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This turns out to have a really boring answer. We can find the two circular orbits by finding the maxima and minima of the effective potential, and we get: $$ r = \frac{L^2}{2M}\left(1 \pm \sqrt{1 - \frac{12M^2}{L^2}}\right) \tag{1} $$ where the $+$ gives the outer stable orbit and the $-$ gives the inner unstable orbit. Note that both orbits exist only ...


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The Quora quote is not useful because it does not tell you how to do any calculations. You need websites which explain how to calculate radius and period of orbit. eg http://www.physicsclassroom.com/calcpad/circgrav http://www.physicsclassroom.com/class/circles/Lesson-4/Mathematics-of-Satellite-Motion A more advanced webpage for calculating general (non-...


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More of an alternative suggestion than an answer to your parameterized ellipse approach. I've recreationally programmed a few similar many-body simulations myself (e.g., click my profile, then my homepage, then click on that voronoi link -- the points are moving under a mutual force law). Just model the force (gravitational in your case) directly, and the ...


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The local escape velocity is $$v_{esc} = \sqrt{\text{2 G M}/\text{r}}$$ At infinty you observe that velocity slower by a factor of $$1-\text{r}_s/\text{r}$$ so at infinity you observe $$\text{v}_{esc} = \sqrt{\text{2 G M}/\text{r}} \cdot (1-\text{r}_s/\text{r})$$ because of gravitational length contraction radial to the mass and gravitational time ...


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You may have mis-interpreted the question slightly. The question means to imply an instantaneous perturbation of a circular orbit (an impulse -- momenum kick -- for example) gives an elliptical orbit. Your second approach implies you are adding a simple harmonic motion to the ellipse, or perhaps that you are applying a time-dependent SHM external force to ...


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Short answer: Use the radial acceleration component $a_\mathrm r$ of the total $\bf a$ by the central force and then deduce the SHM equation using the properties of circular orbital motion (effective potential energy curves can also be used) that $\dot r= 0$ and keeping in mind the fact that SHM would only occur when the circular orbit is stable. Here the ...


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If you use Newton (left), there should be no perihelion shift. If you use Einstein (right), there is one: The equations of motion are not so different, if you differentiate by proper time the only difference between Newton and Schwarzschild is the red term in $$\color{black}{\ddot{r}(t) = -\frac{G\cdot M}{r(t)^2} + r(t)\cdot \dot{\theta}(t)^2} \color{red}...


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To begin with, it is important to clarify the scenario. When two masses are attracted to each other, regardless of their magnitudes, the gravitational force (and hence acceleration) vector acting on each is in the direction of the other. If two masses were sitting motionless in a vacuum, they would accelerate toward each other in a straight line until such ...


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In the Earth-Moon system, the mass of the Moon is sufficiently large and it is sufficiently close to raise tides in the matter of Earth. Because of this tidal friction can occur which dissipates energy. Enormous amount of power are involved in this process (of $\approx$ TW) and part of this causes the Moon to constantly being promoted to a slightly higher ...


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According to Newton's theory of gravity, the orbit would be an ellipse, which could take the form of a perfect circle. However, Einstein's theory of General Relativity tells us that this elliptical orbit would very gradually decay due to the emission of gravitational waves, and perhaps also precess. The Wikipedia page on the two-body problem in general ...


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It will depend on the initial conditions and I personally think that all you will need to answer your question can be found in any book on classical mechanics ("Kepler problem") or on Wikipedia


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The stable orbits around a star are given by the Kepler's laws oft planetary motion. In general these are ellipses with the center star in one of the two foci. Circular orbits are the special case when there is only one focus. For a orbit with a given radius, there is only one speed which allows a circular orbit. If you have a starting condition where the ...



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