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Here's how I demonstrated this concept to my son when he was younger. Take a plastic bottle and put some pebbles or little toys in it. Then toss it in the air and catch it. If you look in mid-flight, you can see the little toys just floating around inside the bottle. But they're still at 1G. And when you look at this example, it's totally obvious what's ...

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Your problem is that you are trying to do this in Cartesian coordinates (x,y). This makes the math much harder than it needs to be. It is more natural, when dealing with circular motion, to use polar coordinates - we express a position relative to the origin by its distance ($r$) and the angle relative to some reference axis ($\theta$). The relationship ...

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I suggest that it doesn't make much sense to say that the planets orbit the barycenter of the solar system. Beware that you are going very much against the grain of the best models of the solar system in writing that. All three of the leading ephemeris models (JPL's Development Ephemeris, the Russian Institute for Applied Astronomy's Ephemerides of the ...

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You seem to be groping towards the fact that the gravitational three-body problem is, in general, not solvable. We can get away with saying "the sun is at one focus of an elliptical orbit" in the solar system because the sun is so much larger than anything else around. The sun is 1000 times more massive than Jupiter, so the sun-Jupiter barycenter is about ...

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Ellipses, parabolas and hyperbolas are both: Defined as the conic sections you speak of: work out the intersection between the cone $\vec{R}.(\cos\phi\,\hat{X} + \sin\phi\,\hat{Z}) = \sin\theta\,|\vec{R}|$ and the plane $z=const$ where $\theta$ is your polar angle and $\phi$ the angle between the cone's axis of symmetry and the slicing plane and you'll ...

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The polar angle of the plane (the angle of the plane with respect to the symmetry axis) should relate to the eccentricity of the orbit. For $90^o$, the section is a circle and the eccentricity is zero. For an angle between the $90^o$ and the angle of the cone, you will get an ellipse with an eccentricity $0<\epsilon<1$. In both of these cases, the ...

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The Earth and the Moon revolve about their barycenter (center of mass of the Earth/Moon system), which is inside the Earth and is about 4,670 Km from the center of the Earth on a line connecting the center of the Earth with the center of the Moon (http://en.wikipedia.org/wiki/Barycentric_coordinates_(astronomy)). The apocenter of the Moon is its center when ...

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Since a new moon has to have the sun shine on the entire surface facing away from the Earth, the time between 2 new moons is the time for the Earth Moon system to complete a synodic period. The period of two masses around the common center of mass is always the same.

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This circle is the set of locations where the gravity of the sun and earth cancel out. Uhm. No. They are neither co-linear nor the same magnitude. To get a zero "force" at the Lagrange points you have to work in a rotating frame of reference, implying a centrifugal pseudo-force which closes the triangle. But the points on the circle you defined other ...

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At L4 and L5, the object would stay in the same relative position because it is in the same orbit around the Sun as Earth. Any other points on the circle would not be in the same orbit as Earth. They would be in orbits of differing inclination and so, in the course of the orbit of that object about the Sun, it would necessarily leave those points. The L4 and ...

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