# Tag Info

49

Still new at stellarium but here are some quick capture gif lasting one month. Sorry about the quality- limited to 256 colors for smaller gifs. Date on lower left corner. By the way the sun is of course the brightest and i use it as reference for recording (start record when sun is in frame then stop when it appears again in the same position which is ...

42

You are correct. The Earth would always appear in approximately the same location in the sky, when viewed from a point on the lunar surface. And it would be seen to spin, the continents coming in and out of view over the course of an Earth day (24 hours). The sun would make it's way across the sky, from one horizon to the other over a period of about two ...

20

To add to Mark Bailey's answer; the Earth would indeed hang in the sky and rotate, but it would also wax and wane over the course of a lunar day (27.3 Earth-days). Starting at lunar dawn, the Earth would be half-full. The Earth would then wane (more shadow) towards lunar noon. At lunar noon, the Earth would be all in shadow (New Earth) and quite close to ...

17

No. These words are English, not Greek. "Periapsis" means the point on the orbit when the two bodies are at their closest. It doesn't matter if this good Greek or bad Greek, it is correct English. Apsis actually derives from "arch" and indicates the two points where the orbital curve is most "arched": ie the points of greatest ...

16

What exactly are "non-Keplerian" orbits? Strictly speaking, no orbits are in perfect accordance with Kepler’s laws. Kepler’s laws aren’t really “laws” in terms of physical laws, but are instead trends that Kepler noticed and calculated using astronomical observations of the planets. Kepler’s laws are very accurate for planetary orbits since he ...

16

Both ellipticity $f$ (also called flattening) and eccentricity $e$ are measures of how elongated an ellipse is, based on the semi-major axis $a$ and the semi-minor axis $b$ (figure from wikipedia). They are defined respectively as $$f=\frac{a-b}{a}$$ and $$e=\sqrt{1-\frac{b^2}{a^2}}$$ For a circle, $a=b$, which implies that $f=e=0$. In modern orbital ...

15

Well done you. I double checked the calculations and couldn't fault what you had done. So I contacted the lead author of the paper about it and here is the response: "After checking the numbers in our paper, I found an error: we actually used a mass of 1.0 MSun for J1407 in our simulations, instead of the 0.9 MSun as stated. This accounts for the difference ...

15

What exactly are "non-Keplerian" orbits? Orbits that don't follow Kepler's laws. Strictly speaking, all orbits are non-Keplerian. In practice, one can model some orbits as basically Keplerian, but with perturbations. Sun synchronous satellites are one example of orbits that are close to Keplerian, but not quite so. The Earth's equatorial bulge ...

13

I'm assuming this is Pluto's equatorial plane. That is incorrect. Charon's right ascension of ascending node with respect to Pluto's equator is undefined. Seemingly paradoxically, it is well defined with respect to Earth's equatorial plane, and since that has become the universal plane of reference, that is what is used. Charon and Pluto are tidally locked ...

13

I'm not sure what the focus is on Sgr A*? Only the stars that are very close to the Galactic center can be said to be "orbiting Sgr A*", the rest of the stars in the Galaxy orbit in the the entire Galactic potential, to which Sgr A* is a minor contributor. The stars near Sgr A* have a wide variety of orbital eccentricities. This is shown clearly in ...

12

To expand a little more, yes the Earth would hang in the same spot in the sky, moving around in a small circle as the moon rotated around it over the course of each of its 28 day orbits. It would have phases, full Earth when the moon is between it and the sun, new Earth when the Earth is between the moon and the sun, and wax and wane between these two points....

12

Well yes, because Brown & Batygin (2021) say the perihelion would be $300^{+85}_{-60}$ au, so there is roughly a 50% chance that the perihelion is smaller than 300 au according to their work. The quoted error bars represent their estimate of the 16th to 84th percentiles of a probability distribution, so conceivably the perihelion could even be a little ...

11

The vis-viva equation doesn't care about eccentricity, so no, you can't get it from just the semimajor axis. However, given the radial distance and the velocity vector, and the gravitational parameter, this is how I would go about calculating orbital eccentricity: Orbital Elements Givens: Parameter Symbol gravitational parameter $\mu$ radial distance ...

9

You are correct that the IAU definition of "clearing the orbit" has the problem of being not explicitly quantified. And a complete clearing was obviously never the intention behind the definition. I like this statement by Steven Soter: The IAU definition of a planet as a heliocentric body that "has cleared the neighborhood around its orbit” is problematic....

9

Ellipses have a "long radius" called the "semi-major-axis" which is the length from the centre to the ellipse measured along the long axis. And a "semi-minor-axis" which is measured along the short axis. Call the semi-major-axis "a" and the semi-minor-axis "b". Ellipses also have foci: which is where the ...

7

I like to classify solutions of the problem of the time evolution of the complete initial state of a set of objects at some epoch time, where the objects are subject to Newtonian gravitation into two main groups. One approach is to use orbital elements of some sort. The other is to use a numerical initial value problem solver, aka a numerical integrator. The ...

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

6

I am assuming that the question you want answered is how to calculate the elevation of an orbit above a reference plane given the orbital inclination with this plane. If so, please update your question to reflect this, heeding the advice given in the comments. Kepler's first law tells us that planets move in elliptical orbits, which we can define as follows,...

6

Planets don't orbit stars. Stars don't orbit planets. Whenever there are two bodies bound by gravity, they are both orbiting their common center of mass. For example, both the Earth and the Moon orbit their common center of mass - but that's pretty close to the center of the Earth actually, so it seems like the Moon orbits the Earth. For a star to seem to ...

6

The IAU gives no definition, it leaves the term somewhat vague. Various calculations of the "planetariness" of solar system bodies use slightly different notions of the "neighbourhood" Soter's $\mu$ considers bodies to be in the neighbourhood if they share a common radial distance (ie the orbits cross each other) and the orbital periods differ by less than ...

6

A disclaimer first - I am not an astronomer, but I am a Greek with some ancient Greek language knowledge. "Periapsis" is definitely ancient Greek and it derives from peri+apsi (περί+αψη). Apsi is the noun of the verb "άπτομαι" which means touch something. Having said that, this word is only being used in astronomy and there is no literal ...

5

Perigee is the Earth-specific name for periapsis. People use longitude (which is a composite angle rather than an angle) because this solves the problems of circular and equatorial orbits. The reason you need to use an epoch time to specify the Moon's argument of perigee and longitude of ascending node is because the Moon's orbit about the Earth precesses. ...

5

A little reading and effort is recommended and the Wikipedia page is straight forward enough, if you're not completely math-phobic. But if you want an answer, Rob Jeffries 9,500 km is right, however, that's center point to center point. Surface to surface, that would only be less than 2,000 km, which would be crazy close. That close the Moon would take ...

5

You can't without assuming something about the overall velocity. The radial velocity is one component of a velocity vector; you are missing the other two components, which could in principle be anything. However, you could assume that as open clusters are mostly quite young and members of the Galactic disc population, that they are moving in the disc on ...

5

I think E and M are in degrees in your code, but the trig functions in the math library in C# expect an argument in radians. One way to handle this is to define constants radians = Pi/180 and degrees = 180/pi then always do Mathf.Sin(M*radians). You will also need to do this for the lines float xx=. Alternatively, you could work entirely in radians. ...

5

I agree that the common explanation of the longitude of periapsis, as the question gave it, appears at first sight to make no sense -- because it constructs a sum of two angles in two different planes, and that seems at first sight to lack geometrical meaning. In fact the fault, if any, is in the explanation rather than in the geometry. It is not so ...

5

I believe you are right, this is extracted from the "Explantory supplement to the Astronomical Almanac (2006?)" (although it might date back to the 1960s...) For example a similar PDF document gives that citation. The supplement is published in a separate binding from the main astronomical almanc and a new version is published each year. Finding ...

5

But where is this orbit centered? The true anomaly is the angle as measured from the central body between periapsis passage and the object's current location. The orbit is an ellipse with one of the two foci at the central body. This concept is central to Kepler's laws and Newtonian mechanics. The eccentric anomaly is the angle as measured from the center ...

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