moon both orbit their common center of mass. Since the Earth has a greater mass, this point is much closer to the Earth than to the moon.

If you don't ignore the motion of the bodies around the sun, you are going to have a much more complicated set of equations than you are probably prepared to deal with. I would ignore the presence of the sun. Furthermore, I would make the additional simplifying assumption that the Moon orbits the Earth, rather than both orbiting their common center of mass, and also that the orbit is a perfect circle, rather than an ellipse. I think you will still end up with a pretty good number.

The force required to keep any body moving in a circle is mv^2/r. I would try to equate this to the gravitational force the Earth exerts on the moon.

Dear Sir,

I know what you are saying; I truly do. However, my purpose here is to suggest that the relative mass values of the earth and moon, currently about 81.3 to 1 respectively, may be RADICALLY WRONG. This is why I wish to treat the system as one not in equilibrium, to see if this is indeed the case. My problem is that I do not know what Equation is valid here. All I do assume is that Mass x Acceleration = Mass x Acceleration, is wrongly applied to the earth and moon. In effect, I am looking for someone who is somewhat of an expert in phasing mathematically the correct relationships of a non-equilibrium system.

Sincerely

Keith.

Dear Rap

Could the relations in a three body problem as you put it be radically different than in a two body problem? That is, could the relative mass values of the earth and moon be very different than 81.3 to 1 in a three body problem than when treated as a two body problem with the equation Acceleration x mass = Acceleration x mass?

Moreover, do you know the equation of the three body problem that would be of help here?

Sincerely

Keith

Alright, you can go ahead and do the calculation, but bear in mind that the whole world scientific community has been looking at this for over 300 years. I am pretty skeptical that there is has been a major error in the celestial dynamics of the Earth-moon system for that length of time, with so many physicists looking at the equations.

Brandon

Brandon,

I tell you the truth, the theory of an earth centred universe was dominant for at least 1500 years, and the epicyles of Ptolemy and Copernicus for about the same. Both were proven false in time.

People can get things completely wrong and hold to their wrong views for millennia. History proves the truth of this.

Sincerely

Keith.

Keith,

All of that happened before physics was invented. Newton published his Principia in the late 17th century. This is quite different. Now people just look at the equations. While it is theoretically possible, it seems unlikely that the most common problem in celestial mechanics has been wrong under the noses of centuries of scientists.

Brandon

I think this has all been worked out in detail, that is unless you consider the Apollo landings fiction. From the mirror/lazar measurements it is known that the moon is moving away from the earth about 5 cm per year indicating a barely unstable orbit, and that the baracenter (the center of common mass) is about 900 miles inside the surface of the earth.

A good discussion of the mathematics of this subject can be found at Orbital Mechanics (Click on the blue type to link)

As for the three body problem, Lagrange demonstrated that there were no analytical solutions except for his five special cases. Numerical methods have been developed and these numerical methods do an excellent job (they worked for the tour of the solar system).

Wikipedia gives a good description of the geophysical properties of the Earth and the Moon

Rap

How precise do you want to get? The Sun is probably treated as a point mass as are Earth and the Moon.

If the whole system is in equilibrium you can easily do calcs using a centre of mass between the earth and the moon, then supposing this centre is consistent around the Sun as a point source.

Its a bit more complicated than that. First the path of the Earth around the Sun is elipitical, the Earth/Moon pair are closest to the Sun in January.

Second an orbiting body produces a tidal force on another body. So in the Earth / Moon example the Moon is slowing down the Earth rotation, and the Earth Moon is expanding every year. When the dinosaurs where alive the Moon was 20 times closer to the Earth! For the Earth/Moon to Sun bodies the Sun has a fluid surface, the Sun rotates every 4.5 days at its poles and every 6 days at its equatator.

Third although you can calculate an elipitical path of the centre of suspension of the Earth/Moon around the Sun using Keplar's laws, the fact the Earth is heavier than the Moon and the curvature of Space time being a 1/d^2 calculation will produce error into your model.

Fourth the mass of the Earth, Moon and Sun is changing every day. The Sun radiates a fantastic amount of energy every second, and E = mc^2, so the Sun's gravitational potential is decreasing every day (although this amount is tiny expressed as a percentage against its overall mass). All three are being struck by meteors every day, increasing their mass. The Moon has lost all its atomsphere, whereas the Earth is probably just heavy enough for its gravity to trap a large percentage of the atomic (i.e un-paired) Hydrogen in our atomsphere, else like Mars we'd be venting atomsphere every year until life couldn't be supported.

Fourth as the Sun is almost the centre of our Solar System's point of suspension, given all the eliptically orbiting planets change this point.

Fifth our Solar system is just one of hundrerd of millions orbiting around our galaxy's hub, which is in turn orbiting a glactic cluster's hub, which is in turn orbiting a super clusters hub, which is in turn orbiting a super thread's hub.

So once again how much precision do you require? It's all varying with time. Imagine watching with time set to 1 second = 10 million years. A typical Sun would last for 20 minutes before it would have radiated probably 60% of its mass as energy. And this is happening across the entire Universe as we ponder. Perhaps this wash of energy everywhere, plus the reminant momentum of the big bang's inflationary epoch accounts for both dark matter and dark energy.

Point,

What does the term "equilibrium" mean to you?

The way this term is being used in this thread doesn't square with the term used in normal science. I must admit I don't understand what you are trying to describe

Before you define this key term in your hypothesis, this conversation doesn't make any sense.