# Tag Info

30

C means "comet": it has a coma which means that volatiles are being released due to solar heating. Other possible letters are "A" for asteroid, "P" for (short) periodic comet, "D" for disappeared comet, "I" for interstellar object and "SN" for supernova. 2020 is the year of discovery F represent ...

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Velocity is a form of kinetic energy, while height within a gravity well is a form of potential energy. For an orbiting body, conservation of energy will keep the total energy constant. So as a planet moves away from the parent star, it loses velocity and gains potential energy. As it moves closer, it trades the potential energy back for velocity. The point ...

9

The formula is Kepler's equation, but to understand it you need to know three values: $M$ is the "Mean Anomaly". It increases linearly from 0 to 360 over the period of one orbit, measured from periapse to periapse. So if a planet has a period of 100 days, then the mean anomaly at day 0 is 0, at day 50 it 180degrees, at day 25 it is 90 degrees. This is only ...

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I want to point out that the Moon's orbit isn't circular now. A 0.055 mean eccentricity isn't that circular. But, onto your question. I think you're making a bad assumption on the "must have started very elliptic. Individual objects that are ejected from a planet need to follow their orbital path. So any object flung from the earth would need ...

7

Both formulae are correct. The discrepancy is because the formula from the eccentric anomaly article uses the centre of the ellipse as the origin, but the formula from the Kepler's equation article uses a focus of the ellipse (i.e, the central gravitating body, eg the Sun) as the origin. Note that $c = ae$ is the distance from the ellipse centre to a focus

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I know that quoting Wikipedia is frowned upon here, but as there has been no other answer posted, this is what the Wikipedia article on Kepler has to say about the matter: He then set about calculating the entire orbit of Mars, using the geometrical rate law and assuming an egg-shaped ovoid orbit. After approximately 40 failed attempts, in early 1605 he ...

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You may want to use perturbation theory. This only gives you an approximate answer, but allows for analytic treatment. Your force is considered a small perturbation to the Keplerian elliptic orbit and the resulting equations of motion are expanded in powers of $K$. For linear perturbation theory, only terms linear in $K$ are retained. This simply leads to ...

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The reason is that gravity is a radial force, and so conserves angular momentum $\mathbf{L} = \mathbf{r}\times m\mathbf{v}$, where $\mathbf{r}$ is the vector to the planet from the star and $\mathbf{v}$ is its velocity. First, remember that the cross product of two vectors $\mathbf{a}$ and $\mathbf{b}$ has a magnitude that's equal to the area of a ...

5

Not sure if you're looking for a more mathematical answer or just the "why", but to answer the why, I'll start with some history on this. Everyone who worked out a model for the Solar System, from Aristotle to Copernicus, liked circles. Even though Copernicus correctly reasoned that the Earth moved around the Sun and not the Sun around the Earth, he ...

5

To find the width (semi-major axis, see Wikipedia) of an orbit, you only need two of the following: perihelion, aphelion, and eccentricity, represented by the variables $P, A,$ and $e$, respectively, and $a$ is the semi-major axis. If you are given perihelion and aphelion, you can just take the average of the two: $a=\dfrac{A+P}{2}$ If you are given ...

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One of the answers to the researchgate question What are the equation of motion for elliptical restricted three body problem? says: https://www.sciencedirect.com/search?qs=equation%20of%20motion%20for%20elliptical%20restricted%20three%20body%20problem&show=25&sortBy=relevance which returns a large number of papers, most paywalled. The answer also ...

4

The Moon has a "Roche lobe", where the Moon's gravity dominates, and all you need for the situation you describe is that the Roche lobe lies physically outside the Moon. That will be true, because the Lagrange points are on the outside edge of that Roche lobe, and it's outside the Moon. The Moon's Roche lobe is shown here: http://hyperphysics.phy-astr.gsu....

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The answer is yes. Stellar collisions are events that occur mainly in Globular clusters due to their high stellar density. The likelihood of these events is highly dependent on the stellar density. As we know the galactic centers concentrate large amounts of matter and, of course, thousads of stars which are mainly in the old red main sequence. However, ...

4

Just to provide an analytical formula for @uhoh's correct time-averaged distance, here the derivation of $\langle r\rangle_t=1+\epsilon^2/2$: $$a=1 \qquad c=e\qquad b=\sqrt{1-e^2}\\ \vec{r}=(\cos \beta-e,\sqrt{1-e^2}\sin \beta)\\ \vec{r'}=(-\sin \beta,\sqrt{1-e^2}\cos \beta)\\ |r|^2=\cos^2 \beta -2e\cos \beta+e^2+\sin^2\beta-e^2\sin^2\beta=1-2e\cos \beta+e^... 4 The maths says that the semi-major axis is not a good measure of average distance for high eccentricity (elliptical) orbits. There are basically two ways to measure this : (1) an average over the entire orbit on a purely geometric basis, and (2) the average over time. These give quite different results - qualitatively different. Average distance on ... 3 For bodies small enough to be non-spherical, the mass is relatively low (as an example, Rosetta's bounces took a long time - it touched the surface at 15:34, 17:25 and 17:32 GMT comet time) but the centre of mass is still the point the lander will orbit. So a chaotic orbit is not a problem here What will be problems are: trying to arrange for the lander's ... 3 We know that two celestial bodies rotate around a center of mass. "We" do not know that. Rhetorical question: Pick up an apple and drop it at high noon. Does it fall towards the solar system barycenter? The answer is of course "No." It falls away from the barycenter, towards the Earth. I'll steal a graph from an answer I wrote to a question at the sister ... 3 Orbital eccentricities cannot be normally distributed around zero, since the minimum eccentricity is zero. Thus, in the same way that the Maxwell-Boltzmann distribution for particle speeds is zero at zero speed and peaks at finite speed, then no asteroids have exactly zero eccentricity and we would expect a peak at higher eccentricities. The analogy can be ... 2 A high density of stars definitely increases the chances of collisions, however, the high velocity of a halo star orbiting wouldn't increase its chances. Since the halo star is traveling very quickly, the halo star would only spend a small amount of time near the galactic core. Additionally, it would have less time to get deflected, gravitationally, by other ... 2 You may take a look at the lates parametrization file by JPL-NAIF for the precession, nutation and pole orientation of the largest known bodies. Although, for the large time scales you are asking, I expect you will need to propagate the data and make your own wild guess, or dig into appropiate literature about solar system physics. 2 No. While you're on that planet, you are also in the same orbit around the star as the planet is. As long as you're in the same orbit as your vehicle (planet, spacecraft), you don't experience any acceleration relative to the vehicle. Whether you're on the planet, or you're just tagging along on the same orbit, is the same thing. Similar question: in a ... 2 The gist of this, is that your assumption is incorrect. It's the semi-major axis that defines the period, not the average distance. Newton worked this out when he invented calculus and derived Kepler's laws. (Look up Derivation of Kepler's laws for some explanations). Here's the Wikipedia version of the Math. I should add that Kepler's three laws, ... 2 This was intended to be a supplemental answer to StephenG's answer. However, there appears to be a problem with the expression for time-averaged distance in that answer. I think it's great to seek a mathematical expression, but it should be confirmed numerically. I did a quick numerical double check and verified those general trends, but there may still be ... 2 My answer won't be complete, from lack of time and resources, but I still wanted to share some interesting aspects here that could be helpful. The difficulty in aswering this question revolves around the complex and irregular shapes involved here. Also, finding a "best-fit" ellipse for comparison is not as easy as it seems, because it depends on what and ... 2 You can get a position and velocity snapshot easily using the Python package Skyfield. It downloads a JPL Development Ephemeris and then interpolates the positions for whatever time you choose for your snapshot. You can also use JPL Horizons to produce one or thousands of positions of solar system bodies. This answer describes how to set up Horizons to ... 2 Earth's Eccentricity varies between about 0.000055 and 0.0679. This is the first Milankovich cycle. When Earth' eccentricity is at its lowest, it is lower than that of Neptune. The Earth's orbit approximates an ellipse. Eccentricity measures the departure of this ellipse from circularity. The shape of the Earth's orbit varies between nearly circular (with ... 1 tl;dr: The synodic period would be the same. For this object it will be about:$$\left( \frac{1}{365.25} - \frac{1}{557} \right)^{-1} \approx 1061 \ days. You can confirm that by reading this new ArXiv preprint about Roadster: The random walk of cars and their collision probabilities with planets. In the 2nd paragraph of Section 3 it says: The ...

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Well, I don't really think the GI Hypothesis can answer the orbital plan of the moon, I would say that the circular orbit of the moon today regarding the GIH is also due to other variables, like Jupiter, the Sun, etc... If the GIH is true, then the moon should still be on a quite elliptical orbit still, since the gravitational forces of other celestial ...

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