I've had a read of a few pages linked in other threads, but I can't quite follow. So I tried to have a go at my own pace.

To start off, I simplified things into a one dimensional situation, considering only the points on the Earth-Moon line (L1-3).

Now, the frame of reference is of course non-inertial, centred on the barycentre, rotating with angular speed w. The positive direction is moonward. The position of Earth relative to the barycentre is R1 and the position of the moon is R2.

Now, a third orbiter lies somewhere along the line at position r relative to the barycentre. The forces acting on it are the gravity from Earth, the gravity from the moon and the inertial force.

F = -GM2m/(R1+r)² + GM2m/(R2-r)² + mw²r

In general GM=µ. So the acceleration is simply

r'' = -µ2/(R1+r)² + µ2/(R2-r)² + w²r

We're looking for equilibrium points, which are stationary points on the potential energy versus displacement graph. In other words, a LaGrange point will happen when the rate of change of potential energy with respect to displacement is zero. dU/dr is in fact the force. So, in fact, the LaGrange points are where r''=0. The kind of equilibrium point is determined by the direction of the line as it crosses the axis. If it is going from negative to positive, it is a stable equilibrium position. If it is going from positive to negative it is an unstable equilibrium position.

I plotted r'' on the hyperquack and got a curve that only once touched the touched the x axis. It did so in a position which would correspond with L1, except the lines was going from negative to positive when in fact that implies L1 is stable, which it is not.

Clearly something is wrong with the equation. Anyone?

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Freedom For Fission A breath of fresh Iodine-131

I think your equation is okay, assuming you didn't mean to use the same mass for the Earth and Moon. :wink: I don't think it's "general" though, in the sense that it applies for all three points. The directions of your forces imply a position between the two other bodies. Try writing your equation assuming a position beyond the Earth (or Moon) and see if it gives you the other point(s). As for it being stable, I'm not sure but that may have to do with the simplified frame you've defined.

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Wayne
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"Epistemology models ontology" - John Polkinghorne

You need to use a few restrictions in order to solve the three body problem. The primary and the secondary masses must be “large”, the secondary's orbit around the primary must be circular, and the third mass – the satellite – must be neglectable compared to the other two.

The Three Body Problem
Lagrange points
A simple derivation of L1

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"The moment we want to believe something, we suddenly see all the arguments for it, and become blind to the arguments against it." -George Bernard Shaw

The signs in front of the three terms are consistent with L1, but not L2 and L3. For L2, both the first two terms would be negative. L3 similarly but it should fall out of the L2 calculation too.
For your equation, for L1, that is backwards. In order for a point between the Earth and moon to be stable, it would have to experience a positive acceleration if r is less, and a negative acceleration if r is greater--which would tend to restore it to its previous position. So, if r'' goes from positive to negative, it is a stable point.

So I'll need to go with the full vector equation.

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Freedom For Fission A breath of fresh Iodine-131

Or just change the signs of the terms, if you're only interested in finding L2 and L3 right now.

Here's another attempt at an equation:

r'' = -µ1(r-R1)/|r-R1|³ - µ2(r-R2)/|r-R2|³ + w²r

I'm going to have to wait until tomorrow to get access to the University computer software to plot it.

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Freedom For Fission A breath of fresh Iodine-131

Does anyone have experience with Mathcad Professional? I want to try to plot the potential energy surface graph.

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Freedom For Fission A breath of fresh Iodine-131

Okay, so I came up with the potential energy eqn

U = -µ1/|r-R1| - µ2/|r-R2| + ½w²r²

The surface plot I got showed essentially a honkin' potential energy well around the barycentre with no sign of LaGrange points.

I tried changing the sign on the centrifugal energy term to -ve but interestingly, that simply flipped the well around to become a maximum as though the centrifugal energy was the only thing that mattered.

Standby One!

Ah. Entered the wrong value for w. No that's it's corrected, something changed. Instead, it looks like a plane with a small but sharp, almost singular, potential energy well in the middle.

But there's no sign of the usual LaGrange points.

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Freedom For Fission A breath of fresh Iodine-131

So you left mass out just to simplify--still that's not "potential" energy.

Regardless, are you setting R1 to zero? If not, why not? Doesn't that effect the value of r? What values are you using for the u?

Shouldn't w be the angular velocity of m2, which you can solve for?

Okay, so specific potential energy!

R1 is about 3×10^6m. I simply used the centre of mass principle. µ1R1 = µ2R2

µ=GM in general so µ1 = 3×10^14 and µ2 = 3×10^12

w = 2pi/(29×24×3600s)

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Freedom For Fission A breath of fresh Iodine-131

So it's the distance of the Earth from the barycenter? That's about 4.7x10^6m, idnit? Then R2 is a negative 4x10^8m, eh?
Those are the masses of the Earth and moon times G? I get 4x10^14 (N m^2/kg) and 5x10^12
It's closer to 27 days than 29.

If you're using mathcad, you might as well use accurate values for the parameters.

I defined the x component of R1 to be -ve and the other one to be positive.

Yeah that's it for µ1. I remembered that the moon's mass is approximately a hundredth of the Earth.

I'll improve as soon as I get hold of the data.

What about the specific potential energy eqn? Is that right?

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Freedom For Fission A breath of fresh Iodine-131

Is "-ve" a shorthand for "negative"?

Yes.

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Freedom For Fission A breath of fresh Iodine-131

So you're allowing r to vary but not w? That's not realistic, is it?

Why not? Since we're Newtonian, both bodies would orbit round the barycentre at the same rate, wouldn't they?

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Freedom For Fission A breath of fresh Iodine-131

r is the radius of the orbit of the third body, right?

What would 1/2 w^2 r^2 represent if r were allowed to vary, but w was not?

No, no. r is the radius vector of the satellite relative to the barycentre. w is the angular rotation of the frame of reference, not specifically the satellite.

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Freedom For Fission A breath of fresh Iodine-131