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A good starting point would be <insert name of some scientist from long ago> planetary equations of motion. For example, there are Lagrange's planetary equations (sometimes called the Lagrange-Laplace planetary equations), Gauss' planetary equations, Delaunay's planetary equations, Hill's planetary equations, and several more. The common theme amongst ...

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It's mere coincidence. In about 10,000 years, perihelion will coincide with the northern hemisphere summer solstice. There are two precessions of interest here. One is the 26,000 year period axial precession of the Earth's rotation axis, caused mostly by the Moon and the Sun; the other is the 112,000 year period apsidal precession of the Earth's orbit, ...

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You must solve the Kepler's equation: $M = E - e sin E$ (where $M$ is the mean anomaly, $e$ the eccentricity of the orbit, $E$ the eccentric anomaly). As you can see, this is a transcendental function, so you will probably need to iterate through a calculator, in order to solve it. Many of useful formulas to understand the parameters dependence are ...

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I'm not sure how much this will help, but I can give you this passage from Wikipedia: Under ideal conditions of a perfectly spherical central body, and zero perturbations, all orbital elements, with the exception of the Mean anomaly are constants, and Mean anomaly changes linearly with time[dubious – discuss], scaled by the Mean motion, ...

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I do not know where that formula was first published in full, but Oppenheim at least does something very close to it. First, keep in mind some of the relevant symbols in Oppenheim, though they're rather standard: $$\def\anode{☊}\begin{eqnarray*} k &=& \small\sqrt{G} = \small\text{Gaussian gravitational constant}\\ \anode &=& ... 1 If you want to project onto the celestial sphere. as you say, then you want to convert to right ascension (\alpha) and declination (\delta). The equations of transformation are here 2 Since I don't have Walter's book, I'm uncertain as the context of the derivation of the equation you quote. Therefore, I've simply re-derived it here; apologies if there's some repetition of things you already know, but perhaps it'll be useful for anyone else reading this regardless. Constants of Motion The Schwarzschild solution is the unique nontrivial ... 0 No special techniques are required that I am aware of, but several observations must be taken to make a reasonable estimate of the orbit. Once a sufficient number of observations are made so that the orbital elements can be determined, then anything that has an elliptical orbit (eccentricity >0 but >1) can be assumed to be orbiting as part of the solar ... 2 The really simple explanation is that Venus' orbital period is 225 days. Earth's is 365 days. That ratio is 13:8 (well, more accurately 12.9777777777..:8) so you can see that your complete cycle is 13 orbits of Venus to 8 of Earth, but within that cycle our view of Venus travelling round the sun gives the pretty pattern as shown in your image. This YouTube ... 1 1) It is the mass internal to the orbit that determines the orbital speeds. These stars are so close to the center that the amount of bulge star mass contained within the orbit would be insignificant compared to the BH. 2) If the BH disappeared these stars would go into highly radial orbits out to much greater distances. That distance would be about where ... 5 Let the two bodies involved have masses m_1, m_2. Start with Newton's second law$$F = ma$$where a is acceleration. The gravitational force on body 2 from body 1 is given by$$F_{21} = \frac{G m_1 m_2}{|r_{21}|^3}r_{21} where $r_{21}$ is the relative position vector for the two bodies in question. The force on body 1 from body two is of course ...

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