# Longitude of the Sun

1. Oct 14, 2008

### Philosophaie

How do you know where the Longitude of the Sun begins?

2. Oct 14, 2008

### mgb_phys

Do you mean how do we decide where longitude ON the sun starts?
It's tricky, first of all there are no fixed features to use as a Greenwhich meridian and secondly it varies with latitude - since the sun is a fluid the rotation speed is different at different distances from the equator.

So for long term data we use "Carrington rotation number" which is based on counting the sun's rotations from earth and using some aribtrary start time.

3. Oct 14, 2008

### D H

Staff Emeritus
Assuming the OP truly did mean longitude OF the sun, that term is a reference to the ecliptic coordinate system. The zero point in this system is the location of the Sun as seen from the Earth at the Vernal Equinox.

4. Oct 14, 2008

### Philosophaie

The earth @ Vernal Equinox is at an Argument of Perihelion of 102.96519141545deg away from the Longitude of the Sun = 0.

5. Oct 14, 2008

### D H

Staff Emeritus
You are confusing your orbital elements. The angular components for some orbit at some epoch time:
• Inclination, $i$. The angle between the orbital plane and the reference x-y plane. The inclination takes the same same as the inner product of the angular momentum and the reference z-axis.

• Longitude of ascending node, $\Omega[/tex]. The angle between the reference x-axis and the point where the orbit crosses the reference x-y plane with a positive z component of velocity. Note that longitude of ascending node doesn't make sense if the inclination is zero, in which case the longitude of ascending node is arbitrarily defined to be zero. • Argument of periapsis, [itex]\omega$. The angle in the orbital plane distended with respect to the central mass from the longitude of ascending node point to the periapsis point, with the angular momentum vector defining the sense of the rotation. The argument of periapsis doesn't make sense if the eccentricity is zero, in which case the argument of periapsis is arbitrarily defined to be zero.

• True anomaly, $\nu$. The angle in the orbital plane distended with respect to the central mass from the periapsis point to the object's location at the epoch time, with the angular momentum vector once again defining the sense of the rotation.

Some derived items:
• Longitude of periapsis, $\bar{\omega}[/tex], defined as [itex]\bar{\omega} = \Omega + \omega$. This is the longitude at which periapsis would occur if the inclination was zero.

• True longitude, $l$, defined as $l = \bar{\omega}+\nu = \Omega + \omega + \nu$. This is the longitude at which the body would be located at the epoch time if the inclination was zero.

So, what about the Earth? It has zero inclination in the ecliptic coordinate system, so the true longitude of the Earth is the sum of the argument of perihelion and the true anomaly. You forgot the true anomaly.

6. Oct 14, 2008

### Philosophaie

Isn't the true anomaly = 0 at the vernal equinox? So the distance from Longitude of the Sun is equal to Argument of Perihelion away at this juncture.

Does anyone know why there is an angle at Earth's Periapsis?

Last edited: Oct 14, 2008
7. Oct 14, 2008

### D H

Staff Emeritus
No. Did you read my post on the various orbital elements? I'll try again.

The seasons result from the Earth's tilt with respect to the Earth's orbital plane, not from how close the Earth is to the Sun. Think about it this way: If the proximity to the sun dictated the seasons, it would be fall worldwide right now because we are less than three months from perihelion. Right now it is indeed fall in the northern hemisphere but it is springtime in the southern hemisphere.

True anomaly is measured from perihelion passage, not from the vernal equinox. Read post #5, or even better, read a tutorial on orbital elements, or even better, read a book.

Perihelion passage occurs when the distance between the Earth and the Sun reaches a minimum. This currently happens around January 3 or so. Since the Earth and Moon are also orbiting one another, the exact timing of perihelion passage depends in part on the orientation of the Earth and Moon with respect to the Sun.

There is no reason whatsoever for the time of perihelion passage to be related to the minimum sun angle in some hemisphere. That perihelion passage right now is close to December solstice is a chance circumstance. A very chance circumstance: perihelion passage advances about 1 day every 58 years or so, marching completely around the calendar in about 21,000 years. We merely happen to live in a period where perihelion passage is fairly close in time to the December solstice.

Aside: Have you fixed your bug in calculating Julian date?

8. Oct 14, 2008

### Philosophaie

Yes, that was the whole problem. On Jan 3,2008 @6:01:20PM, the True Anomaly is aprox=0 and is at the vernal equinox. Does that mean you can balance an egg on it's end?

My next project is 3D model of the Solar System. Any suggestions on how to go about making parametric equations for graphing the data would be appreciated.

9. Oct 14, 2008

### D H

Staff Emeritus
No!
The true anomaly is approximately zero because on that date the earth was at perihelion. One more time: True anomaly is measured from perihelion, by definition, so there is no balancing on eggs here.

It also has absolutely nothing to do with the vernal equinox. The vernal equinox is a point in time, not space, and it occurs around March 20th. It occurs when the sun as viewed from the Earth crosses the projection of the Earth's equatorial plane out into space. The location of the Sun at this point in time with respect to the Earth basically defines one of the principal axes, the x axis, for the two commonly used astronomical coordinate systems.

Before you proceed to any further projects you really should get a better understanding of astronomical coordinate systems and of planetary motion.

10. Oct 15, 2008

### Philosophaie

Sorry for the confusion about the vernal equinox. My main question from this discussion is why the periasis does not occur at the longitude of the sun equal to Zero and the rest adjusted from that?

Last edited: Oct 15, 2008
11. Oct 17, 2008

### D H

Staff Emeritus
Philosophaie,

There is absolutely no reason to expect that Earth's perihelion will occur when the longitude of the Sun is equal to zero. They are very different things. Perihelion passage occurs once per anomalistic year, when the Earth is closest to the Sun. The longitude of the Sun equaling zero occurs once per tropical year, when the Sun as viewed from the Earth crosses the Earth's equatorial plane heading northward.

Note well: perihelion passage occurs once per anomalistic year while longitude of the Sun equalling zero occurs once per [/b]tropical[/b] year. They don't even occur at the same frequency. If they happen to coincide during some year, they will not coincide the next.

I've already introduced the concepts of the anomalistic year and the tropical year. There is one more definition of a "year": The time it takes the Earth to complete one orbit around the Sun with respect to an inertial reference frame, or to use a slightly outdated terminology, with respect to the fixed stars. This is the sidereal year.

Our calendar is based on the tropical year because we want the calendar to by in synch with seasons. The Earth's axial tilt (technically, the obliquity of the ecliptic) is what determines the seasons. The distant stars (the sidereal year) obviously have no impact on the seasons. Less obviously, the proximity of the Earth to the Sun (the anomalistic year) plays a much lesser role in determining climate than does the axial tilt.

That proximity to the Sun has only a secondary effect on climate would not be true if the Earth's axial tilt was much smaller than it is and the eccentricity of the Earth's orbit was much larger. Playing "what-if", if the axial tilt was near zero but the eccentricity remained small (~0.0167), the weather would be much more uniform year-round because that small eccentricity results in only a 7% variation in radiation intensity over the course of a year. If the eccentricity was also significantly greater, the "what-if" Earth would once again have seasons, but in this case the seasons would be dictated by the anomalistic year.

Back to the real world. Why are there three different "years"? First, the sidereal year versus the anomalistic year. First a bit on Keplerian orbits and orbital elements. Orbital elements comprise the semi-major axis of the orbit, the orbit's eccentricity, three angles that describe the shape and orientation of the orbit, plus the true (or mean or eccentric) anomaly. Keplerian orbits occur only in fiction -- and in introductory physics classes. The Earth and Sun would orbit their center of mass in a Keplerian orbit if the only objects in the universe were the Earth and the Sun, and if Newton's law of gravity perfectly described reality, and if both had spherical mass distributions.

In a true Keplerian orbit, only one of the six orbital elements changes with time: the anomaly. The remaining five elements are constant. The anomalistic year and the sidereal year would be equal to one another in this fictional universe. Our solar system has multiple planets and Newton's law of gravitation is only approximately correct. Those other planets, particularly Jupiter, perturb the Earth's orbit around the Sun. The Earth's perihelion passage does not occur at exactly the same place every year with respect to the fixed stars. It instead precesses (called anomalistic precession) a tiny bit. That Newton's law of gravity is not quite correct adds a tiny bit to this precession. The end result: The anomalistic year is about 4.7039 minutes longer than the sidereal year.

The tropical year differs from the sidereal year because the Earth isn't a perfect sphere. The Earth has an equatorial bulge because the Earth is rotating about its axis. The Moon, the Sun, and the other planets "grab" this bulge gravitationally, resulting in a torque on the Earth. The Earth wobbles a bit (called precession of the equinoxes). The Earth's rotation axis in turn rotates about another axis, completing a revolution in about 26,000 years. The end result: The tropical year is about 20.3628 minutes shorter than the sidereal year.