# Tag Info

46

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 ...

41

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 ...

19

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 ...

14

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 ...

11

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 ...

10

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....

8

Short answer: because Neptune's argument of the perifocus, the reference point for the true and mean anomalies jumps around a lot because Neptune's orbit is nearly circular. If you run HORIZONS w/ the following parameters: you'll notice Neptune's mean anomaly also decreases at times. Both of these occur because Neptune's argument of the perifocus (from ...

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

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 ...

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

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 ...

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 ...

4

So, basically you're trying to merge orbital inclination to Kepler's laws. The simplest way is to take the measured orbital inclination of the planet, which is constant, and apply the Pythagorean theorem to any given location and that gives you 3 dimensions of distance from the 2 dimensions defined in Kepler's laws. That's probably what Kepler did. (see ...

4

As the commenter states, $e$ is indeed called the orbital eccentricity. If you add a radial scale length (e.g., semi-major axis) to both of those values the $1-e$ describes the periapsis (closest approach of orbit) and the $1+e$ apoapsis (furthest) of an elliptical orbit. They don't have a specific special name, as they are dimensionless measures, but can be ...

4

There's no reason why 2 bodies of equal mass couldn't have elliptical orbits around each other. There's an example of that here . The simple way to think about this is, if two bodies of similar mass approach each other, one of two things can happen, they either have sufficient velocity to pass each other with some hyperbolic curving of both objects ...

4

Not exactly. During either of the equinoxes there is a moment when the line between center of earth and the sun aligns with the equator. This doesn't coincide with prime meridian in any way though; it may happen at any meridian whatsoever that happens to coincide with the line. Of course if it happens the sun is in zenith above prime meridian at that ...

4

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. ...

4

The primary driver of ice ages over the history of the Earth is much more Earth science than astronomy. The modern ice age cycle (last 2.6 million years or so) is only possible with relatively low levels of CO2 in the atmosphere and land masses near the North Pole where glaciers can form. Other factors in the modern ice age period are thought to be the ...

4

Unless I'm missing something we know very little about the orbital planes of individual systems. Most exoplanetary systems (all of the transiting ones, whereas it is just a strong bias for the doppler-discovered planets) are discovered because they have high orbital inclination. This means that the planet's orbit takes it across the stellar disc as we view ...

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 ...

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

Normally three planets would not line up along a simple line like that. Planets have orbits with different inclinations, so at best they'd be in the same plane. Working that out requires a detailed calculation of their motions. The simplest approximation is to actually assume everything has the same inclination and treat the orbits as circles (which they'...

4

"US Government sensors is a euphemism for "nuclear weapon warning early warning satellites", which is why we won't know details of the actual sensor: It is a military satellite and secret. This particular event occurred in the Bismark sea, North of Papua New Guinea, it probably could have been seen from Papua or parts of Indonesia, if the sky was clear (...

4

The AAS Nova article cites Rawls et al. 2016, who analyze light curves and spectra of the eclipsing binary KIC 9246715 to estimate its physical properties. Besides the stellar masses and radii in the article, their Table 2 lists orbital semimajor axis a = 211 R☉ = 147 million km = 0.98 au, and orbital eccentricity e = 0.356. During the 171-day period, ...

4

The year length depends on the distance between the planet's centre & the Sun's centre, not the Sun's surface. So if the Sun merely expands, Earth at 1 AU will still take a standard year to perform 1 orbit. However, when the Sun becomes a red giant, it won't just expand. As Wikipedia mentions, red giants shed a considerable amount of mass in the form of ...

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