# Tag Info

8

First, don't think that equinox is when the day and night are the same. It is the moment, when the sun's declination is zero, or when the sun is directly over the equator. Let me explain ... Earth is always rotating around the Sun, so declination of the Sun is always changing (except at the solstices when it stops for a minute and goes in the other way). As @...

7

As you said, the Sun takes a year to cycle the Zodiac, and it takes a year between two spring or autumn equinoxes. Both statements are true, but "year" in them means two different things. It takes a sidereal year to cycle the Zodiac, but it takes a tropic year to cycle between equinoxes. Since the tropic year is about 20 minutes shorter, the equinoctial ...

7

The position of the ecliptic does change, due to perturbation of other planets, but only very slowly. The invariant plane is the plane of angular momentum of the whole solar system. This is dominated by the orbital momentum of the planets, particularly Jupiter, and since other stars are much too far away to perturb the solar system, it is invariant with ...

6

The ecliptic was understood to be the path of the sun relative to the sky. Ptolemy believed the sky rotated daily around the Earth. All the planets, including the Moon and the Sun, shared in this motion. However, the planets had secondary motions, relative to the sky. And in particular relative to the stars that they thought were fixed to the sky. This ...

6

I wanted to ask this question even though I realized the correct geometry in the middle of typing it. So I will answer this myself. First of all, the sub-lunar point never exceeds 28.545 N/S. If you're standing at the sub-lunar point, the Moon will be directly overhead. The Moon never wanders outside this latitude bracket. But that does not mean the Moon's ...

6

Edit: Getting the exact value winds up being simple, but it was only later in the day that the answer occurred to me. All we need to do is ask for the altitude and azimuth of the Ecliptic’s south pole! The ecliptic itself will be highest above our horizon in exactly that direction, at an altitude that is exactly 90° above its pole. Thus the answer is given ...

5

Two further points to make are that: (1) because of the way we define sunrise and sunset as being when the entire Sun is below the horizon not just the centre of the Sun. (2) When the Sun is close to the horizon the Sun’s ray are refracted (bent) by the Earth’s atmosphere so that the Sun appears to be slightly higher in the sky that it would be if the ...

5

The easiest graph to consider is the graph of day-length. If the day-length varies from 8 to 16 hours, there can only be two equinoxes: At the time of equinox, the day length changes at the fastest rate, by a few minutes in temperate regions, compared to the solstices, which change by a few seconds every day. So, the equinox doesn't exist perfectly, the ...

5

The moon's orbit around the Earth is inclined at about 5° to the ecliptic So as the Earth progresses in its own orbit around the sun, the moon's altitude also changes During summers the part of the orbit on the other side of the Earth(night side) is below the ecliptic. While during winters it is the upper part which is on the night side Source

4

Mg I does indeed mean neutral magnesium atoms, but May and his collaborators weren't observing magnesium atoms orbiting in the inner Solar System -- they were seeing absorption by magnesium atoms in the Sun's atmosphere. In the optical and near-infrared, the zodiacal light is sunlight scattered/reflected by interplanetary dust particles. Not surprisingly, ...

4

Earth is a special case since the equatorial and ecliptic coordinate systems are defined in terms of its own rotation and orbit. Earth's north pole vector in equatorial coordinates is $$\vec N_{\oplus,eq} = (0, 0, 1)$$ To transform this to ecliptic coordinates, we rotate about the $x$ axis by the obliquity $\varepsilon$ = 23.44$^\circ$ and get \vec N_{\...

4

Both the equator and the ecliptic are great circles on the celestial sphere. The appearance of each on a map depends on the map projection. In an equirectangular projection centered on the equator, the equator (brown) is a straight line, and the ecliptic (blue) is approximately sinusoidal. If the same projection is centered on the ecliptic instead, the ...

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By inspection of a star charting app, I'd say the circles lie in the celestial equator, ie declination equal 0 degrees. Stars above the circles are to the north and vice versa.

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The equator (Earth's rotation plane) and ecliptic (Earth's solar orbit plane) cross at the vernal equinox ♈, the zero point of both right ascension and ecliptic longitude. Relative to the ecliptic, the Earth's north pole is tilted toward ecliptic longitude λ=90°, where the Sun appears at the June solstice. Constellations in that general direction are highest ...

3

The tidal field of the Galaxy does lead to the oscillation of the plane of very wide binaries. The mechanism of this oscillation is identical to the Kozai-Lidov mechanism (the only difference is that in the case of KL oscillations the tidal field is generated by the averaged orbit of a tertiary stellar companion). However, if you run the numbers, the ...

3

I cant help you with skyfield, but i usually use JPL Horizons Web interface. No installation required, you can also print it in a text file if you want: https://ssd.jpl.nasa.gov/horizons.cgi Otherwise i found the documentation for skyfield: https://rhodesmill.org/skyfield/toc.html And if nothing of that works, i made a little astropy script for you from ...

3

Maps of the sky are being used to actually get to know the sky. Thus you take the map and hold it above your head (as opposed to a map of the Earth's surface where you look down on Earth). Thus East and West are switched between that two types of maps, so that directions match actual directions when viewing the map.

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HIP 76880 = $\kappa$ Librae (V=4.72) has an Ecliptic latitude of -0.019 degrees. This is the winner amongst all stars in the Hipparco/Tycho catalogue with a Hipparcos magnitude <6. If you mean physical distance, rather than angular distance, then the sine of the ecliptic latitude must be multiplied by a distance estimate for the star. This cannot be ...

2

Regulus ($\alpha$ Leo) is the one to beat, with ecliptic latitude $\beta$ = +0.46$^\circ$ and apparent V magnitude +1.4. Some dimmer stars are closer to the ecliptic, e.g. $\alpha$ Lib ($\beta$=+0.33$^\circ$, V=+2.7), $\delta$ Gem ($\beta$=-0.18$^\circ$, V=+3.5), and $\delta$ Cnc ($\beta$=+0.08$^\circ$, V=+3.9). For more suggestions, poke around in ...

2

The nodes are the two points on the orbit of the moon at which the plane of the orbit crosses the elliptic. The position of the nodes change in a predictable way due to the perturbation by the sun (and the other planets). Since the orbit is constantly changing, the position of the longitude. The node is not the position where the crossing last occurred. ...

2

As you mentioned, the Earth's rotational axis undergoes a slow precession. This results from tidal torques caused largely by the Sun and the Moon. There would be no precession if the axial tilt with respect to the orbital angular momentum vectors was zero or if the Earth was spherical. The Sun is almost spherical it's oblate to about $10^{-6}$, and is ...

2

Not within plus/minus a few hundred years at least! The Moon's orbital plane around the Earth (viewed in an inertial frame) rocks back and forth a bit (Lunar Nodal Precession with a period of about 18.6 years), but that's not going to help here. You can play with the numbers in this program. import numpy as np import matplotlib.pyplot as plt from skyfield....

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I don't think the question was clearly phrased. The question that I will answer is how did the ancients see the invisible line known as the ecliptic. The problem is similar in that it is easy to tell how far north and south you are but difficult to tell how far east and west you are. You can just keep track of where the heart of the sun rises on the ...

2

The stars are not visible during the day, but they are still there (we just can't see them because of the brightness of the sun). We can't see the stars, but we can calculate where they would be if we could see them. Over the course of the year the sun appears to move relative to the stars. This is actually due to the orbit of the Earth around the sun. ...

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